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Lemma 2. For \( x \neq 0,\frac{{\partial }^{2}}{\partial {x}_{j}^{2}}\left| x\right| = {\left| x\right| }^{-1} - {x}_{j}^{2}{\left| x\right| }^{-3} \) . | Proof.\n\n\[ \frac{{\partial }^{2}}{\partial {x}_{j}^{2}}\left| x\right| = \frac{\partial }{\partial {x}_{j}}\left\lbrack {{x}_{j}{\left| x\right| }^{-1}}\right\rbrack = {\left| x\right| }^{-1} + {x}_{j}\left( {-1}\right) {\left| x\right| }^{-2}{x}_{j}{\left| x\right| }^{-1} \]\n\n\[ = {\left| x\right| }^{-1} - {x}_{j}... | Yes |
Lemma 3. For \( x \neq 0 \), and \( g \in {C}^{2}\left( {0,\infty }\right) \) , \[ {\Delta g}\left( \left| x\right| \right) = {g}^{\prime \prime }\left( \left| x\right| \right) + \left( {n - 1}\right) {\left| x\right| }^{-1}{g}^{\prime }\left( x\right) \] | Proof. \[ \frac{\partial }{\partial {x}_{j}}g\left( \left| x\right| \right) = {g}^{\prime }\left( \left| x\right| \right) \frac{\partial }{\partial {x}_{j}}\left| x\right| = {g}^{\prime }\left( \left| x\right| \right) {x}_{j}{\left| x\right| }^{-1} \] \[ \frac{{\partial }^{2}}{\partial {x}_{j}^{2}}g\left( \left| x\righ... | Yes |
Find a fundamental solution of the operator \( A \) defined (for \( n = 1 \) ) by the equation\n\n\[ \n{A\phi } = {\phi }^{\prime \prime } + {2a}{\phi }^{\prime } + {b\phi }\;\left( {\phi \in \mathcal{D}}\right)\n\] | We seek a distribution \( T \) such that \( {AT} = \delta \) . Let us look for a regular distribution, \( T = \widetilde{f} \) . Using the definition of derivatives of distributions, we have\n\n\[ \n\left( {A\widetilde{f}}\right) \left( \phi \right) = \widetilde{f}\left( {{\phi }^{\prime \prime } - {2a}{\phi }^{\prime ... | Yes |
Theorem 5. Consider the operator\n\n\[ \nA = \mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}\left( x\right) \frac{{d}^{j}}{d{x}^{j}} \n\]\n\nin which \( {c}_{j} \in {C}^{\infty }\left( \mathbb{R}\right) \) and \( {c}_{m}\left( x\right) \neq 0 \) for all \( x \) . This operator has a fundamental solution that is a regular di... | Proof. We find a function \( f \) defined on \( \lbrack 0,\infty ) \) such that\n\n\[ \n\text{(i)}\;\mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}\left( x\right) {f}^{\left( j\right) }\left( x\right) = 0 \n\]\n\n(ii) \( \;{c}_{m - 1}\left( 0\right) {f}^{\left( m - 1\right) }\left( 0\right) = 1 \)\n\n(iii) \( \;{c}_{j}\left... | Yes |
Lemma 1. There is a function \( f \in {C}^{\infty }\left( \mathbb{R}\right) \) such that \( 0 \leq f \leq 1 \) , \( f\left( x\right) = 0 \) on \( ( - \infty ,0\rbrack \), and \( f\left( x\right) = 1 \) on \( \lbrack 1,\infty ) \) . | Proof. Define\n\n\[ g\left( x\right) = \left\{ \begin{array}{ll} \exp \left\lbrack {{x}^{2}/\left( {{x}^{2} - 1}\right) }\right\rbrack & \left| x\right| < 1 \\ 0 & \left| x\right| \geq 1 \end{array}\right. \]\n\nand\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} g\left( {x - 1}\right) & x \leq 1 \\ 1 & \text{ other... | Yes |
Lemma 2. If \( {x}_{0} \in {\mathbb{R}}^{n} \) and \( \rho > r > 0 \), then there is a test function\n\n\( \phi \) such that\n\n(i) \( 0 \leq \phi \leq 1 \)\n\n(ii) \( \phi \left( x\right) = 1 \) if \( \left| {x - {x}_{0}}\right| \leq r \)\n\n(iii) \( \phi \left( x\right) = 0 \) if \( \left| {x - {x}_{0}}\right| \geq \... | Proof. Use the function \( f \) from the preceding lemma, and define\n\n\[ \phi \left( x\right) = 1 - f\left( {a{\left| x - {x}_{0}\right| }^{2} - b}\right) \]\n\nwith \( a = {\left( {\rho }^{2} - {r}^{2}\right) }^{-1} \) and \( b = {r}^{2}a \) . If \( \left| {x - {x}_{0}}\right| \leq r \), then \( a{\left| x - {x}_{0}... | Yes |
