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Proposition 3.8 Let \( \\left( {x}_{n}\\right) \) be a sequence in \( E \) converging weakly to \( x \) . Then, for all \( T \\in L\\left( E\\right) \), the sequence \( \\left( {T{x}_{n}}\\right) \) converges weakly to \( {Tx} \) .
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Proof. For every \( y \\in E \) ,\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow + \\infty }}\\left( {T{x}_{n} \\mid y}\\right) = \\mathop{\\lim }\\limits_{{n \\rightarrow + \\infty }}\\left( {{x}_{n} \\mid {T}^{ * }y}\\right) = \\left( {x \\mid {T}^{ * }y}\\right) = \\left( {{Tx} \\mid y}\\right) .\n\]
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Yes
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Proposition 4.1 An orthogonal family that does not include the zero vector is free.
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Proof. Let \( J \) be a finite subset of \( I \) and let \( {\left( {\lambda }_{j}\right) }_{j \in J} \) be elements of \( \mathbb{K} \) such that \( \mathop{\sum }\limits_{{j \in J}}{\lambda }_{j}{X}_{j} = 0 \) . Then\n\n\[ \n{\begin{Vmatrix}\mathop{\sum }\limits_{{j \in J}}{\lambda }_{j}{X}_{j}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{j \in J}}{\left| {\lambda }_{j}\right| }^{2}{\begin{Vmatrix}{X}_{j}\end{Vmatrix}}^{2} = 0 \n\]\n\nwhich clearly implies that \( {\lambda }_{j} = 0 \) for all \( j \in J \) .
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Yes
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Proposition 4.2 Let \( {\left\{ {e}_{j}\right\} }_{j \in J} \) be a finite orthonormal family in \( E \), spanning the vector subspace \( F \) . For every \( x \in E \), the orthogonal projection \( {P}_{F}\left( x\right) \) of \( x \) onto \( F \) is given by\n\n\[ \n{P}_{F}\left( x\right) = \mathop{\sum }\limits_{{j \in J}}\left( {x \mid {e}_{j}}\right) {e}_{j} \n\]\n\nAs a consequence,\n\n\[ \n\parallel x{\parallel }^{2} = {\begin{Vmatrix}x - \mathop{\sum }\limits_{{j \in J}}\left( x \mid {e}_{j}\right) {e}_{j}\end{Vmatrix}}^{2} + \mathop{\sum }\limits_{{j \in J}}{\left| \left( x \mid {e}_{j}\right) \right| }^{2}. \n\]
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Proof. To prove the first statement, it is enough to show that the vector \( y = \mathop{\sum }\limits_{{j \in J}}\left( {x \mid {e}_{j}\right) {e}_{j} \) satisfies the conditions characterizing \( {P}_{F}\left( x\right) \) (see Proposition 2.3 and the remark on page 107). Now, it is clear that \( y \in F \) and that \( \left( {x - y \mid {e}_{j}}\right) = 0 \) for all \( j \in J \), which implies \( x - y \in {F}^{ \bot } \) . The rest of the theorem follows immediately from the Pythagorean Theorem.
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Yes
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Theorem 4.5 Let \( {\left( {e}_{i}\right) }_{i \in I} \) be a Hilbert basis of \( E \) . For any element \( x \) of \( E \) , \[ x = \mathop{\sum }\limits_{{i \in I}}\left( {x \mid {e}_{i}}\right) {e}_{i} \]
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Proof. By Proposition 4.2, we know that, for any finite subset \( J \) of \( I \) , \[ {\begin{Vmatrix}x - \mathop{\sum }\limits_{{j \in J}}\left( x \mid {e}_{j}\right) {e}_{j}\end{Vmatrix}}^{2} = \parallel x{\parallel }^{2} - \mathop{\sum }\limits_{{j \in J}}{\left| \left( x \mid {e}_{j}\right) \right| }^{2}. \] Now just apply the definitions and property 2 of Theorem 4.4.
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No
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Proposition 4.6 (Schmidt orthonormalization process) Suppose that \( N \in \{ 1,2,3,\ldots \} \cup \{ + \infty \} \) and let \( {\left( {f}_{n}\right) }_{0 \leq n < N} \) be a free family in \( E \) . There exists an orthonormal family \( {\left( {e}_{n}\right) }_{0 \leq n < N} \) of \( E \) such that, for each nonnegative integer \( n < N \), the families \( {\left( {e}_{p}\right) }_{0 \leq p \leq n} \) and \( {\left( {f}_{p}\right) }_{0 \leq p \leq n} \) span the same vector subspace of \( E \) .
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Such a family can be constructed by setting\n\n\[ \n{e}_{0} = \frac{1}{\begin{Vmatrix}{f}_{0}\end{Vmatrix}}{f}_{0} \n\] \n\nand, for \( 0 \leq n < N - 1 \) , \n\n\[ \n{x}_{n + 1} = {f}_{n + 1} - {P}_{n}{f}_{n + 1}\;\text{ and }\;{e}_{n + 1} = \frac{1}{\begin{Vmatrix}{x}_{n + 1}\end{Vmatrix}}{x}_{n + 1}, \n\] \n\nwhere \( {P}_{n} \) is the orthogonal projection onto the span of the family \( {\left( {f}_{p}\right) }_{0 \leq p \leq n} \) .\n\nProof. We show that the sequence \( {\left( {e}_{n}\right) }_{n \in \mathbb{N}} \) defined in the statement satisfies the desired conditions. First, since the family \( \left( {f}_{n}\right) \) is assumed to be free, it is clear that \( {x}_{n} \neq 0 \) for all \( n \), and so that \( {e}_{n} \) is defined for all \( n \) . Let \( {E}_{n} \) and \( {F}_{n} \) be the vector subspaces of \( E \) spanned by, respectively, \( {\left( {e}_{p}\right) }_{0 \leq p \leq n} \) and \( {\left( {f}_{p}\right) }_{0 \leq p \leq n} \) . Trivially, \( {E}_{0} = {F}_{0} \) . Suppose that \( {E}_{n} = {F}_{n} \) for \( n < N - 1 \) . Clearly \( {e}_{n + 1} \in {F}_{n + 1} \), so \( {E}_{n + 1} \subset {F}_{n + 1} \) . Moreover \( {f}_{n + 1} \in {E}_{n + 1} \), which shows the reverse inclusion. Hence, \( {E}_{n} = {F}_{n} \) for all \( 0 \leq n < N \) . At the same time, for each \( n \geq 1 \) the vector \( {e}_{n + 1} \) is, by construction, orthogonal to \( {F}_{n} \) and thus to \( {E}_{n} \) . Therefore the family \( {\left( {e}_{n}\right) }_{0 \leq n < N} \) is orthonormal.
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Yes
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Corollary 4.7 A scalar product space is separable if and only if it has a countable Hilbert basis.
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Proof. According to Proposition 2.6 on page 10, the condition is sufficient. By the same proposition, separability implies the existence of a free and fundamental family \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) . Applying the Schmidt orthonormalization process to the family \( \left( {f}_{n}\right) \) we obtain a family \( \left( {e}_{n}\right) \) that is a Hilbert basis.
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Yes
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Proposition 1.4 If \( f \in {L}^{1} \cap {L}^{\infty } \), then \( f \in {L}^{p} \) for every \( p \in \left( {1,\infty }\right) \), and\n\n\[ \parallel f{\parallel }_{p} \leq \parallel f{\parallel }_{1}^{1/p}\parallel f{\parallel }_{\infty }^{1 - 1/p}. \]\n\nIn addition, if \( 1 \leq p < \infty ,{L}^{1} \cap {L}^{\infty } \) is dense in \( {L}^{p} \) .
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Proof. If \( f \in {L}^{\infty } \) and \( 1 < p < \infty \), we clearly have \( {\left| f\right| }^{p} \leq \left| f\right| \parallel f{\parallel }_{\infty }^{p - 1} \) \( m \) -almost everywhere, which proves the first assertion of the proposition.\n\nNow suppose that \( 1 \leq p < \infty \) and that \( f \in {L}^{p} \) . Since \( {\left| f\right| }^{p} \) is a positive integrable function, there exists an increasing sequence \( {\left( {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) of positive, integrable, piecewise constant functions that converges almost everywhere to \( {\left| f\right| }^{p} \) . Set\n\n\[ \alpha \left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) /\left| {f\left( x\right) }\right| & \text{ if }f\left( x\right) \neq 0 \\ 0 & \text{ if }f\left( x\right) = 0 \end{array}\right. \]\n\nThen the sequence \( \left( {\alpha {\varphi }_{n}^{1/p}}\right) \) is a sequence in \( {L}^{1} \cap {L}^{\infty } \) that converges almost everywhere to \( f \), being bounded above in absolute value by \( \left| f\right| \) . By the Dominated Convergence Theorem, this sequence converges to \( f \) in \( {L}^{p} \) .
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Yes
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Lemma 1.5 For each nonnegative real a, define a map \( {\Pi }_{a} : \mathbb{K} \rightarrow \mathbb{K} \) by setting \( {\Pi }_{0}\left( x\right) = 0 \) and\n\n\[ \n{\Pi }_{a}\left( x\right) = \frac{ax}{\max \left( {a,\left| x\right| }\right) }\;\text{ if }a > 0.\n\]\n\nThen, for every \( x \in \mathbb{K} \), we have \( \left| {{\Pi }_{a}\left( x\right) }\right| \leq \min \left( {a,\left| x\right| }\right) \) and, if \( \left| x\right| \leq a \), then \( {\Pi }_{a}\left( x\right) = x \) . Moreover,\n\n\[ \n\left| {{\Pi }_{a}\left( x\right) - {\Pi }_{a}\left( y\right) }\right| \leq \left| {x - y}\right| \;\text{ for all }x, y \in \mathbb{K}.\n\]
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Proof. It is clear that \( {\Pi }_{a} \) is exactly the projection map from the canonical euclidean space \( \mathbb{R} \) (or the canonical hermitian space \( \mathbb{C} \), as the case may be) onto \( \bar{B}\left( {0, a}\right) \) . The claims made are then obvious; the last of them can be seen as a particular case of Proposition 2.2 on page 106.
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No
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Theorem 1.6 If \( m \) is a Radon measure, the space \( {C}_{c}\left( X\right) \) is dense in \( {L}^{p}\left( m\right) \) for \( 1 \leq p < + \infty \) .
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Proof. The case \( p = 1 \) was proved in Chapter 2, Proposition 3.5 on page 70. Suppose \( 1 < p < \infty \) . By Proposition 1.4, it suffices to approximate \( f \in \) \( {L}^{1} \cap {L}^{\infty } \) in the sense of the \( \parallel \cdot {\parallel }_{p} \) norm. Thus, fix \( f \in {L}^{1} \cap {L}^{\infty } \) and let \( \left( {\varphi }_{n}\right) \) be a sequence in \( {C}_{c}\left( X\right) \) that converges to \( f \) in \( {L}^{1} \) . Set \( {\psi }_{n} = {\Pi }_{\parallel f{\parallel }_{\infty }}\left( {\varphi }_{n}\right) \) , using the notation of Lemma 1.5. Then \( {\psi }_{n} \in {C}_{c}\left( X\right) \) and\n\n\[ \n{\left| f - {\psi }_{n}\right| }^{p} \leq \left| {f - {\psi }_{n}}\right| {\left( 2\parallel f{\parallel }_{\infty }\right) }^{p - 1}.\n\]\n\nNow \( f = {\Pi }_{\parallel f{\parallel }_{\infty }}\left( f\right) \) and so, since by Lemma 1.5 the maps \( {\Pi }_{a} \) are contracting,\n\n\[ \n{\left| f - {\psi }_{n}\right| }^{p} \leq \left| {f - {\varphi }_{n}}\right| {\left( 2\parallel f{\parallel }_{\infty }\right) }^{p - 1},\n\]\n\nwhich proves the result.
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Yes
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Corollary 1.7 If \( m \) is a Radon measure, the space \( {L}^{p}\left( m\right) \) is separable for \( 1 \leq p < \infty \) .
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Proof. Let \( \left( {K}_{n}\right) \) be a sequence of compact sets exhausting \( X \) . Since\n\n\[ \n{C}_{c}\left( X\right) = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{C}_{{K}_{n}}\left( X\right) \n\]\n\nit suffices, by the preceding theorem, to show that each \( {C}_{{K}_{n}}\left( X\right) \) is separable with respect to the \( \parallel \cdot {\parallel }_{p} \) norm. But \( {C}_{{K}_{n}}\left( X\right) \) is separable with respect to the uniform norm \( \parallel \cdot \parallel \), and \( \parallel f{\parallel }_{p} \leq \parallel f\parallel m{\left( {K}_{n}\right) }^{1/p} \) for every \( f \in {C}_{{K}_{n}}\left( X\right) \) . This proves the result.
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Yes
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Lemma 2.3 If \( T \in {\left( {L}^{p}\right) }^{\prime } \), there exists a measurable function \( g \) such that, for all \( f \in {L}^{p} \), \[ {fg} \in {L}^{1}\;\text{ and }\;{Tf} = \int {fgdm} \]
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Proof. For \( f \in {L}_{\mathbb{R}}^{p} \), set \( {T}_{1}f = \operatorname{Re}\left( {Tf}\right) \) and \( {T}_{2}f = \operatorname{Im}\left( {Tf}\right) \). Then \( {T}_{1} \) and \( {T}_{2} \) belong to \( {\left( {L}_{\mathbb{R}}^{p}\right) }^{\prime } \). If Lemma 2.3 is true in the real case, we can apply it to \( {T}_{1} \) and \( {T}_{2} \) to obtain real functions \( {g}_{1} \) and \( {g}_{2} \), and clearly the function \( g = {g}_{1} + i{g}_{2} \) works for \( T \). Therefore we can suppose we are in the real case.\n\nIn this case \( T \) can be written as the difference of two continuous and positive linear forms on \( {L}_{\mathbb{R}}^{p} \) (apply Remark 2 on page 88 to the lattice \( {L}_{\mathbb{R}}^{p} \)). So we can in fact suppose that \( T \) is a positive continuous linear form on \( {L}_{\mathbb{R}}^{p} \), and we do so.\n\nSince the measure \( m \) is \( \sigma \)-finite, there exists a countable partition \( \left( {K}_{n}\right) \) of \( X \) consisting of elements of \( \mathcal{F} \) of finite measure. For each integer \( n \), let \( {m}_{n} \) be the restriction of \( m \) to \( {K}_{n} \). If \( f \in {L}_{\mathbb{R}}^{p}\left( {m}_{n}\right) \), denote by \( \widetilde{f} \) the extension of \( f \) to \( X \) taking the value 0 on \( X \smallsetminus {K}_{n} \). The linear form on \( {L}_{\mathbb{R}}^{p}\left( {m}_{n}\right) \) defined by \( f \mapsto T\left( \widetilde{f}\right) \) then satisfies on \( {K}_{n} \) the hypotheses of Lemma 2.2. Therefore there is a positive measurable function \( {g}_{n} \) on \( {K}_{n} \) such that, for all \( f \in {L}_{\mathbb{R}}^{p}\left( {m}_{n}\right) \)\n\n\[ f{g}_{n} \in {L}_{\mathbb{R}}^{1}\left( {m}_{n}\right) \;\text{ and }\;T\left( \widetilde{f}\right) = \int f{g}_{n}d{m}_{n}. \]\n\nNow let \( g \) be the measurable function on \( X \) whose restriction to each \( {K}_{n} \) is \( {g}_{n} \). If \( f \in {L}_{\mathbb{R}}^{p}\left( m\right) \) and \( f \geq 0 \), we have \( f = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}f{1}_{{K}_{n}} \), the series being convergent in \( {L}_{\mathbb{R}}^{p}\left( m\right) \). By the continuity of \( T \) and monotone convergence in the integral, we deduce that\n\n\[ {Tf} = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}T\left( {f{1}_{{K}_{n}}}\right) = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\int }_{{K}_{n}}{fgdm} = \int {fgdm}. \]\n\nThus \( g \) satisfies the necessary conditions.
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Yes
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Lemma 2.4 With the notation of Lemma 2.3, we have \( g \in {L}^{{p}^{\prime }} \) and \( \parallel g{\parallel }_{{p}^{\prime }} \leq \parallel T{\parallel }_{p}^{\prime } \), where \( \parallel \cdot {\parallel }_{p}^{\prime } \) is the norm in \( {\left( {L}^{p}\right) }^{\prime } \) .
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Proof. Since the measure \( m \) is \( \sigma \) -finite, there exists an increasing sequence \( \left( {A}_{n}\right) \) of elements of \( \mathcal{F} \) of finite measure that cover \( X \) . 1. Case \( p = 1 \) . Suppose the conclusion of the lemma is false. Then the set \( \left\{ {\left| g\right| > \parallel T{\parallel }_{1}^{\prime }}\right\} \) has positive measure, so there exists \( \varepsilon > 0 \) such that the set \( A = \left\{ {\left| g\right| > \parallel T{\parallel }_{1}^{\prime } + \varepsilon }\right\} \) has positive measure. Let \( \alpha \) be the function that equals \( \left| g\right| /g \) on \( \{ g \neq 0\} \) and 1 on \( \{ g = 0\} \) . Then, on the one hand, \[ T\left( {\alpha {1}_{A \cap {A}_{n}}}\right) = {\int }_{A \cap {A}_{n}}\left| g\right| {dm} \geq \left( {\parallel T{\parallel }_{1}^{\prime } + \varepsilon }\right) m\left( {A \cap {A}_{n}}\right) \] and, on the other, \[ T\left( {\alpha {1}_{A \cap {A}_{n}}}\right) \leq \parallel T{\parallel }_{1}^{\prime }m\left( {A \cap {A}_{n}}\right) \] There certainly exists an integer \( n \) for which \( m\left( {A \cap {A}_{n}}\right) > 0 \), so we deduce that \( \parallel T{\parallel }_{1}^{\prime } + \varepsilon \leq \parallel T{\parallel }_{1}^{\prime } \), which is absurd. 2. Case \( 1 < p < \infty \) . Define \( \alpha \) as in the preceding case and, for \( n \in \mathbb{N} \), set \( {B}_{n} = {A}_{n} \cap \{ \left| g\right| \leq n\} \) and \( {f}_{n} = {1}_{{B}_{n}}\alpha {\left| g\right| }^{{p}^{\prime } - 1} \) . Then, for every \( n \) , \[ {\int }_{{B}_{n}}{\left| g\right| }^{{p}^{\prime }}{dm} = T{f}_{n} \leq \parallel T{\parallel }_{p}^{\prime }{\left( {\int }_{{B}_{n}}{\left| g\right| }^{{p}^{\prime }}dm\right) }^{1/p}, \] so \[ {\left( {\int }_{{B}_{n}}{\left| g\right| }^{{p}^{\prime }}dm\right) }^{1/{p}^{\prime }} \leq \parallel T{\parallel }_{p}^{\prime } \] whence we deduce the result by making \( n \) approach infinity.
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Yes
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If \( 1 \leq p \leq \infty \), the family \( {\left( {\tau }_{a}\right) }_{a \in {\mathbb{R}}^{d}} \) forms an abelian group of isometries of \( {L}^{p} \).
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The first assertion follows immediately from the remarks preceding the theorem (in particular, from the translation invariance of \( \lambda \) ).
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No
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Proposition 3.2 Let \( p,{p}^{\prime } \in \left\lbrack {1,\infty }\right\rbrack \) be conjugate exponents and suppose \( f \in {L}^{p} \) and \( g \in {L}^{{p}^{\prime }} \) . Then \( f * g \) is uniformly continuous and bounded, and\n\n\[ \parallel f * g{\parallel }_{\infty } \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{{p}^{\prime }} \]
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Proof. The Hölder inequality yields\n\n\[ \left| {\left( {f * g}\right) \left( x\right) - \left( {f * g}\right) \left( {x}^{\prime }\right) }\right| \leq {\begin{Vmatrix}{\tau }_{x}\check{f} - {\tau }_{{x}^{\prime }}\check{f}\end{Vmatrix}}_{p}\parallel g{\parallel }_{{p}^{\prime }}\;\text{ for all }x,{x}^{\prime } \in {\mathbb{R}}^{d}; \]\n\nthe uniform continuity of \( f * g \) if \( p < \infty \) follows because of Proposition 3.1. If \( p = \infty \), we have \( {p}^{\prime } = 1 \) and the property remains true since \( f * g = g * f \) .\n\nWe also have \( \parallel f * g{\parallel }_{\infty } \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{{p}^{\prime }} \), by the Hölder inequality and the fact that \( {\begin{Vmatrix}{\tau }_{x}\check{f}\end{Vmatrix}}_{p} = \parallel f{\parallel }_{p} \) for every \( x \) . This implies, in particular, that the bilinear map \( \left( {f, g}\right) \mapsto f * g \) is continuous as a map from \( {L}^{p} \times {L}^{{p}^{\prime }} \) to \( {C}_{b}\left( {\mathbb{R}}^{d}\right) \) with the uniform norm.
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Yes
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Proposition 3.3 If \( f \) and \( g \) are convolvable equivalence classes of functions,
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\[ \operatorname{Supp}\left( {f * g}\right) \subset \overline{\operatorname{Supp}f + \operatorname{Supp}g}. \] In particular, if \( f \) or \( g \) has compact support, we have \[ \operatorname{Supp}\left( {f * g}\right) \subset \operatorname{Supp}f + \operatorname{Supp}g \] (since, if \( F \) is closed and \( K \) is compact, \( F + K \) is closed). Thus, the convolution of two classes of functions with compact support has compact support.
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No
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Proposition 3.5 Let \( p, q, r \in \left\lbrack {1, + \infty }\right\rbrack \) be such that \( 1/p + 1/q + 1/r \geq 2 \) . If \( f \in {L}^{p}, g \in {L}^{q} \), and \( h \in {L}^{r} \), then \( f * \left( {g * h}\right) \) and \( \left( {f * g}\right) * h \) are well defined and belong to \( {L}^{s} \), where \( s \) is given by \( 1/s = 1/p + 1/q + 1/r - 2 \) . In addition, \[ f * \left( {g * h}\right) = \left( {f * g}\right) * h. \]
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Proof. That \( f * \left( {g * h}\right) \) and \( \left( {f * g}\right) * h \) are well defined and belong to \( {L}^{s} \) follows from Theorem 3.4. Next, \[ \left( {f * \left( {g * h}\right) }\right) \left( x\right) = \iint f\left( {x - y}\right) g\left( {y - z}\right) h\left( z\right) {dydz} \] \[ = \iint f\left( {x - y - z}\right) g\left( y\right) h\left( z\right) {dydz} = \left( {\left( {f * g}\right) * h}\right) \left( x\right) , \] which concludes the proof. (As an exercise, the reader might justify these formal calculations, especially the use of Fubini's Theorem.)
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No
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Corollary 3.6 The operations + and \( * \) make \( {L}^{1} \) into a commutative ring.
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Proof. The convolution product is commutative and, by Theorem 3.4, \( {L}^{1} \) is closed under it. Proposition 3.5 says it is also associative. The rest is obvious.
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No
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Proposition 3.7 Suppose \( p \in \lbrack 1,\infty ) \) and let \( {\left( {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) be a Dirac sequence. If \( f \in {L}^{p} \), then\n\n\[ f * {\varphi }_{n} \in {L}^{p}\;\text{ and }\;{\begin{Vmatrix}f * {\varphi }_{n}\end{Vmatrix}}_{p} \leq \parallel f{\parallel }_{p}\;\text{ for every }n \in \mathbb{N}, \]\n\nand\n\n\[ \mathop{\lim }\limits_{{n \rightarrow + \infty }}f * {\varphi }_{n} = f\;\text{ in }{L}^{p} \]
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Proof. That \( f * {\varphi }_{n} \in {L}^{p} \) and \( {\begin{Vmatrix}f * {\varphi }_{n}\end{Vmatrix}}_{p} \leq \parallel f{\parallel }_{p} \) follows from Theorem 3.4. Further, for almost every \( x \) ,\n\n\[ \left| {f\left( x\right) - \left( {f * {\varphi }_{n}}\right) \left( x\right) }\right| \leq \int \left| {f\left( x\right) - f\left( {x - y}\right) }\right| {\varphi }_{n}\left( y\right) {dy} \]\n\n\[ \leq {\left( \int {\left| f\left( x\right) - f\left( x - y\right) \right| }^{p}{\varphi }_{n}\left( y\right) dy\right) }^{1/p} \]\n\nthe latter inequality being a consequence of Hölder's inequality applied to the measure \( {\varphi }_{n}\left( y\right) {dy} \) . We deduce that\n\n\[ {\begin{Vmatrix}f - f * {\varphi }_{n}\end{Vmatrix}}_{p}^{p} \leq \int {\begin{Vmatrix}f - {\tau }_{y}f\end{Vmatrix}}_{p}^{p}{\varphi }_{n}\left( y\right) {dy}. \]\n\nNow, for every \( \varepsilon > 0 \), we can write\n\n\[ \int {\begin{Vmatrix}f - {\tau }_{y}f\end{Vmatrix}}_{p}^{p}{\varphi }_{n}\left( y\right) {dy} \leq \mathop{\sup }\limits_{{\left| y\right| < \varepsilon }}{\begin{Vmatrix}f - {\tau }_{y}f\end{Vmatrix}}_{p}^{p} + {\left( 2\parallel f{\parallel }_{p}\right) }^{p}{\int }_{\{ \left| y\right| > \varepsilon \} }{\varphi }_{n}\left( y\right) {dy} \]\n\nby breaking \( {\mathbb{R}}^{d} \) into the disjoint union of \( \{ \left| y\right| \leq \varepsilon \} \) and \( \{ \left| y\right| > \varepsilon \} \) . It follows that\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}f - f * {\varphi }_{n}\end{Vmatrix}}_{p} \leq \mathop{\sup }\limits_{{\left| y\right| \leq \varepsilon }}{\begin{Vmatrix}f - {\tau }_{y}f\end{Vmatrix}}_{p} \]\n\nNow it suffices to apply Proposition 3.1.
