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Proposition 1.12 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then for every \( \epsilon > 0 \) :\n\n(i) There exists a set of finite measure \( B \) (a ball, for example) such that\n\n\[{\int }_{{B}^{c}}\left| f\right| < \epsilon\]\n\n(ii) There is a \( \delta > 0 \) such that\n\n\[{\int }_{E}\left| f\rig...
Proof. By replacing \( f \) with \( \left| f\right| \) we may assume without loss of generality that \( f \geq 0 \) .\n\nFor the first part, let \( {B}_{N} \) denote the ball of radius \( N \) centered at the origin, and note that if \( {f}_{N}\left( x\right) = f\left( x\right) {\chi }_{{B}_{N}}\left( x\right) \), then...
Yes
Theorem 1.13 Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of measurable functions such that \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) a.e. \( x \), as \( n \) tends to infinity. If \( \left| {{f}_{n}\left( x\right) }\right| \leq g\left( x\right) \), where \( g \) is integrable, then\n\n\[ \n\int ...
Proof. For each \( N \geq 0 \) let \( {E}_{N} = \{ x : \left| x\right| \leq N, g\left( x\right) \leq N\} \) . Given \( \epsilon > 0 \), we may argue as in the first part of the previous lemma, to see that there exists \( N \) so that \( {\int }_{{E}_{N}^{c}}g < \epsilon \) . Then the functions \( {f}_{n}{\chi }_{{E}_{N...
Yes
Theorem 2.4 The following families of functions are dense in \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) :\n\n(i) The simple functions.\n\n(ii) The step functions.\n\n(iii) The continuous functions of compact support.
Proof. Let \( f \) be an integrable function on \( {\mathbb{R}}^{d} \) . First, we may assume that \( f \) is real-valued, because we may approximate its real and imaginary parts independently. If this is the case, we may then write \( f = {f}^{ + } - {f}^{ - } \) , where \( {f}^{ + },{f}^{ - } \geq 0 \), and it now su...
Yes
Proposition 2.5 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\begin{Vmatrix}{f}_{h} - f\end{Vmatrix}}_{{L}^{1}} \rightarrow 0\;\text{ as }h \rightarrow 0.\n\]
The proof is a simple consequence of the approximation of integrable functions by continuous functions of compact support as given in Theorem 2.4. In fact for any \( \epsilon > 0 \), we can find such a function \( g \) so that \( \parallel f - g\parallel < \epsilon \) . Now\n\n\[ \n{f}_{h} - f = \left( {{g}_{h} - g}\ri...
Yes
Theorem 3.1 Suppose \( f\left( {x, y}\right) \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is integrable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by \( {\int }_{{\mathbb{R}}...
We first note that we may assume that \( f \) is real-valued, since the theorem then applies to the real and imaginary parts of a complex-valued function. The proof of Fubini's theorem which we give next consists of a sequence of six steps. We begin by letting \( \mathcal{F} \) denote the set of integrable functions on...
No
Theorem 3.2 Suppose \( f\left( {x, y}\right) \) is a non-negative measurable function on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) . Then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) :\n\n(i) The slice \( {f}^{y} \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \) .\n\n(ii) The function defined by ...
Proof of Theorem 3.2. Consider the truncations\n\n\[ \n{f}_{k}\left( {x, y}\right) = \left\{ \begin{matrix} f\left( {x, y}\right) & \text{ if }\left| \left( {x, y}\right) \right| < k\text{ and }f\left( {x, y}\right) < k, \\ 0 & \text{ otherwise. } \end{matrix}\right. \n\]\n\nEach \( {f}_{k} \) is integrable, and by par...
Yes
Corollary 3.3 If \( E \) is a measurable set in \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \), then for almost every \( y \in {\mathbb{R}}^{{d}_{2}} \) the slice\n\n\[ \n{E}^{y} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : \left( {x, y}\right) \in E}\right\} \n\]\n\nis a measurable subset of \( {\mathbb{R}}^{...
This is an immediate consequence of the first part of Theorem 3.2 applied to the function \( {\chi }_{E} \) . Clearly a symmetric result holds for the \( x \) -slices in \( {\mathbb{R}}^{{d}_{2}} \) .
Yes
Proposition 3.4 If \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \), and \( {m}_{ * }\left( {E}_{2}\right) > 0 \), then \( {E}_{1} \) is measurable.
Proof. By Corollary 3.3, we know that for a.e. \( y \in {\mathbb{R}}^{{d}_{2}} \), the slice function\n\n\[ \n{\left( {\chi }_{{E}_{1} \times {E}_{2}}\right) }^{y}\left( x\right) = {\chi }_{{E}_{1}}\left( x\right) {\chi }_{{E}_{2}}\left( y\right) \n\]\n\nis measurable as a function of \( x \) . In fact, we claim that t...
Yes
Lemma 3.5 If \( {E}_{1} \subset {\mathbb{R}}^{{d}_{1}} \) and \( {E}_{2} \subset {\mathbb{R}}^{{d}_{2}} \), then\n\n\[ \n{m}_{ * }\left( {{E}_{1} \times {E}_{2}}\right) \leq {m}_{ * }\left( {E}_{1}\right) {m}_{ * }\left( {E}_{2}\right) \n\]\n\nwith the understanding that if one of the sets \( {E}_{j} \) has exterior me...
Proof. Let \( \epsilon > 0 \) . By definition, we can find cubes \( {\left\{ {Q}_{k}\right\} }_{k = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{1}} \) and \( {\left\{ {Q}_{\ell }^{\prime }\right\} }_{\ell = 1}^{\infty } \) in \( {\mathbb{R}}^{{d}_{2}} \) such that\n\n\[ \n{E}_{1} \subset \mathop{\bigcup }\limits_{{k = 1}}...
Yes
Proposition 3.6 Suppose \( {E}_{1} \) and \( {E}_{2} \) are measurable subsets of \( {\mathbb{R}}^{{d}_{1}} \) and \( {\mathbb{R}}^{{d}_{2}} \), respectively. Then \( E = {E}_{1} \times {E}_{2} \) is a measurable subset of \( {\mathbb{R}}^{d} \) . Moreover, \[ m\left( E\right) = m\left( {E}_{1}\right) m\left( {E}_{2}\r...
Proof. It suffices to prove that \( E \) is measurable, because then the assertion about \( m\left( E\right) \) follows from Corollary 3.3. Since each set \( {E}_{j} \) is measurable, there exist sets \( {G}_{j} \subset {\mathbb{R}}^{{d}_{j}} \) of type \( {G}_{\delta } \), with \( {G}_{j} \supset {E}_{j} \) and \( {m}...
Yes
Corollary 3.7 Suppose \( f \) is a measurable function on \( {\mathbb{R}}^{{d}_{1}} \) . Then the function \( \widetilde{f} \) defined by \( \widetilde{f}\left( {x, y}\right) = f\left( x\right) \) is measurable on \( {\mathbb{R}}^{{d}_{1}} \times {\mathbb{R}}^{{d}_{2}} \) .
Proof. To see this, we may assume that \( f \) is real-valued, and recall first that if \( a \in \mathbb{R} \) and \( {E}_{1} = \left\{ {x \in {\mathbb{R}}^{{d}_{1}} : f\left( x\right) < a}\right\} \), then \( {E}_{1} \) is measurable by definition. Since\n\n\[ \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{{d}_{1}} \...
Yes
Corollary 3.8 Suppose \( f\left( x\right) \) is a non-negative function on \( {\mathbb{R}}^{d} \), and let\n\n\[ \mathcal{A} = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{d} \times \mathbb{R} : 0 \leq y \leq f\left( x\right) }\right\} \]\n\nThen:\n\n(i) \( f \) is measurable on \( {\mathbb{R}}^{d} \) if and only if...
Proof. If \( f \) is measurable on \( {\mathbb{R}}^{d} \), then the previous proposition guarantees that the function\n\n\[ F\left( {x, y}\right) = y - f\left( x\right) \]\nis measurable on \( {\mathbb{R}}^{d + 1} \), so we immediately see that \( \mathcal{A} = \{ y \geq 0\} \cap \{ F \leq \) \( 0\} \) is measurable.\n...