Theorem 2. Let \( \operatorname{supp}\left( T\right) \) denote the intersection of all closed sets having property (3). Then \( \operatorname{supp}\left( T\right) \) is the smallest closed set having property (3). | Proof. Let \( \mathcal{F} \) be the family of all closed sets \( F \) having property (3). Then\n\n\[ \operatorname{supp}\left( T\right) = \bigcap \{ F : F \in \mathcal{F}\} \]\n\nBeing an intersection of closed sets, \( \operatorname{supp}\left( T\right) \) is itself closed. The only question is whether it has propert... | Yes |
Each distribution having a compact support has an extension to \( \mathcal{E} \) that is continuous. | Let \( T \) be a distribution for which \( \operatorname{supp}\left( T\right) \) is compact. By the theorem on partitions of unity, there is a test function \( \psi \) such that \( \psi \left( x\right) = 1 \) on a neighborhood of \( \operatorname{supp}\left( T\right) \) . Define \( \bar{T} \) on \( \mathcal{E} \) by th... | Yes |
Theorem 3. Each distribution having a compact support has an extension to \( \mathcal{E} \) that is continuous. | Proof. Let \( T \) be a distribution for which \( \operatorname{supp}\left( T\right) \) is compact. By the theorem on partitions of unity, there is a test function \( \psi \) such that \( \psi \left( x\right) = 1 \) on a neighborhood of \( \operatorname{supp}\left( T\right) \) . Define \( \bar{T} \) on \( \mathcal{E} \... | Yes |
Theorem 3. Each distribution having a compact support has an extension to \( \mathcal{E} \) that is continuous. | Proof. Let \( T \) be a distribution for which \( \operatorname{supp}\left( T\right) \) is compact. By the theorem on partitions of unity, there is a test function \( \psi \) such that \( \psi \left( x\right) = 1 \) on a neighborhood of \( \operatorname{supp}\left( T\right) \) . Define \( \bar{T} \) on \( \mathcal{E} \... | Yes |
Theorem 4. Each continuous linear functional on \( \mathcal{E} \) is an extension of some distribution having compact support. | Proof. Let \( L \) be a continuous linear functional on \( \mathcal{E} \). Let \( T = L \mid \mathfrak{D} \), which denotes the restriction of \( L \) to \( \mathbf{D} \). It is easily seen that \( T \) is a distribution. In order to prove that the support of \( T \) is compact, suppose otherwise. Then for each \( k \)... | Yes |
Theorem 5. If \( T \) is a distribution with compact support and if \( \phi \in \mathcal{E} \), then \( T * \phi \in \mathcal{E} \). | Proof. See [Ru1], Theorem 6.35, page 159. | No |
Theorem 2. Let \( E \) denote the translation operator, defined by\n\n\( \left( {{E}_{y}f}\right) \left( x\right) = f\left( {x - y}\right) \) . Then we have \( \widehat{{E}_{y}f} = {e}_{-y}\widehat{f} \) and \( \widehat{{e}_{y}f} = {E}_{y}\widehat{f} \) . | Proof. We verify the first equation and leave the second to the problems. We have\n\n\[ \widehat{{E}_{y}f}\left( x\right) = \int f\left( {u - y}\right) {e}^{-{2\pi ixu}}{du} = \int f\left( v\right) {e}^{-{2\pi ix}\left( {v + y}\right) }{dv} \]\n\n\[ = {e}^{-{2\pi ixy}}\int f\left( v\right) {e}^{-{2\pi ixv}}{dv} = {e}_{... | No |
Theorem 3. If \( f \) and \( g \) belong to \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \), then the same is true of \( f * g \), and \[ \parallel f * g{\parallel }_{1} \leq \parallel f{\parallel }_{1} \cdot \parallel g{\parallel }_{1} \] | Proof. ([Smi]) Define a function \( h \) on \( {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \) by the equation \[ h\left( {x, y}\right) = g\left( {x - y}\right) \] Let us prove that \( h \) is measurable. It is not enough to observe that the map \( \left( {x, y}\right) \mapsto x - y \) is continuous and that \( g \) is mea... | Yes |