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Yes
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Proposition 1.2 Suppose \( T \in L\left( E\right) \) . The limit \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} \) exists and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} = \mathop{\inf }\limits_{{n \in {\mathbb{N}}^{ * }}}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n}. \]
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Proof\n\n1. Set \( a = \mathop{\inf }\limits_{{n \in {\mathbb{N}}^{ * }}}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} \) . Certainly we have\n\n\[ a \leq \mathop{\liminf }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} \]\n\nTake \( \varepsilon > 0 \) and let \( {n}_{0} \in {\mathbb{N}}^{ * } \) be such that \( {\begin{Vmatrix}{T}^{{n}_{0}}\end{Vmatrix}}^{1/{n}_{0}} \leq a + \varepsilon \) . Given \( n \in {\mathbb{N}}^{ * } \), we can write, by dividing with remainder, \( n = p\left( n\right) {n}_{0} + q\left( n\right) \) , with \( p\left( n\right) \in \mathbb{N}, q\left( n\right) \in \mathbb{N} \) and \( 0 \leq q\left( n\right) < {n}_{0} \) . Thus\n\n\[ \begin{Vmatrix}{T}^{n}\end{Vmatrix} \leq {\begin{Vmatrix}{T}^{{n}_{0}}\end{Vmatrix}}^{p\left( n\right) }\parallel T{\parallel }^{q\left( n\right) }.\]\n\nSince \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}q\left( n\right) /n = 0 \) and \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}p\left( n\right) /n = 1/{n}_{0} \), we deduce\n\nthat\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} \leq {\begin{Vmatrix}{T}^{{n}_{0}}\end{Vmatrix}}^{1/{n}_{0}} \leq a + \varepsilon . \]\n\nThis holds for all \( \varepsilon > 0 \), so \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} = a \) .\n\n2. The map \( \lambda \mapsto \left( {{\lambda I} - T}\right) \) from \( \mathbb{K} \) to \( L\left( E\right) \) is clearly continuous. Therefore, by Proposition 1.1, \( \rho \left( T\right) \) is open and \( \sigma \left( T\right) \) is closed. All that remains to show is that \( \sigma \left( T\right) \) is bounded by \( r\left( T\right) \).\n\n3. Take \( \lambda \in \mathbb{K} \) such that \( \left| \lambda \right| > r\left( T\right) \), and consider \( r \in \left( {r\left( T\right) ,\left| \lambda \right| }\right) \) . Since \( r > r\left( T\right) \), there exists an integer \( {n}_{0} \in {\mathbb{N}}^{ * } \) such that\n\n\[ \begin{Vmatrix}{T}^{n}\end{Vmatrix} \leq {r}^{n}\;\text{ for all }n \geq {n}_{0}. \]\n\nThe series \( \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\lambda }^{-n - 1}{T}^{n} \) converges absolutely in \( L\left( E\right) \) (since \( r < \left| \lambda \right| \) ) and it is easy to see that\n\n\[ \left( {{\lambda I} - T}\right) \left( {\mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\lambda }^{-n - 1}{T}^{n}}\right) = \left( {\mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\lambda }^{-n - 1}{T}^{n}}\right) \left( {{\lambda I} - T}\right) = I \]\n\nand so that \( \lambda \in \rho \left( T\right) \) . Since this holds for all \( \left| \lambda \right| > r\left( T\right) \), the proof is complete.
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Yes
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Proposition 1.3 Suppose \( T \in L\left( E\right) \) . For all \( \lambda ,\mu \in \rho \left( T\right) \), we have\n\n\[ R\left( {\lambda, T}\right) - R\left( {\mu, T}\right) = \left( {\mu - \lambda }\right) R\left( {\lambda, T}\right) R\left( {\mu, T}\right) = \left( {\mu - \lambda }\right) R\left( {\mu, T}\right) R\left( {\lambda, T}\right) . \]\n\n(This is called the resolvent equation.) Moreover, the map \( \lambda \mapsto R\left( {\lambda, T}\right) \) from the open subset \( \rho \left( T\right) \) of \( \mathbb{K} \) to \( L\left( E\right) \) is differentiable and\n\n\[ \frac{d}{d\lambda }R\left( {\lambda, T}\right) = - {\left( R\left( \lambda, T\right) \right) }^{2} \]
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Proof. First,\n\n\[ R\left( {\lambda, T}\right) - R\left( {\mu, T}\right) = R\left( {\lambda, T}\right) \left( {\left( {{\mu I} - T}\right) - \left( {{\lambda I} - T}\right) }\right) R\left( {\mu, T}\right) \]\n\n\[ = \left( {\mu - \lambda }\right) R\left( {\lambda, T}\right) R\left( {\mu, T}\right) \]\n\nwhich proves the resolvent equation. In particular,\n\n\[ \frac{1}{h}\left( {R\left( {\lambda + h, T}\right) - R\left( {\lambda, T}\right) }\right) = - R\left( {\lambda, T}\right) R\left( {\lambda + h, T}\right) ,\]\nwith \( h \in {\mathbb{K}}^{ * } \) and \( \lambda ,\lambda + h \in \rho \left( T\right) \) . By the continuity of the map \( \lambda \mapsto R\left( {\lambda, T}\right) \) (an immediate consequence of Proposition 1.1) and the continuity of the product in \( L\left( E\right) \), we obtain\n\n\[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {R\left( {\lambda + h, T}\right) - R\left( {\lambda, T}\right) }\right) = - {\left( R\left( \lambda, T\right) \right) }^{2},\]\n\nwhich concludes the proof.
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Yes
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Theorem 1.5 (spectral image) If \( T \in L\left( E\right) \) and \( P \in \mathbb{K}\left\lbrack X\right\rbrack \), we have\n\n\[ P\left( {\sigma \left( T\right) }\right) \subset \sigma \left( {P\left( T\right) }\right) \]\n\nwith equality if \( \mathbb{K} = \mathbb{C} \) .
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Proof\n\n1. Take \( \lambda \in \mathbb{K} \) . Since \( \lambda \) is a root of the polynomial \( P - P\left( \lambda \right) \), there exists a polynomial \( {Q}_{\lambda } \in \mathbb{K}\left\lbrack X\right\rbrack \) such that \( P - P\left( \lambda \right) = \left( {X - \lambda }\right) {Q}_{\lambda } \) . Then\n\n\[ P\left( T\right) - P\left( \lambda \right) I = \left( {T - {\lambda I}}\right) {Q}_{\lambda }\left( T\right) = {Q}_{\lambda }\left( T\right) \left( {T - {\lambda I}}\right) . \]\n\nSuppose that \( P\left( \lambda \right) \notin \sigma \left( {P\left( T\right) }\right) \), and set \( S = {\left( P\left( \lambda \right) I - P\left( T\right) \right) }^{-1} \) . Then\n\n\[ \left( {{\lambda I} - T}\right) {Q}_{\lambda }\left( T\right) S = S{Q}_{\lambda }\left( T\right) \left( {{\lambda I} - T}\right) = I, \]\n\nshowing that \( \left( {{\lambda I} - T}\right) \) is invertible, with inverse \( S{Q}_{\lambda }\left( T\right) = {Q}_{\lambda }\left( T\right) S \) ; thus \( \lambda \notin \sigma \left( T\right) \) . Thus \( \lambda \in \sigma \left( T\right) \) implies \( P\left( \lambda \right) \in \sigma \left( {P\left( T\right) }\right) \), which is to say \( P\left( {\sigma \left( T\right) }\right) \subset \sigma \left( {P\left( T\right) }\right) \) .\n\n2. Suppose that \( \mathbb{K} = \mathbb{C} \) and that \( P \) has degree at least 1 (if \( P \) is constant, the result is trivial). Take \( \mu \in \sigma \left( {P\left( T\right) }\right) \) . Write the polynomial \( P - \mu \) as a product of factors of degree 1 :\n\n\[ P - \mu = C\left( {X - {\lambda }_{1}}\right) \ldots \left( {X - {\lambda }_{n}}\right) \]\n\nwith \( C \neq 0 \) . Then\n\n\[ P\left( T\right) - {\mu I} = C\left( {T - {\lambda }_{1}I}\right) \ldots \left( {T - {\lambda }_{n}I}\right) . \]\n\nSince, by assumption, \( P\left( T\right) - {\mu I} \) is not invertible, one of the factors \( T - {\lambda }_{j}I \) is not invertible. Then, for this value of \( j \), we have \( {\lambda }_{j} \in \sigma \left( T\right) \) . Since \( P\left( {\lambda }_{j}\right) = \mu \), this shows that \( \mu \in P\left( {\sigma \left( T\right) }\right) \) .
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Yes
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Proposition 2.1 Suppose \( T \in L\left( E\right) \) . Then:\n\ni. \( \ker T = {\left( \operatorname{im}{T}^{ * }\right) }^{ \bot } \) .\n\nii. \( \overline{\operatorname{im}T} = {\left( \ker {T}^{ * }\right) }^{ \bot } \) .\n\niii. \( T \) is invertible if and only if \( {T}^{ * } \) is, and in this case\n\n\[{\left( {T}^{ * }\right) }^{-1} = {\left( {T}^{-1}\right) }^{ * }\]
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Proof. For \( x \in E \), we have \( x \in \ker T \) if and only if\n\n\[ \left( {{Tx} \mid y}\right) = \left( {x \mid {T}^{ * }y}\right) = 0\;\text{ for all }y \in E,\]\n\nwhich proves the first assertion. The second is a consequence of the first, in view of Corollary 2.7 on page 108 and of the equality \( {T}^{* * } = T \) . Finally, if \( T \) is invertible, we have \( T{T}^{-1} = {T}^{-1}T = I \) and, by Proposition 3.3 on page 112, \( {\left( {T}^{-1}\right) }^{ * }{T}^{ * } = {T}^{ * }{\left( {T}^{-1}\right) }^{ * } = I \) . Therefore \( {T}^{ * } \) is invertible and \( {\left( {T}^{ * }\right) }^{-1} = {\left( {T}^{-1}\right) }^{ * }. \)
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Yes
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Proposition 2.3 The spectral radius and the norm of a hermitian operator on \( E \) coincide.
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Proof. If \( T \) is hermitian, Proposition 3.4 on page 113 says that \( \begin{Vmatrix}{T}^{2}\end{Vmatrix} = \) \( \parallel T{\parallel }^{2} \) . Iterating this property, which we can do because the square of a hermitian operator is hermitian, we obtain\n\n\[\n\begin{Vmatrix}{T}^{{2}^{n}}\end{Vmatrix} = \parallel T{\parallel }^{{2}^{n}}\;\text{ for all }n \in \mathbb{N}.\n\]\n\nWe conclude that\n\n\[\nr\left( T\right) = \mathop{\lim }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}{T}^{{2}^{n}}\end{Vmatrix}}^{{2}^{-n}} = \parallel T\parallel\n\]\n\nsince the limit of the sequence \( {\left( {\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n}\right) }_{n \in \mathbb{N}} \) equals the limit of any of its subsequences.
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Yes
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Proposition 2.5 Every hermitian operator \( T \) on \( E \) has the following properties:\n\ni. The eigenvalues of \( T \) are real.\n\nii. For every \( \lambda \in \mathbb{C} \), we have \( \overline{\operatorname{im}\left( {{\lambda I} - T}\right) } = {\left( \ker \left( \bar{\lambda }I - T\right) \right) }^{ \bot } \) .\n\niii. The eigenspaces of \( T \) associated with distinct eigenvalues are orthogonal.
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Proof. Suppose that \( \lambda \) is an eigenvalue of \( T \), and let \( x \in E \) be an associated nonzero eigenvector, so that \( {Tx} = {\lambda x} \) and \( x \neq 0 \) . Then\n\n\[ \lambda \parallel x{\parallel }^{2} = \left( {{\lambda x} \mid x}\right) = \left( {{Tx} \mid x}\right) .\n\]\n\nSince the operator \( T \) is selfadjoint, we have \( \left( {{Tx} \mid x}\right) \in \mathbb{R} \) and so \( \lambda \in \mathbb{R} \) , which proves the first part of the proposition.\n\nThe second part is an immediate consequence of the equality \( \overline{\operatorname{im}S} = \) \( {\left( \ker {S}^{ * }\right) }^{ \bot } \), valid for all \( S \in L\left( E\right) \) by Proposition 2.1.\n\nFinally, if \( \lambda \) and \( \mu \) are distinct eigenvalues of \( T \) and if \( x \) and \( y \) are corresponding eigenvectors, we have\n\n\[ \lambda \left( {x \mid y}\right) = \left( {{Tx} \mid y}\right) = \left( {x \mid {Ty}}\right) = \mu \left( {x \mid y}\right) ,\n\]\n\nsince \( \mu \in \mathbb{R} \) . Therefore \( \left( {x \mid y}\right) = 0 \) .
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Yes
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Proposition 2.8 For every \( P \in \mathbb{C}\left\lbrack X\right\rbrack \), we have \( {\left( P\left( T\right) \right) }^{ * } = \bar{P}\left( T\right) \) and \[ \parallel P\left( T\right) \parallel = \mathop{\max }\limits_{{t \in \sigma \left( T\right) }}\left| {P\left( t\right) }\right| \]
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Proof. The first assertion is an immediate consequence of the fact that \( T \) is hermitian (see Proposition 3.3 on page 112). Next, for \( P \in \mathbb{C}\left\lbrack X\right\rbrack \) , Proposition 3.4 on page 113 gives \[ \parallel P\left( T\right) \parallel = {\begin{Vmatrix}P\left( T\right) P{\left( T\right) }^{ * }\end{Vmatrix}}^{1/2} = {\begin{Vmatrix}{\left| P\right| }^{2}\left( T\right) \end{Vmatrix}}^{1/2}. \] But, since \( {\left| P\right| }^{2}\left( T\right) \) is positive hermitian, \[ \begin{Vmatrix}{{\left| P\right| }^{2}\left( T\right) }\end{Vmatrix} = \max \sigma \left( {{\left| P\right| }^{2}\left( T\right) }\right) \] by Corollary 2.7. By the Spectral Image Theorem (page 194), we have \( \sigma \left( {{\left| P\right| }^{2}\left( T\right) }\right) = {\left| P\right| }^{2}\left( {\sigma \left( T\right) }\right) \), so \[ \parallel P\left( T\right) \parallel = {\left( \mathop{\max }\limits_{{t \in \sigma \left( T\right) }}{\left| P\right| }^{2}\left( t\right) \right) }^{1/2} = \mathop{\max }\limits_{{t \in \sigma \left( T\right) }}\left| {P\left( t\right) }\right| \] which concludes the proof.
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Yes
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Corollary 2.10 Let \( f \) be a continuous function from \( \sigma \left( T\right) \) to \( \mathbb{C} \) . The operator \( f\left( T\right) \) is hermitian if and only if \( f \) is real-valued. It is positive hermitian if and only if \( f \geq 0 \) .
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Proof. The first assertion follows from part ii of Theorem 2.9. The second follows from part iii of the same theorem and from Corollary 2.7.
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No
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Proposition 1.1 \( \mathcal{K}\left( {E, F}\right) \) is a vector subspace of \( L\left( {E, F}\right) \) .
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Proof. Consider compact operators \( T \) and \( S \) from \( E \) to \( F \) and elements \( \lambda ,\mu \in \mathbb{K} \) . Then\n\n\[ \left( {{\lambda T} + {\mu S}}\right) \left( {\bar{B}\left( E\right) }\right) \subset \lambda \overline{T\left( {\bar{B}\left( E\right) }\right) } + \mu \overline{S\left( {\bar{B}\left( E\right) }\right) }.\]\n\nBut, if \( {K}_{1} \) and \( {K}_{2} \) are compact sets in \( F \), the set \( \lambda {K}_{1} + \mu {K}_{2} \), being the image of the compact \( {K}_{1} \times {K}_{2} \) under the continuous map \( \left( {x, y}\right) \mapsto {\lambda x} + {\mu y} \) , is also compact.
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Yes
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Proposition 1.2 Let \( R \) be a compact operator from \( E \) to \( F \) . If \( {E}_{1} \) and \( {F}_{1} \) are normed spaces and if \( T \in L\left( {{E}_{1}, E}\right) \) and \( S \in L\left( {F,{F}_{1}}\right) \) are arbitrary, the composition SRT is a compact operator from \( {E}_{1} \) to \( {F}_{1} \) .
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Proof. Indeed,\n\n\[ \n{SRT}\left( {\bar{B}\left( {E}_{1}\right) }\right) \subset \parallel T\parallel S\left( \overline{R\left( {\bar{B}\left( E\right) }\right) }\right) .\n\]\n\nSince a continuous image of a compact set is compact, the result follows.
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Yes
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Proposition 1.4 If \( F \) is complete, the limit in \( L\left( {E, F}\right) \) of every convergent sequence of compact operators from \( E \) to \( F \) is a compact operator.
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Proof. Let \( {\left( {T}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence of compact operators from \( E \) to \( F \) that converges to \( T \) in \( L\left( {E, F}\right) \) . By Theorem 3.3 on page 14, it suffices to show that \( T\left( {\bar{B}\left( E\right) }\right) \) is precompact. Choose \( \varepsilon > 0 \) and let \( n \in \mathbb{N} \) be such that \( \begin{Vmatrix}{T - {T}_{n}}\end{Vmatrix} \leq \varepsilon /3 \) . We can cover \( {T}_{n}\left( {\bar{B}\left( E\right) }\right) \) with a finite number \( k \) of balls \( B\left( {{T}_{n}{f}_{j},\varepsilon /3}\right) \), where \( {f}_{1},\ldots ,{f}_{k} \in \bar{B}\left( E\right) \) . Suppose \( f \in \bar{B}\left( E\right) \) and let \( j \leq k \) be such that \( \begin{Vmatrix}{{T}_{n}f - {T}_{n}{f}_{j}}\end{Vmatrix} < \varepsilon /3 \) . By the triangle inequality, \( \begin{Vmatrix}{{Tf} - T{f}_{j}}\end{Vmatrix} < \varepsilon \) . Therefore\n\n\[ T\left( {\bar{B}\left( E\right) }\right) \subset \mathop{\bigcup }\limits_{{j = 1}}^{k}B\left( {T{f}_{j},\varepsilon }\right) \]\n\nand \( T\left( {\bar{B}\left( E\right) }\right) \) is precompact.
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Yes
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Proposition 1.4 If \( F \) is complete, the limit in \( L\left( {E, F}\right) \) of every convergent sequence of compact operators from \( E \) to \( F \) is a compact operator.
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Proof. Let \( {\left( {T}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence of compact operators from \( E \) to \( F \) that converges to \( T \) in \( L\left( {E, F}\right) \) . By Theorem 3.3 on page 14, it suffices to show that \( T\left( {\bar{B}\left( E\right) }\right) \) is precompact. Choose \( \varepsilon > 0 \) and let \( n \in \mathbb{N} \) be such that \( \begin{Vmatrix}{T - {T}_{n}}\end{Vmatrix} \leq \varepsilon /3 \) . We can cover \( {T}_{n}\left( {\bar{B}\left( E\right) }\right) \) with a finite number \( k \) of balls \( B\left( {{T}_{n}{f}_{j},\varepsilon /3}\right) \), where \( {f}_{1},\ldots ,{f}_{k} \in \bar{B}\left( E\right) \) . Suppose \( f \in \bar{B}\left( E\right) \) and let \( j \leq k \) be such that \( \begin{Vmatrix}{{T}_{n}f - {T}_{n}{f}_{j}}\end{Vmatrix} < \varepsilon /3 \) . By the triangle inequality, \( \begin{Vmatrix}{{Tf} - T{f}_{j}}\end{Vmatrix} < \varepsilon \) . Therefore\n\n\[ T\left( {\bar{B}\left( E\right) }\right) \subset \mathop{\bigcup }\limits_{{j = 1}}^{k}B\left( {T{f}_{j},\varepsilon }\right) \]\n\nand \( T\left( {\bar{B}\left( E\right) }\right) \) is precompact.
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Yes
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Corollary 1.5 If \( F \) is complete, every limit in \( L\left( {E, F}\right) \) of finite-rank operators is a compact operator.
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This provides a frequently useful criterion for proving that an operator is compact.
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No
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Lemma 1.7 If \( F \) is a proper closed subspace of a normed vector space \( G \), there exists \( u \in G \) such that \( \parallel u\parallel = 1 \) and \( d\left( {u, F}\right) \geq \frac{1}{2} \) .
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Proof. Take \( v \in G \smallsetminus F \) and set \( \delta = d\left( {v, F}\right) > 0 \) . Certainly there exists \( w \in F \) such that \( \parallel v - w\parallel < {2\delta } \) . Then the point \( u = \parallel v - w{\parallel }^{-1}\left( {v - w}\right) \) works: if \( z \in F \), we have\n\n\[ \parallel u - z\parallel = \parallel v - w{\parallel }^{-1}\parallel v - w - \parallel v - w\parallel z\parallel \geq \frac{1}{2\delta }\delta = \frac{1}{2}, \]\n\nproving the lemma.
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Yes
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Lemma 2.1 Let \( S \) be a compact selfadjoint operator on a scalar product space \( F \) not equal to \( \{ 0\} \) . Then \( S \) has at least one eigenvalue and\n\n\[ \max \{ \left| \lambda \right| : \lambda \in \operatorname{ev}\left( S\right) \} = \parallel S\parallel \]
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Proof. Clearly, if \( \lambda \) is an eigenvalue of \( S \), then \( \left| \lambda \right| \leq \parallel S\parallel \) . On the other hand, we know from the remark following Theorem 2.6 on page 203 that there exists a spectral value \( \lambda \) of \( S \) such that \( \left| \lambda \right| = \mathop{\sup }\limits_{{\parallel x\parallel = 1}}\left| \left( {{Sx} \mid x}\right) \right| \) , which equals \( \parallel S\parallel \) by Proposition 3.5 on page 114 (whose proof did not use the completeness of \( E \) ). We can assume \( S \neq 0 \) (else the result is trivial), so \( \lambda \) is nonzero and must be an eigenvalue, by Theorem 1.8 on page 219.
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Yes
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Theorem 2.2 Let \( \Lambda \) be the set of eigenvalues of \( T \) . Write \( {\Lambda }^{ * } = \Lambda \smallsetminus \{ 0\} \) and, for each eigenvalue \( \lambda \), let \( {E}_{\lambda } \) be the eigenspace of \( T \) associated with \( \lambda \) .\n\n- \( \Lambda \) is a countable, infinite, bounded subset of \( \mathbb{R} \) whose only cluster point is 0 .\n\n- The eigenspace associated with any nonzero eigenvalue of \( T \) has finite dimension.\n\n– Eigenspaces of T associated with distinct eigenvalues are orthogonal.\n\n- For each nonzero eigenvalue \( \lambda \) of \( T \), let \( {P}_{\lambda } \) be the orthogonal projection operator onto \( {E}_{\lambda } \) . Then\n\n\[ T = \mathop{\sum }\limits_{{\lambda \in {\Lambda }^{ * }}}\lambda {P}_{\lambda } \]\n\nin the sense of a summable family in \( L\left( E\right) \) .