Yes
Proposition 3.9 If \( f \) is a measurable function on \( {\mathbb{R}}^{d} \), then the function \( \widetilde{f}\left( {x, y}\right) = f\left( {x - y}\right) \) is measurable on \( {\mathbb{R}}^{d} \times {\mathbb{R}}^{d} \) .
By picking \( E = \left\{ {z \in {\mathbb{R}}^{d} : f\left( z\right) < a}\right\} \), we see that it suffices to prove that whenever \( E \) is a measurable subset of \( {\mathbb{R}}^{d} \), then \( \widetilde{E} = \{ \left( {x, y}\right) : x - y \in \) \( E\} \) is a measurable subset of \( {\mathbb{R}}^{d} \times {\m...
Yes
Proposition 4.1 Suppose \( f \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then \( \widehat{f} \) defined by (14) is continuous and bounded on \( {\mathbb{R}}^{d} \) .
In fact, since \( \left| {f\left( x\right) {e}^{-{2\pi ix} \cdot \xi }}\right| = \left| {f\left( x\right) }\right| \), the integral representing \( \widehat{f} \) converges for each \( \xi \) and \( \mathop{\sup }\limits_{{\xi \in {\mathbb{R}}^{d}}}\left| {\widehat{f}\left( \xi \right) }\right| \leq {\int }_{{\mathbb{R...
Yes
Corollary 4.3 Suppose \( \widehat{f}\left( \xi \right) = 0 \) for all \( \xi \) . Then \( f = 0 \) a.e.
The proof of the theorem requires only that we adapt the earlier arguments carried out for Schwartz functions in Chapter 5 of Book I to the present context. We begin with the \
No
Lemma 4.4 Suppose \( f \) and \( g \) belong to \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) . Then\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}\widehat{f}\left( \xi \right) g\left( \xi \right) {d\xi } = {\int }_{{\mathbb{R}}^{d}}f\left( y\right) \widehat{g}\left( y\right) {dy}.\n\]
Note that both integrals converge in view of the proposition above. Consider the function \( F\left( {\xi, y}\right) = g\left( \xi \right) f\left( y\right) {e}^{-{2\pi i\xi } \cdot y} \) defined for \( \left( {\xi, y}\right) \in {\mathbb{R}}^{d} \times \) \( {\mathbb{R}}^{d} = {\mathbb{R}}^{2d} \) . It is measurable as...
Yes
Theorem 1.1 Suppose \( f \) is integrable on \( {\mathbb{R}}^{d} \) . Then:\n\n(i) \( {f}^{ * } \) is measurable.\n\n(ii) \( {f}^{ * }\left( x\right) < \infty \) for a.e. \( x \) .\n\n(iii) \( {f}^{ * } \) satisfies\n\n(1)\n\n\[ m\left( \left\{ {x \in {\mathbb{R}}^{d} : {f}^{ * }\left( x\right) > \alpha }\right\} \righ...
Before we come to the proof we want to clarify the nature of the main conclusion (iii). As we shall observe, one has that \( {f}^{ * }\left( x\right) \geq \left| {f\left( x\right) }\right| \) for a.e. \( x \) ; the effect of (iii) is that, broadly speaking, \( {f}^{ * } \) is not much larger than \( \left| f\right| \) ...
No
Lemma 1.2 Suppose \( \mathcal{B} = \left\{ {{B}_{1},{B}_{2},\ldots ,{B}_{N}}\right\} \) is a finite collection of open balls in \( {\mathbb{R}}^{d} \) . Then there exists a disjoint sub-collection \( {B}_{{i}_{1}},{B}_{{i}_{2}},\ldots ,{B}_{{i}_{k}} \) of \( \mathcal{B} \) that satisfies\n\n\[ m\left( {\mathop{\bigcup ...
Proof. The argument we give is constructive and relies on the following simple observation: Suppose \( B \) and \( {B}^{\prime } \) are a pair of balls that intersect, with the radius of \( {B}^{\prime } \) being not greater than that of \( B \) . Then \( {B}^{\prime } \) is contained in the ball \( \widetilde{B} \) th...
Yes
Theorem 1.3 If \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} = f\left( x\right) \;\text{ for a.e. }x. \]
Proof. It suffices to show that for each \( \alpha > 0 \) the set\n\n\[ {E}_{\alpha } = \left\{ {x : \mathop{\limsup }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\left| {\frac{1}{m\left( B\right) }{\int }_{B}f\left( y\right) {dy} - f\left( x\right) }\right| > {2\alpha }}\right\} \]\n\nhas measure...
Yes
Corollary 1.6 If \( f \) is locally integrable on \( {\mathbb{R}}^{d} \), then almost every point belongs to the Lebesgue set of \( f \) .
Proof. An application of Theorem 1.4 to the function \( \left| {f\left( y\right) - r}\right| \) shows that for each rational \( r \), there exists a set \( {E}_{r} \) of measure zero, such that\n\n\[ \mathop{\lim }\limits_{\substack{{m\left( B\right) \rightarrow 0} \\ {x \in B} }}\frac{1}{m\left( B\right) }{\int }_{B}\...
Yes
Corollary 1.7 Suppose \( f \) is locally integrable on \( {\mathbb{R}}^{d} \). If \( \left\{ {U}_{\alpha }\right\} \) shrinks regularly to \( \bar{x} \), then\n\n\[\n\mathop{\lim }\limits_{\substack{{m\left( {U}_{\alpha }\right) \rightarrow 0} \\ {x \in {U}_{\alpha }} }}\frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{...
The proof is immediate once we observe that if \( \bar{x} \in B \) with \( {U}_{\alpha } \subset B \) and \( m\left( {U}_{\alpha }\right) \geq {cm}\left( B\right) \), then\n\n\[\n\frac{1}{m\left( {U}_{\alpha }\right) }{\int }_{{U}_{\alpha }}\left| {f\left( y\right) - f\left( \bar{x}\right) }\right| {dy} \leq \frac{1}{{...
Yes
Theorem 2.1 If \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity and \( f \) is integrable on \( {\mathbb{R}}^{d} \), then\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ as }\delta \rightarrow 0 \]\n\nfor every \( x \) in the Lebesgue ...
Since the integral of each kernel \( {K}_{\delta } \) is equal to 1, we may write\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = \int \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {K}_{\delta }\left( y\right) {dy}. \]\n\nConsequently,\n\n\[ \left| {\left( {f * {K}_...
Yes
Lemma 2.2 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \), and that \( x \) is a point of the Lebesgue set of \( f \) . Let\n\n\[ \mathcal{A}\left( r\right) = \frac{1}{{r}^{d}}{\int }_{\left| y\right| \leq r}\left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy},\;\text{ whenever }r > 0. \]\n\nThe...
Proof. The continuity of \( \mathcal{A}\left( r\right) \) follows by invoking the absolute continuity in Proposition 1.12 of Chapter 2.\n\nThe fact that \( \mathcal{A}\left( r\right) \) tends to 0 as \( r \) tends to 0 follows since \( x \) belongs to the Lebesgue set of \( f \), and the measure of a ball of radius \( ...
Yes
Theorem 2.3 Suppose that \( f \) is integrable on \( {\mathbb{R}}^{d} \) and that \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is an approximation to the identity. Then, for each \( \delta > 0 \), the convolution\n\n\[\n\left( {f * {K}_{\delta }}\right) \left( x\right) = {\int }_{{\mathbb{R}}^{d}}f\left( {x - y}...
The proof is merely a repetition in a more general context of the argument in the special case where \( {K}_{\delta }\left( x\right) = {\delta }^{-d/2}{e}^{-\pi {\left| x\right| }^{2}/\delta } \) given in Section \( {4}^{ * } \) , Chapter 2, and so will not be repeated.