Theorem 4. If \( f \) and \( g \) belong to \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \), then\n\n\[ \overset{⏜}{f * g} = \widehat{f}\widehat{g} \] | Proof. We use the Fubini Theorem again:\n\n\[ \widehat{f * g}\left( x\right) = \int {e}_{-x}\left( y\right) \left( {f * g}\right) \left( y\right) {dy} = \int {e}_{-x}\left( y\right) \int f\left( u\right) g\left( {y - u}\right) {dudy} \]\n\n\[ = \iint {e}_{-x}\left( {u + y - u}\right) f\left( u\right) g\left( {y - u}\ri... | Yes |
Example 1. The Gaussian function \( \phi \) defined by\n\n\[ \phi \left( x\right) = {e}^{-{\left| x\right| }^{2}} \]\n\nbelongs to \( S \) . | It is easily seen, with the aid of Leibniz’s formula, that if \( \phi \in \mathcal{S} \), then \( P \cdot \phi \in \mathcal{S} \) for any polynomial \( P \), and \( {D}^{\alpha }\phi \in \mathcal{S} \) for any multi-index \( \alpha \) . | No |
Lemma 1. If \( P \) is a polynomial, then the mapping \( \phi \mapsto P \cdot \phi \) is linear and continuous from \( \mathbf{S} \) into \( \mathbf{S} \) . | Proof. Let \( {\phi }_{j} \rightarrow 0 \) . We ask whether \( Q \cdot {D}^{\beta }\left( {P \cdot {\phi }_{j}}\right) \rightarrow 0 \) uniformly for each polynomial \( Q \) and multi-index \( \beta \) . By using the Leibniz formula, this expression can be exhibited as a sum of terms \( {Q}_{\gamma } \cdot {D}^{\alpha ... | No |
Lemma 2. If \( g \in \mathcal{S} \), then the mapping \( \phi \mapsto {g\phi } \) is linear and continuous from \( \mathbf{S} \) into \( \mathbf{S} \) | Proof. This is left to the problems. | No |
Lemma 3. For any multi-index \( \alpha \), the mapping \( \phi \mapsto {D}^{\alpha }\phi \) is linear and continuous from \( \mathbf{S} \) into \( \mathbf{S} \) . | Proof. This is left to the problems. | No |
Lemma 4. The function \( {e}_{y} \) defined by \( {e}_{y}\left( x\right) = {e}^{2\pi ixy} \) obeys the equation \( P\left( D\right) {e}_{y} = P\left( {2\pi iy}\right) {e}_{y} \) for any polynomial \( P \) . | Proof. It suffices to deal with the case of one monomial and establish that \( {D}^{\alpha }{e}_{y} = {\left( 2\pi iy\right) }^{\alpha }{e}_{y} \) . We have\n\n\[ \frac{\partial }{\partial {x}_{j}}{e}_{y}\left( x\right) = \frac{\partial }{\partial {x}_{j}}{e}^{{2\pi i}\left( {{y}_{1}{x}_{1} + \cdots + {y}_{n}{x}_{n}}\r... | Yes |
Theorem 1. If \( \phi \in \mathcal{S} \), and if \( P \) is a polynomial, then \( {\left\lbrack P\left( D/\left( 2\pi i\right) \right) \phi \right\rbrack }^{ \land } = P \cdot \widehat{\phi } \) . Equivalently, \( {\left\lbrack P\left( D\right) \phi \right\rbrack }^{ \land } = {P}^{ + }\widehat{\phi } \), where \( {P}^... | Proof. We have to show that\n\n\[{\left\lbrack \sum {c}_{\alpha }{\left( \frac{D}{2\pi i}\right) }^{\alpha }\phi \right\rbrack }^{ \land }\left( y\right) = \sum {c}_{\alpha }{y}^{\alpha }\widehat{\phi }\left( y\right)\]\n\nSince the Fourier map \( f \mapsto \widehat{f} \) is linear, it suffices to prove that\n\n\[{\lef... | Yes |
Theorem 2. If \( \phi \in \mathcal{S} \) and \( P \) is a polynomial, then \( P\left( {-D/\left( {2\pi i}\right) }\right) \widehat{\phi } = \) \( \widehat{P\phi } \) . Equivalently, \( P\left( D\right) \widehat{\phi } = \widehat{{P}^{ * }\phi } \), where \( {P}^{ * }\left( y\right) = P\left( {-{2\pi iy}}\right) \) . | Proof. We insert the variables, and interpret \( P\left( D\right) \) as differentiating with respect to the variable \( x \) . Thus, with the help of Lemma 4, we have\n\n\[ \left\lbrack {P\left( D\right) \widehat{\phi }}\right\rbrack \left( x\right) = P\left( D\right) \int {e}_{-x}\left( y\right) \phi \left( y\right) {... | Yes |