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## Proof\n\n1. That all eigenvalues are real and that eigenspaces associated with distinct eigenvalues are orthogonal comes from parts i and iii of Proposition 2.5 on page 203, whose proof did not use the completeness of \( E \) . That eigenspaces associated with nonzero eigenvalues are finite-dimensional comes from Theorem 1.8 on page 219.\n\n2. We prove that \( {\Lambda }^{ * } \) is infinite. By Lemma 2.1, there exists an eigenvalue \( \lambda \) of \( T \) such that \( \left| \lambda \right| = \parallel T\parallel \) . Since \( T \) is nonzero (recall that \( T \) has infinite rank), we deduce that \( \lambda \neq 0 \) and so that \( {\Lambda }^{ * } \) is nonempty. Suppose that \( T \) has finitely many nonzero eigenvalues: \( {\Lambda }^{ * } = \left\{ {{\lambda }_{1},\ldots ,{\lambda }_{k}}\right\} \) . Set \( G = {\bigoplus }_{j = 1}^{k}{E}_{{\lambda }_{j}} \) and \( F = {G}^{ \bot } \) . Since \( G \) is finite-dimensional, \( E = F \oplus G \) (once more by the remark following Corollary 2.4 on page 107). It is clear that \( T\left( G\right) \subset G \) . Since \( T \) is selfadjoint, we quickly deduce that \( T\left( F\right) \subset F \) . The operator \( T \) therefore induces an operator \( {T}_{F} \) from \( F \) to itself, and we easily check that \( {T}_{F} \) is compact, because \( F \) is closed. Naturally, \( {T}_{F} \) is a selfadjoint operator on \( F \), and it is nonzero \( \left( {{T}_{F} = 0}\right. \) would imply im \( T \subset G \), contradicting the fact that \( T \) has infinite rank). By Lemma 2.1, \( {T}_{F} \) has a nonzero eigenvalue \( \lambda \) . We see then that \( \lambda \) is a nonzero eigenvalue of \( T \) distinct from all the \( {\lambda }_{j} \), for \( 1 \leq j \leq k \), since one of its associated eigenvectors lies in \( F \) and thus not in \( G \) . This is a contradiction. It follows that \( {\Lambda }^{ * } \) is infinite and, by Theorem 1.8 on page \( {219},\Lambda \) is countable and has 0 as its only cluster point.\n\n3. Let \( J \) be a finite subset of \( {\Lambda }^{ * } \) and put \( {G}_{J} = {\bigoplus }_{\lambda \in J}{E}_{\lambda } \) and \( {F}_{J} = \) \( {G}_{J}^{ \bot } \) . Arguing as above and using Lemma 2.1, we see that \( T \) induces on \( {F}_{J} \) a compact selfadjoint operator \( {T}_{{F}_{J}} \) whose norm equals \( \begin{Vmatrix}{T}_{{F}_{J}}\end{Vmatrix} = \) \( \mathop{\max }\limits_{{\lambda \in \operatorname{ev}\left( {T}_{{F}_{J}}\right) }}\left| \lambda \right| \) . Now observe that, as before, every eigenvalue \( \lambda \) of \( {T}_{{F}_{J}} \) is an eigenvalue of \( T \) (this is clear) but does not belong to \( J \) , since, by construction, \( {F}_{J} \) intersects trivially all the eigenspaces \( {E}_{\mu } \) , for \( \mu \in J \) . Therefore \( \operatorname{ev}\left( {T}_{{F}_{J}}\right) \subset \Lambda \small
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Yes
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Corollary 2.3 In the notation of Theorem 2.2,\n\n\\[ \n\\overline{\\operatorname{im}T} = \\overline{{\\bigoplus }_{\\lambda \\in {\\Lambda }^{ * }}{E}_{\\lambda }} \n\\]
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Proof. We know that \\( {Tx} = \\mathop{\\sum }\\limits_{{\\lambda \\in {\\Lambda }^{ * }}}\\lambda {P}_{\\lambda }x \\) for every \\( x \\in E \\) . It follows that\n\n\\[ \n\\operatorname{im}T \\subset \\overline{{\\bigoplus }_{\\lambda \\in {\\Lambda }^{ * }}{E}_{\\lambda }},\\;\\text{ and hence }\\;\\overline{\\operatorname{im}T} \\subset \\overline{{\\bigoplus }_{\\lambda \\in {\\Lambda }^{ * }}{E}_{\\lambda }}.\n\\]\n\nOn the other hand, if \\( \\lambda \\in {\\Lambda }^{ * } \\), we clearly have \\( {E}_{\\lambda } \\subset \\operatorname{im}T \\), proving the reverse inclusion.
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Yes
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Corollary 2.6 Suppose that \( E \) is complete. Let \( {P}_{0} \) be the operator of orthogonal projection onto \( {E}_{0} = \ker T \) . Then\n\n\[ x = \mathop{\sum }\limits_{{\lambda \in \Lambda }}{P}_{\lambda }x\;\text{ for all }x \in E \]\n\nand\n\n\[ E = \overline{{\bigoplus }_{\lambda \in \Lambda }{E}_{\lambda }} \]
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Proof. Since \( T \) is selfadjoint, we have \( {E}_{0} = \ker T = {\overline{\operatorname{im}T}}^{ \bot } \) . Therefore, if \( E \) is complete, \( E = {E}_{0} \oplus \overline{\operatorname{im}T} \) by Corollary 2.4 on page 107 .\n\nIf, moreover, \( E \) is separable, so is \( \ker T \) . Thus \( \ker T \) has a countable Hilbert basis, by Corollary 4.7 on page 129. Taking the union of such a basis with the Hilbert basis of \( \overline{\operatorname{im}T} \) given by Corollary 2.4, we obtain the following diagonalization result:\n\nCorollary 2.7 If \( E \) is a separable Hilbert space, it has a Hilbert basis consisting of eigenvectors of \( T \) .\n\nThis is still true if \( E \) is an arbitrary Hilbert space, but then we have to use the axiom of choice in order to guarantee the existence of a Hilbert basis for \( \ker T \) and so for \( E \) (see Exercise 11 on page 133).
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Yes
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Corollary 2.6 Suppose that \( E \) is complete. Let \( {P}_{0} \) be the operator of orthogonal projection onto \( {E}_{0} = \ker T \) . Then\n\n\[ x = \mathop{\sum }\limits_{{\lambda \in \Lambda }}{P}_{\lambda }x\;\text{ for all }x \in E \]\n\nand\n\n\[ E = \overline{{\bigoplus }_{\lambda \in \Lambda }{E}_{\lambda }} \]
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Proof. Since \( T \) is selfadjoint, we have \( {E}_{0} = \ker T = {\overline{\operatorname{im}T}}^{ \bot } \) . Therefore, if \( E \) is complete, \( E = {E}_{0} \oplus \overline{\operatorname{im}T} \) by Corollary 2.4 on page 107 .
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No
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Proposition 2.8 With the notation and hypotheses above,\n\n\[ \iint {\left| K\left( x, y\right) \right| }^{2}{dm}\left( x\right) {dm}\left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}^{2} = \mathop{\sum }\limits_{\substack{{\lambda \in \mathrm{{ev}}\left( T\right) } \\ {\lambda \neq 0} }}{d}_{\lambda }{\lambda }^{2}. \]
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Proof. Take \( u \in \ker T \) . For almost every \( y \), the function \( {K}_{y} : x \mapsto K\left( {x, y}\right) \) lies in \( E \) and\n\n\[ \left( {{K}_{y} \mid u}\right) = \int K\left( {x, y}\right) \bar{u}\left( x\right) {dm}\left( x\right) = \overline{Tu}\left( y\right) = 0. \]\n\nThe second of these equalities is true for almost every \( y \) : more precisely, for every \( y \) not in a subset \( {A}_{u} \) of \( X \) of measure zero, and which a priori may depend on \( u \) . But, since \( E \) is separable, \( \ker T \) is also separable. Let \( {\left( {u}_{n}\right) }_{n \in \mathbb{N}} \) be a dense subset of \( \ker T \) . Then, for every \( y \) not belonging to the set \( A = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{{u}_{n}} \) of measure zero, we have \( \left( {{K}_{y} \mid {u}_{n}}\right) = 0 \) for every \( n \in \mathbb{N} \) and, because of denseness, \( \left( {{K}_{y} \mid u}\right) = 0 \) for every \( u \in \ker T \) . It follows that \( {K}_{y} \in {\left( \ker T\right) }^{ \bot } = \overline{\operatorname{im}T} \) for almost every \( y \) . At the same time, for each \( n \in \mathbb{N} \),\n\n\[ \left( {{K}_{y} \mid {f}_{n}}\right) = \overline{T{f}_{n}}\left( y\right) = \overline{{\mu }_{n}{f}_{n}}\left( y\right) = {\mu }_{n}\overline{{f}_{n}}\left( y\right) \]\n\n\( \left( *\right) \)\n\nfor almost every \( y \) . We then deduce from the Bessel equality that, for almost every \( y \),\n\n\[ \int {\left| K\left( x, y\right) \right| }^{2}{dm}\left( x\right) = {\begin{Vmatrix}{K}_{y}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}^{2}{\left| {f}_{n}\left( y\right) \right| }^{2}. \]\n\n\( \left( {* * }\right) \)\n\nNow just integrate with respect to \( y \) to obtain the desired result.
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Yes
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Proposition 2.9 We have\n\n\[ K\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}{f}_{n}\left( x\right) \overline{{f}_{n}}\left( y\right) \]\n\nthe series being convergent in \( {L}^{2}\left( {m \times m}\right) \) .
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Proof. Set \( {K}_{N}\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 0}}^{N}{\mu }_{n}{f}_{n}\left( x\right) \overline{{f}_{n}}\left( y\right) \) . By equality (*) above, for almost every \( y \), we have\n\n\[ {K}_{y} = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}\overline{{f}_{n}}\left( y\right) {f}_{n} \]\n\nin the sense of convergence in \( {L}^{2}\left( m\right) \) . Thus, still by Bessel’s equality,\n\n\[ \int {\left| K\left( x, y\right) - {K}_{N}\left( x, y\right) \right| }^{2}{dm}\left( x\right) = {\begin{Vmatrix}{\left( {K}_{N}\right) }_{y} - {K}_{y}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{n = N + 1}}^{{+\infty }}{\mu }_{n}^{2}{\left| {f}_{n}\left( y\right) \right| }^{2}. \]\n\nWe deduce, integrating this equality with respect to \( y \), that\n\n\[ {\begin{Vmatrix}K - {K}_{N}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{n = N + 1}}^{{+\infty }}{\mu }_{n}^{2} \]\n\nwhere \( \parallel \cdot \parallel \) represents the norm in \( {L}^{2}\left( {m \times m}\right) \) . This proves the result.
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Yes
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Proposition 2.10 Suppose that \( \Phi : x \mapsto \int {\left| K\left( x, y\right) \right| }^{2}{dm}\left( y\right) \) belongs to \( {L}^{\infty }\left( m\right) \) . Then, for every \( n \in \mathbb{N} \), we have \( {f}_{n} \in {L}^{\infty }\left( m\right) \) and \[ f = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}\left( {f \mid {f}_{n}}\right) {f}_{n}\;\text{ for every }f \in \operatorname{im}T \] the convergence of the series taking place in \( {L}^{\infty }\left( m\right) \) .
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Proof. For every \( n \in \mathbb{N} \) we have \[ {f}_{n}\left( x\right) = \frac{1}{{\mu }_{n}}\int K\left( {x, y}\right) {f}_{n}\left( y\right) {dm}\left( y\right) . \] Therefore \( {f}_{n} \in {L}^{\infty }\left( m\right) \) and \( {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{\infty } \leq {\mu }_{n}^{-1}\sqrt{L} \), where \( L = \parallel \Phi {\parallel }_{\infty } \) . We show that the series \( \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}\left( {f \mid {f}_{n}}\right) {f}_{n} \) satisfies Cauchy’s criterion on \( {L}^{\infty }\left( m\right) \) . Let \( f = {Tg} \) be an element of \( \operatorname{im}T \) . For every \( n \in \mathbb{N},\left( {f \mid {f}_{n}}\right) = \left( {{Tg} \mid {f}_{n}}\right) = \left( {g \mid T{f}_{n}}\right) = {\mu }_{n}\left( {g \mid {f}_{n}}\right) \) . If \( k \leq l \), \[ \left| {\mathop{\sum }\limits_{{n = k}}^{l}\left( {f \mid {f}_{n}}\right) {f}_{n}\left( x\right) }\right| = \left| {\mathop{\sum }\limits_{{n = k}}^{l}{\mu }_{n}\left( {g \mid {f}_{n}}\right) {f}_{n}\left( x\right) }\right| \] \[ \leq {\left( \mathop{\sum }\limits_{{n = k}}^{l}{\left| \left( g|{f}_{n}\right) \right| }^{2}\right) }^{1/2}{\left( \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}^{2}{\left| {f}_{n}\left( x\right) \right| }^{2}\right) }^{1/2}, \] by the Schwarz inequality. Now, by an earlier calculation (see equality \( \left( {* * }\right) \) on page 240), we have \[ \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}^{2}{\left| {f}_{n}\left( x\right) \right| }^{2} = \int {\left| K\left( x, y\right) \right| }^{2}{dm}\left( y\right) \leq L \] which finally implies that \[ {\begin{Vmatrix}\mathop{\sum }\limits_{{n = k}}^{l}\left( f \mid {f}_{n}\right) {f}_{n}\end{Vmatrix}}_{\infty } \leq {\left( \mathop{\sum }\limits_{{n = k}}^{l}{\left| \left( g \mid {f}_{n}\right) \right| }^{2}\right) }^{1/2}{L}^{1/2} \] which proves the result, since the series \( \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\left| \left( g \mid {f}_{n}\right) \right| }^{2} \) converges by the Bessel inequality and so satisfies Cauchy's criterion.
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Yes
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Proposition 1.1 (Leibniz’s formula) Suppose \( f, g \in {\mathcal{E}}^{m}\left( \Omega \right) \) . For each multiindex \( p \) such that \( \left| p\right| \leq m \) ,
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\[ {D}^{p}\left( {fg}\right) = \mathop{\sum }\limits_{{q \leq p}}\left( \begin{array}{l} p \\ q \end{array}\right) {D}^{p - q}f{D}^{q}g. \]
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Yes
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Proposition 1.2 Assume \( \varphi \in {\mathcal{D}}^{m} \), for some \( m \in \mathbb{N} \) . For every integer \( n \geq 1 \), the convolution \( \varphi * {\chi }_{n} \) belongs to \( \mathcal{D} \) and \[ \mathop{\lim }\limits_{{n \rightarrow + \infty }}\varphi * {\chi }_{n} = \varphi \;\text{ in }{\mathcal{D}}^{m} \]
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Proof. Since the functions \( \varphi \) and \( {\chi }_{n} \) have compact support, so does \( \varphi * {\chi }_{n} \) . More precisely, \[ \operatorname{Supp}\left( {\varphi * {\chi }_{n}}\right) \subset \operatorname{Supp}\varphi + \operatorname{Supp}{\chi }_{n} \subset \operatorname{Supp}\varphi + \bar{B}\left( {0,1/n}\right) \subset \operatorname{Supp}\varphi + \bar{B}\left( {0,1}\right) . \] At the same time, a classical theorem about differentiation under the integral sign easily implies, on the one hand, that \( \varphi * {\chi }_{n} \) is of class \( {C}^{\infty } \) and so \( \varphi * {\chi }_{n} \in \mathcal{D} \), and, on the other, that \( {D}^{p}\left( {\varphi * {\chi }_{n}}\right) = \left( {{D}^{p}\varphi }\right) * {\chi }_{n} \) for \( \left| p\right| \leq m \) . Now, since the support of \( {\chi }_{n} \) is contained in \( \bar{B}\left( {0,1/n}\right) \) and \( \int {\chi }_{n}\left( y\right) {dy} = 1 \) , we get \[ \left( {{D}^{p}\varphi }\right) * {\chi }_{n}\left( x\right) - \left( {{D}^{p}\varphi }\right) \left( x\right) = {\int }_{\left| y\right| \leq 1/n}\left( {{D}^{p}\varphi \left( {x - y}\right) - {D}^{p}\varphi \left( x\right) }\right) {\chi }_{n}\left( y\right) {dy} \] and \[ \mathop{\sup }\limits_{{x \in {\mathbb{R}}^{d}}}\left| {\left( {{D}^{p}\varphi }\right) * {\chi }_{n}\left( x\right) - \left( {{D}^{p}\varphi }\right) \left( x\right) }\right| \leq \mathop{\sup }\limits_{\substack{{x, z \in {\mathbb{R}}^{d}} \\ {\left| {z - x}\right| \leq 1/n} }}\left| {{D}^{p}\varphi \left( z\right) - {D}^{p}\varphi \left( x\right) }\right| . \] Since \( {D}^{p}\varphi \) is uniformly continuous (being continuous and having compact support), we deduce that the sequence \( {\left( {D}^{p}\left( \varphi * {\chi }_{n}\right) \right) }_{n \in \mathbb{N}} \) converges uniformly to \( {D}^{p}\varphi \) .
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Yes
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Corollary 1.3 For every \( n \in \mathbb{N} \), the space \( \mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{D}}^{m}\left( \Omega \right) \) . In particular, \( \mathcal{D}\left( \Omega \right) \) is dense in \( {C}_{c}\left( \Omega \right) \) .
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Proof. If \( \varphi \in {\mathcal{D}}^{m}\left( \Omega \right) \), we can consider \( \varphi \) as an element of \( {\mathcal{D}}^{m} \) (by extending it with the value 0 on \( {\mathbb{R}}^{d} \smallsetminus \Omega \) ). Now\n\n\[ \operatorname{Supp}\left( {\varphi * {\chi }_{n}}\right) \subset \operatorname{Supp}\varphi + \bar{B}\left( {0,1/n}\right) \]\n\ntherefore \( \operatorname{Supp}\left( {\varphi * {\chi }_{n}}\right) \subset \Omega \) for \( n \) large enough \( - \) say \( n > 1/d\left( {\operatorname{Supp}\varphi ,{\mathbb{R}}^{d} \smallsetminus \Omega }\right) \) . Then, by the preceding proposition, \( \varphi * {\chi }_{n} \) belongs to \( \mathcal{D}\left( \Omega \right) \) for \( n \) large enough, and \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}\varphi * {\chi }_{n} = \varphi \) in \( {\mathcal{D}}^{m}\left( \Omega \right) \) .
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Yes
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Proposition 1.4 If \( K \) is a compact subset of \( {\mathbb{R}}^{d} \) and \( {O}_{1},\ldots ,{O}_{n} \) are open sets in \( {\mathbb{R}}^{d} \) such that \( K \subset \mathop{\bigcup }\limits_{{j = 1}}^{n}{O}_{j} \), there exist functions \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) in \( \mathcal{D} \) such that\n\n\[ 0 \leq {\varphi }_{j} \leq 1\;\text{ and }\;\operatorname{Supp}{\varphi }_{j} \subset {O}_{j}\;\text{ for }j \in \{ 1,\ldots, n\} \]\n\nand such that \( \mathop{\sum }\limits_{{j = 1}}^{n}{\varphi }_{j}\left( x\right) = 1 \) for every \( x \in K \) .
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Proof. Set \( d = d\left( {K,{\mathbb{R}}^{d} \smallsetminus O}\right) \), with \( O = \mathop{\bigcup }\limits_{{j = 1}}^{n}{O}_{j} \) (the metric being the canonical euclidean metric in \( \left. {\mathbb{R}}^{d}\right) \) . Set \( {K}^{\prime } = \{ x : d\left( {x, K}\right) \leq d/2\} \) . The set \( {K}^{\prime } \) is compact and, since \( d > 0 \), \n\n\[ {\mathring{K}}^{\prime } \supset \{ x : d\left( {x, K}\right) < d/2\} \supset K. \]\n\nThus \( K \subset {\mathring{K}}^{\prime } \subset {K}^{\prime } \subset O \) . By Proposition 1.8 on page 53, there exist functions \( {h}_{1},\ldots ,{h}_{n} \) in \( {C}_{c} \) such that\n\n\[ 0 \leq {h}_{j} \leq 1\;\text{ and }\;\operatorname{Supp}{h}_{j} \subset {O}_{j}\;\text{ for }j \in \{ 1,\ldots, n\} ,\]\n\nand such that \( \mathop{\sum }\limits_{{j = 1}}^{n}{h}_{j}\left( x\right) = 1 \) for every \( x \in {K}^{\prime } \) . Define \( \delta = d\left( {K,{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) \), \( {\eta }_{j} = d\left( {\operatorname{Supp}{h}_{j},{\mathbb{R}}^{d} \smallsetminus {O}_{j}}\right) \) for \( 1 \leq j \leq n \), and\n\n\[ \varepsilon = \frac{1}{2}\min \left( {\delta ,{\eta }_{1},\ldots ,{\eta }_{n}}\right) \]\n\nLet \( \chi \) be the function defined on page 261 and let \( u \) be defined by\n\n\[ u\left( x\right) = {\varepsilon }^{-d}\chi \left( {x/\varepsilon }\right) \]\n\nThen \( u \in \mathcal{D}, u \geq 0,\int u\left( x\right) {dx} = 1 \), and \( \operatorname{Supp}u = \bar{B}\left( {0,\varepsilon }\right) \) .\n\nFor \( 1 \leq j \leq n \), set \( {\varphi }_{j} = {h}_{j} * u \) . Then \( {\varphi }_{j} \) is of class \( {C}^{\infty } \) (this follows immediately from the theorem on differentiation under the integral sign) and\n\n\[ \operatorname{Supp}{\varphi }_{j} \subset \operatorname{Supp}{h}_{j} + \bar{B}\left( {0,\varepsilon }\right) \subset {O}_{j}. \]\n\nIn particular, \( {\varphi }_{j} \in \mathcal{D} \) . Moreover, \( 0 \leq {\varphi }_{j} \leq 1 \) . Finally, if \( x \in K \) and \( y \in \bar{B}\left( {0,\varepsilon }\right) \), we have \( x - y \in {K}^{\prime } \) and so\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{h}_{j}\left( {x - y}\right) u\left( y\right) = u\left( y\right) \]\n\nIntegrating we obtain\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{\varphi }_{j}\left( x\right) = \int u\left( y\right) {dy} = 1\;\text{ for all }x \in K. \]
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Yes
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Proposition 1.5 The space \( \mathcal{D}\left( \Omega \right) \) is dense in \( \mathcal{E}\left( \Omega \right) \) and in \( {\mathcal{E}}^{m}\left( \Omega \right) \), for every \( m \in \mathbb{N} \) .
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Proof. Let \( {\left( {K}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence of compact subsets of \( \Omega \) exhausting \( \Omega \) . By the previous proposition, there exists, for every integer \( n \in \mathbb{N} \), an element \( {\varphi }_{n} \in \mathcal{D}\left( \Omega \right) \) such that\n\n\[ 0 \leq {\varphi }_{n} \leq 1,\;{\varphi }_{n} = 1\text{ on }{K}_{n},\;\operatorname{Supp}{\varphi }_{n} \subset {K}_{n + 1}. \]\n\nIf \( f \in \mathcal{E}\left( \Omega \right) \), we have \( f{\varphi }_{n} \in \mathcal{D}\left( \Omega \right) \) for every \( n \in \mathbb{N} \) . If \( K \) is a compact subset of \( \Omega \), there exists \( N \in \mathbb{N} \) such that \( K \subset {\mathring{K}}_{N} \) (see Proposition 1.6 on page 52); thus, for every \( n \geq N \) and every \( p \in {\mathbb{N}}^{d} \), we have \( {D}^{p}\left( {f{\varphi }_{n}}\right) = {D}^{p}f \) on \( K \) . By the definition of convergence in \( \mathcal{E}\left( \Omega \right) \), we deduce that \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}\left( {f{\varphi }_{n}}\right) = f \) in \( \mathcal{E}\left( \Omega \right) \) .\n\nUsing the same reasoning, one shows that \( {\mathcal{D}}^{m}\left( \Omega \right) \) is dense in \( {\mathcal{E}}^{m}\left( \Omega \right) \) . Moreover, as we saw in Corollary \( {1.3},\mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{D}}^{m}\left( \Omega \right) \) . Thus every element of \( {\mathcal{D}}^{m}\left( \Omega \right) \) is the limit of a sequence of elements of \( \mathcal{D}\left( \Omega \right) \) in the sense of convergence in \( {\mathcal{E}}^{m}\left( \Omega \right) \) (since the canonical injection from \( {\mathcal{D}}^{m}\left( \Omega \right) \) into \( {\mathcal{E}}^{m}\left( \Omega \right) \) is continuous: see page 260). This implies, finally, that \( \mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{E}}^{m}\left( \Omega \right) \) (because \( {\mathcal{E}}^{m}\left( \Omega \right) \) is a metric space: see Exercise 7 below).
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Yes
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Proposition 2.1 Let \( T \) be a linear form on \( \mathcal{D}\left( \Omega \right) \). Then \( T \) is a distribution on \( \Omega \) if and only if, for every compact \( K \) in \( \Omega \), there exist \( m \in \mathbb{N} \) and \( C \geq 0 \) such that \[ \left| {T\left( \varphi \right) }\right| \leq C\parallel \varphi {\parallel }^{\left( m\right) }\;\text{ for all }\varphi \in {\mathcal{D}}_{K}\left( \Omega \right) . \]
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Proof. The \
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No
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Proposition 2.2 Two locally integrable functions on \( \Omega \) define the same distribution if and only if they coincide almost everywhere.