No
Theorem 3.1 A curve parametrized by \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is rectifiable if and only if both \( x\left( t\right) \) and \( y\left( t\right) \) are of bounded variation.
The proof is immediate once we observe that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \), then\n\n\[ F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) = \left( {x\left( {t}_{j}\right) - x\left( {t}_{j - 1}\right) }\right) + i\left( {y\left( {t}_{j}\right) - y\left( {t}_{j - 1}\right) }\right) ,\...
Yes
Lemma 3.2 Suppose \( F \) is real-valued and of bounded variation on \( \left\lbrack {a, b}\right\rbrack \) . Then for all \( a \leq x \leq b \) one has \[ F\left( x\right) - F\left( a\right) = {P}_{F}\left( {a, x}\right) - {N}_{F}\left( {a, x}\right) , \] and \[ {T}_{F}\left( {a, x}\right) = {P}_{F}\left( {a, x}\right...
Proof. Given \( \epsilon > 0 \) there exists a partition \( a = {t}_{0} < \cdots < {t}_{N} = x \) of \( \left\lbrack {a, x}\right\rbrack \), such that \[ \left| {{P}_{F} - \mathop{\sum }\limits_{\left( +\right) }F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| < \epsilon \text{ and }\left| {{N}_{F} - \matho...
Yes
Theorem 3.3 A real-valued function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) is of bounded variation if and only if \( F \) is the difference of two increasing bounded functions.
Proof. Clearly, if \( F = {F}_{1} - {F}_{2} \), where each \( {F}_{j} \) is bounded and increasing, then \( F \) is of bounded variation.\n\nConversely, suppose \( F \) is of bounded variation. Then, we let \( {F}_{1}\left( x\right) = \) \( {P}_{F}\left( {a, x}\right) + F\left( a\right) \) and \( {F}_{2}\left( x\right)...
Yes
Theorem 3.4 If \( F \) is of bounded variation on \( \left\lbrack {a, b}\right\rbrack \), then \( F \) is differentiable almost everywhere.
In other words, the quotient\n\n\[\n\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{F\left( {x + h}\right) - F\left( x\right) }{h}\n\]\n\nexists for almost every \( x \in \left\lbrack {a, b}\right\rbrack \) . By the previous result, it suffices to consider the case when \( F \) is increasing. In fact, we shall first als...
No
Lemma 3.5 Suppose \( G \) is real-valued and continuous on \( \mathbb{R} \) . Let \( E \) be the set of points \( x \) such that\n\n\[ G\left( {x + h}\right) > G\left( x\right) \;\text{ for some }h = {h}_{x} > 0. \]\n\nIf \( E \) is non-empty, then it must be open, and hence can be written as a countable disjoint union...
Proof. Since \( G \) is continuous, it is clear that \( E \) is open whenever it is non-empty and can therefore be written as a disjoint union of countably many open intervals (Theorem 1.3 in Chapter 1). If \( \left( {{a}_{k},{b}_{k}}\right) \) denotes a finite interval in this decomposition, then \( {a}_{k} \notin E \...
Yes
If \( F \) is increasing and continuous, then \( {F}^{\prime } \) exists almost everywhere. Moreover \( {F}^{\prime } \) is measurable, non-negative, and\n\n\[{\int }_{a}^{b}{F}^{\prime }\left( x\right) {dx} \leq F\left( b\right) - F\left( a\right)\]
Proof. For \( n \geq 1 \), we consider the quotient\n\n\[{G}_{n}\left( x\right) = \frac{F\left( {x + 1/n}\right) - F\left( x\right) }{1/n}.\].\n\nBy the previous theorem, we have that \( {G}_{n}\left( x\right) \rightarrow {F}^{\prime }\left( x\right) \) for a.e. \( x \), which shows in particular that \( {F}^{\prime } ...
Yes
Theorem 3.8 If \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \), then \( {F}^{\prime }\left( x\right) \) exists almost everywhere. Moreover, if \( {F}^{\prime }\left( x\right) = 0 \) for a.e. \( x \), then \( F \) is constant.
Since an absolutely continuous function is the difference of two continuous monotonic functions, as we have seen above, the existence of \( {F}^{\prime }\left( x\right) \) for a.e. \( x \) follows from what we have already proved. To prove that \( {F}^{\prime }\left( x\right) = 0 \) a.e. implies \( F \) is constant req...
No
Corollary 3.10 We can arrange the choice of the balls so that\n\n\[ m\left( {E - \mathop{\bigcup }\limits_{{i = 1}}^{N}{B}_{i}}\right) < {2\delta } \]
In fact, let \( \mathcal{O} \) be an open set, with \( \mathcal{O} \supset E \) and \( m\left( {\mathcal{O} - E}\right) < \delta \) . Since we are dealing with a Vitali covering of \( E \), we can restrict all of our choices above to balls contained in \( \mathcal{O} \) . If we do this, then \( \left( {E - \mathop{\big...
Yes
Theorem 3.11 Suppose \( F \) is absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then \( {F}^{\prime } \) exists almost everywhere and is integrable. Moreover,\n\n\[ F\left( x\right) - F\left( a\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy},\;\text{ for all }a \leq x \leq b. \]
Proof. Since we know that a real-valued absolutely continuous function is the difference of two continuous increasing functions, Corollary 3.7 shows that \( {F}^{\prime } \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . Now let \( G\left( x\right) = {\int }_{a}^{x}{F}^{\prime }\left( y\right) {dy} \) . Then...
Yes
Lemma 3.12 A bounded increasing function \( F \) on \( \left\lbrack {a, b}\right\rbrack \) has at most countably many discontinuities.
Proof. If \( F \) is discontinuous at \( x \), we may choose a rational number \( {r}_{x} \) so that \( F\left( {x}^{ - }\right) < {r}_{x} < F\left( {x}^{ + }\right) \) . If \( f \) is discontinuous at \( x \) and \( z \) with \( x < z \), we must have \( F\left( {x}^{ + }\right) \leq F\left( {z}^{ - }\right) \), hence...
Yes
Lemma 3.13 If \( F \) is increasing and bounded on \( \left\lbrack {a, b}\right\rbrack \), then:\n\n(i) \( J\left( x\right) \) is discontinuous precisely at the points \( \left\{ {x}_{n}\right\} \) and has a jump at \( {x}_{n} \) equal to that of \( F \) .\n\n(ii) The difference \( F\left( x\right) - J\left( x\right) \...
Proof. If \( x \neq {x}_{n} \) for all \( n \), each \( {j}_{n} \) is continuous at \( x \), and since the series converges uniformly, \( J \) must be continuous at \( x \) . If \( x = {x}_{N} \) for some \( N \), then we write\n\n\[ J\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{\alpha }_{n}{j}_{n}\left( x\rig...
Yes
Theorem 3.14 If \( J \) is the jump function considered above, then \( {J}^{\prime }\left( x\right) \) exists and vanishes almost everywhere.
Proof. Given any \( \epsilon > 0 \), we note that the set \( E \) of those \( x \) where\n\n(10)\n\n\[ \mathop{\limsup }\limits_{{h \rightarrow 0}}\frac{J\left( {x + h}\right) - J\left( x\right) }{h} > \epsilon \]\n\nis a measurable set. (The proof of this little fact is outlined in Exercise 14 below.) Suppose \( \delt...
No
Theorem 4.1 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right) \) is a curve defined for \( a \leq t \leq b \) . If both \( x\left( t\right) \) and \( y\left( t\right) \) are absolutely continuous, then the curve is rectifiable, and if \( L \) denotes its length, we have\n\n\[ L = {\int }_{a}^{b}{\left( {x}...
Note that if \( F\left( t\right) = x\left( t\right) + {iy}\left( t\right) \) is absolutely continuous then it is automatically of bounded variation, and hence the curve is rectifiable. The identity (12) is an immediate consequence of the proposition below, which can be viewed as a more precise version of Corollary 3.7 ...