Theorem 2. If \( \phi \in \mathcal{S} \) and \( P \) is a polynomial, then \( P\left( {-D/\left( {2\pi i}\right) }\right) \widehat{\phi } = \) \( \widehat{P\phi } \) . Equivalently, \( P\left( D\right) \widehat{\phi } = \widehat{{P}^{ * }\phi } \), where \( {P}^{ * }\left( y\right) = P\left( {-{2\pi iy}}\right) \) . | Proof. We insert the variables, and interpret \( P\left( D\right) \) as differentiating with respect to the variable \( x \) . Thus, with the help of Lemma 4, we have\n\n\[ \left\lbrack {P\left( D\right) \widehat{\phi }}\right\rbrack \left( x\right) = P\left( D\right) \int {e}_{-x}\left( y\right) \phi \left( y\right) {... | Yes |
Theorem 4. The mapping \( \phi \mapsto \widehat{\phi } \) is continuous and linear from \( \mathcal{S} \) into S. | Proof. First we must prove that \( \widehat{\phi } \in \mathcal{S} \) when \( \phi \in \mathcal{S} \) . It is to be shown that \( \widehat{\phi } \) is a \( {C}^{\infty } \) -function and that \( P \cdot {D}^{\alpha }\widehat{\phi } \) is bounded for each polynomial \( P \) and for each multi-index \( \alpha \) . In Th... | Yes |
Theorem 5. Poisson Summation Formula. If \( f \in C\left( {\mathbb{R}}^{n}\right) \) and if\n\n\[ \mathop{\sup }\limits_{x}\left( {\left| {f\left( x\right) }\right| + \left| {\widehat{f}\left( x\right) }\right| }\right) {\left( 1 + \left| x\right| \right) }^{n + \varepsilon } < \infty \]\n\nfor some \( \varepsilon > 0 ... | Proof. Let \( c \) equal the supremum in the hypotheses. Then for \( \parallel x{\parallel }_{\infty } \leq 1 \) and \( \nu \neq 0 \) we have\n\n\[ \left| {f\left( {x + \nu }\right) }\right| \leq c{\left( 1 + \left| x + \nu \right| \right) }^{-n - \varepsilon } \leq c{\left( 1 + \parallel x + \nu {\parallel }_{\infty }... | Yes |
Theorem 1. The function \( \theta \) defined on \( {\mathbb{R}}^{n} \) by \( \theta \left( x\right) = {e}^{-\pi {x}^{2}} \) is a fixed point of the Fourier transform. Thus, \( \widehat{\theta } = \theta \) . | Proof. First observe that the notation is \[ {x}^{2} = {xx} = x \cdot x = \langle x, x\rangle = \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}^{2} = {\left| x\right| }^{2} \] We prove our result first when \( n = 1 \) and then derive the general case. Define, for \( x \in \mathbb{R} \), the analogous function \( \psi \left... | Yes |
Theorem 2. First Inversion Theorem. If \( \phi \in \mathcal{S}\left( {\mathbb{R}}^{n}\right) \), then\n\n\[ \phi \left( x\right) = {\int }_{{\mathbb{R}}^{n}}\widehat{\phi } \cdot {e}_{x} = {\int }_{{\mathbb{R}}^{n}}\widehat{\phi }\left( y\right) {e}^{2\pi iyx}{dy} \] | Proof. We use the conjurer’s tricks of smoke and mirrors. Let \( \theta \) be the function in the preceding theorem, and put \( g\left( x\right) = \theta \left( {x/\lambda }\right) \) . Then \( \widehat{g}\left( y\right) = {\lambda }^{n}\widehat{\theta }\left( {\lambda y}\right) \) . (Problem 8 in Section 6.1, page 293... | No |
Theorem 3. The Fourier transform operator \( \mathcal{F} \) from \( \mathcal{S}\left( {\mathbb{R}}^{n}\right) \) to \( \mathcal{S}\left( {\mathbb{R}}^{n}\right) \) is a continuous linear bijection, and \( {\mathcal{F}}^{-1} = {\mathcal{F}}^{3} \) . | Proof. The continuity and linearity of \( \mathcal{F} \) were established by Theorem 4 in Section 6.2, page 297. The fact that \( \mathcal{F} \) is surjective is established by writing the basic inversion formula from the preceding theorem as \[ \phi \left( x\right) = \int \widehat{\phi }\left( y\right) {e}_{x}\left( y... | Yes |