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Proof. Take \( f \in {L}_{\text{loc }}^{1}\left( \Omega \right) \) such that \( \left\lbrack f\right\rbrack = 0 \) . Because \( \mathcal{D}\left( \Omega \right) \) is dense in \( {C}_{c}\left( \Omega \right) = {\mathcal{D}}^{0}\left( \Omega \right) \) (Corollary 1.3), we see that\n\n\[ \n{\int }_{\Omega }g\left( x\right) f\left( x\right) {dx} = 0\;\text{ for all }g \in {C}_{c}\left( \Omega \right) .\n\]\n\nThus, for every \( g \in {C}_{c}^{\mathbb{R}}\left( \Omega \right) \) ,\n\n\[ \n{\int }_{\Omega }g\left( x\right) {\left( \operatorname{Re}f\left( x\right) \right) }^{ + }{dx} = {\int }_{\Omega }g\left( x\right) {\left( \operatorname{Re}f\left( x\right) \right) }^{ - }{dx}\n\]\n\n\[ \n{\int }_{\Omega }g\left( x\right) {\left( \operatorname{Im}f\left( x\right) \right) }^{ + }{dx} = {\int }_{\Omega }g\left( x\right) {\left( \operatorname{Im}f\left( x\right) \right) }^{ - }{dx}.\n\]\n\nBy the uniqueness part of the Radon-Riesz Theorem (page 69), these equalities are valid for any positive Borel function \( g \) . Applying them to the characteristic functions of the sets \( \{ \operatorname{Re}f > 0\} ,\{ \operatorname{Re}f < 0\} ,\{ \operatorname{Im}f > 0\} \) and \( \{ \operatorname{Im}f < 0\} \), we deduce that \( f = 0 \) almost everywhere.\n\nThus, the map that associates to each \( f \in {L}_{\text{loc }}^{1}\left( \Omega \right) \) the distribution \( \left\lbrack f\right\rbrack \in \) \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) is injective. By identifying \( f \) with \( \left\lbrack f\right\rbrack \), we can write \( {L}_{\mathrm{{loc}}}^{1}\left( \Omega \right) \subset {\mathcal{D}}^{\prime }\left( \Omega \right) \) . It is in this sense that distributions are \
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Yes
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Proposition 2.3 Every positive distribution has order 0.
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Proof. Let \( T \) be a positive distribution on \( \Omega \) . Let \( K \) be a compact subset of \( \Omega \) and let \( \rho \in \mathcal{D}\left( \Omega \right) \) be such that \( 0 \leq \rho \leq 1 \) and \( \rho = 1 \) on \( K \) . For every \( \varphi \in {\mathcal{D}}_{K}\left( \Omega \right) \), we have \( \left| \varphi \right| = \left| {\varphi \rho }\right| \leq \parallel \varphi \parallel \rho \), where \( \parallel \varphi \parallel \) denotes the uniform norm of \( \varphi \) . If \( \varphi \) is real-valued, this means that\n\n\[ - \parallel \varphi \parallel \rho \leq \varphi \leq \parallel \varphi \parallel \rho \]\n\nWe then deduce from the linearity and the positivity of \( T \) that \( \left| {T\left( \varphi \right) }\right| \leq \) \( \parallel \varphi \parallel T\left( \rho \right) \) . When we no longer assume \( \varphi \) to be real-valued, the decomposition \( \varphi = \operatorname{Re}\varphi + i\operatorname{Im}\varphi \) leads to the inequality \( \left| {T\left( \varphi \right) }\right| \leq {2T}\left( \rho \right) \parallel \varphi \parallel \), which proves that \( T \) has order 0 .
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Yes
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For every \( \varphi \in \mathcal{D}\left( \mathbb{R}\right) \), the limit\n\n\[ \langle T,\varphi \rangle = \mathop{\lim }\limits_{{\varepsilon \rightarrow {0}^{ + }}}\left( {{\int }_{\varepsilon }^{+\infty }\frac{\varphi \left( x\right) }{x}{dx} + \varphi \left( 0\right) \log \varepsilon }\right) \]\n\nexists. The linear form \( T \) thus defined is a distribution of order 1 on \( \mathbb{R} \), called the finite part of \( Y\left( x\right) /x \) and denoted by \( \mathrm{{fp}}\left( {Y\left( x\right) /x}\right) \).
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Proof. Take \( \varphi \in \mathcal{D}\left( \mathbb{R}\right) \) and \( A > 0 \) such that \( \operatorname{Supp}\varphi \subset \left\lbrack {-A, A}\right\rbrack \). Then\n\n\[ {\int }_{\varepsilon }^{+\infty }\frac{\varphi \left( x\right) }{x}{dx} = {\int }_{\varepsilon }^{A}\frac{\varphi \left( x\right) - \varphi \left( 0\right) }{x}{dx} + \varphi \left( 0\right) \log A - \varphi \left( 0\right) \log \varepsilon . \]\n\nThus\n\n\[ \mathop{\lim }\limits_{{\varepsilon \rightarrow {0}^{ + }}}\left( {{\int }_{\varepsilon }^{+\infty }\frac{\varphi \left( x\right) }{x}{dx} + \varphi \left( 0\right) \log \varepsilon }\right) = {\int }_{0}^{A}\frac{\varphi \left( x\right) - \varphi \left( 0\right) }{x}{dx} + \varphi \left( 0\right) \log A, \]\n\nand this expression is bounded in absolute value by \( \parallel \varphi {\parallel }^{\left( 1\right) }\max \left( {A,\left| {\log A}\right| }\right) \), by the Mean Value Theorem. It follows that \( \operatorname{fp}\left( {Y\left( x\right) /x}\right) \) is indeed a distribution of order at most 1.
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Yes
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For every \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{2}\right) \), the limit\n\n\[ \langle T,\varphi \rangle = \mathop{\lim }\limits_{{\varepsilon \rightarrow {0}^{ + }}}\left( {{\iint }_{\{ r \geq \varepsilon \} }{r}^{-4}{\varphi dxdy} - {\pi \varphi }\left( {0,0}\right) {\varepsilon }^{-2} + \frac{\pi }{2}{\Delta \varphi }\left( {0,0}\right) \log \varepsilon }\right) \]\n\nexists. The linear form \( T \) thus defined is a distribution of order 3 on \( {\mathbb{R}}^{2} \).
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Summary of proof. Take \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{2}\right) \) and \( A > 0 \) such that \( \operatorname{Supp}\varphi \subset \) \( \bar{B}\left( {0, A}\right) \) . A quick calculation shows that\n\n\[ {\iint }_{\{ r \geq \varepsilon \} }\frac{\varphi }{{r}^{4}}\;{dx}\;{dy} \]\n\n\[ = {\iint }_{\{ A \geq r \geq \varepsilon \} }{r}^{-4}\left( {\varphi \left( {x, y}\right) - \varphi \left( {0,0}\right) - x\frac{\partial \varphi }{\partial x}\left( {0,0}\right) - y\frac{\partial \varphi }{\partial y}\left( {0,0}\right) }\right. \]\n\n\[ - \left( {\frac{{x}^{2}}{2}\;\frac{{\partial }^{2}\varphi }{\partial {x}^{2}}(0,0) + {xy}\;\frac{{\partial }^{2}\varphi }{\partial x\partial y}(0,0) + \frac{{y}^{2}}{2}\;\frac{{\partial }^{2}\varphi }{\partial {y}^{2}}(0,0)}\right) )\;{dx}\;{dy} \]\n\n\[ - {\pi \varphi }\left( {0,0}\right) \left( {{A}^{-2} - {\varepsilon }^{-2}}\right) + \frac{\pi }{2}{\Delta \varphi }\left( {0,0}\right) \log \frac{A}{\varepsilon }. \]\n\nWe then deduce from Taylor's formula that the limit given in the statement of the proposition exists and is bounded in absolute value by \( {C}_{A}\parallel \varphi {\parallel }^{\left( 3\right) } \) , with \( {C}_{A} > 0 \) . Therefore the distribution thus defined has order at most 3 . It remains to show that it is not of order less than 3.
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Yes
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Proposition 3.1 Let \( T \) be a distribution on \( \Omega \) and suppose \( m \in \mathbb{N} \) . A necessary and sufficient condition for \( T \) to have order at most \( m \) is that \( T \) can be extended to a continuous linear form on \( {\mathcal{D}}^{m}\left( \Omega \right) \) . The extension is then unique.
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Proof. Suppose that \( T \) has order at most \( m \) . Property \( \left( *\right) \) on page 268 then implies that \( T \) is continuous (and even uniformly continuous) on the space \( \mathcal{D}\left( \Omega \right) \) regarded, topologically speaking, as a subspace of \( {\mathcal{D}}^{m}\left( \Omega \right) \) . Since \( \mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{D}}^{m}\left( \Omega \right) \) by Corollary 1.3, we can apply the theorem of extension of continuous linear forms. This theorem applies a priori to continuous linear forms on normed spaces, but we can reduce the problem to that situation by considering the normed spaces \( {\mathcal{D}}_{K}^{m}\left( \Omega \right) \) . Similarly, since \( \mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{D}}^{m}\left( \Omega \right) \), this extension is unique.\n\nIn the other direction, it is clear from the definitions that the restriction of a continuous linear form on \( {\mathcal{D}}^{m}\left( \Omega \right) \) to \( \mathcal{D}\left( \Omega \right) \) is a distribution of order at most \( m \) .
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Yes
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Proposition 3.4 Every distribution \( T \) with compact support in \( \Omega \) has finite order. More precisely, there exists an integer \( m \in \mathbb{N} \) and a constant \( C \geq 0 \) such that\n\n\[ \n\\left| {\\langle T,\\varphi \\rangle }\\right| \\leq C\\parallel \\varphi {\\parallel }^{\\left( m\\right) }\\;\\text{ for all }\\varphi \\in \\mathcal{D}\\left( \\Omega \\right) .\n\]
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Proof. Let \( K \) be the support of \( T \) and let \( {K}^{\\prime },{K}^{\\prime \\prime } \) be compact sets such that\n\n\[ \nK \\subset {\\mathring{K}}^{\\prime } \\subset {K}^{\\prime } \\subset {\\mathring{K}}^{\\prime \\prime } \\subset {K}^{\\prime \\prime } \\subset \\Omega .\n\]\n\nBy Proposition 2.1, there exists a constant \( C \\geq 0 \) and an integer \( m \\in \\mathbb{N} \) such that\n\n\[ \n\\left| {\\langle T,\\varphi \\rangle }\\right| \\leq C\\parallel \\varphi {\\parallel }^{\\left( m\\right) }\\;\\text{ for all }\\varphi \\in {\\mathcal{D}}_{{K}^{\\prime \\prime }}\\left( \\Omega \\right) .\n\]\n\nBy Proposition 1.4, there exists \( \\psi \\in \\mathcal{D} \) such that \( 0 \\leq \\psi \\leq 1,\\psi = 1 \) on \( {K}^{\\prime } \) and \( \\operatorname{Supp}\\psi \\subset {\\mathring{K}}^{\\prime \\prime } \) . If \( \\varphi \\in \\mathcal{D}\\left( \\Omega \\right) \), then \( {\\varphi \\psi } \\in {\\mathcal{D}}_{{K}^{\\prime \\prime }}\\left( \\Omega \\right) \) and\n\n\[ \n\\operatorname{Supp}\\left( {\\varphi - {\\varphi \\psi }}\\right) \\subset \\Omega \\smallsetminus {\\mathring{K}}^{\\prime } \\subset \\Omega \\smallsetminus K\n\]\n\nSince the compact \( K \) is the support of \( T \), it follows that there is a positive constant \( {C}^{\\prime } \) depending only on \( C, m \) and \( \\psi \), and such that\n\n\[ \n\\left| {\\langle T,\\varphi \\rangle }\\right| = \\left| {\\langle T,{\\varphi \\psi }\\rangle }\\right| \\leq C\\parallel {\\varphi \\psi }{\\parallel }^{\\left( m\\right) } \\leq {C}^{\\prime }\\parallel \\varphi {\\parallel }^{\\left( m\\right) },\n\]\n\nthe last inequality being a consequence of Leibniz's formula.
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Yes
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Lemma 1.1 Suppose \( \alpha \in \mathcal{E}\left( \Omega \right) \) . The map \( \varphi \mapsto {\alpha \varphi } \) from \( \mathcal{D}\left( \Omega \right) \) to \( \mathcal{D}\left( \Omega \right) \) is continuous. Likewise, if \( \alpha \in {\mathcal{E}}^{m}\left( \Omega \right) \), with \( m \in \mathbb{N} \), the map \( \varphi \mapsto {\alpha \varphi } \) from \( {\mathcal{D}}^{m}\left( \Omega \right) \) to \( {\mathcal{D}}^{m}\left( \Omega \right) \) is continuous.
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In other words, if \( {\left( {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) is a sequence in \( \mathcal{D}\left( \Omega \right) \) or \( {\mathcal{D}}^{m}\left( \Omega \right) \) converging to 0 in \( \mathcal{D}\left( \Omega \right) \) or \( {\mathcal{D}}^{m}\left( \Omega \right) \), respectively, the same is true about the sequence \( {\left( \alpha {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) .\n\nProof. The lemma follows immediately from Leibniz's formula (page 258) and from the fact that, if \( {\varphi }_{n} \in \mathcal{D}\left( \Omega \right) \), the support of \( \alpha {\varphi }_{n} \) is contained in the support of \( {\varphi }_{n} \) .
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Yes
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Proposition 1.3 With the notation introduced in Definition 1.2, we have\n\n\[ \operatorname{Supp}\left( {\alpha T}\right) \subset \operatorname{Supp}\alpha \cap \operatorname{Supp}T \]
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Proof. The second claim is obvious. To show the first, take \( \varphi \in \mathcal{D}\left( \Omega \right) \) . If \( \operatorname{Supp}\varphi \subset \Omega \smallsetminus \operatorname{Supp}\alpha \), then \( {\alpha \varphi } = 0 \), so \( \langle {\alpha T},\varphi \rangle = 0 \) . It follows that \( \Omega \smallsetminus \operatorname{Supp}\alpha \) is contained in \( \Omega \smallsetminus \operatorname{Supp}\left( {\alpha T}\right) \), so \( \operatorname{Supp}\left( {\alpha T}\right) \subset \operatorname{Supp}\alpha \) .\n\nNow if \( \operatorname{Supp}\varphi \subset \Omega \smallsetminus \operatorname{Supp}T \), then\n\n\[ \operatorname{Supp}{\alpha \varphi } \subset \operatorname{Supp}\varphi \subset \Omega \smallsetminus \operatorname{Supp}T, \]\n\nwhich implies that \( \langle {\alpha T},\varphi \rangle = 0 \) . Therefore \( \Omega \smallsetminus \operatorname{Supp}T \) is contained in \( \Omega \smallsetminus \) \( \operatorname{Supp}\left( {\alpha T}\right) \), so \( \operatorname{Supp}\left( {\alpha T}\right) \subset \operatorname{Supp}T \) . The result follows.
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Yes
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Proposition 1.4 For every \( S \in {\mathcal{D}}^{\prime }\left( \mathbb{R}\right) \), there exists \( T \in {\mathcal{D}}^{\prime }\left( \mathbb{R}\right) \) such that \( {xT} = S \) . If \( {T}_{0} \) is such that \( x{T}_{0} = S \), the set of solutions of the equation \( {xT} = S \) equals \( \left\{ {{T}_{0} + {C\delta } : C \in \mathbb{C}}\right\} \) .
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Proof. Take \( \chi \in \mathcal{D}\left( \mathbb{R}\right) \) such that \( \chi \left( 0\right) = 1 \) . To each \( \varphi \in \mathcal{D}\left( \mathbb{R}\right) \) we associate \( \widetilde{\varphi } \), defined by\n\n\[ \widetilde{\varphi }\left( x\right) = {\int }_{0}^{1}\left( {{\varphi }^{\prime }\left( {tx}\right) - \varphi \left( 0\right) {\chi }^{\prime }\left( {tx}\right) }\right) {dt}. \]\n\nOne easily checks that \( \widetilde{\varphi } \in \mathcal{D}\left( \mathbb{R}\right) \) and that the map \( \varphi \mapsto \widetilde{\varphi } \) from \( \mathcal{D}\left( \mathbb{R}\right) \) to \( \mathcal{D}\left( \mathbb{R}\right) \) is continuous. Moreover, if \( x \in {\mathbb{R}}^{ * },\widetilde{\varphi }\left( x\right) = \left( {\varphi \left( x\right) - \varphi \left( 0\right) \chi \left( x\right) }\right) /x \) . Now put\n\n\[ \langle T,\varphi \rangle = \langle S,\widetilde{\varphi }\rangle \;\text{ for all }\varphi \in \mathcal{D}\left( \mathbb{R}\right) . \]\n\nSince \( \varphi \mapsto \widetilde{\varphi } \) is continuous, \( T \) belongs to \( {\mathcal{D}}^{\prime }\left( \mathbb{R}\right) \) ; since \( \widetilde{x\varphi } = \varphi \), we get \( {xT} = S \) .\n\nNow take \( T \in {\mathcal{D}}^{\prime }\left( \mathbb{R}\right) \) with \( {xT} = 0 \) . If \( \varphi \in \mathcal{D}\left( \mathbb{R}\right) \), we have\n\n\[ 0 = \langle {xT},\widetilde{\varphi }\rangle = \langle T,\varphi - \varphi \left( 0\right) \chi \rangle = \langle T,\varphi \rangle - \langle T,\chi \rangle \langle \delta ,\varphi \rangle . \]\n\nIt follows that \( T = \langle T,\chi \rangle \delta \) .
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Yes
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Proposition 1.5 Suppose \( T \in {\mathcal{D}}^{\prime }\left( \mathbb{R}\right) \) . Then \( {xT} = 1 \) if and only if there exists \( C \in \mathbb{C} \) such that \( T = \operatorname{pv}\left( {1/x}\right) + {C\delta } \) .
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Proof. By Proposition 1.4, it suffices to show that \( x\mathrm{{pv}}\left( {1/x}\right) = 1 \) . To do this, take \( \varphi \in \mathcal{D}\left( \mathbb{R}\right) \) . By definition,\n\n\[ \langle x\operatorname{pv}\left( {1/x}\right) ,\varphi \rangle = \langle \operatorname{pv}\left( {1/x}\right) ,{x\varphi }\rangle = \mathop{\lim }\limits_{{\varepsilon \rightarrow {0}^{ + }}}{\int }_{\{ \left| x\right| > \varepsilon \} }\left( {1/x}\right) {x\varphi }\left( x\right) {dx} \]\n\n\[ = \int \varphi \left( x\right) {dx} = \langle \left\lbrack 1\right\rbrack ,\varphi \rangle \]\n\n as we wished to show.
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Yes
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Proposition 2.1 Suppose \( m \in \mathbb{N} \) . For every \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) \), we have \( {D}^{p}T \in {\mathcal{D}}^{\prime m + \left| p\right| }\left( \Omega \right) \) and \[ \left\langle {{D}^{p}T,\varphi }\right\rangle = {\left( -1\right) }^{\left| p\right| }\left\langle {T,{D}^{p}\varphi }\right\rangle \;\text{ for all }\varphi \in {\mathcal{D}}^{m + \left| p\right| }\left( \Omega \right) .
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Proof. By Fubini’s Theorem, we can reduce to the case \( d = 1 \), to which we apply the classical theorem of integration by parts, taking into account that the support of \( \varphi \) is a compact subset of \( \Omega \), so that the \
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No
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Proposition 2.2 Let \( m \in \mathbb{N} \) and \( p \in {\mathbb{N}}^{d} \) satisfy \( \left| p\right| \leq m \) . If \( f \in {\mathcal{E}}^{m}\left( \Omega \right) \) , then\n\n\[ \n{D}^{p}\left( \left\lbrack f\right\rbrack \right) = \left\lbrack {{D}^{p}f}\right\rbrack \n\]\n\nIn this equality, the first \( {D}^{p} \) denotes differentiation in the sense of distributions as defined above, and the second denotes ordinary differentiation in the sense of functions.
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The proposition is easily obtained by induction on \( \left| p\right| \) starting from the case \( \left| p\right| = 1 \), which is a consequence of the following lemma.\n\nLemma 2.3 (Integration by parts) If \( f \in {\mathcal{E}}^{1}\left( \Omega \right) \) and \( \varphi \in {\mathcal{D}}^{1}\left( \Omega \right) \) , then, for every \( j \in \{ 1,\ldots, d\} \) ,\n\n\[ \n{\int }_{\Omega }{D}_{j}{f\varphi dx} = - {\int }_{\Omega }f{D}_{j}{\varphi dx} \n\]\n\nProof. By Fubini’s Theorem, we can reduce to the case \( d = 1 \), to which we apply the classical theorem of integration by parts, taking into account that the support of \( \varphi \) is a compact subset of \( \Omega \), so that the \
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Yes
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Lemma 2.3 (Integration by parts) If \( f \in {\mathcal{E}}^{1}\left( \Omega \right) \) and \( \varphi \in {\mathcal{D}}^{1}\left( \Omega \right) \) , then, for every \( j \in \{ 1,\ldots, d\} \) , \[ {\int }_{\Omega }{D}_{j}{f\varphi dx} = - {\int }_{\Omega }f{D}_{j}{\varphi dx} \]
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Proof. By Fubini’s Theorem, we can reduce to the case \( d = 1 \), to which we apply the classical theorem of integration by parts, taking into account that the support of \( \varphi \) is a compact subset of \( \Omega \), so that the \
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No
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Proposition 2.5 (Leibniz’s formula) Consider \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) ,\alpha \in \mathcal{E}\left( \Omega \right) \) , and \( p \in {\mathbb{N}}^{d}. \) Then\n\n\[ \n{D}^{p}\left( {\alpha T}\right) = \mathop{\sum }\limits_{{q \leq p}}\left( \begin{array}{l} p \\ q \end{array}\right) {D}^{p - q}\alpha {D}^{q}T.\n\]\n\nThis formula remains true for \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) \) and \( \alpha \in {\mathcal{E}}^{m + \left| p\right| }\left( \Omega \right) \) .
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Proof. This is obvious if \( \left| p\right| = 0 \) . Consider the case \( \left| p\right| = 1 \) . If \( j \in \{ 1,\ldots, d\} \) , we have\n\n\[ \n\left\langle {{D}_{j}\left( {\alpha T}\right) ,\varphi }\right\rangle = - \left\langle {{\alpha T},{D}_{j}\varphi }\right\rangle = - \left\langle {T,\alpha {D}_{j}\varphi }\right\rangle = - \left\langle {T,{D}_{j}\left( {\alpha \varphi }\right) }\right\rangle + \left\langle {T,\left( {{D}_{j}\alpha }\right) \varphi }\right\rangle ,\n\]\n\nso that\n\n\[ \n\left\langle {{D}_{j}\left( {\alpha T}\right) ,\varphi }\right\rangle = \left\langle {\left( {{D}_{j}\alpha }\right) T + \alpha {D}_{j}T,\varphi }\right\rangle\n\]\n\nThus \( {D}_{j}\left( {\alpha T}\right) = \alpha {D}_{j}T + \left( {{D}_{j}\alpha }\right) T \) . From here the formula can be extended by induction on \( \left| p\right| \) as in the case of functions.
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Yes
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Theorem 2.7 Let \( \Omega \) be a connected open subset of \( {\mathbb{R}}^{d} \) and suppose that \( T \) is a distribution on \( \Omega \) such that \( {D}_{j}T = 0 \) for every \( j \in \{ 1,\ldots, d\} \) . Then \( T = C \) for some \( C \in \mathbb{C} \) .
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Proof. By the preceding theorem, there exists \( f \in {\mathcal{E}}^{1}\left( \Omega \right) \) such that \( T = \left\lbrack f\right\rbrack \) and \( {D}_{j}f = 0 \) in the ordinary sense, for all \( j \in \{ 1,\ldots, d\} \) . The result follows.
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No
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Theorem 2.8 Suppose that \( \Omega \) is an open interval in \( \mathbb{R} \) and that \( \alpha \in \Omega \) . Let \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) and \( f \in {L}_{\mathrm{{loc}}}^{1}\left( \Omega \right) \) . The following properties are equivalent:\n\ni. \( {T}^{\prime } = \left\lbrack f\right\rbrack \) .\n\nii. There exists \( C \in \mathbb{C} \) such that \( T = \left\lbrack F\right\rbrack \), with \( F\left( x\right) = C + {\int }_{\alpha }^{x}f\left( t\right) {dt} \) .
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Proof. Suppose \( \Omega = \left( {a, b}\right) \) . Take \( f \in {L}_{\text{loc }}^{1}\left( \left( {a, b}\right) \right) \) and let \( F\left( x\right) = {\int }_{\alpha }^{x}f\left( t\right) {dt} \) . Then, for every \( \varphi \in \mathcal{D}\left( \left( {a, b}\right) \right) \) ,\n\n\[ \left\langle {{\left\lbrack F\right\rbrack }^{\prime },\varphi }\right\rangle = - {\int }_{a}^{b}{\varphi }^{\prime }\left( x\right) \left( {{\int }_{\alpha }^{x}f\left( t\right) {dt}}\right) {dx}. \]\n\nTherefore, by Fubini's Theorem,\n\n\[ \left\langle {{\left\lbrack F\right\rbrack }^{\prime },\varphi }\right\rangle = {\iint }_{\{ a \leq x \leq t \leq \alpha \} }{\varphi }^{\prime }\left( x\right) f\left( t\right) {dtdx} - {\iint }_{\{ \alpha \leq t \leq x \leq b\} }{\varphi }^{\prime }\left( x\right) f\left( t\right) {dtdx} \]\n\n\[ = {\int }_{a}^{\alpha }\varphi \left( t\right) f\left( t\right) {dt} + {\int }_{\alpha }^{b}\varphi \left( t\right) f\left( t\right) {dt} = \langle \left\lbrack f\right\rbrack ,\varphi \rangle . \]\n\nThus \( {\left\lbrack F\right\rbrack }^{\prime } = \left\lbrack f\right\rbrack \) and the desired result follows from the uniqueness theorem proved earlier (Theorem 2.7).