Yes
Proposition 4.2 Suppose \( F \) is complex-valued and absolutely continuous on \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[{T}_{F}\left( {a, b}\right) = {\int }_{a}^{b}\left| {{F}^{\prime }\left( t\right) }\right| {dt}\]
In fact, because of Theorem 3.11, for any partition \( a = {t}_{0} < {t}_{1} < \cdots < \) \( {t}_{N} = b \) of \( \left\lbrack {a, b}\right\rbrack \), we have\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{N}\left| {F\left( {t}_{j}\right) - F\left( {t}_{j - 1}\right) }\right| = \mathop{\sum }\limits_{{j = 1}}^{N}\left| {{\int...
Yes
Theorem 4.3 Suppose \( \left( {x\left( t\right), y\left( t\right) }\right), a \leq t \leq b \), is a rectifiable curve that has length \( L \) . Consider the arc-length parametrization \( \widetilde{z}\left( s\right) = \left( {\widetilde{x}\left( s\right) ,\widetilde{y}\left( s\right) }\right) \) described above. Then ...
Proof. We noted that \( \left| {\widetilde{z}\left( {s}_{1}\right) - \widetilde{z}\left( {s}_{2}\right) }\right| \leq \left| {{s}_{1} - {s}_{2}}\right| \), so it follows immediately that \( \widetilde{z}\left( s\right) \) is absolutely continuous, hence differentiable almost everywhere. Moreover, this inequality also p...
Yes
Theorem 4.4 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a quasi-simple curve. The Minkowski content of \( \Gamma \) exists if and only if \( \Gamma \) is rectifiable. When this is the case and \( L \) is the length of the curve, then \( \mathcal{M}\left( \Gamma \right) = L \) .
To prove the theorem, we also consider for any compact set \( K \)\n\n\[ \n{\mathcal{M}}^{ * }\left( K\right) = \mathop{\limsup }\limits_{{\delta \rightarrow 0}}\frac{m\left( {K}^{\delta }\right) }{2\delta }\;\text{ and }\;{\mathcal{M}}_{ * }\left( K\right) = \mathop{\liminf }\limits_{{\delta \rightarrow 0}}\frac{m\lef...
No
Proposition 4.5 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a quasi-simple curve. If \( {\mathcal{M}}_{ * }\left( \Gamma \right) < \infty \), then the curve is rectifiable, and if \( L \) denotes its length, then \[ L \leq {\mathcal{M}}_{ * }\left( \Gamma \right) \]
The proof depends on the following simple observation. Lemma 4.6 If \( \Gamma
No
Lemma 4.6 If \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is any curve, and \( \Delta = \left| {z\left( b\right) - z\left( a\right) }\right| \) is the distance between its end-points, then \( m\left( {\Gamma }^{\delta }\right) \geq {2\delta \Delta } \) .
Proof. Since the distance function and the Lebesgue measure are invariant under translations and rotations (see Section 3 in Chapter 1 and Problem 4 in Chapter 2) we may transform the situation by an appropriate composition of these motions. Therefore we may assume that the end-points of the curve have been placed on t...
Yes
Proposition 4.7 Suppose \( \Gamma = \{ z\left( t\right), a \leq t \leq b\} \) is a rectifiable curve with length \( L \) . Then\n\n\[{\mathcal{M}}^{ * }\left( \Gamma \right) \leq L\]
The quantities \( {\mathcal{M}}^{ * }\left( \Gamma \right) \) and \( L \) are of course independent of the parametrization used; since the curve is rectifiable, it will be convenient to use the arclength parametrization. Thus we write the curve as \( z\left( s\right) = \left( {x\left( s\right), y\left( s\right) }\right...
Yes
Proposition 1.1 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) has the following properties:\n\n(i) \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is a vector space.\n\n(ii) \( f\left( x\right) \overline{g\left( x\right) } \) is integrable whenever \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), and the Cauch...
Proof. If \( f, g \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then since \( \left| {f\left( x\right) + g\left( x\right) }\right| \leq 2\max \left( {\left| {f\left( x\right) }\right| ,\left| {g\left( x\right) }\right| }\right) \), we have\n\n\[{\left| f\left( x\right) + g\left( x\right) \right| }^{2} \leq 4\left( {{\l...
Yes
Theorem 1.3 The space \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) is separable, in the sense that there exists a countable collection \( \left\{ {f}_{k}\right\} \) of elements in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) such that their linear combinations are dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) .
Proof. Consider the family of functions of the form \( r{\chi }_{R}\left( x\right) \), where \( r \) is a complex number with rational real and imaginary parts, and \( R \) is a rectangle in \( {\mathbb{R}}^{d} \) with rational coordinates. We claim that finite linear combinations of these type of functions are dense i...
Yes
Proposition 2.1 If \( f \bot g \), then \( \parallel f + g{\parallel }^{2} = \parallel f{\parallel }^{2} + \parallel g{\parallel }^{2} \) .
Proof. It suffices to note that \( \left( {f, g}\right) = 0 \) implies \( \left( {g, f}\right) = 0 \), and therefore\n\n\[ \parallel f + g{\parallel }^{2} = \left( {f + g, f + g}\right) = \parallel f{\parallel }^{2} + \left( {f, g}\right) + \left( {g, f}\right) + \parallel g{\parallel }^{2} \]\n\n\[ = \parallel f{\para...
Yes
Proposition 2.2 If \( {\left\{ {e}_{k}\right\} }_{k = 1}^{\infty } \) is orthonormal, and \( f = \sum {a}_{k}{e}_{k} \in \mathcal{H} \) where the sum is finite, then\n\n\[ \parallel f{\parallel }^{2} = \sum {\left| {a}_{k}\right| }^{2} \]
The proof is a simple application of the Pythagorean theorem.
No
Theorem 2.4 Any Hilbert space has an orthonormal basis.
The first step in the proof of this fact is to recall that (by definition) a Hilbert space \( \mathcal{H} \) is separable. Hence, we may choose a countable collection of elements \( \mathcal{F} = \left\{ {h}_{k}\right\} \) in \( \mathcal{H} \) so that finite linear combinations of elements in \( \mathcal{F} \) are dens...
Yes
Corollary 2.5 Any two infinite-dimensional Hilbert spaces are unitarily equivalent.
Proof. If \( \mathcal{H} \) and \( {\mathcal{H}}^{\prime } \) are two infinite-dimensional Hilbert spaces, we may select for each an orthonormal basis, say\n\n\[ \left\{ {{e}_{1},{e}_{2},\ldots }\right\} \subset \mathcal{H}\;\text{ and }\;\left\{ {{e}_{1}^{\prime },{e}_{2}^{\prime },\ldots }\right\} \subset {\mathcal{H...
Yes
Proposition 2.7 Suppose we are given a pre-Hilbert space \( {\mathcal{H}}_{0} \) with inner product \( {\left( \cdot , \cdot \right) }_{0} \) . Then we can find a Hilbert space \( \mathcal{H} \) with inner product \( \left( {\cdot , \cdot }\right) \) such that\n\n(i) \( {\mathcal{H}}_{0} \subset \mathcal{H} \) .\n\n(ii...
A Hilbert space satisfying properties like \( \mathcal{H} \) in the above proposition is called a completion of \( {\mathcal{H}}_{0} \) . We shall only sketch the construction of \( \mathcal{H} \), since it follows closely Cantor’s familiar method of obtaining the real numbers as the completion of the rationals in term...
Yes
Theorem 3.1 Suppose \( f \) is integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) .\n\n(i) If \( {a}_{n} = 0 \) for all \( n \), then \( f\left( x\right) = 0 \) for a.e. \( x \) .\n\n(ii) \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{r}^{\left| n\right| }{e}^{inx} \) tends to \( f\left( x\right) \...
Proof. The first conclusion is an immediate consequence of the second. To prove the latter we recall the identity\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{r}^{\left| n\right| }{e}^{iny} = {P}_{r}\left( y\right) = \frac{1 - {r}^{2}}{1 - {2r}\cos y + {r}^{2}} \]\n\nfor the Poisson kernel; see Book I, Chapt...