Theorem 4. Second Inversion Theorem. If \( f \) and \( \widehat{f} \) belong to \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \), then for almost all \( x \) , \[ f\left( x\right) = {\int }_{{\mathbb{R}}^{n}}\widehat{f}\left( y\right) {e}^{2\pi ixy}{dy} \] | Proof. Assume that \( f \) and \( \widehat{f} \) are in \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \). Let \( \phi \in \mathcal{S} \). Then by Theorem 2, \( \phi \left( x\right) = \int {e}_{x}\widehat{\phi } \). By Problem 13 in Section 6.2, page 300, \( \int \widehat{f}\phi = \int f\widehat{\phi } \). Hence if we put \(... | Yes |
Lemma 1. If \( f \) and \( g \) belong to the Schwartz space \( \mathcal{S}\left( {\mathbb{R}}^{n}\right) \), then \( f * g \) also belongs to \( \mathcal{S}\left( {\mathbb{R}}^{n}\right) \), and furthermore, \( \widehat{fg} = \widehat{f} * \widehat{g} \) . | Proof. Since \( f \) and \( g \) belong to \( \mathcal{S} \), so does \( {fg} \) by Lemma 2 in Section 6.2, page 295. By Theorem 4 of Section 6.2, page 297, \( \widehat{f},\widehat{g} \), and \( \widehat{fg} \) belong to S. Consequently, \( \widehat{f}\widehat{g} \) belongs to \( \mathbf{S} \) . By Theorem 4 of Section... | Yes |
Lemma 2. If \( f \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \) and \( \phi \in \mathcal{D}\left( {\mathbb{R}}^{n}\right) \), then \( f * \phi \in {C}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . | Proof. By the theorem in Section 5.5, page 271,\n\n\[ \n{D}^{\alpha }\left( {T * \phi }\right) = T * {D}^{\alpha }\phi \;\left( {T \in {\mathcal{D}}^{\prime },\phi \in \mathcal{D}}\right) \n\]\n\nIn particular, for \( f \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \), \n\n(1)\n\n\[ \n{D}^{\alpha }\left( {f * \phi }\right)... | Yes |
Lemma 3. The translation operator \( {E}_{x} \) has the following continuity property: If \( 1 \leq p < \infty \) and \( f \in {L}^{p}\left( {\mathbb{R}}^{n}\right) \), then the mapping \( x \mapsto {E}_{x}f \) is continuous from \( {\mathbb{R}}^{n} \) to \( {L}^{p}\left( {\mathbb{R}}^{n}\right) \) . | Proof. The continuous functions with compact support form a dense set in \( {L}^{p} \), if \( 1 \leq p < \infty \) . Hence, if \( \varepsilon > 0 \), then there exists such a continuous function \( h \) for which \( \parallel f - h{\parallel }_{p} \leq \varepsilon \) . Let the support of \( h \) be contained in the bal... | Yes |
Theorem 1. If \( f \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \), then \( f * {\rho }_{k} \rightarrow f \) in the metric of \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \) . | Proof. Since \( \int {\rho }_{k} = 1 \), \[ \left( {f * {\rho }_{k}}\right) \left( x\right) - f\left( x\right) = \int \left\lbrack {f\left( {x - z}\right) - f\left( x\right) }\right\rbrack {\rho }_{k}\left( z\right) {dz} \] Hence by Fubini's Theorem (Chapter 8, page 426) \[ \int \left| {f * {\rho }_{k} - f}\right| \leq... | Yes |
Theorem 2. The space of test functions \( \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) is a dense subspace of \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \). | Proof. Let \( f \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \), and let \( \varepsilon > 0 \) . We wish to find an element of \( \mathcal{D}\left( {R}^{n}\right) \) within distance \( \varepsilon \) of \( f \) . The function \( f * {\rho }_{k} \) from the preceding theorem would be a candidate, but it need not have compa... | No |
Let \( n = 1 \) and \( D = \frac{d}{dx} \). If \( P \) is a polynomial, say \( P\left( \lambda \right) = \mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}{\lambda }^{j} \), then \( P\left( D\right) \) is a linear differential operator with constant coefficients: | \[ P\left( D\right) = \mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}{D}^{j} = \mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}{\left( 2\pi i\right) }^{j}{\left( \frac{D}{2\pi i}\right) }^{j} \] Consider the ordinary differential equation \[ P\left( D\right) u = g\; - \infty < x < \infty \] in which \( g \) is given and is assume... | Yes |