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Yes
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Theorem 2.9 Suppose that \( \Omega \) is an open interval in \( \mathbb{R} \), and that \( T \in \) \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) . If there exists an increasing function \( \alpha \) on \( \Omega \) such that \( T = \left\lbrack \alpha \right\rbrack \) , then \( {T}^{\prime } = {d\alpha } \) and therefore \( {T}^{\prime } \) is positive.\n\nConversely, if \( {T}^{\prime } \) is positive, there exists an increasing function \( \alpha \) on \( \Omega \) and a constant \( C \in \mathbb{R} \) such that \( T = \left\lbrack {\alpha + {iC}}\right\rbrack \) .
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Proof. Set \( \Omega = \left( {a, b}\right) \) . Let \( \alpha \) be an increasing function on \( \left( {a, b}\right) \) . Take \( \varphi \in \) \( \mathcal{D}\left( \Omega \right) \) and let \( c, d \) be such that \( a < c < d < b \) and the support of \( \varphi \) is contained is \( \left\lbrack {c, d}\right\rbrack \) . For \( n \in {\mathbb{N}}^{ * } \) and \( k \in \{ 0,\ldots, n\} \), set\n\n\[{x}_{k} = c + k\frac{d - c}{n}.\n\]\n\nThen, by definition,\n\n\[ \int {\varphi d\alpha } = {\int }_{c}^{d}\varphi \left( x\right) {d\alpha }\left( x\right) = \mathop{\lim }\limits_{{n \rightarrow + \infty }}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}\varphi \left( {x}_{k}\right) \left( {\alpha \left( {x}_{k + 1}\right) - \alpha \left( {x}_{k}\right) }\right) .\n\]\n\nWe perform a summation by parts. Since \( \varphi \left( c\right) = \varphi \left( d\right) = 0 \), we have\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}\varphi \left( {x}_{k}\right) \left( {\alpha \left( {x}_{k + 1}\right) - \alpha \left( {x}_{k}\right) }\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\alpha \left( {x}_{k}\right) \left( {\varphi \left( {x}_{k - 1}\right) - \varphi \left( {x}_{k}\right) }\right) .\n\]\n\nConsequently,\n\n\[ \int {\varphi d\alpha } = - \mathop{\lim }\limits_{{n \rightarrow + \infty }}{\int }_{c}^{d}{\varphi }^{\prime }\left( x\right) \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{1}_{\left\lbrack {x}_{k - 1},{x}_{k}\right) }\left( x\right) \alpha \left( {x}_{k}\right) }\right) {dx}.\n\]\n\nUsing the Dominated Convergence Theorem, we obtain\n\n\[ \int {\varphi d\alpha } = - \int {\varphi }^{\prime }\left( x\right) \alpha \left( {x}_{ + }\right) {dx}\n\]\n\n(Recall that \( \alpha \left( {x}_{ + }\right) \) denotes the limit from the right of the function \( \alpha \) at \( x \) .) Now, \( \alpha \left( {x}_{ + }\right) = \alpha \left( x\right) \) except at a set of points \( x \) that is countable, and so of Lebesgue measure zero (see Exercise 6 on page 5). Therefore\n\n\[ \int {\varphi d\alpha } = - \int {\varphi }^{\prime }\left( x\right) \alpha \left( x\right) {dx}\n\]\n\nso \( {d\alpha } = {\left\lbrack \alpha \right\rbrack }^{\prime } \) . This proves the first part of the theorem.\n\nNow suppose that \( {T}^{\prime } \) is positive. By Proposition 2.3 on page 270, \( {T}^{\prime } \) is a positive Radon measure on \( \Omega \) . By Theorem 3.8 on page 73 (applied to \( \Omega \) rather than \( \mathbb{R} \) ), there exists an increasing function \( \alpha \) such that \( {T}^{\prime } = {d\alpha } \) (we may assume \( \alpha \) is right continuous). Then, by the first part of this proof, \( {T}^{\prime } = {\left\lbrack \alpha \right\rbrack }^{\prime } \) . Now it suffices to apply the uniqueness theorem (Theorem 2.7) to obtain \( T = \left\lbrack {\alpha + C}\right\rbrack \), for \( C \in \mathbb{C} \) . The desired result follows by replacing \( \alpha \) with \( \alpha + \operatorname{Re}C \) and \( C \) with \( \operatorname{Im}C \) .
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Yes
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Theorem 2.10 Suppose that \( \Omega \) is an open subset of \( \mathbb{R} \) and that \( f \) is a function on \( \Omega \) for which there exist points \( {x}_{1} < \cdots < {x}_{n} \) in \( \Omega \) satisfying these conditions:\n\n- \( f \) is of class \( {C}^{1} \) on \( \Omega \smallsetminus \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) .\n\n- For every \( j \in \{ 1,\ldots, n\}, f \) has right and left limits at \( {x}_{j} \), which we denote by \( f\left( {x}_{j + }\right) \) and \( f\left( {x}_{j - }\right) \), respectively.\n\n- The ordinary derivative \( {f}^{\prime } \) of \( f \), defined on \( \Omega \smallsetminus \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \), belongs to \( {L}_{\mathrm{{loc}}}^{1}\left( \Omega \right) \) .\n\nThen\n\n\[ \n{\left\lbrack f\right\rbrack }^{\prime } = \left\lbrack {f}^{\prime }\right\rbrack + \mathop{\sum }\limits_{{j = 1}}^{n}\left( {f\left( {x}_{j + }\right) - f\left( {x}_{j - }\right) }\right) {\delta }_{{x}_{j}}.\n\]
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Proof. Considering separately each of the connected components of \( \Omega \), we can assume that \( \Omega \) is an open interval \( \left( {a, b}\right) \) . Put \( {x}_{0} = a \) and \( {x}_{n + 1} = b \) . Then, if \( \varphi \in \mathcal{D}\left( \Omega \right) \), we have\n\n\[ \n\left\langle {{\left\lbrack f\right\rbrack }^{\prime },\varphi }\right\rangle = - \left\langle {\left\lbrack f\right\rbrack ,{\varphi }^{\prime }}\right\rangle = - \mathop{\sum }\limits_{{j = 0}}^{n}{\int }_{{x}_{j}}^{{x}_{j + 1}}f\left( t\right) {\varphi }^{\prime }\left( t\right) {dt}\n\]\n\nor yet, integrating by parts (and setting \( \varphi \left( a\right) = \varphi \left( b\right) = 0 \) ),\n\n\[ \n\left\langle {{\left\lbrack f\right\rbrack }^{\prime },\varphi }\right\rangle = \mathop{\sum }\limits_{{j = 0}}^{n}\left( {{\int }_{{x}_{j}}^{{x}_{j + 1}}\varphi \left( t\right) {f}^{\prime }\left( t\right) {dt} + f\left( {x}_{j + }\right) \varphi \left( {x}_{j}\right) - f\left( {\left( {x}_{j + 1}\right) }_{ - }\right) \varphi \left( {x}_{j + 1}\right) }\right)\n\]\n\n\[ \n= \int \varphi \left( t\right) {f}^{\prime }\left( t\right) {dt} + \mathop{\sum }\limits_{{j = 1}}^{n}\varphi \left( {x}_{j}\right) \left( {f\left( {x}_{j + }\right) - f\left( {x}_{j - }\right) }\right) ,\n\]\n\nwhich concludes the proof.
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Yes
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Theorem 2.12 Suppose that \( d \geq 2 \) and, if \( \left( {{x}_{2},\ldots ,{x}_{d}}\right) \in {\mathbb{R}}^{d - 1} \), write\n\n\[{\Omega }_{{x}_{2},\ldots ,{x}_{d}} = \left\{ {{x}_{1} \in \mathbb{R} : \left( {{x}_{1},{x}_{2},\ldots ,{x}_{d}}\right) \in \Omega }\right\} .\n\]\n\nLet \( f \in {L}_{\text{loc }}^{1}\left( \Omega \right) \) satisfy the following conditions:\n\n- For almost every \( \left( {{x}_{2},\ldots ,{x}_{d}}\right) \in {\mathbb{R}}^{d - 1} \), the map on \( {\Omega }_{{x}_{2},\ldots ,{x}_{d}} \) defined by\n\n\[{x}_{1} \mapsto f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{d}}\right)\n\]\n\nis continuous on \( {\Omega }_{{x}_{2},\ldots ,{x}_{d}} \) and of class \( {C}^{1} \) except at finitely many points of \( {\Omega }_{{x}_{2},\ldots ,{x}_{d}} \) .\n\n- The ordinary partial derivative \( \partial f/\partial {x}_{1} \), defined almost everywhere on \( \Omega \), is an element of \( {L}_{\mathrm{{loc}}}^{1}\left( \Omega \right) \) .\n\nThen\n\n\[{D}_{1}\left\lbrack f\right\rbrack = \left\lbrack \frac{\partial f}{\partial {x}_{1}}\right\rbrack .\n\]\n\nOf course, an analogous result holds if we replace the subscript 1 by any \( j \in \{ 2,\ldots, d\} \) .
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Proof. Argue as in the proof of Theorem 2.10 and apply Fubini’s Theorem.
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No
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Theorem 3.1 Let \( P\left( X\right) = \mathop{\sum }\limits_{{j = 0}}^{m}{a}_{j}{X}^{j} \), where \( m \in {\mathbb{N}}^{ * },{a}_{1},\ldots ,{a}_{m} \in \mathbb{C} \) , and \( {a}_{m} \neq 0 \) . Let \( \varphi \) be the solution on \( \mathbb{R} \) of the differential equation\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{m}{a}_{j}{\varphi }^{\left( j\right) } = 0 \]\n\nsuch that \( {\varphi }^{\left( m - 1\right) }\left( 0\right) = 1 \) and \( {\varphi }^{\left( j\right) }\left( 0\right) = 0 \) for every \( j \leq m - 2 \) . Then \( E = \) \( \left( {1/{a}_{m}}\right) {Y\varphi } \) is a fundamental solution of \( P\left( D\right) \) .
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Proof. As a particular case of Example 4 on page 301, we have\n\n\[ {\left\lbrack Y\varphi \right\rbrack }^{\left( m\right) } = \left\lbrack {Y{\varphi }^{\left( m\right) }}\right\rbrack + \delta \]\n\n\[ {\left\lbrack Y\varphi \right\rbrack }^{\left( k\right) } = \left\lbrack {Y{\varphi }^{\left( k\right) }}\right\rbrack \text{ for all }k \leq m - 1, \]\n\nso that\n\n\[ P\left( D\right) E = {a}_{m}^{-1}\mathop{\sum }\limits_{{j = 0}}^{m}{a}_{j}{\left\lbrack Y\varphi \right\rbrack }^{\left( j\right) } = {a}_{m}^{-1}\left\lbrack {\mathop{\sum }\limits_{{j = 0}}^{m}{a}_{j}Y{\varphi }^{\left( j\right) }}\right\rbrack + \delta = \delta . \]
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Yes
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Theorem 3.4 In \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{2}\right) \), \[ \frac{\partial }{\partial \bar{z}}\left( \frac{1}{\pi z}\right) = \delta \]
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Proof. We follow a method analogous to the one used in the proof of Theorem 3.2. For \( \varepsilon > 0 \), put \[ {f}_{\varepsilon }\left( {x, y}\right) = \left\{ \begin{array}{ll} 1/z & \text{ if }\left| z\right| > \varepsilon \\ \bar{z}/{\varepsilon }^{2} & \text{ if }\left| z\right| \leq \varepsilon \end{array}\right. \] Then \( {f}_{\varepsilon } \) is continuous on \( {\mathbb{R}}^{2} \) and, by Theorem 2.12, \[ \frac{\partial }{\partial x}\left\lbrack {f}_{\varepsilon }\right\rbrack = \left\lbrack {g}_{1,\varepsilon }\right\rbrack ,\;\frac{\partial }{\partial y}\left\lbrack {f}_{\varepsilon }\right\rbrack = \left\lbrack {g}_{2,\varepsilon }\right\rbrack \] with \[ {g}_{1,\varepsilon }\left( {x, y}\right) = \left\{ {\begin{aligned} - \frac{1}{{z}^{2}} & \text{ if }\left| z\right| > \varepsilon , \\ \frac{1}{{\varepsilon }^{2}} & \text{ if }\left| z\right| < \varepsilon ; \end{aligned}\;{g}_{2,\varepsilon }\left( {x, y}\right) = \left\{ \begin{array}{ll} - \frac{i}{{z}^{2}} & \text{ if }\left| z\right| > \varepsilon , \\ - \frac{i}{{\varepsilon }^{2}} & \text{ if }\left| z\right| < \varepsilon . \end{array}\right. }\right. \] Thus \[ \frac{\partial }{\partial \bar{z}}\left\lbrack {f}_{\varepsilon }\right\rbrack = \frac{1}{{\varepsilon }^{2}}{1}_{\bar{B}\left( {0,\varepsilon }\right) } \] which tends to \( {\pi \delta } \) in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{2}\right) \) when \( \varepsilon \) tends to 0 . We have \( \left| {{f}_{\varepsilon }\left( {x, y}\right) }\right| \leq 1/\left| z\right| \) , so the Dominated Convergence Theorem implies that \( \left\lbrack {f}_{\varepsilon }\right\rbrack \) tends to \( \left\lbrack f\right\rbrack \) in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{2}\right) \), with \( f\left( {x, y}\right) = 1/z \) . Therefore \[ \frac{\partial }{\partial \bar{z}}\left\lbrack f\right\rbrack = \mathop{\lim }\limits_{{\varepsilon \rightarrow 0}}\frac{\partial }{\partial \bar{z}}\left\lbrack {f}_{\varepsilon }\right\rbrack = {\pi \delta } \] whence the result.
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Yes
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Theorem 1.1 (Differentiation inside the brackets) Let \( m \in \mathbb{N} \) and \( r \in \mathbb{N} \) . If \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) \) and \( \varphi \in {\mathcal{D}}^{m + r}\left( {\Omega \times {\Omega }^{\prime }}\right) \), the map on \( {\Omega }^{\prime } \) defined by\n\n\[ y \mapsto \langle T,\varphi \left( {\cdot, y}\right) \rangle \]\n\n\( \left( *\right) \)\n\nbelongs to \( {\mathcal{D}}^{r}\left( {\Omega }^{\prime }\right) \) and, for every multiindex \( p \in {\mathbb{N}}^{{d}^{\prime }} \) of length at most \( r \) ,\n\n\[ \frac{{\partial }^{\left| p\right| }}{\partial {y}^{p}}\langle T,\varphi \left( {\cdot, y}\right) \rangle = \left\langle {T,\frac{{\partial }^{\left| p\right| }}{\partial {y}^{p}}\varphi \left( {\cdot, y}\right) }\right\rangle \]\n\n\( \left( {* * }\right) \)\n\nfor every \( y \in {\Omega }^{\prime } \) .
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Proof. We carry out the proof in the case \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) ,\varphi \in {\mathcal{D}}^{m + r}\left( {\Omega \times {\Omega }^{\prime }}\right) \) . The other case is very similar.\n\nCase \( r = 0 \) . Take \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) \) and \( \varphi \in {\mathcal{D}}^{m}\left( {\Omega \times {\Omega }^{\prime }}\right) \), and let \( K \) and \( {K}^{\prime } \) be compact subsets of \( \Omega \) and \( {\Omega }^{\prime } \), respectively, satisfying \( \operatorname{Supp}\varphi \subset \) \( K \times {K}^{\prime } \) . Since, for every multiindex \( p \) of length at most \( m \), the function \( \left( {{\partial }^{\left| p\right| }\varphi }\right) /\left( {\partial {x}^{p}}\right) \) is uniformly continuous (being continuous and having compact support), and since all the functions \( \varphi \left( {\cdot, y}\right) \), with \( y \in {\Omega }^{\prime } \), are supported within the same compact \( K \), we see that, if \( {\left( {y}_{n}\right) }_{n \in \mathbb{N}} \) is a sequence in \( {\Omega }^{\prime } \) converging to \( y \in {\Omega }^{\prime } \), the sequence of functions \( {\left( \varphi \left( \cdot ,{y}_{n}\right) \right) }_{n \in \mathbb{N}} \) converges to \( \varphi \left( {\cdot, y}\right) \) in \( {\mathcal{D}}^{m}\left( \Omega \right) \), so the sequence \( {\left( \left\langle T,\varphi \left( \cdot ,{y}_{n}\right) \right\rangle \right) }_{n \in \mathbb{N}} \) converges to \( \langle T,\varphi \left( {\cdot, y}\right) \rangle \) . We deduce that the map \( y \mapsto \langle T,\varphi \left( {\cdot, y}\right) \rangle \) is continuous on \( {\Omega }^{\prime } \) . Since its support is compact (being contained in \( {K}^{\prime } \) ), this map does belong to \( {C}_{c}\left( {\Omega }^{\prime }\right) = {\mathcal{D}}^{0}\left( {\Omega }^{\prime }\right) \) .\n\nCase \( r = 1 \) . Take \( T \in {\mathcal{D}}^{\prime m}\left( \Omega \right) \) and \( \varphi \in {\mathcal{D}}^{m + 1}\left( {\Omega \times {\Omega }^{\prime }}\right) \), and again let \( K \) and \( {K}^{\prime } \) be compact subsets of \( \Omega \) and \( {\Omega }^{\prime } \), respectively, satisfying \( \operatorname{Supp}\varphi \subset \) \( K \times {K}^{\prime } \) . For \( 1 \leq j \leq {d}^{\prime } \), let \( {e}_{j} \) be the \( j \) -th vector of the canonical basis of \( {\mathbb{R}}^{{d}^{\prime }} \) . Take \( y \in {\Omega }^{\prime } \) . If \( x \in \Omega \) and \( t \neq 0 \), we have\n\n\[ \left| {\frac{\varphi \left( {x, y + t{e}_{j}}\right) - \varphi \left( {x, y}\right) }{t} - \frac{\partial \varphi }{\partial {y}_{j}}\left( {x, y}\right) }\right| \leq \mathop{\sup }\limits_{{{t}^{\prime } \in \left\lbrack {0, t}\right\rbrack }}\left| {\frac{\partial \varphi }{\partial {y}_{j}}\left( {x, y + {t}^{\prime }{e}_{j}}\right) - \frac{\partial \varphi }{\partial {y}_{j}}\left( {x, y}\right) }\right| \]
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Yes
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Theorem 1.2 The vector space \( \mathcal{D}\left( \Omega \right) \otimes \mathcal{D}\left( {\Omega }^{\prime }\right) \) spanned by the functions\n\n\[ f \otimes g : \left( {x, y}\right) \mapsto f\left( x\right) g\left( y\right) \]\n\nwith \( f \in \mathcal{D}\left( \Omega \right) \) and \( g \in \mathcal{D}\left( {\Omega }^{\prime }\right) \), is dense in \( \mathcal{D}\left( {\Omega \times {\Omega }^{\prime }}\right) \) .
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Proof. We use a lemma that allows us to approximate the convolution by means of a \
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No
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Lemma 1.3 Suppose \( \varphi ,\psi \in \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) . For \( \varepsilon > 0 \) and \( x \in {\mathbb{R}}^{n} \), set\n\n\[ \n{g}_{\varepsilon }\left( x\right) = {\varepsilon }^{n}\mathop{\sum }\limits_{{\nu \in {\mathbb{Z}}^{n}}}\varphi \left( {x - {\varepsilon \nu }}\right) \psi \left( {\varepsilon \nu }\right) \n\]\n\nThen \( {g}_{\varepsilon } \in \mathcal{D}\left( {\mathbb{R}}^{n}\right) ,\operatorname{Supp}{g}_{\varepsilon } \subset \operatorname{Supp}\varphi + \operatorname{Supp}\psi \), and\n\n\[ \n\mathop{\lim }\limits_{{\varepsilon \rightarrow 0}}{g}_{\varepsilon } = \varphi * \psi \n\]\n\nin \( \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) .
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Proof. The function \( {g}_{\varepsilon } \) is defined by a finite sum whose number of terms depends only on \( \varepsilon \) (since \( \psi \) has compact support). Since each of these terms is an element of \( \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) and is supported within Supp \( \varphi + \operatorname{Supp}\psi \), the same holds for \( {g}_{\varepsilon } \) . At the same time, for every \( p \in {\mathbb{N}}^{d} \) ,\n\n\[ \n{D}^{p}{g}_{\varepsilon }\left( x\right) = {\varepsilon }^{n}\mathop{\sum }\limits_{{\nu \in {\mathbb{Z}}^{n}}}{D}^{p}\varphi \left( {x - {\varepsilon \nu }}\right) \psi \left( {\varepsilon \nu }\right) . \n\]\n\nThus the result will be proved if we show that \( {g}_{\varepsilon } \) converges uniformly to \( \varphi * \psi \) (for then we will be able to apply the same result to \( {D}^{p}\varphi \) and \( \psi \) instead of \( \varphi \) and \( \psi \) ).\n\nDenote by \( \parallel \cdot \parallel \) the uniform norm on \( {\mathbb{R}}^{n} \) and set \( N = \mathop{\max }\limits_{{x \in \operatorname{Supp}\psi }}\parallel x\parallel \) . By the Mean Value Theorem, there exists \( C > 0 \) such that, for every \( x, y,{y}^{\prime } \in {\mathbb{R}}^{n}, \)\n\n\[ \n\left| {\varphi \left( {x - y}\right) \psi \left( y\right) - \varphi \left( {x - {y}^{\prime }}\right) \psi \left( {y}^{\prime }\right) }\right| \leq C\begin{Vmatrix}{y - {y}^{\prime }}\end{Vmatrix}. \n\]\n\nFor \( \nu \in {\mathbb{Z}}^{n} \), set\n\n\[ \n{Q}_{\nu }^{\varepsilon } = \mathop{\prod }\limits_{{j = 1}}^{n}\left\lbrack {{\nu }_{j}\varepsilon ,\left( {{\nu }_{j} + 1}\right) \varepsilon }\right) \n\]\n\nThen\n\n\[ \n\varphi * \psi \left( x\right) = \mathop{\sum }\limits_{{\parallel \nu \parallel \leq \left( {N/\varepsilon }\right) + 1}}{\int }_{{Q}_{\nu }^{\varepsilon }}\varphi \left( {x - y}\right) \psi \left( y\right) {dy} \n\]\nso that\n\n\[ \n\left| {\varphi * \psi \left( x\right) - {g}_{\varepsilon }\left( x\right) }\right| \leq \mathop{\sum }\limits_{{\parallel \nu \parallel \leq \left( {N/\varepsilon }\right) + 1}}{\int }_{{Q}_{\nu }^{\varepsilon }}\left| {\varphi \left( {x - y}\right) \psi \left( y\right) - \varphi \left( {x - {\nu \varepsilon }}\right) \psi \left( {\nu \varepsilon }\right) }\right| {dy} \n\]\n\n\[ \n\leq C{\varepsilon }^{n + 1}{\left( 2\left( \frac{N}{\varepsilon } + 1\right) + 1\right) }^{n} \leq {C}^{\prime }\varepsilon \;\left( {\text{ for }\varepsilon < 1}\right) , \n\]\n\nproving the result.