Yes
Theorem 3.2 Suppose \( f \in {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) . Then:\n\n(i) We have Parseval's relation\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left| {a}_{n}\right| }^{2} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left| f\left( x\right) \right| }^{2}{dx} \]\n\n(ii) The map...
To apply the previous results, we let \( \mathcal{H} = {L}^{2}\left( \left\lbrack {-\pi ,\pi }\right\rbrack \right) \) with inner product \( \left( {f, g}\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \overline{g\left( x\right) }{dx} \), and take the orthonormal set \( {\left\{ {e}_{k}\right\} }_{k = 1...
Yes
Theorem 3.3 A bounded holomorphic function \( F\left( {r{e}^{i\theta }}\right) \) on the unit disc has radial limits at almost every \( \theta \) .
Proof. We know that \( F\left( z\right) \) has a power series expansion \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \) in \( \mathbb{D} \) that converges absolutely and uniformly whenever \( z = r{e}^{i\theta } \) and \( r < 1 \) . In fact, for \( r < 1 \) the series \( \mathop{\sum }\limits_{{n = 0}}^{\...
Yes
Proposition 4.2 If \( \mathcal{S} \) is a closed subspace of a Hilbert space \( \mathcal{H} \), then\n\n\[ \mathcal{H} = \mathcal{S} \oplus {\mathcal{S}}^{ \bot } \]\n\nThe notation in the proposition means that every \( f \in \mathcal{H} \) can be written uniquely as \( f = g + h \), where \( g \in \mathcal{S} \) and ...
The proof of the proposition relies on the previous lemma giving the closest element of \( f \) in \( \mathcal{S} \) . In fact, for any \( f \in \mathcal{H} \), we choose \( {g}_{0} \) as in the lemma and write\n\n\[ f = {g}_{0} + \left( {f - {g}_{0}}\right) \]\n\nBy construction \( {g}_{0} \in \mathcal{S} \), and the ...
Yes
Lemma 5.1 \( \parallel T\parallel = \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g\parallel \leq 1\} \)
Proof. If \( \parallel T\parallel \leq M \), the Cauchy-Schwarz inequality gives\n\n\[ \left| \left( {{Tf}, g}\right) \right| \leq M\;\text{ whenever }\parallel f\parallel \leq 1\text{ and }\parallel g\parallel \leq 1 \]\n\nthus \( \sup \{ \left| \left( {{Tf}, g}\right) \right| : \parallel f\parallel \leq 1,\parallel g...
Yes
Proposition 5.2 A linear operator \( T : {\mathcal{H}}_{1} \rightarrow {\mathcal{H}}_{2} \) is bounded if and only if it is continuous.
Proof. If \( T \) is bounded, then \( {\begin{Vmatrix}T\left( f\right) - T\left( {f}_{n}\right) \end{Vmatrix}}_{{\mathcal{H}}_{2}} \leq M{\begin{Vmatrix}f - {f}_{n}\end{Vmatrix}}_{{\mathcal{H}}_{1}} \) , hence \( T \) is continuous. Conversely, suppose that \( T \) is continuous but not bounded. Then for each \( n \) t...
Yes
Proposition 5.4 Let \( T : \mathcal{H} \rightarrow \mathcal{H} \) be a bounded linear transformation. There exists a unique bounded linear transformation \( {T}^{ * } \) on \( \mathcal{H} \) so that:\n\n(i) \( \left( {{Tf}, g}\right) = \left( {f,{T}^{ * }g}\right) \) ,\n\n(ii) \( \parallel T\parallel = \begin{Vmatrix}{...
To prove the existence of an operator satisfying (i) above, we observe that for each fixed \( g \in \mathcal{H} \), the linear functional \( \ell = {\ell }_{g} \), defined by\n\n\[ \ell \left( f\right) = \left( {{Tf}, g}\right) \]\n\nis bounded. Indeed, since \( T \) is bounded one has \( \parallel {Tf}\parallel \leq M...
Yes
Proposition 5.5 Let \( T \) be a Hilbert-Schmidt operator on \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) with kernel \( K \).\n\n(i) If \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then for almost every \( x \) the function \( y \mapsto K\left( {x, y}\right) f\left( y\right) \) is integrable.\n\n(ii) The operator...
Proof. By Fubini’s theorem we know that for almost every \( x \), the function \( y \mapsto {\left| K\left( x, y\right) \right| }^{2} \) is integrable. Then, part (i) follows directly from an application of the Cauchy-Schwarz inequality.\n\nFor (ii), we make use again of the Cauchy-Schwarz inequality as follows\n\n\[ \...
Yes
Proposition 6.1 Suppose \( T \) is a bounded linear operator on \( \mathcal{H} \). (i) If \( S \) is compact on \( \mathcal{H} \), then \( {ST} \) and \( {TS} \) are also compact.
Proof. Part (i) is immediate.
No
Lemma 6.3 Suppose \( T \) is a bounded symmetric linear operator on a Hilbert space \( \mathcal{H} \). (i) If \( \lambda \) is an eigenvalue of \( T \), then \( \lambda \) is real. (ii) If \( {f}_{1} \) and \( {f}_{2} \) are eigenvectors corresponding to two distinct eigenvalues, then \( {f}_{1} \) and \( {f}_{2} \) ar...
Proof. To prove (i), we first choose a non-zero eigenvector \( f \) such that \( T\left( f\right) = {\lambda f} \). Since \( T \) is symmetric (that is, \( T = {T}^{ * } \)), we find that \[ \lambda \left( {f, f}\right) = \left( {{Tf}, f}\right) = \left( {f,{Tf}}\right) = \left( {f,{\lambda f}}\right) = \bar{\lambda }\...
Yes
Lemma 6.5 Suppose \( T \neq 0 \) is compact and symmetric. Then either \( \parallel T\parallel \) or \( - \parallel T\parallel \) is an eigenvalue of \( T \) .
Proof. By the observation (7) made earlier, either\n\n\[ \parallel T\parallel = \sup \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} \;\text{ or }\; - \parallel T\parallel = \inf \{ \left( {{Tf}, f}\right) : \parallel f\parallel = 1\} . \]\n\nWe assume the first case, that is,\n\n\[ \lambda = \parallel T\parall...
Yes
Theorem 1.1 The Fourier transform \( {\mathcal{F}}_{0} \), initially defined on \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) , has a (unique) extension \( \mathcal{F} \) to a unitary mapping of \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) to itself. In particular,\n\n\[ \parallel \mathcal{F}\left( f\right) {\parallel...
The extension \( \mathcal{F} \) will be given by a limiting process: if \( \left\{ {f}_{n}\right\} \) is a sequence in the Schwartz space that converges to \( f \) in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \), then \( \left\{ {{\mathcal{F}}_{0}\left( {f}_{n}\right) }\right\} \) will converge to an element in \( {L}^{...
Yes
Lemma 1.2 The space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) is dense in \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) . In other words, given any \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \), there exists a sequence \( \left\{ {f}_{n}\right\} \subset \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) such that\n\n...
For the proof of the lemma, we fix \( f \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and \( \epsilon > 0 \) . Then, for each \( M > 0 \), we define\n\n\[ \n{g}_{M}\left( x\right) = \left\{ \begin{matrix} f\left( x\right) & \text{ if }\left| x\right| \leq M\text{ and }\left| {f\left( x\right) }\right| \leq M, \\ 0 & \te...
No
Lemma 1.3 Let \( {\mathcal{H}}_{1} \) and \( {\mathcal{H}}_{2} \) denote Hilbert spaces with norms \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \), respectively. Suppose \( \mathcal{S} \) is a dense subspace of \( {\mathcal{H}}_{1} \) and \( {T}_{0} : \mathcal{S} \rightarrow {\mathcal{...
Proof. Given \( f \in {\mathcal{H}}_{1} \), let \( \left\{ {f}_{n}\right\} \) be a sequence in \( \mathcal{S} \) that converges to \( f \) , and define \[ T\left( f\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{T}_{0}\left( {f}_{n}\right) \] where the limit is taken in \( {\mathcal{H}}_{2} \) . To see that \...