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Yes
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Proposition 1.4 Suppose \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) and \( S \in {\mathcal{D}}^{\prime }\left( {\Omega }^{\prime }\right) \). There exists a unique distribution on \( \Omega \times {\Omega }^{\prime } \), denoted \( T \otimes S \) and called the tensor product of \( T \) and \( S \), such that\n\n\[ \langle T \otimes S,\varphi \otimes \psi \rangle = \langle T,\varphi \rangle \langle S,\psi \rangle \]\n\nfor all \( \varphi \in \mathcal{D}\left( \Omega \right) \) and \( \psi \in \mathcal{D}\left( {\Omega }^{\prime }\right) \). Moreover, for every \( \varphi \in \mathcal{D}\left( {\Omega \times {\Omega }^{\prime }}\right) \),\n\n\[ \langle T \otimes S,\varphi \rangle = \left\langle {{T}_{x},\left\langle {{S}_{y},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle = \left\langle {{S}_{y},\left\langle {{T}_{x},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle . \]
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Proof. Uniqueness follows immediately from Theorem 1.2. For existence, consider the linear map on \( \mathcal{D}\left( {\Omega \times {\Omega }^{\prime }}\right) \) defined by\n\n\[ \varphi \mapsto \left\langle {{S}_{y},\left\langle {{T}_{x},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle \]\n\n(*) \n\nThis map is well defined, by Theorem 1.1. Let \( {K}_{1} \) be a compact subset of \( \Omega \times {\Omega }^{\prime } \), and let \( K \) and \( {K}^{\prime } \) be compact subsets in \( \Omega \) and \( {\Omega }^{\prime } \), respectively, such that \( {K}_{1} \subset K \times {K}^{\prime } \). Take \( m,{m}^{\prime } \in \mathbb{N} \) and \( C,{C}^{\prime } > 0 \) such that\n\n\[ \left| {\langle T,\varphi \rangle }\right| \leq C\parallel \varphi {\parallel }^{\left( m\right) }\;\text{ for all }\varphi \in {\mathcal{D}}_{K}\left( \Omega \right) \]\n\nand\n\n\[ \left| {\langle S,\varphi \rangle }\right| \leq {C}^{\prime }\parallel \varphi {\parallel }^{\left( {m}^{\prime }\right) }\;\text{ for all }\varphi \in {\mathcal{D}}_{{K}^{\prime }}\left( {\Omega }^{\prime }\right) \]\n\n(see Proposition 2.1 on page 268). Then, again by Theorem 1.1, there exists a constant \( {C}^{\prime \prime } \geq 0 \) such that\n\n\[ \left| \left\langle {{S}_{y},\left\langle {{T}_{x},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle \right| \leq {C}^{\prime \prime }\parallel \varphi {\parallel }^{\left( m + {m}^{\prime }\right) }\;\text{ for all }\varphi \in {\mathcal{D}}_{{K}_{1}}\left( {\Omega \times {\Omega }^{\prime }}\right) . \]\n\nThus, the linear map defined in \( \left( *\right) \) is indeed a distribution on \( \Omega \times {\Omega }^{\prime } \) satisfying the indicated condition, namely\n\n\[ \left\langle {{S}_{y},\left\langle {{T}_{x},\varphi \left( x\right) \psi \left( y\right) }\right\rangle }\right\rangle = \langle T,\varphi \rangle \langle S,\psi \rangle \]\n\nfor all \( \varphi \in \mathcal{D}\left( \Omega \right) \) and \( \psi \in \mathcal{D}\left( {\Omega }^{\prime }\right) \). One argues likewise for the expression \( \left\langle {{T}_{x},\left\langle {{S}_{y},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle \), interchanging the roles of \( x \) and \( y \).
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Yes
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Proposition 1.5 Suppose \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) and \( S \in {\mathcal{D}}^{\prime }\left( {\Omega }^{\prime }\right) \) . Then:\n\ni. \( \operatorname{Supp}\left( {T \otimes S}\right) = \left( {\operatorname{Supp}T}\right) \times \left( {\operatorname{Supp}S}\right) \) .\n\nii. For any \( p \in {\mathbb{N}}^{d} \) and \( q \in {\mathbb{N}}^{{d}^{\prime }} \), \[ {\partial }_{x}^{p}{\partial }_{y}^{q}\left( {T \otimes S}\right) = \left( {{\partial }_{x}^{p}T}\right) \otimes \left( {{\partial }_{y}^{q}S}\right) \]
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Proof. If \( \varphi \) is supported within \( \left( {\Omega \smallsetminus \operatorname{Supp}T}\right) \times {\Omega }^{\prime } \), the support of \( \varphi \left( {\cdot, y}\right) \) , for every \( y \in {\Omega }^{\prime } \), is contained in \( \Omega \smallsetminus \operatorname{Supp}T \) . Therefore \[ \langle T \otimes S,\varphi \rangle = \left\langle {{S}_{y},\left\langle {{T}_{x},\varphi \left( {x, y}\right) }\right\rangle }\right\rangle = 0. \] It follows that the support of \( T \otimes S \) is contained in Supp \( T \times {\Omega }^{\prime } \) ; similarly, it is contained in \( \Omega \times \operatorname{Supp}S \), and so also in the intersection of these two sets, which is \( \operatorname{Supp}T \times \operatorname{Supp}S \) .\n\nConversely, if \( \left( {x, y}\right) \in \operatorname{Supp}T \times \operatorname{Supp}S \) and if \( \left( {x, y}\right) \notin \operatorname{Supp}\left( {T \otimes S}\right) \), let \( O \) denote the complement of the support of \( T \otimes S \) in \( \Omega \times {\Omega }^{\prime } \) . Then there exist open sets \( {O}_{1} \) and \( {O}_{2} \) containing \( x \) and \( y \), respectively, and such that \( O \supset {O}_{1} \times {O}_{2} \) . By the definition of \( x \) and \( y \), there exist \( \varphi \in \mathcal{D}\left( {O}_{1}\right) \) and \( \psi \in \mathcal{D}\left( {O}_{2}\right) \) such that \( \langle T,\varphi \rangle \neq 0 \) and \( \langle S,\psi \rangle \neq 0 \) . But then \( \varphi \otimes \psi \in \mathcal{D}\left( O\right) \) and \( \langle T \otimes S,\varphi \otimes \psi \rangle \neq 0 \), which contradicts the definition of \( O \) . Therefore \( \operatorname{Supp}T \times \operatorname{Supp}S \subset \operatorname{Supp}\left( {T \otimes S}\right) \), and the first assertion of the theorem is proved.\n\nNext, if \( \varphi \in \mathcal{D}\left( \Omega \right) \) and \( \psi \in \mathcal{D}\left( {\Omega }^{\prime }\right) \), \[ \left\langle {{\partial }_{x}^{p}{\partial }_{y}^{q}\left( {T \otimes S}\right) ,\;\varphi \otimes \psi }\right\rangle = {\left( -1\right) }^{\left| p\right| + \left| q\right| }\left\langle {T \otimes S,\;\left( {{\partial }_{x}^{p}\varphi }\right) \otimes \left( {{\partial }_{y}^{q}\psi }\right) }\right\rangle \] \[ = {\left( -1\right) }^{\left| p\right| + \left| q\right| }\langle T,{\partial }_{x}^{p}\varphi \rangle \langle S,{\partial }_{y}^{q}\psi \rangle \] \[ = \left\langle {\left( {{\partial }_{x}^{p}T}\right) \otimes \left( {{\partial }_{y}^{q}S}\right) ,\varphi \otimes \psi }\right\rangle . \] Now just apply the denseness theorem (Theorem 1.2) to obtain the second part of the theorem.
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Yes
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Proposition 2.2 If \( T, S \in {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \), then \( T * S \in {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) and\n\n\[ \operatorname{Supp}\left( {T * S}\right) \subset \operatorname{Supp}T + \operatorname{Supp}S \]
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Proof. Let \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) . If \( x \notin \operatorname{Supp}\varphi - \operatorname{Supp}S \), then\n\n\[ \operatorname{Supp}\varphi \left( {x + \cdot }\right) \cap \operatorname{Supp}S = \left( {\operatorname{Supp}\varphi - x}\right) \cap \operatorname{Supp}S = \varnothing ,\]\n\nso \( \left\langle {{S}_{y},\varphi \left( {x + y}\right) }\right\rangle = 0 \) . Thus, \( \operatorname{Supp}\left\langle {{S}_{y},\varphi \left( {\cdot + y}\right) }\right\rangle \subset \operatorname{Supp}\varphi - \operatorname{Supp}S \) . It follows that, if the support of \( T \) does not intersect \( \operatorname{Supp}\varphi - \operatorname{Supp}S \), we have \( \langle T * S,\varphi \rangle = 0 \) and therefore \( \operatorname{Supp}\left( {T * S}\right) \subset \operatorname{Supp}T + \operatorname{Supp}S \) . The rest of the proposition follows immediately from the results proved in Section 1.
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Yes
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Proposition 2.5 Let \( \Omega \) be open in \( {\mathbb{R}}^{d} \) . Let \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) and \( \varphi \in \mathcal{E}\left( \Omega \right) \) be such that \( \operatorname{Supp}T \cap \operatorname{Supp}\varphi \) is compact. Then, if \( \rho \in \mathcal{D}\left( \Omega \right) \) is a function taking the value 1 on an open set containing \( \operatorname{Supp}T \cap \operatorname{Supp}\varphi \), the value of \( \langle T,{\rho \varphi }\rangle \) does not depend on \( \rho \) .
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Proof. Take \( \rho \in \mathcal{D}\left( \Omega \right) \) such that \( \rho = 0 \) on an open set containing \( \operatorname{Supp}T \cap \) Supp \( \varphi \) . Then the support of \( \rho \) is contained in the complement of Supp \( T \cap \) Supp \( \varphi \), and therefore\n\n\[ \operatorname{Supp}{\rho \varphi } \subset \operatorname{Supp}\varphi \cap \left( {{\mathbb{R}}^{d} \smallsetminus \left( {\operatorname{Supp}T \cap \operatorname{Supp}\varphi }\right) }\right) = \operatorname{Supp}\varphi \cap \left( {{\mathbb{R}}^{d} \smallsetminus \operatorname{Supp}T}\right) ,\]\n\nwhich implies that \( \langle T,{\rho \varphi }\rangle = 0 \) .\n\nConsequently, if \( \rho \) and \( \widetilde{\rho } \) are functions in \( \mathcal{D}\left( \Omega \right) \) that coincide on an open set containing \( \operatorname{Supp}T \cap \operatorname{Supp}\varphi \), we have \( \langle T,{\rho \varphi }\rangle = \langle T,\widetilde{\rho }\varphi \rangle \) .
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Yes
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Proposition 2.6 Let \( \left( {{T}_{1},\ldots ,{T}_{n}}\right) \) be a family of distributions on \( {\mathbb{R}}^{d} \) satisfying condition \( \left( \mathrm{C}\right) \) . 1. If \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \), we define a function \( \widehat{\varphi } \) on \( {\left( {\mathbb{R}}^{d}\right) }^{n} \) by \[ \widehat{\varphi }\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \varphi \left( {{x}^{1} + \cdots + {x}^{n}}\right) . \] Then \( \widehat{\varphi } \in \mathcal{E}\left( {\left( {\mathbb{R}}^{d}\right) }^{n}\right) \) and \( \operatorname{Supp}\left( {{T}_{1} \otimes \cdots \otimes {T}_{n}}\right) \cap \operatorname{Supp}\widehat{\varphi } \) is compact. The map defined on \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) by \[ \varphi \mapsto \left\langle {{T}_{1} \otimes \cdots \otimes {T}_{n},\widehat{\varphi }}\right\rangle \] is a distribution on \( {\mathbb{R}}^{d} \), denoted \( {T}_{1} * \cdots * {T}_{n} \) and called the convolution of \( {T}_{1},\ldots ,{T}_{n} \) .
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Proof. Let \( \Omega \) be a bounded open set in \( {\mathbb{R}}^{d} \) and \( \varphi \) an element of \( \mathcal{D}\left( \Omega \right) \) . We know that \( \operatorname{Supp}\left( {{T}_{1} \otimes \cdots \otimes {T}_{n}}\right) = \operatorname{Supp}{T}_{1} \times \cdots \times \operatorname{Supp}{T}_{n} \), so \( \operatorname{Supp}\left( {{T}_{1} \otimes \cdots \otimes {T}_{n}}\right) \cap \operatorname{Supp}\widehat{\varphi } \) \[ \subset \left\{ {\left( {{x}^{1},\ldots ,{x}^{n}}\right) \in \operatorname{Supp}{T}_{1} \times \cdots \times \operatorname{Supp}{T}_{n} : {x}^{1} + \cdots + {x}^{n} \in \bar{\Omega }}\right\} . \] By condition (C), we deduce that \( \operatorname{Supp}\left( {{T}_{1} \otimes \cdots \otimes {T}_{n}}\right) \cap \operatorname{Supp}\widehat{\varphi } \) is a compact subset of \( {\left( {\mathbb{R}}^{d}\right) }^{n} \) contained in a compact \( {K}_{\Omega } \) that depends only on \( \Omega \), not on \( \varphi \) . Thus, \( \left\langle {{T}_{1} \otimes \cdots \otimes {T}_{n},\widehat{\varphi }}\right\rangle \) is well defined and coincides with \( \left\langle {{T}_{1} \otimes \cdots \otimes {T}_{n}}\right. \) , \( \left. {\left. {\left( {{\rho }_{l} \otimes \cdots \otimes {\rho }_{l}}\right) \widehat{\varphi }}\right\rangle \text{ if }{K}_{\Omega } \subset {\left( B\left( 0, l\right) \right) }^{n}\text{. Now,}}\right\rangle \) \[ \left\langle {{T}_{1} \otimes \cdots \otimes {T}_{n},\left( {{\rho }_{l} \otimes \cdots \otimes {\rho }_{l}}\right) \widehat{\varphi }}\right\rangle = \left\langle {{\rho }_{l}{T}_{1} \otimes \cdots \otimes {\rho }_{l}{T}_{n},\widehat{\varphi }}\right\rangle \] \[ = \left\langle {\left( {{\rho }_{l}{T}_{1}}\right) * \cdots * \left( {{\rho }_{l}{T}_{n}}\right) ,\varphi }\right\rangle . \] This shows that \( {T}_{1} * \cdots * {T}_{n} \)
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Yes
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Proposition 2.8 (Continuity) Let \( {\left( {T}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) , and let \( T, S \) belong to \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) . Suppose that the sequence \( {\left( {T}_{n}\right) }_{n \in \mathbb{N}} \) converges to \( T \) in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \), that there exists a closed set \( F \) in \( {\mathbb{R}}^{d} \) such that \( \operatorname{Supp}{T}_{n} \subset F \) for all \( n \in \mathbb{N} \), and that \( \left( {F,\operatorname{Supp}S}\right) \) satisfies (C). Then \[ \mathop{\lim }\limits_{{n \rightarrow + \infty }}{T}_{n} * S = T * S \] in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) .
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Proof. Take \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) . As above, write \( \widehat{\varphi }\left( {x, y}\right) = \varphi \left( {x + y}\right) \) . Since the family \( \left( {F,\operatorname{Supp}S}\right) \) satisfies (C), the intersection \( \operatorname{Supp}\widehat{\varphi } \cap \left( {F \times \operatorname{Supp}S}\right) \) is compact. Let \( \rho \in \mathcal{D}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) \) satisfy \( \rho = 1 \) on an open set that contains this compact. Then, by definition, \[ \left\langle {{T}_{n} * S,\varphi }\right\rangle = \left\langle {{\left( {T}_{n}\right) }_{x},\left\langle {{S}_{y},\rho \left( {x, y}\right) \widehat{\varphi }\left( {x, y}\right) }\right\rangle }\right\rangle . \] Since the map \( x \mapsto \left\langle {{S}_{y},\rho \left( {x, y}\right) \widehat{\varphi }\left( {x, y}\right) }\right\rangle \) belongs to \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \), we deduce that \[ \mathop{\lim }\limits_{{n \rightarrow + \infty }}\left\langle {{T}_{n} * S,\varphi }\right\rangle = \left\langle {{T}_{x},\left\langle {{S}_{y},\rho \left( {x, y}\right) \widehat{\varphi }\left( {x, y}\right) }\right\rangle }\right\rangle = \langle T * S,\varphi \rangle \] which is the desired result.
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Yes
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Proposition 2.9 Suppose \( \\left( {T, S}\\right) \) satisfies property \( \\left( \\mathrm{C}\\right) \) . Then, for every \( \\varphi \\in \\mathcal{D}\\left( {\\mathbb{R}}^{d}\\right) \), the function \( \\widetilde{\\varphi } \) on \( {\\mathbb{R}}^{d} \) defined by\n\n\[ \n\\widetilde{\\varphi }\\left( x\\right) = \\left\\langle {{S}_{y},\\varphi \\left( {x + y}\\right) }\\right\\rangle \n\]\n\nbelongs to \( \\mathcal{E}\\left( {\\mathbb{R}}^{d}\\right) \), the intersection \( \\operatorname{Supp}\\widetilde{\\varphi } \\cap \\operatorname{Supp}T \) is compact, and\n\n\[ \n\\langle T * S,\\varphi \\rangle = \\langle T,\\widetilde{\\varphi }\\rangle = \\left\\langle {{T}_{x},\\left\\langle {{S}_{y},\\varphi \\left( {x + y}\\right) }\\right\\rangle }\\right\\rangle .\n\]
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Proof. Put \( K = \\{ \\left( {x, y}\\right) \\in \\operatorname{Supp}T \\times \\operatorname{Supp}S : x + y \\in \\operatorname{Supp}\\varphi \\} \) . Then the support of \( \\widetilde{\\varphi } \) is contained in Supp \( \\varphi \) -Supp \( S \) and \( \\left( {\\operatorname{Supp}\\varphi \\text{-Supp}S}\\right) \\cap \\operatorname{Supp}T \) is the projection of \( K \) on the first factor. Therefore \( \\operatorname{Supp}\\widetilde{\\varphi } \\cap \\operatorname{Supp}T \) is compact. At the same time, if \( {\\rho }_{l} \\in \\mathcal{D}\\left( {\\mathbb{R}}^{d}\\right) \) satisfies \( {\\rho }_{l} = 1 \) on \( B\\left( {0, l}\\right) \), the function\n\n\[ \n{\\rho }_{l}\\widetilde{\\varphi } : x \\mapsto \\left\\langle {{S}_{y},{\\rho }_{l}\\left( x\\right) \\varphi \\left( {x + y}\\right) }\\right\\rangle \n\]\n\nbelongs to \( \\mathcal{D}\\left( {\\mathbb{R}}^{d}\\right) \), by Theorem 1.1. Therefore \( \\widetilde{\\varphi } \) is of class \( {C}^{\\infty } \) on \( B\\left( {0, l}\\right) \) for every \( l > 0 \), which is to say that \( \\widetilde{\\varphi } \\in \\mathcal{E}\\left( {\\mathbb{R}}^{d}\\right) \) .\n\nAt the same time, by Proposition 2.8,\n\n\[ \n\\langle T * S,\\varphi \\rangle = \\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\mathop{\\lim }\\limits_{{{l}^{\\prime } \\rightarrow + \\infty }}\\left\\langle {{\\rho }_{l}T * {\\rho }_{{l}^{\\prime }}S,\\varphi }\\right\\rangle \n\]\n\n\[ \n= \\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\mathop{\\lim }\\limits_{{{l}^{\\prime } \\rightarrow + \\infty }}\\left\\langle {{T}_{x},{\\rho }_{l}\\left( x\\right) \\left\\langle {{S}_{y},{\\rho }_{{l}^{\\prime }}\\left( y\\right) \\varphi \\left( {x + y}\\right) }\\right\\rangle }\\right\\rangle .\n\]\n\nNow, if \( B\\left( {0,{l}^{\\prime }}\\right) \\supset \\operatorname{Supp}\\varphi - \\operatorname{Supp}{\\rho }_{l} \), we have\n\n\[ \n\\operatorname{Supp}\\left( {\\varphi \\left( {x + \\cdot }\\right) }\\right) \\subset B\\left( {0,{l}^{\\prime }}\\right) \\;\\text{ for every }x \\in \\operatorname{Supp}{\\rho }_{l}.\n\]\n\nTherefore \( {\\rho }_{{l}^{\\prime }}\\left( y\\right) \\varphi \\left( {x + y}\\right) = \\varphi \\left( {x + y}\\right) \) . We deduce that\n\n\[ \n\\langle T * S,\\varphi \\rangle = \\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\left\\langle {{T}_{x},{\\rho }_{l}\\left( x\\right) \\left\\langle {{S}_{y},\\varphi \\left( {x + y}\\right) }\\right\\rangle }\\right\\rangle .\n\]\n\nBy definition, if \( B\\left( {0, l}\\right) \\supset \\operatorname{Supp}\\widetilde{\\varphi } \\cap \\operatorname{Supp}T \), then\n\n\[ \n\\left\\langle {{T}_{x},{\\rho }_{l}\\left( x\\right) \\left\\langle {{S}_{y},\\varphi \\left( {x + y}\\right) }\\right\\rangle }\\right\\rangle = \\left\\langle {{T}_{x},\\left\\langle {{S}_{y},\\varphi \\left( {x + y}\\right) }\\right\\rangle }\\right\\rangle ,\n\]\n\nwhich proves the result.
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Yes
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Corollary 2.10 Let \( f \) and \( g \) be elements of \( {L}_{\mathrm{{loc}}}^{1}\left( {\mathbb{R}}^{d}\right) \) whose supports satisfy condition (C). Then \( f \) and \( g \) are convolvable in the sense of the definition on page 171; moreover \( f * g \in {L}_{\mathrm{{loc}}}^{1}\left( {\mathbb{R}}^{d}\right) \) and \[ \left\lbrack f\right\rbrack * \left\lbrack g\right\rbrack = \left\lbrack {f * g}\right\rbrack \]
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Proof. For every \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) , \[ \iint \left| {f\left( {x - y}\right) }\right| \left| {g\left( y\right) }\right| \left| {\varphi \left( x\right) }\right| {dxdy} = \iint \left| {f\left( x\right) }\right| \left| {g\left( y\right) }\right| \left| {\varphi \left( {x + y}\right) }\right| {dxdy} \] (because Lebesgue measure is invariant under translations); the term on the right is finite because the supports of \( f \) and \( g \) satisfy condition (C). This proves that \( f \) and \( g \) are convolvable and that \( f * g \in {L}_{\text{loc }}^{1}\left( {\mathbb{R}}^{d}\right) \) . Moreover, if \( \varphi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \), we have \[ \langle \left\lbrack {f * g}\right\rbrack ,\varphi \rangle = \int f\left( x\right) \left( {\int g\left( y\right) \varphi \left( {x + y}\right) {dy}}\right) {dx} \] by Fubini’s Theorem, and this quantity equals \( \langle \left\lbrack f\right\rbrack * \left\lbrack g\right\rbrack ,\varphi \rangle \) by Proposition 2.9.
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Yes
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Proposition 2.11 (Associativity) Let \( \left( {{T}_{1},{T}_{2},{T}_{3}}\right) \) be a family of distributions on \( {\mathbb{R}}^{d} \) satisfying (C). The distributions \( \left( {{T}_{1} * {T}_{2}}\right) * {T}_{3} \) and \( {T}_{1} * \left( {{T}_{2} * {T}_{3}}\right) \) are well-defined and coincide.
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Proof. By property 1 on page 326, the distributions \( {T}_{1} * {T}_{2} \) and \( {T}_{2} * {T}_{3} \) are well defined and, by Proposition 2.7,\n\n\( \operatorname{Supp}\left( {{T}_{1} * {T}_{2}}\right) \subset \operatorname{Supp}{T}_{1} + \operatorname{Supp}{T}_{2},\;\operatorname{Supp}\left( {{T}_{2} * {T}_{3}}\right) \subset \operatorname{Supp}{T}_{2} + \operatorname{Supp}{T}_{3}. \)\n\nIt follows then from properties 1 and 7 on page 326 that the distributions \( \left( {{T}_{1} * {T}_{2}}\right) * {T}_{3} \) and \( {T}_{1} * \left( {{T}_{2} * {T}_{3}}\right) \) are well defined.\n\nIn view of this we obtain, by applying Proposition 2.8 several times,\n\n\[ \left( {{T}_{1} * {T}_{2}}\right) * {T}_{3} = \mathop{\lim }\limits_{{{l}_{1} \rightarrow + \infty }}\mathop{\lim }\limits_{{{l}_{2} \rightarrow + \infty }}\mathop{\lim }\limits_{{{l}_{3} \rightarrow + \infty }}\left( {{\rho }_{{l}_{1}}{T}_{1} * {\rho }_{{l}_{2}}{T}_{2}}\right) * {\rho }_{{l}_{3}}{T}_{3} \]\n\nwhere, for \( l > 0 \), we have \( {\rho }_{l} \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) and \( {\rho }_{l} = 1 \) on \( B\left( {0, l}\right) \) . Because the convolution product is associative in \( {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) (Proposition 2.2), we get\n\n\[ \left( {{\rho }_{{l}_{1}}{T}_{1} * {\rho }_{{l}_{2}}{T}_{2}}\right) * {\rho }_{{l}_{3}}{T}_{3} = {\rho }_{{l}_{1}}{T}_{1} * \left( {{\rho }_{{l}_{2}}{T}_{2} * {\rho }_{{l}_{3}}{T}_{3}}\right) . \]\n\nThen it suffices to use Proposition 2.8 several times again to obtain\n\n\[ \left( {{T}_{1} * {T}_{2}}\right) * {T}_{3} = {T}_{1} * \left( {{T}_{2} * {T}_{3}}\right) \]
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Yes
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Proposition 2.12 If \( \\left( {{T}_{1},\\ldots ,{T}_{n}}\\right) \) satisfies condition (C), we have\n\n\[ \n{D}_{j}\\left( {{T}_{1} * \\cdots * {T}_{n}}\\right) = {T}_{1} * \\cdots * {T}_{k - 1} * {D}_{j}{T}_{k} * {T}_{k + 1} * \\cdots * {T}_{n} \n\]\n\nfor all \( j \\in \\{ 1,\\ldots, d\\} \) and \( k \\in \\{ 1,\\ldots, n\\} \) . This remains so if we replace \( {D}_{j} \) by an arbitrary differential operator of the form \( P\\left( D\\right) \) .
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Proof. Note first that \( \\operatorname{Supp}{D}_{j}{T}_{k} \\subset \\operatorname{Supp}{T}_{k} \), so the two sides in the equality above are well defined (see property 1 on page 326). By associativity and commutativity, it suffices to show that, if \( \\left( {T, S}\\right) \) satisfies \( \\left( \\mathrm{C}\\right) \), then \( {D}_{j}\\left( {T * S}\\right) = \\left( {{D}_{j}T}\\right) * S \) . We already know this is so wher. \( T \) and \( S \) have compact support (Proposition 2.3 on page 325). The general case follows by passing to the limit, using Proposition 2.8 and the continuity in \( {\\mathcal{D}}^{\\prime }\\left( {\\mathbb{R}}^{d}\\right) \) of the map \( T \\mapsto {D}_{j}T \) (as well as the formula for the derivative of a product):\n\n\[ \n{D}_{j}\\left( {T * S}\\right) = \\mathop{\\lim }\\limits_{{{l}^{\\prime } \\rightarrow + \\infty }}\\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}{D}_{j}\\left( {{\\rho }_{l}T * {\\rho }_{{l}^{\\prime }}S}\\right) \n\]\n\n\[ \n= \\mathop{\\lim }\\limits_{{{l}^{\\prime } \\rightarrow + \\infty }}\\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\left( {{\\rho }_{l}{D}_{j}T * {\\rho }_{{l}^{\\prime }}S}\\right) + \\mathop{\\lim }\\limits_{{{l}^{\\prime } \\rightarrow + \\infty }}\\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\left( {\\left( {{D}_{j}{\\rho }_{l}}\\right) T * {\\rho }_{{l}^{\\prime }}S}\\right) \n\]\n\n\[ \n= {D}_{j}T * S \n\]\n\nwhere the latter equality comes from the fact that \( \\mathop{\\lim }\\limits_{{l \\rightarrow + \\infty }}\\left( {{D}_{j}{\\rho }_{l}}\\right) T = 0 \) .
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Yes
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Consider \( T \in {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) and \( f \in \mathcal{E}\left( {\mathbb{R}}^{d}\right) \), and suppose \( \left( {T, f}\right) \) satisfies condition (C). Then \( T * f \in \mathcal{E}\left( {\mathbb{R}}^{d}\right) \) and, for all \( x \in {\mathbb{R}}^{d} \), the intersection \( \operatorname{Supp}f\left( {x - \cdot }\right) \cap \operatorname{Supp}T \) is compact and \[ T * f\left( x\right) = \left\langle {{T}_{y}, f\left( {x - y}\right) }\right\rangle .
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Proof. For each \( l > 0 \), we again fix an element \( {\rho }_{l} \) of \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) equal to 1 on \( B\left( {0, l}\right) \) . Take \( T \in {\mathcal{D}}^{\prime m}\left( {\mathbb{R}}^{d}\right) \) and \( f \in {\mathcal{E}}^{m + r}\left( {\mathbb{R}}^{d}\right) \) (or \( T \in {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) and \( f \in \mathcal{E}\left( {\mathbb{R}}^{d}\right) ) \), and suppose that \( \left( {T, f}\right) \) satisfies (C). For every compact subset \( K \) of \( {\mathbb{R}}^{d} \), the set \[ \widetilde{K} = \{ \left( {x, y}\right) \in \operatorname{Supp}f \times \operatorname{Supp}T : x + y \in K\} \] is a compact subset of \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} \) . Denote by \( {K}^{\prime } \) its projection on the second factor; then \( {K}^{\prime } \) is compact in \( {\mathbb{R}}^{d} \) . For every \( x \in K \) , \[ \operatorname{Supp}\left( {f\left( {x - \cdot }\right) }\right) \cap \operatorname{Supp}T = \left( {x - \operatorname{Supp}f}\right) \cap \operatorname{Supp}T \] \[ = \{ y \in \operatorname{Supp}T : \exists z \in \operatorname{Supp}f\text{ such that }y + z = x\} \] \[ \subset {K}^{\prime }\text{.} \] Now take \( l > \mathop{\max }\limits_{{x \in K}}\left| x\right| + \mathop{\max }\limits_{{y \in {K}^{\prime }}}\left| y\right| \) . For every \( x \in K \), the function \( y \mapsto {\rho }_{l}\left( y\right) {\rho }_{l}\left( {x - y}\right) \) equals 1 on an open that contains \( {K}^{\prime } \), so \[ \left\langle {{T}_{y}, f\left( {x - y}\right) }\right\rangle = \left\langle {{T}_{y},{\rho }_{l}\left( y\right) {\rho }_{l}\left( {x - y}\right) f\left( {x - y}\right) }\right\rangle \;\text{ for all }x \in K. \] \( \left( *\right) \) Since the function \( \left( {x, y}\right) \mapsto {\rho }_{l}\left( y\right) {\rho }_{l}\left( {x - y}\right) f\left( {x - y}\right) \) lies in \( {\mathcal{D}}^{m + r}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) \) (or \( \mathcal{D}\left( {{\mathbb{R}}^{d} \times {\mathbb{R}}^{d}}\right) \), as the case may be), we deduce from Theorem 1.1 that the function \( x \mapsto \left\langle {{T}_{y}, f\left( {x - y}\right) }\right\rangle \) is of class \( {C}^{r} \) (or \( {C}^{\infty } \) ) in \( \widetilde{K} \) . This reasoning is valid for every compact subset \( K \) of \( {\mathbb{R}}^{d} \), so the function belongs to \( {\mathcal{E}}^{r}\left( {\mathbb{R}}^{d}\right) \) (or \( \mathcal{E}\left( {\mathbb{R}}^{d}\right) \) ).
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Yes
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Proposition 2.14 For every open \( \Omega \) in \( {\mathbb{R}}^{d} \), the set \( \mathcal{D}\left( \Omega \right) \) is dense in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) . In other words, every distribution on \( \Omega \) is the limit in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) of a sequence of elements of \( \mathcal{D}\left( \Omega \right) \) .
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Proof. Let \( \Omega \) be open in \( {\mathbb{R}}^{d} \), and let \( {\left( {K}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence of compact sets exhausting \( \Omega \) . For every \( n \in \mathbb{N} \), take \( {\varphi }_{n} \in \mathcal{D}\left( \Omega \right) \) such that \( {\varphi }_{n} = 1 \) on \( {K}_{n} \) . Also let \( {\left( {\chi }_{n}\right) }_{n \in \mathbb{N}} \) be a smoothing sequence in \( {\mathbb{R}}^{d} \) and \( {\left( {\chi }_{{p}_{n}}\right) }_{n \in \mathbb{N}} \) a subsequence such that \( \operatorname{Supp}{\varphi }_{n} + \operatorname{Supp}{\chi }_{{p}_{n}} \subset \Omega \) for every \( n \in \mathbb{N} \) .\n\nTake \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) and write \( {\psi }_{n} = \left( {{\varphi }_{n}T}\right) * {\chi }_{{p}_{n}} \) for every integer \( n \in \mathbb{N} \) . (The distribution \( {\varphi }_{n}T \) has compact support in \( \Omega \) and so can be identified with a distribution on \( {\mathbb{R}}^{d} \) with compact support, as explained on page 283; see particularly Equation \( \left( *\right) \) on that page. Thus the convolution product \( \left( {{\varphi }_{n}T}\right) * {\chi }_{{p}_{n}} \) does make sense.) By Proposition 2.13 and our assumption on the supports of \( {\varphi }_{n} \) and \( {\chi }_{{p}_{n}} \), we have \( {\psi }_{n} \in \mathcal{D}\left( \Omega \right) \) . We will show that the sequence \( {\left( {\psi }_{n}\right) }_{n \in \mathbb{N}} \) converges in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) to \( T \), and this will prove the proposition.\n\nTo do this, take \( \varphi \in \mathcal{D}\left( \Omega \right) \) . By definition,\n\n\[ \left\langle {{\psi }_{n},\varphi }\right\rangle = \left\langle {{T}_{x},{\varphi }_{n}\left( x\right) \int {\chi }_{{p}_{n}}\left( y\right) \varphi \left( {x + y}\right) {dy}}\right\rangle = \left\langle {T,{\varphi }_{n}\left( {\varphi * {\check{\chi }}_{{p}_{n}}}\right) }\right\rangle . \]\n\nNow, for \( n \) large enough, \( \operatorname{Supp}\left( {\varphi * {\check{\chi }}_{{p}_{n}}}\right) \subset \operatorname{Supp}\varphi - \operatorname{Supp}{\chi }_{{p}_{n}} \subset {K}_{n} \), so \( {\varphi }_{n}\left( {\varphi * {\check{\chi }}_{{p}_{n}}}\right) = \varphi * {\check{\chi }}_{{p}_{n}} \), whence\n\n\[ \left\langle {{\psi }_{n},\varphi }\right\rangle = \left\langle {T,\varphi * {\check{\chi }}_{{p}_{n}}}\right\rangle \]\n\nThe sequence \( {\left( \varphi * {\check{\chi }}_{{p}_{n}}\right) }_{n \in \mathbb{N}} \) converges to \( \varphi \) in \( \mathcal{D}\left( \Omega \right) \), by Proposition 1.2 of page 261 applied to every \( m \in \mathbb{N} \), since \( {\left( {\check{\chi }}_{n}\right) }_{n \in \mathbb{N}} \) is also a smoothing sequence. Therefore the sequence \( {\left( {\psi }_{n}\right) }_{n \in \mathbb{N}} \) converges to \( T \) in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) .
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Yes
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Proposition 3.1 If \( T \in {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \), then\n\n\[ T = \frac{1}{{s}_{d}}\mathop{\sum }\limits_{{j = 1}}^{d}\left( \frac{{x}_{j}}{{r}^{d}}\right) * {D}_{j}T \]\n\nwhere \( r = \left| x\right| \) and \( {s}_{d} \) is the area of the unit sphere in \( {\mathbb{R}}^{d} \) .
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Proof. Let \( E \) be the fundamental solution of the Laplacian given in Theorem 3.2 on page 308. A simple calculation using Theorem 2.12 on page 301 shows that\n\n\[ {D}_{j}E = \frac{1}{{s}_{d}}\frac{{x}_{j}}{{r}^{d}}\;\text{ for all }j \in \{ 1,\ldots, d\} \]\n\nand this in any dimension \( d \) . At the same time, \( {\Delta E} * T = T \), since \( {\Delta E} = \delta \) . Since \( T \) has compact support (so that \( \left( {E, T}\right) \) satisfies \( \left( \mathrm{C}\right) \) ), we deduce from Proposition 2.12 that\n\n\[ {\Delta E} * T = \Delta \left( {E * T}\right) = \mathop{\sum }\limits_{{j = 1}}^{d}{D}_{j}^{2}\left( {E * T}\right) = \mathop{\sum }\limits_{{j = 1}}^{d}\left( {{D}_{j}E}\right) * \left( {{D}_{j}T}\right) . \]\n\nTherefore,\n\n\[ T = \mathop{\sum }\limits_{{j = 1}}^{d}\left( {{D}_{j}E}\right) * \left( {{D}_{j}T}\right) \]\n\nwhich yields the result.
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Yes
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Proposition 3.2 The norm \( \parallel \cdot {\parallel }_{1, p} \) makes \( {W}^{1, p}\left( {\mathbb{R}}^{d}\right) \) into a Banach space.
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Proof. Let \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) be a Cauchy sequence in \( {W}^{1, p}\left( {\mathbb{R}}^{d}\right) \) . Since the space \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \) is complete, the sequences \( \left( {f}_{n}\right) ,\left( {{D}_{1}{f}_{n}}\right) ,\ldots ,\left( {{D}_{d}{f}_{n}}\right) \), which are clearly Cauchy sequences in \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \), converge in \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \) . Let \( f,{g}_{1},\ldots ,{g}_{d} \) be their limits in \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \) . Since \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) is contained in \( {L}^{{p}^{\prime }}\left( {\mathbb{R}}^{d}\right) \) (where \( {p}^{\prime } \) is the exponent conjugate to \( p \) ), we deduce easily from Hölder’s inequality\n\nthat the same sequences also converge in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) . Since the operators \( {D}_{j} \) are continuous in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \), we deduce that\n\n\[ \n{D}_{j}f = \mathop{\lim }\limits_{{n \rightarrow + \infty }}{D}_{j}{f}_{n} = {g}_{j}\;\text{ for }1 \leq j \leq d, \n\]\n\nby the uniqueness of the limit in \( {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) . This shows that \( f \in {W}^{1, p}\left( {\mathbb{R}}^{d}\right) \) and that the sequence \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) converges to \( f \) in \( {W}^{1, p}\left( {\mathbb{R}}^{d}\right) \) .
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Yes
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Theorem 3.4 Let \( T \) be a distribution on an open set \( \Omega \) in \( {\mathbb{R}}^{d} \) . Suppose that \( p \in \left\lbrack {1,\infty }\right\rbrack \) and that \( {D}_{j}T \in {L}_{\text{loc }}^{p}\left( \Omega \right) \) for every \( j \in \{ 1,\ldots, d\} \) . - If \( p \leq d \), then \( T \in {L}_{\text{loc }}^{r}\left( \Omega \right) \) for every \( r \in \lbrack p,{pd}/\left( {d - p}\right) ) \) . - If \( p > d \), then \( T \in C\left( \Omega \right) \) . If \( p = d \), we interpret \( {pd}/\left( {d - p}\right) \) as \( \infty \) .
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Proof. Let \( K \) be a compact subset of \( \Omega \) and \( {K}^{\prime } \) a compact subset of \( \Omega \) whose interior contains \( K \) . Let \( \varphi \in \mathcal{D}\left( \Omega \right) \) be such that \( \varphi = 1 \) on \( {K}^{\prime } \) . Put \( \mathfrak{d} = d\left( {K,\Omega \smallsetminus {\mathring{K}}^{\prime }}\right) > 0 \) and let \( \gamma \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) be such that \( \gamma = 1 \) in a neighborhood of 0 and Supp \( \gamma \subset B\left( {0,\mathfrak{d}/2}\right) \) . If \( E \) is the fundamental solution of the Laplacian provided by Theorem 3.2 on page 308 , we saw in the proof of Theorem 3.3 that there exist \( \eta ,{\eta }_{1},\ldots ,{\eta }_{d} \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) such that \[ \Delta \left( {\gamma E}\right) = \eta + \delta ,\;{D}_{j}\left( {\gamma E}\right) = {\eta }_{j} + \frac{1}{{s}_{d}}\gamma \frac{{x}_{j}}{{r}^{d}}\;\text{ for all }j \in \{ 1,\ldots, d\} . \] Using formula \( \left( {* * }\right) \) from the previous page and replacing \( T \) by \( {\varphi T} \) (considered as a distribution on \( {\mathbb{R}}^{d} \) : see page 283), we obtain \[ {\varphi T} = - \left( {\varphi T}\right) * \eta + \mathop{\sum }\limits_{{j = 1}}^{d}{D}_{j}\left( {\varphi T}\right) * {\eta }_{j} \] \[ + \frac{1}{{s}_{d}}\mathop{\sum }\limits_{{j = 1}}^{d}\left( {\left( {{D}_{j}\varphi }\right) T}\right) * \left( {\gamma \frac{{x}_{j}}{{r}^{d}}}\right) + \frac{1}{{s}_{d}}\mathop{\sum }\limits_{{j = 1}}^{d}\left( {\varphi {D}_{j}T}\right) * \left( {\gamma \frac{{x}_{j}}{{r}^{d}}}\right) . \] By Proposition 2.13, \( \left( {\varphi T}\right) * \eta \) and the \( {D}_{j}\left( {\varphi T}\right) * {\eta }_{j} \), for every \( j \), belong to \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) . At the same time, \[ \operatorname{Supp}\left( {\left( {{D}_{j}\varphi }\right) T * \left( {\gamma \frac{{x}_{j}}{{r}^{d}}}\right) }\right) \subset \left( {{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) + \bar{B}\left( {0,\mathfrak{d}/2}\right) \subset \left( {{\mathbb{R}}^{d} \smallsetminus K}\right) . \] Finally, \( \varphi {D}_{j}T \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) . We then apply Young’s inequality (Theorem 3.4 on page 172) as in the proof of Theorem 3.3. We conclude that \[ {\Omega }_{K} = {\mathbb{R}}^{d} \smallsetminus \left( {\left( {{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) + \bar{B}\left( {0,\mathfrak{d}/2}\right) }\right) \] is an open set satisfying \( K \subset {\Omega }_{K} \subset {K}^{\prime } \) and that the restriction of \( T \) to \( {\Omega }_{K} \) belongs to \( {L}^{r}\left( {\Omega }_{K}\right) \) if \( p \leq d \) and \( r \in \lbrack p,{pd}/\left( {p - d}\right) ) \) and to \( C\left( {\Omega }_{K}\right) \) if \( p > d \) . Since this happens for every compact \( K \), the theorem is proved.
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Yes
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Theorem 3.5 Any differential operator with constant coefficients having a fundamental solution whose restriction to \( {\mathbb{R}}^{d} \smallsetminus \{ 0\} \) is a function of class \( {C}^{\infty } \) is hypoelliptic.
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Proof. The proof is analogous to that of Theorem 3.4. Let \( \Omega \) be an open subset of \( {\mathbb{R}}^{d} \) and let \( K,{K}^{\prime } \) be compact subsets of \( \Omega \) such that \( K \subset {K}^{\prime } \) . Write \( \mathfrak{d} = d\left( {K,{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) \) . Let \( \varphi \in \mathcal{D}\left( \Omega \right) \) have the value 1 on \( {K}^{\prime } \) and let \( \gamma \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) have the value 1 on a neighborhood of 0 ; assume also that Supp \( \gamma \subset \widehat{B}\left( {0,\mathfrak{d}/2}\right) \) . Finally, set\n\n\[ \n{\Omega }_{K} = {\mathbb{R}}^{d} \smallsetminus \left( {\left( {{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) + \bar{B}\left( {0,\mathfrak{d}/2}\right) }\right) .\n\]\n\nThen\n\n\[ \nK \subset {\Omega }_{K} \subset {K}^{\prime }\n\]\n\nConsider a differential operator \( P\left( D\right) \) having a fundamental solution \( E \) of class \( {C}^{\infty } \) on \( {\mathbb{R}}^{d} \smallsetminus \{ 0\} \) . Let \( T \in {\mathcal{D}}^{\prime }\left( \Omega \right) \) be such that \( f = P\left( D\right) T \in \mathcal{E}\left( \Omega \right) \) . By Leibniz's formula,\n\n\[ \nP\left( D\right) \left( {\gamma E}\right) = \delta + \eta \;\text{ with }\eta \in \mathcal{D}\left( {\mathbb{R}}^{d}\right)\n\]\n\nand\n\n\[ \nP\left( D\right) \left( {\varphi T}\right) = {\varphi f} + S\text{ with }\operatorname{Supp}S \subset \left( {{\mathbb{R}}^{d} \smallsetminus {\mathring{K}}^{\prime }}\right) .\n\]\n\nThen\n\n\[ \nP\left( D\right) \left( {{\gamma E} * {\varphi T}}\right) = {\varphi T} + {\varphi T} * \eta = {\gamma E} * {\varphi f} + {\gamma E} * S,\n\]\n\nthat is,\n\n\[ \n{\varphi T} = - \left( {\varphi T}\right) * \eta + {\gamma E} * {\varphi f} + {\gamma E} * S.\n\]\n\nSince \( \eta \) and \( {\varphi f} \) belong to \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \), we deduce from Proposition 2.13 that \( - \left( {\varphi T}\right) * \eta + {\gamma E} * {\varphi f} \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) . On the other hand, \( \operatorname{Supp}\left( {{\gamma E} * S}\right) \subset {\mathbb{R}}^{d} \smallsetminus {\Omega }_{K} \) . We deduce that the restriction of \( T \) to \( {\Omega }_{K} \) is of class \( {C}^{\infty } \) . Since \( K \) is arbitrary and \( {\Omega }_{K} \supset K \), this implies that \( T \in \mathcal{E}\left( \Omega \right) \) .
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Yes
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Consider a linear differential operator \( P\left( D\right) \) with constant coefficients, a fundamental solution \( E \) of \( P\left( D\right) \), and \( S \in {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \). The distribution \( {T}_{0} = E * S \) satisfies \( P\left( D\right) {T}_{0} = S \). Moreover, the set of solutions \( T \in {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) \) of the equation \( P\left( D\right) T = S \) equals \( \left\{ {T = {T}_{0} + U : U \in {\mathcal{D}}^{\prime }\left( {\mathbb{R}}^{d}\right) }\right. \) such that \( \left. {P\left( D\right) U = 0}\right\} \).
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If \( S \in {\mathcal{E}}^{\prime }\left( {\mathbb{R}}^{d}\right) \), Proposition 2.12 yields \( P\left( D\right) \left( {E * S}\right) = P\left( D\right) E * S = \delta * S = S \). Set \( U = T - E * S \). Clearly, \( P\left( D\right) T = S \) if and only if \( P\left( D\right) U = 0 \).
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Yes
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Proposition 1.2 Suppose that \( \Omega = \left( {a, b}\right) \), with \( - \infty \leq a < b \leq + \infty \) . Every element \( f \) of \( {H}^{1}\left( \Omega \right) \) has a continuous representative on \( \Omega \) (still denoted by \( f \) ) that has finite limits at \( a \) and \( b \) . Moreover, if \( a = - \infty \), we have \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = 0 \) ; similarly, if \( b = + \infty \), we have \( \mathop{\lim }\limits_{{x \rightarrow b}}f\left( x\right) = 0 \) .
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Proof. By Theorem 2.8 on page 297, every element of \( {H}^{1}\left( \Omega \right) \) has a continuous representative \( f \) satisfying, for \( \alpha \in \Omega \) ,\n\n\[ f\left( t\right) = f\left( \alpha \right) + {\int }_{\alpha }^{t}{f}^{\prime }\left( u\right) {du}\;\text{ for all }t \in \Omega . \]\n\n(*) \n\nIf, for example, \( b < + \infty \), then \( {L}^{2}\left( \left( {\alpha, b}\right) \right) \subset {L}^{1}\left( \left( {\alpha, b}\right) \right) \), so \( {f}^{\prime } \in {L}^{1}\left( \left( {\alpha, b}\right) \right) \) . Therefore \( f \) does have a finite limit at \( b \) . Similarly, \( f \) certainly has a finite limit at \( a \) if \( a > - \infty \) .\n\nNow suppose that \( b = + \infty \) . Multiplying equality \( \left( *\right) \) by \( {f}^{\prime }\left( t\right) \) and integrating the resulting equality between \( \alpha \) and \( x \), we conclude that, if \( x > \alpha \),\n\n\[ {\int }_{\alpha }^{x}f\left( t\right) {f}^{\prime }\left( t\right) {dt} = f\left( \alpha \right) \left( {f\left( x\right) - f\left( \alpha \right) }\right) + {\int }_{\alpha }^{x}\left( {{\int }_{\alpha }^{t}{f}^{\prime }\left( u\right) {du}}\right) {f}^{\prime }\left( t\right) {dt}. \]\n\nBy Fubini's Theorem,\n\n\[ {\int }_{\alpha }^{x}\left( {{\int }_{\alpha }^{t}{f}^{\prime }\left( u\right) \;{du}}\right) {f}^{\prime }\left( t\right) \;{dt} = {\iint }_{{\left\lbrack \alpha, x\right\rbrack }^{2}}{1}_{\{ u \leq t\} }{f}^{\prime }\left( u\right) {f}^{\prime }\left( t\right) \;{du}\;{dt} \]\n\n\[ = \frac{1}{2}{\iint }_{{\left\lbrack \alpha, x\right\rbrack }^{2}}{f}^{\prime }\left( u\right) {f}^{\prime }\left( t\right) {dudt} = \frac{1}{2}{\left( f\left( x\right) - f\left( \alpha \right) \right) }^{2}. \]\n\nWe deduce that\n\n\[ {f}^{2}\left( x\right) = {f}^{2}\left( \alpha \right) + 2{\int }_{\alpha }^{x}f\left( t\right) {f}^{\prime }\left( t\right) {dt}. \]\n\n\( \left( {* * }\right) \)\n\nSince the two functions \( f \) and \( {f}^{\prime } \) belong to \( {L}^{2}\left( \Omega \right) \), their product \( f{f}^{\prime } \) belongs to \( {L}^{1}\left( \Omega \right) \) by the Schwarz inequality; therefore, by \( \left( {* * }\right) ,{f}^{2} \) has a finite limit at \( + \infty \) . Since \( {f}^{2} \in {L}^{1}\left( \Omega \right) \), this limit can only be 0 . The reasoning is similar if \( a = - \infty \) .
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Yes
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Proposition 1.3 Suppose that \( \Omega = \left( {a, b}\right) \), with \( - \infty < a < b < + \infty \) . Then \( {C}^{1}\left( \bar{\Omega }\right) \) is a dense subspace of \( {H}^{1}\left( \Omega \right) \) .