Yes
Theorem 2.1 The elements \( F \) in \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \) are exactly the functions given by (6), with \( {\widehat{F}}_{0} \in {L}^{2}\left( {0,\infty }\right) \) . Moreover \[ \parallel F{\parallel }_{{H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) } = {\begin{Vmatrix}{\widehat{F}}_{0}\end{Vma...
The crucial point in the proof of the theorem is the following fact. For any fixed strictly positive \( y \), we let \( {\widehat{F}}_{y}\left( \xi \right) \) denote the Fourier transform of the \( {L}^{2} \) function \( F\left( {x + {iy}}\right), x \in \mathbb{R} \) . Then for any pair of choices of \( y \) , \( {y}_{...
Yes
Lemma 2.2 If \( F \) belongs to \( {H}^{2}\left( {\mathbb{R}}_{ + }^{2}\right) \), then \( F \) is bounded in any proper half-plane \( \{ z = x + {iy}, y \geq \delta \} \), where \( \delta > 0 \) .
To prove this we exploit the mean-value property of holomorphic functions. This property may be stated in two alternative ways. First, in terms of averages over circles,\n\n(8)\n\n\[ F\left( \zeta \right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( {\zeta + r{e}^{i\theta }}\right) {d\theta }\;\text{ if }0 < r \leq \del...
Yes
Lemma 3.1 The space \( {C}_{0}^{\infty }\left( \Omega \right) \) is dense in \( {L}^{2}\left( \Omega \right) \) in the norm \( \parallel \cdot {\parallel }_{{L}^{2}\left( \Omega \right) } \) .
The proof is essentially a repetition of that of Lemma 1.2. We take the precaution of modifying the definition of \( {g}_{M} \) given there to be: \( {g}_{M}\left( x\right) = \) \( f\left( x\right) \) if \( \left| x\right| \leq M, d\left( {x,{\Omega }^{c}}\right) \geq 1/M \) and \( \left| {f\left( x\right) }\right| \le...
Yes
Lemma 3.4 Suppose \( P\left( z\right) = {z}^{m} + \cdots + {a}_{1}z + {a}_{0} \) is a polynonial of degree \( m \) with leading coefficient 1 . If \( F \) is a holomorphic function on \( \mathbb{C} \) , then\n\n\[ \n{\left| F\left( 0\right) \right| }^{2} \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| P\left( {e}^{i\the...
Proof. The lemma is a consequence of the special case when \( P = 1 \)\n\n(16)\n\n\[ \n{\left| F\left( 0\right) \right| }^{2} \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }{\int }_{0}^{2\pi }{\left| F\left( {e}^{i\theta }\right) \right| }^{2}{d\theta }\n\]\n\nThis assertion follows directly from the mean-value identity (8) in...
No
Proposition 4.1 Suppose there exists a function \( u \in {C}^{2}\left( \bar{\Omega }\right) \) that minimizes \( \mathcal{D}\left( U\right) \) among all \( U \in {C}^{2}\left( \bar{\Omega }\right) \) with \( {\left. U\right| }_{\partial \Omega } = f \) . Then \( u \) is harmonic in \( \Omega \) .
Proof. For functions \( F \) and \( G \) in \( {C}^{2}\left( \bar{\Omega }\right) \) define the following inner-product\n\n\[ \langle F, G\rangle = {\int }_{\Omega }\left( {\frac{\partial F}{\partial {x}_{1}}\overline{\frac{\partial G}{\partial {x}_{1}}} + \frac{\partial F}{\partial {x}_{2}}\overline{\frac{\partial G}{...
Yes
Theorem 4.3 Any weakly harmonic function \( u \) in \( \Omega \) can be corrected on a set of measure zero so that the resulting function is harmonic in \( \Omega \) .
The above statement says that for a given weakly harmonic function \( u \) there exists a harmonic function \( \widetilde{u} \), so that \( \widetilde{u}\left( x\right) = u\left( x\right) \) for a.e. \( x \in \Omega \) . Notice since \( \widetilde{u} \) is necessarily continuous it is uniquely determined by \( u \) .
No
Corollary 4.4 Suppose \( \Omega \) is a bounded open set, and let \( \partial \Omega = \bar{\Omega } - \Omega \) denote its boundary. Assume that \( u \) is continuous in \( \bar{\Omega } \) and is harmonic in \( \Omega \) . Then\n\n\[ \mathop{\max }\limits_{{x \in \bar{\Omega }}}\left| {u\left( x\right) }\right| = \ma...
Proof. Since the sets \( \bar{\Omega } \) and \( \partial \Omega \) are compact and \( u \) is continuous, the two maxima above are clearly attained. We suppose that \( \mathop{\max }\limits_{{x \in \bar{\Omega }}}\left| {u\left( x\right) }\right| \) is attained at an interior point \( {x}_{0} \in \Omega \), for otherw...
Yes
Lemma 4.5 We have the identity\n\n\[ \n{\int }_{B}\left( {v\bigtriangleup u - u\bigtriangleup v}\right) {\eta dx} = {\int }_{B}u\left( {\nabla v \cdot \nabla \eta }\right) - v\left( {\nabla u \cdot \nabla \eta }\right) {dx}. \n\]
Here \( \nabla u \) is the gradient of \( u \), that is, \( \nabla u = \left( {\frac{\partial u}{\partial {x}_{1}},\frac{\partial u}{\partial {x}_{2}},\ldots ,\frac{\partial u}{\partial {x}_{d}}}\right) \) and\n\n\[ \n\nabla v \cdot \nabla \eta = \mathop{\sum }\limits_{{j = 1}}^{d}\frac{\partial v}{\partial {x}_{j}}\fr...
Yes
Corollary 4.8 Suppose \( \left\{ {u}_{n}\right\} \) is a sequence of harmonic functions in \( \Omega \) that converges to a function \( u \) uniformly on compact subsets of \( \Omega \) as \( n \rightarrow \infty \) . Then \( u \) is also harmonic.
The first of these corollaries was already proved as a consequence of (26). For the second, we use the fact that each \( {u}_{n} \) satisfies the mean-value property\n\n\[ \n{u}_{n}\left( {x}_{0}\right) = \frac{1}{m\left( B\right) }{\int }_{B}{u}_{n}\left( x\right) {dx} \n\]\n\nwhenever \( B \) is a ball with center at...
No
Lemma 4.9 Let \( \Omega \) be an open bounded set in \( {\mathbb{R}}^{d} \). Suppose \( v \) belongs to \( {C}^{1}\left( \bar{\Omega }\right) \) and \( v \) vanishes on \( \partial \Omega \). Then\n\n\[ \n{\int }_{\Omega }{\left| v\left( x\right) \right| }^{2}{dx} \leq {c}_{\Omega }{\int }_{\Omega }{\left| \nabla v\lef...
Proof. This conclusion could in fact be deduced from the considerations given in Lemma 3.3. We prefer to prove this easy version separately to highlight a simple idea that we shall also use later. It should be noted that the argument yields the estimate \( {c}_{\Omega } \leq d{\left( \Omega \right) }^{2} \), where \( d...
Yes
Lemma 4.11 Let \( f \) be a continuous function on a compact subset \( \Gamma \) of \( {\mathbb{R}}^{d} \). Then there exists a function \( G \) on \( {\mathbb{R}}^{d} \) that is continuous, and so that \( {\left. G\right| }_{\partial \Gamma } = f \)
Proof. We begin with the observation that if \( {K}_{0} \) and \( {K}_{1} \) are two disjoint compact sets, there exists a continuous function \( 0 \leq g\left( x\right) \leq 1 \) on \( {\mathbb{R}}^{d} \) which takes the value 0 on \( {K}_{0} \) and 1 on \( {K}_{1} \). Indeed, if \( d\left( {x,\Omega }\right) \) denot...
Yes
Theorem 1.2 If \( {\mu }_{ * } \) is a metric exterior measure on a metric space \( X \) , then the Borel sets in \( X \) are measurable. Hence \( {\mu }_{ * } \) restricted to \( {\mathcal{B}}_{X} \) is a measure.
Proof. By the definition of \( {\mathcal{B}}_{X} \) it suffices to prove that closed sets in \( X \) are Carathéodory measurable. Therefore, let \( F \) denote a closed set and \( A \) a subset of \( X \) with \( {\mu }_{ * }\left( A\right) < \infty \) . For each \( n > 0 \), let\n\n\[ \n{A}_{n} = \left\{ {x \in {F}^{c...
Yes
Lemma 1.4 If \( {\mu }_{0} \) is a premeasure on an algebra \( \mathcal{A} \), define \( {\mu }_{ * } \) on any subset \( E \) of \( X \) by\n\n\[ \n{\mu }_{ * }\left( E\right) = \inf \left\{ {\mathop{\sum }\limits_{{j = 1}}^{\infty }{\mu }_{0}\left( {E}_{j}\right) : E \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty...
Proof. Proving that \( {\mu }_{ * } \) is an exterior measure presents no difficulty. To see why the restriction of \( {\mu }_{ * } \) to \( \mathcal{A} \) coincides with \( {\mu }_{0} \), suppose that \( E \in \mathcal{A} \) . Clearly, one always has \( {\mu }_{ * }\left( E\right) \leq {\mu }_{0}\left( E\right) \) sin...
Yes
Theorem 1.5 Suppose that \( \mathcal{A} \) is an algebra of sets in \( X,{\mu }_{0} \) a premeasure on \( \mathcal{A} \), and \( \mathcal{M} \) the \( \sigma \) -algebra generated by \( \mathcal{A} \) . Then there exists a measure \( \mu \) on \( \mathcal{M} \) that extends \( {\mu }_{0} \) .
Proof. The exterior measure \( {\mu }_{ * } \) induced by \( {\mu }_{0} \) defines a measure \( \mu \) on the \( \sigma \) -algebra of Carathéodory measurable sets. Therefore, by the result in the previous lemma, \( \mu \) is also a measure on \( \mathcal{M} \) that extends \( {\mu }_{0} \) . (We should observe that in...
Yes
Proposition 3.2 If \( E \) is an arbitrary measurable set in \( X \), then the conclusion of Proposition 3.1 are still valid except that we only assert that \( {E}^{{x}_{2}} \) is \( {\mu }_{1} \) -measurable and \( {\mu }_{1}\left( {E}^{{x}_{2}}\right) \) is defined for almost every \( {x}_{2} \in {X}_{2} \) .
Proof. Consider first the case when \( E \) is a set of measure zero. Then we know by Proposition 1.6 that there is a set \( F \in {\mathcal{A}}_{\sigma \delta } \) such that \( E \subset F \) and \( \left( {{\mu }_{1} \times {\mu }_{2}}\right) \left( F\right) = 0 \) . Since \( {E}^{{x}_{2}} \subset {F}^{{x}_{2}} \) fo...
Yes
Theorem 3.3 In the setting above, suppose \( f\left( {{x}_{1},{x}_{2}}\right) \) is an integrable function on \( \left( {{X}_{1} \times {X}_{2},{\mu }_{1} \times {\mu }_{2}}\right) \) . (i) For almost every \( {x}_{2} \in {X}_{2} \), the slice \( {f}^{{x}_{2}}\left( {x}_{1}\right) = f\left( {{x}_{1},{x}_{2}}\right) \) ...
Proof. Note that if the desired conclusions hold for finitely many functions, they also hold for their linear combinations. In particular it suffices to assume that \( f \) is non-negative. When \( f = {\chi }_{E} \), where \( E \) is a set of finite measure, what we wish to prove is contained in Proposition 3.2. Hence...
Yes
Proposition 4.1 The total variation \( \left| \nu \right| \) of a signed measure \( \nu \) is itself a (positive) measure that satisfies \( \nu \leq \left| \nu \right| \) .
Proof. Suppose \( {\left\{ {E}_{j}\right\} }_{j = 1}^{\infty } \) is a countable collection of disjoints sets in \( \mathcal{M} \), and let \( E = \bigcup {E}_{j} \) . It suffices to prove:\n\n(11)\n\n\[ \sum \left| \nu \right| \left( {E}_{j}\right) \leq \left| \nu \right| \left( E\right) \;\text{ and }\;\left| \nu \ri...
Yes
Proposition 4.2 The assertion (14) implies (12). Conversely, if \( \left| \nu \right| \) is a finite measure, then (12) implies (14).
That (12) is a consequence of (14) is obvious because \( \mu \left( E\right) = 0 \) gives \( \left| {\nu \left( E\right) }\right| < \epsilon \) for every \( \epsilon > 0 \) . To prove the converse, it suffices to consider the case when \( \nu \) is positive, upon replacing \( \nu \) by \( \left| \nu \right| \) . We the...
Yes
Theorem 4.3 Suppose \( \mu \) is a \( \sigma \) -finite positive measure on the measure space \( \left( {X,\mathcal{M}}\right) \) and \( \nu \) a \( \sigma \) -finite signed measure on \( \mathcal{M} \) . Then there exist unique signed measures \( {\nu }_{a} \) and \( {\nu }_{s} \) on \( \mathcal{M} \) such that \( {\n...
We start with the case when both \( \nu \) and \( \mu \) are positive and finite. Let \( \rho = \nu + \mu \), and consider the transformation on \( {L}^{2}\left( {X,\rho }\right) \) defined by\n\n\[ \ell \left( \psi \right) = {\int }_{X}\psi \left( x\right) {d\nu }\left( x\right) \]\n\nThe mapping \( \ell \) defines a ...
Yes
Lemma 5.2 The following relations hold among the subspaces \( S,{S}_{ * } \) , and \( \overline{{S}_{1}} \) .\n\n(i) \( S = {S}_{ * } \) .\n\n(ii) The orthogonal complement of \( \overline{{S}_{1}} \) is \( S \) .
Proof. First, since \( T \) is an isometry, we have that \( \left( {{Tf},{Tg}}\right) = \left( {f, g}\right) \) for all \( f, g \in \mathcal{H} \), and thus \( {T}^{ * }T = I \) . (See Exercise 22 in Chapter 4.) So if \( {Tf} = f \) then \( {T}^{ * }{Tf} = {T}^{ * }f \), which means that \( f = {T}^{ * }f \) . To prove...
No
Theorem 6.1 Suppose \( T \) is a bounded symmetric operator on a Hilbert space \( \mathcal{H} \) . Then there exists a spectral resolution \( \{ E\left( \lambda \right) \} \) such that\n\n\[ T = {\int }_{{a}^{ - }}^{b}{\lambda dE}\left( \lambda \right) \]\n\nin the sense that for every \( f, g \in \mathcal{H} \)\n\n(32...
The result encompasses the spectral theorem for compact symmetric operators \( T \) in the following sense. Let \( \left\{ {\varphi }_{k}\right\} \) be an orthonormal basis of eigenvectors of \( T \) with corresponding eigenvalues \( {\lambda }_{k} \), as guaranteed by Theorem 6.2 in Chapter 4 . In this case, we take t...
No
Proposition 6.2 Suppose \( T \) is symmetric. Then \( \parallel T\parallel \leq M \) if and only if \( - {MI} \leq \) \( T \leq {MI} \) .
This is a consequence of (7) in Chapter 4.
No
Proposition 6.4 If \( {T}_{1} \) and \( {T}_{2} \) are positive operators that commute, then \( {T}_{1}{T}_{2} \) is also positive.
Indeed, if \( S \) is a square root of \( {T}_{1} \) given in the previous proposition, then \( {T}_{1}{T}_{2} = \) \( {SS}{T}_{2} = S{T}_{2}S \), and hence \( \left( {{T}_{1}{T}_{2}f, f}\right) = \left( {S{T}_{2}{Sf}, f}\right) = \left( {{T}_{2}{Sf},{Sf}}\right) \), since \( S \) is symmetric, and thus the last term i...