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Proof. Clearly \( {C}^{1}\left( \left\lbrack {a, b}\right\rbrack \right) \) is a subspace of \( {H}^{1}\left( \Omega \right) \) . Consider an element of \( {H}^{1}\left( \Omega \right) \), having a continuous representative \( f \) . By the preceding proposition (and Theorem 2.8 on page 297), \( f \) has a continuous extension to \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ f\left( x\right) = f\left( a\right) + {\int }_{a}^{x}{f}^{\prime }\left( t\right) {dt}\;\text{ for all }x \in \left\lbrack {a, b}\right\rbrack . \]\n\nSince \( {C}_{c}\left( \Omega \right) \) is dense in \( {L}^{2}\left( \Omega \right) \), the derivative \( {f}^{\prime } \) is the limit in \( {L}^{2}\left( \Omega \right) \) of a sequence \( {\left( {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) of elements of \( {C}_{c}\left( \Omega \right) \) . For each \( n \in \mathbb{N} \), set\n\n\[ {f}_{n}\left( x\right) = f\left( a\right) + {\int }_{a}^{x}{\varphi }_{n}\left( t\right) {dt} \]\n\nClearly \( {f}_{n} \in {C}^{1}\left( \left\lbrack {a, b}\right\rbrack \right) \) and, for every \( x \in \left\lbrack {a, b}\right\rbrack \) ,\n\n\[ \left| {f\left( x\right) - {f}_{n}\left( x\right) }\right| \leq {\int }_{a}^{b}\left| {{f}^{\prime }\left( t\right) - {\varphi }_{n}\left( t\right) }\right| {dt} \leq \sqrt{b - a}{\begin{Vmatrix}{f}^{\prime } - {\varphi }_{n}\end{Vmatrix}}_{{L}^{2}}, \]\n\nby the Schwarz inequality. Thus \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) is a sequence in \( {C}^{1}\left( \left\lbrack {a, b}\right\rbrack \right) \) that converges uniformly, and so in \( {L}^{2}\left( \Omega \right) \), to the element \( f \) . Since, in addition, \( {f}_{n}^{\prime } = {\varphi }_{n} \) for every \( n \), which implies that the sequence \( {\left( {f}_{n}^{\prime }\right) }_{n \in \mathbb{N}} \) converges to \( {f}^{\prime } \) in \( {L}^{2}\left( \Omega \right) \), we deduce that \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) converges to \( f \) in \( {H}^{1}\left( \Omega \right) \) .
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Yes
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Proposition 1.5 The spaces \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) and \( {H}_{0}^{1}\left( {\mathbb{R}}^{d}\right) \) coincide.
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Proof. We must show that \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) is dense in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) . Take \( \xi \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) such that \( \xi \left( 0\right) = 1 \) . For each \( n \in {\mathbb{N}}^{ * } \), put \( {\xi }_{n}\left( x\right) = \xi \left( {x/n}\right) \) . If \( f \in {H}^{1}\left( {\mathbb{R}}^{d}\right) \), then \( {\xi }_{n}f \in {H}^{1}\left( {\mathbb{R}}^{d}\right) \) and, by the Dominated Convergence Theorem, the sequence \( {\left( {\xi }_{n}f\right) }_{n \in {\mathbb{N}}^{ * }} \) converges to \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . Moreover, for each \( j \in \{ 1,\ldots, d\} \) ,\n\n\[ \n{D}_{j}\left( {{\xi }_{n}f}\right) \left( x\right) = {\xi }_{n}\left( x\right) {D}_{j}f\left( x\right) + \frac{1}{n}\left( {{D}_{j}\xi }\right) \left( \frac{x}{n}\right) f\left( x\right) ,\n\]\n\nso, again by Dominated Convergence, the sequence \( {\left( {D}_{j}\left( {\xi }_{n}f\right) \right) }_{n \in \mathbb{N}} \) converges to \( {D}_{j}f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . Therefore the sequence \( {\left( {\xi }_{n}f\right) }_{n \in {\mathbb{N}}^{ * }} \) converges to \( f \) in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) . Consequently, the space \( {H}_{c}^{1}\left( {\mathbb{R}}^{d}\right) \) consisting of elements of \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) with compact support is dense in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) .\n\nNow take \( f \in {H}_{c}^{1}\left( {\mathbb{R}}^{d}\right) \) and let \( {\left( {\chi }_{n}\right) }_{n \in \mathbb{N}} \) be a smoothing sequence. Then, for every \( n \in \mathbb{N} \), the convolution \( f * {\chi }_{n} \) is a function of class \( {C}^{\infty } \) on \( {\mathbb{R}}^{d} \) (this follows from the theorems on differentiation under the summation sign) whose support is contained in \( \operatorname{Supp}f + \operatorname{Supp}{\chi }_{n} \) . Therefore \( f * {\chi }_{n} \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) . On the other hand, by Proposition 3.7 on page 174, the sequence \( {\left( {\chi }_{n} * f\right) }_{n \in \mathbb{N}} \) converges to \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . Since, in addition, \( {D}_{j}\left( {f * {\chi }_{n}}\right) = \left( {{D}_{j}f}\right) * {\chi }_{n} \) (by Corollary 2.10 on page 330 and Proposition 2.12 on page 331), we again deduce from Proposition 3.7 on page 174 that the sequence \( {\left( {D}_{j}\left( f * {\chi }_{n}\right) \right) }_{n \in \mathbb{N}} \) converges to \( {D}_{j}f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . It follows that \( {\left( f * {\chi }_{n}\right) }_{n \in \mathbb{N}} \) converges to \( f \) in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \), which concludes the proof.
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Yes
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Proposition 1.6 (Poincaré inequality) If \( \Omega \) is a bounded open set in \( {\mathbb{R}}^{d} \) (more generally, if one of the projections of \( \Omega \) on the coordinate axes is bounded), there exists a constant \( C \geq 0 \) depending only on \( \Omega \) and such that \[ \parallel u{\parallel }_{{L}^{2}\left( \Omega \right) } \leq C{\begin{Vmatrix}\left| \nabla u\right| \end{Vmatrix}}_{{L}^{2}\left( \Omega \right) }\;\text{ for all }u \in {H}_{0}^{1}\left( \Omega \right) . \] If \( \Omega \) is bounded, we can take \( C = d\left( \Omega \right) \) .
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Proof. By denseness, we just have to show the inequality for every \( u \in \) \( \mathcal{D}\left( \Omega \right) \), that is, for every \( u \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) such that \( \operatorname{Supp}u \subset \Omega \) . Suppose for example that the projection on \( \Omega \) onto the first factor is bounded, so there exist real numbers \( A < B \) such that \( \Omega \subset \left\lbrack {A, B}\right\rbrack \times {\mathbb{R}}^{d - 1} \) . Since \( u \) is of class \( {C}^{1} \) , \[ u\left( x\right) = {\int }_{A}^{{x}_{1}}\frac{\partial u}{\partial {x}_{1}}\left( {t,{x}_{2},\ldots ,{x}_{d}}\right) {dt}\;\text{ for all }x \in \Omega . \] It follows, by the Schwarz inequality, that \[ {\left| u\left( x\right) \right| }^{2} \leq \left( {B - A}\right) {\int }_{A}^{B}{\left| \frac{\partial u}{\partial {x}_{1}}\right| }^{2}\left( {t,{x}_{2},\ldots ,{x}_{d}}\right) {dt}\;\text{ for all }x \in \Omega . \] Integrating this inequality over \( \left\lbrack {A, B}\right\rbrack \times {\mathbb{R}}^{d - 1} \) gives \[ \parallel u{\parallel }_{{L}^{2}}^{2} \leq {\left( B - A\right) }^{2}{\begin{Vmatrix}{D}_{1}u\end{Vmatrix}}_{{L}^{2}}^{2} \] Since \( \left| {{D}_{1}u}\right| \leq \left| {\nabla u}\right| \), the result is proved.
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Yes
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Corollary 1.7 Suppose that \( \Omega \) is a bounded open set (more generally, that one of the coordinate projections of \( \Omega \) is bounded). The map \[ u \mapsto \parallel u{\parallel }_{{H}_{0}^{1}\left( \Omega \right) } = \parallel \left| {\nabla u}\right| {\parallel }_{{L}^{2}\left( \Omega \right) } \] is a Hilbert norm on \( {H}_{0}^{1}\left( \Omega \right) \) equivalent to the norm \( \parallel \cdot {\parallel }_{{H}^{1}\left( \Omega \right) } \) .
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Proof. If \( C \) is the constant that appears in the Poincaré inequality, we have, for every \( u \in {H}_{0}^{1}\left( \Omega \right) \) , \[ {\begin{Vmatrix}\left| \nabla u\right| \end{Vmatrix}}_{{L}^{2}\left( \Omega \right) }^{2} \leq \parallel u{\parallel }_{{L}^{2}\left( \Omega \right) }^{2} + {\begin{Vmatrix}\left| \nabla u\right| \end{Vmatrix}}_{{L}^{2}\left( \Omega \right) }^{2} = \parallel u{\parallel }_{{H}^{1}\left( \Omega \right) }^{2} \leq \left( {1 + {C}^{2}}\right) {\begin{Vmatrix}\left| \nabla u\right| \end{Vmatrix}}_{{L}^{2}\left( \Omega \right) }^{2}, \] which proves the equivalence between the two norms. At the same time, we see that the norm \( \parallel u{\parallel }_{{H}_{0}^{1}\left( \Omega \right) } \) is defined by the scalar product \[ {\left( u \mid v\right) }_{{H}_{0}^{1}\left( \Omega \right) } = \mathop{\sum }\limits_{{j = 1}}^{d}{\left( {D}_{j}u \mid {D}_{j}v\right) }_{{L}^{2}\left( \Omega \right) }.\]
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Yes
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Proposition 1.8 For \( f \in {H}_{0}^{1}\left( \Omega \right) \), set\n\n\[ \n\widetilde{f} = \left\{ \begin{array}{ll} f & \text{ on }\Omega , \\ 0 & \text{ on }{\mathbb{R}}^{d} \smallsetminus \Omega . \end{array}\right.\n\]\n\nThen \( \widetilde{f} \in {H}^{1}\left( {\mathbb{R}}^{d}\right) \) and the map that takes \( f \) to \( \widetilde{f} \) is an isometry between \( \left( {{H}_{0}^{1}\left( \Omega \right) ,\parallel \cdot {\parallel }_{{H}^{1}\left( \Omega \right) }}\right) \) and \( \left( {{H}^{1}\left( {\mathbb{R}}^{d}\right) ,\parallel \cdot {\parallel }_{{H}^{1}\left( {\mathbb{R}}^{d}\right) }}\right) \) .
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Proof. If \( f \in \mathcal{D}\left( \Omega \right) \), we clearly have \( \widetilde{f} \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) and \( \widetilde{{D}_{j}f} = {D}_{j}\widetilde{f} \) for \( j \in \{ 1,\ldots, d\} \) . Consequently, the map \( f \mapsto \widetilde{f} \) is an isometry from \( \mathcal{D}\left( \Omega \right) \) , with the norm \( \parallel \cdot {\parallel }_{{H}^{1}\left( \Omega \right) } \), into \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) . Since \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) is complete, the extension theorem says that this isometry extends to an isometry \( f \mapsto \widehat{f} \) from \( \left( {{H}_{0}^{1}\left( \Omega \right) ,\parallel \cdot {\parallel }_{{H}^{1}\left( \Omega \right) }}\right) \) to \( \left( {{H}^{1}\left( {\mathbb{R}}^{d}\right) ,\parallel \cdot {\parallel }_{{H}^{1}\left( {\mathbb{R}}^{d}\right) }}\right) \) . Now, convergence in \( {H}^{1} \) implies convergence in \( {L}^{2} \), so, if \( {\left( {\varphi }_{n}\right) }_{n \in \mathbb{N}} \) is a sequence in \( \mathcal{D}\left( \Omega \right) \) converging to \( f \) in \( {H}_{0}^{1}\left( \Omega \right) \), the sequence \( {\left( {\widetilde{\varphi }}_{n}\right) }_{n \in \mathbb{N}} \) converges to \( \widetilde{f} \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . Since it also converges to \( \widehat{f} \) in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) (by the definition of \( \widehat{f} \) ) and so also in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \), it follows that \( \widetilde{f} = \widehat{f} \), which concludes the proof.
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Yes
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Lemma 1.9 For every \( u \in {H}^{1}\left( {\mathbb{R}}^{d}\right) \) and every \( h \in {\mathbb{R}}^{d} \), \[ {\begin{Vmatrix}{\tau }_{h}u - u\end{Vmatrix}}_{{L}^{2}} \leq \parallel \left| {\nabla u}\right| {\parallel }_{{L}^{2}}\left| h\right| \]
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Proof. By Proposition 1.5, the space \( \mathcal{D}\left( {\mathbb{R}}^{d}\right) \) is dense in \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \). Thus it suffices to prove the property for \( u \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \). If \( u \in \mathcal{D}\left( {\mathbb{R}}^{d}\right) \), we have \[ u\left( {x - h}\right) - u\left( x\right) = - {\int }_{0}^{1}\nabla u\left( {x - {th}}\right) \cdot {hdt} \] thus, by the Schwarz inequality, \[ {\left| {\tau }_{h}u\left( x\right) - u\left( x\right) \right| }^{2} \leq {\left| h\right| }^{2}{\int }_{0}^{1}{\left| \nabla u\right| }^{2}\left( {x - {th}}\right) {dt} \] Now it suffices to integrate this inequality over \( {\mathbb{R}}^{d} \) using the fact that Lebesgue measure is invariant under translations.
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Yes
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Theorem 1.10 (Rellich) If \( \Omega \) is a bounded open set, the canonical injection \( u \mapsto u \) from \( {H}_{0}^{1}\left( \Omega \right) \) into \( {L}^{2}\left( \Omega \right) \) is a compact operator.
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Proof. Since the map \( u \mapsto {u}_{\mid \Omega } \) from \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to \( {L}^{2}\left( \Omega \right) \) is clearly continuous, it suffices to prove that the map \( f \mapsto \widetilde{f} \) from \( {H}_{0}^{1}\left( \Omega \right) \) to \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is a compact operator (where \( \widetilde{f} \) is as in Proposition 1.8). Let \( B \) be the closed unit ball in \( {H}_{0}^{1}\left( \Omega \right) \), and put \( \widetilde{B} = \{ \widetilde{f} : f \in B\} \) . By Proposition 1.8, \( \widetilde{B} \) is contained in the closed unit ball of \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) . We must show that \( \widetilde{B} \) is relatively compact in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . To do this, we use the criterion provided by Theorem 3.8 on page 175 in the case \( p = 2 \) .\n\nProperties \( \mathrm{i} \) and \( \mathrm{{ii}} \) in the statement of that theorem are clearly satisfied since, for every \( f \in B \), we have \( \parallel \widetilde{f}{\parallel }_{{L}^{2}} \leq 1 \) and\n\n\[{\int }_{\{ \left| x\right| > R\} }{\left| \widetilde{f}\left( x\right) \right| }^{2}{dx} = 0\]\n\nfor every \( R > 0 \) such that \( \Omega \subset B\left( {0, R}\right) \) . On the other hand, \( \widetilde{B} \) is contained in the closed unit ball of \( {H}^{1}\left( {\mathbb{R}}^{d}\right) \) ; thus, by Lemma 1.9,\n\n\[{\begin{Vmatrix}{\tau }_{h}\widetilde{f} - \widetilde{f}\end{Vmatrix}}_{{L}^{2}} \leq \left| h\right| \;\text{ for all }f \in B\text{ and }h \in {\mathbb{R}}^{d},\]\n\nwhich proves property iii.
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Yes
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Proposition 2.1 If \( f \in {L}^{2}\left( \Omega \right) \), these statements are equivalent:\n\n- \( u \in {H}_{0}^{1}\left( \Omega \right) \) and \( {\Delta u} = f \) .\n\n- \( u \in {H}_{0}^{1}\left( \Omega \right) \) and \( {\left( u \mid v\right) }_{{H}_{0}^{1}} = - {\left( f \mid v\right) }_{{L}^{2}} \) for all \( v \in {H}_{0}^{1}\left( \Omega \right) \) .
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Proof. If \( f \in {L}^{2}\left( \Omega \right) \) and \( u \in {H}_{0}^{1}\left( \Omega \right) \), we have the following chain of equivalences:\n\n\[ \n{\Delta u} = f \Leftrightarrow \langle {\Delta u},\bar{\varphi }\rangle = {\left( f \mid \varphi \right) }_{{L}^{2}}\;\text{ for all }\varphi \in \mathcal{D}\left( \Omega \right) \n\]\n\n\[ \n\Leftrightarrow \mathop{\sum }\limits_{{j = 1}}^{d}{\left( {D}_{j}u \mid {D}_{j}\varphi \right) }_{{L}^{2}} = - {\left( f \mid \varphi \right) }_{{L}^{2}}\;\text{ for all }\varphi \in \mathcal{D}\left( \Omega \right) \n\]\n\n\[ \n\Leftrightarrow {\left( u \mid \varphi \right) }_{{H}_{0}^{1}} = - {\left( f \mid \varphi \right) }_{{L}^{2}}\;\text{ for all }\varphi \in \mathcal{D}\left( \Omega \right) .\n\]\n\nThe result follows because \( \mathcal{D}\left( \Omega \right) \) is dense in \( {H}_{0}^{1}\left( \Omega \right) \) .
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Yes
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For every \( f \in {L}^{2}\left( \Omega \right) \), the Dirichlet problem on \( \Omega \) with right-hand side \( f \) has a unique solution \( u \in {H}_{0}^{1}\left( \Omega \right) \). The operator \[ {\Delta }^{-1} : {L}^{2}\left( \Omega \right) \rightarrow {H}_{0}^{1}\left( \Omega \right) \] \[ f\; \mapsto \;u \] thus defined is continuous and has norm at most \( C \), the constant that appears in the Poincaré inequality (Proposition 1.6).
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Proof. If \( f \in {L}^{2}\left( \Omega \right) , \) \[ \left| {\left( f \mid v\right) }_{{L}^{2}}\right| \leq C\parallel f{\parallel }_{{L}^{2}}\parallel v{\parallel }_{{H}_{0}^{1}}\;\text{ for all }v \in {H}_{0}^{1}\left( \Omega \right) , \] where \( C \) is the constant in the Poincaré inequality. Thus, the map \( L : v \mapsto \) \( {\left( v \mid f\right) }_{{L}^{2}} \) is a continuous linear form on \( {H}_{0}^{1}\left( \Omega \right) \) of norm at most \( C\parallel f{\parallel }_{{L}^{2}} \) . Therefore the existence and uniqueness of the solution \( u \), together with the inequality \( \parallel u{\parallel }_{{H}_{0}^{1}} \leq C\parallel f{\parallel }_{{L}^{2}} \), follow immediately (in view of the preceding proposition) from an application of Riesz's Theorem (page 111) to the Hilbert space \( {H}_{0}^{1}\left( \Omega \right) \) .
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Yes
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Proposition 2.3 Let \( f \in {L}^{2}\left( \Omega \right) \) . For every \( v \in {H}_{0}^{1}\left( \Omega \right) \), put\n\n\[ \n{J}_{f}\left( v\right) = \frac{1}{2}{\left( \parallel v{\parallel }_{{H}_{0}^{1}}\right) }^{2} + \operatorname{Re}{\left( f \mid v\right) }_{{L}^{2}}.\n\]\n\nThese statements are equivalent:\n\n\( - u \in {H}_{0}^{1}\left( \Omega \right) \) and \( {\Delta u} = f \) .\n\n- \( u \in {H}_{0}^{1}\left( \Omega \right) \) and \( {J}_{f}\left( u\right) = \mathop{\min }\limits_{{v \in {H}_{0}^{1}\left( \Omega \right) }}{J}_{f}\left( v\right) \) .
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Proof. Suppose \( h \in {H}_{0}^{1}\left( \Omega \right) \) . Then\n\n\[ \n{J}_{f}\left( {u + h}\right) = {J}_{f}\left( u\right) + \mathrm{{Re}}\left( {{\left( f \mid h\right) }_{{L}^{2}} + {\left( u \mid h\right) }_{{H}_{0}^{1}}}\right) + \frac{1}{2}{\left( \parallel h{\parallel }_{{H}_{0}^{1}}\right) }^{2}.\n\]\n\nTherefore, by Proposition 2.1, \( {\Delta u} = f \) implies that\n\n\[ \n{J}_{f}\left( {u + h}\right) = {J}_{f}\left( u\right) + \frac{1}{2}{\left( \parallel h{\parallel }_{{H}_{0}^{1}}\right) }^{2}\;\text{ for all }h \in {H}_{0}^{1}\left( \Omega \right) .\n\]\n\nThus \( {J}_{f} \) attains its minimum on \( {H}_{0}^{1}\left( \Omega \right) \) at \( u \) and only at \( u \) .
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Yes
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Proposition 2.4 The operator\n\n\\[ \nT : {H}_{0}^{1}\\left( \\Omega \\right) \\rightarrow {H}_{0}^{1}\\left( \\Omega \\right) \n\\]\n\n\\[ \nv\\; \\mapsto \\;u\\text{ such that }{\\Delta u} = - v \n\\]\nis an injective, compact, positive selfadjoint operator on \\( {H}_{0}^{1}\\left( \\Omega \\right) \\) .
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Proof. Let \\( J : u \\mapsto u \\) be the canonical injection from \\( {H}_{0}^{1}\\left( \\Omega \\right) \\) into \\( {L}^{2}\\left( \\Omega \\right) \\) . Then \\( T = - {\\Delta }^{-1} \\circ J \\), so, by Proposition 1.2 on page 215, \\( T \\) is compact, because \\( {\\Delta }^{-1} \\) is continuous (Theorem 2.2) and \\( J \\) is compact (Rellich’s Theorem, page 355).\n\nOn the other hand, if \\( u \\in {H}_{0}^{1}\\left( \\Omega \\right) \\) and \\( {Tu} = w \\), we have \\( {\\Delta w} = - u \\) . Therefore, by Proposition 2.1,\n\n\\[ \n{\\left( Tu \\mid v\\right) }_{{H}_{0}^{1}} = {\\left( u \\mid v\\right) }_{{L}^{2}}\\;\\text{ for all }u, v \\in {H}_{0}^{1}\\left( \\Omega \\right) , \n\\]\n\nwhich easily implies that \\( T \\) is selfadjoint, positive, and injective.
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Yes
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Proposition 2.6 Suppose \( f \in {L}^{2}\left( \Omega \right) \) . The solution \( u \) of the Dirichlet problem on \( \Omega \) with right-hand side \( f \) is given by\n\n\[ u = - \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\left( f \mid {u}_{n}\right) }_{{L}^{2}}{u}_{n} \]\n\nthe series being convergent in \( {H}_{0}^{1}\left( \Omega \right) \) .
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Proof. By remark 1 above, \( {\left( \sqrt{-{\mu }_{n}}{u}_{n}\right) }_{n \in \mathbb{N}} \) is a Hilbert basis of \( {L}^{2}\left( \Omega \right) \), so\n\n\[ f = - \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}{\left( f \mid {u}_{n}\right) }_{{L}^{2}}{u}_{n} \]\n\nwith convergence in \( {L}^{2}\left( \Omega \right) \) . In particular,\n\n\[ {\left( \parallel f{\parallel }_{{L}^{2}}\right) }^{2} = \mathop{\sum }\limits_{{n = 0}}^{{+\infty }} - {\mu }_{n}{\left| {\left( f \mid {u}_{n}\right) }_{{L}^{2}}\right| }^{2} < + \infty \]\n\nand, since the sequence \( {\left( -{\mu }_{n}\right) }_{n \in \mathbb{N}} \) is increasing and thus bounded below by \( - {\mu }_{0} > 0 \),\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\left| {\left( f \mid {u}_{n}\right) }_{{L}^{2}}\right| }^{2} < + \infty \]\n\nSince the sequence \( {\left( {u}_{n}\right) }_{n \in \mathbb{N}} \) is a Hilbert basis for \( {H}_{0}^{1}\left( \Omega \right) \), it follows that the series\n\n\[ v = - \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\left( f \mid {u}_{n}\right) }_{{L}^{2}}{u}_{n} \]\n\nconverges in \( {H}_{0}^{1}\left( \Omega \right) \) . Since convergence in \( {H}_{0}^{1}\left( \Omega \right) \) implies convergence in \( {L}^{2}\left( \Omega \right) \), which in turn implies convergence in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \), we deduce that, in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \)\n\n\[ {\Delta v} = - \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}{\left( f \mid {u}_{n}\right) }_{{L}^{2}}{u}_{n} \]\n\n(by the definition of the sequence \( {\mu }_{n} \) ). Likewise\n\n\[ f = - \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{\mu }_{n}{\left( f \mid {u}_{n}\right) }_{{L}^{2}}{u}_{n} \]\n\nwith convergence in \( {\mathcal{D}}^{\prime }\left( \Omega \right) \) . It follows that \( {\Delta v} = f \) and so that \( v = u \), the solution of the Dirichlet problem on \( \Omega \) with right-hand side \( f \) .
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Yes
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