Yes
Proposition 6.5 Suppose \( T \) is symmetric and a and \( b \) are given by (33). If \( p\left( t\right) = \) \( \mathop{\sum }\limits_{{k = 0}}^{n}{c}_{k}{t}^{k} \) is a real polynomial which is positive for \( t \in \left\lbrack {a, b}\right\rbrack \), then the operator \( p\left( T\right) = \mathop{\sum }\limits_{{k...
To see this, write \( p\left( t\right) = c\mathop{\prod }\limits_{j}\left( {t - {\rho }_{j}}\right) \mathop{\prod }\limits_{k}\left( {{\rho }_{k}^{\prime } - t}\right) \mathop{\prod }\limits_{\ell }\left( {{\left( t - {\mu }_{\ell }\right) }^{2} + {\nu }_{\ell }}\right) \), where \( c \) is positive and the third facto...
Yes
Corollary 6.6 If \( p\left( t\right) \) is a real polynomial, then\n\n\[ \parallel p\left( T\right) \parallel \leq \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \]
This is an immediate consequence using Proposition 6.2, since \( - M \leq p\left( t\right) \leq M \) , where \( M = \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \), and thus \( - {MI} \leq p\left( T\right) \leq {MI} \).
Yes
Proposition 6.7 Suppose \( \left\{ {T}_{n}\right\} \) is a sequence of positive operators that satisfy \( {T}_{n} \geq {T}_{n + 1} \) for all \( n \) . Then there is a positive operator \( T \), such that \( {T}_{n}f \rightarrow {Tf} \) as \( n \rightarrow \infty \) for every \( f \in \mathcal{H} \) .
Proof. We note that for each fixed \( f \in \mathcal{H} \) the sequence of positive numbers \( \left( {{T}_{n}f, f}\right) \) is decreasing and hence convergent. Now observe that for any positive operator \( S \) with \( \parallel S\parallel \leq M \) we have\n\n(35)\n\n\[ \parallel S\left( f\right) {\parallel }^{2} \l...
Yes
Proposition 6.8 If \( T \) is symmetric, then \( \sigma \left( T\right) \) is a closed subset of the interval \( \left\lbrack {a, b}\right\rbrack \) given by (33).
Note that if \( z \notin \left\lbrack {a, b}\right\rbrack \), the function \( \Phi \left( t\right) = {\left( t - z\right) }^{-1} \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and \( \Phi \left( T\right) \left( {T - {zI}}\right) = \left( {T - {zI}}\right) \Phi \left( T\right) = I \), so \( \Phi \left( T\rig...
Yes
Proposition 6.9 For each \( f \in \mathcal{H} \), the Lebesgue-Stieltjes measure corresponding to \( F\left( \lambda \right) = \left( {E\left( \lambda \right) f, f}\right) \) is supported on \( \sigma \left( T\right) \) .
To prove this, let \( J \) be one of the open intervals in the complement of \( \sigma \left( T\right) \) , \( {x}_{0} \in J \), and \( {J}_{0} \) the sub-interval centered at \( {x}_{0} \) of length \( {2\epsilon } \), with \( \epsilon < \begin{Vmatrix}{\left( T - {x}_{0}I\right) }^{-1}\end{Vmatrix} \) . First note th...
Yes
Property 1 (Monotonicity) If \( {E}_{1} \subset {E}_{2} \), then \( {m}_{\alpha }^{ * }\left( {E}_{1}\right) \leq {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) .
This is straightforward, since any cover of \( {E}_{2} \) is also a cover of \( {E}_{1} \) .
No
Property 2 (Sub-additivity) \( {m}_{\alpha }^{ * }\left( {\mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{\infty }{m}_{\alpha }^{ * }\left( {E}_{j}\right) \) for any countable family \( \left\{ {E}_{j}\right\} \) of sets in \( {\mathbb{R}}^{d} \) .
For the proof, fix \( \delta \), and choose for each \( j \) a cover \( {\left\{ {F}_{j, k}\right\} }_{k = 1}^{\infty } \) of \( {E}_{j} \) by sets of diameter less than \( \delta \) such that \( \mathop{\sum }\limits_{k}{\left( \operatorname{diam}{F}_{j, k}\right) }^{\alpha } \leq {\mathcal{H}}_{\alpha }^{\delta }\lef...
Yes
Property 3 If \( d\left( {{E}_{1},{E}_{2}}\right) > 0 \), then \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) = {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) .
It suffices to prove that \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) \geq {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) since the reverse inequality is guaranteed by sub-additivity. Fix \( \epsilon > 0 \) with \( \epsilon < \) \( d\left( {{E}_{1},{E}_{2}}\right) ...
Yes
Property 5 Hausdorff measure is invariant under translations\n\n\[ \n{m}_{\alpha }\left( {E + h}\right) = {m}_{\alpha }\left( E\right) \;\text{ for all }h \in {\mathbb{R}}^{d}, \n\]\n\nand rotations\n\n\[ \n{m}_{\alpha }\left( {rE}\right) = {m}_{\alpha }\left( E\right) \n\]\n\nwhere \( r \) is a rotation in \( {\mathbb...
These conclusions follow once we observe that the diameter of a set \( S \) is invariant under translations and rotations, and satisfies \( \operatorname{diam}\left( {\lambda S}\right) = \) \( \lambda \operatorname{diam}\left( S\right) \) for \( \lambda > 0 \) .
Yes
Property 6 The quantity \( {m}_{0}\left( E\right) \) counts the number of points in \( E \) , while \( {m}_{1}\left( E\right) = m\left( E\right) \) for all Borel sets \( E \subset \mathbb{R} \) . (Here \( m \) denotes the Lebesgue measure on \( \mathbb{R} \) .)
In fact, note that in one dimension every set of diameter \( \delta \) is contained in an interval of length \( \delta \) (and for an interval its length equals its Lebesgue measure).
No
Property 7 If \( E \) is a Borel subset of \( {\mathbb{R}}^{d} \), then \( {c}_{d}{m}_{d}\left( E\right) = m\left( E\right) \) for some constant \( {c}_{d} \) that depends only on the dimension \( d \) .
The constant \( {c}_{d} \) equals \( m\left( B\right) /{\left( \operatorname{diam}B\right) }^{d} \), for the unit ball \( B \) ; note that this ratio is the same for all balls \( B \) in \( {\mathbb{R}}^{d} \), and so \( {c}_{d} = {v}_{d}/{2}^{d} \) (where \( {v}_{d} \) denotes the volume of the unit ball). The proof o...
Yes
Theorem 2.1 The Cantor set \( \mathcal{C} \) has strict Hausdorff dimension \( \alpha = \) \( \log 2/\log 3 \) .
The inequality\n\n\[ \n{m}_{\alpha }\left( \mathcal{C}\right) \leq 1 \n\]\n\nfollows from the construction of \( \mathcal{C} \) and the definitions. Indeed, recall from Chapter 1 that \( \mathcal{C} = \bigcap {C}_{k} \), where each \( {C}_{k} \) is a finite union of \( {2}^{k} \) intervals of length \( {3}^{-k} \) . Gi...
Yes
Lemma 2.2 Suppose a function \( f \) defined on a compact set \( E \) satisfies a Lipschitz condition with exponent \( \gamma \) . Then\n\n(i) \( {m}_{\beta }\left( {f\left( E\right) }\right) \leq {M}^{\beta }{m}_{\alpha }\left( E\right) \) if \( \beta = \alpha /\gamma \) .\n\n(ii) \( \dim f\left( E\right) \leq \frac{1...
Proof. Suppose \( \left\{ {F}_{k}\right\} \) is a countable family of sets that covers \( E \) . Then \( \left\{ {f\left( {E \cap {F}_{k}}\right) }\right\} \) covers \( f\left( E\right) \) and, moreover, \( f\left( {E \cap {F}_{k}}\right) \) has diameter less than \( M{\left( \operatorname{diam}{F}_{k}\right) }^{\gamma...
Yes