\[ {u}^{\prime }\left( x\right) + {bu}\left( x\right) = {e}^{-\left| x\right| }\;\left( {b > 0,\;b \neq 1}\right) \] | The Fourier transform of the function \( g\left( x\right) = {e}^{-\left| x\right| } \) is \( \widehat{g}\left( t\right) = 2/\left( {1 + 4{\pi }^{2}{t}^{2}}\right) \) (Problem 5 of Section 6.3, page 304). Hence the Fourier transform of Equation (11) is \[ {2\pi it}\;\widehat{u}\left( t\right) + b\;\widehat{u}\left( t\ri... | Yes |
Consider the integral equation\n\n\[ \n{\\int }_{-\\infty }^{\\infty }k\\left( {x - s}\\right) u\\left( s\\right) {ds} = g\\left( x\\right) \n\]\n\nin which \( k \) and \( g \) are given, and \( u \) is an unknown function. We can write\n\n\[ \nu * k = g \n\] | After taking Fourier transforms and using Theorem 4 in Section 6.1 (page 290)\n\nwe have\n\n\[ \n\\widehat{u}\\widehat{k} = \\widehat{g} \n\]\n\nwhence \( \\widehat{u} = \\widehat{g}/\\widehat{k} \) and \( u = {\\mathcal{F}}^{-1}\\left( {\\widehat{g}/\\widehat{k}}\\right) \) . | Yes |
Theorem 1. If \( f \) is the Fourier transform of a positive function in \( {L}^{1}\left( {\mathbb{R}}^{n}\right) \), then for any finite set of points \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \) in \( {\mathbb{R}}^{n} \) the matrix having elements \( f\left( {{x}_{i} - {x}_{j}}\right) \) will be positive definite (and hence... | Proof. Let \( f = \widehat{g} \), where \( g \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \) and \( g\left( x\right) > 0 \) everywhere. The interpolation matrix in question must be shown to be positive definite. This means that \( {u}^{ * }{Au} > 0 \) for all nonzero vectors \( u \) in \( {\mathbb{C}}^{m} \) . We undertak... | No |
Lemma 1. Let \( {\lambda }_{1},\ldots ,{\lambda }_{m} \) be \( m \) distinct complex numbers, and let \( {c}_{1},\ldots ,{c}_{m} \) be complex numbers. If \( \mathop{\sum }\limits_{{j = 1}}^{m}{c}_{j}{e}^{{\lambda }_{j}z} = 0 \) for all \( z \) in a subset of \( \mathbb{C} \) that has an accumulation point, then \( \ma... | Proof. Use induction on \( m \) . If \( m = 1 \), the result is obvious, because \( {e}^{{\lambda }_{1}z} \) is not zero for any \( z \in \mathbb{C} \) . If the lemma has been established for a certain integer \( m - 1 \), then we can prove it for \( m \) as follows. Let \( f\left( z\right) = \mathop{\sum }\limits_{1}^... | Yes |
Lemma 2. Let \( {w}_{1},\ldots ,{w}_{m} \) be \( m \) distinct points in \( {\mathbb{C}}^{n} \). Let \( {c}_{1},\ldots ,{c}_{m} \) be complex numbers. If \( \mathop{\sum }\limits_{{j = 1}}^{m}{c}_{j}{e}^{{w}_{j}x} = 0 \) for all \( x \) in a nonempty open subset of \( {\mathbb{R}}^{n} \), then \( \mathop{\sum }\limits_... | Proof. Let \( \mathcal{O} \) be an open set in \( {\mathbb{R}}^{n} \) having the stated property. Select \( \xi \in \mathcal{O} \) such that the complex inner products \( {w}_{j}\xi \) are all different. This is possible by the following reasoning. The condition on \( \xi \) can be expressed in the form \( {w}_{j}\xi \... | Yes |
Theorem 2. Laurent’s Theorem. Let \( f \) be a function that is analytic inside and on a circle \( C \) in the complex plane, except for having an isolated singularity at the center \( \zeta \) . Then at each point inside \( C \) with the exception of \( \zeta \) we have\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{... | The coefficient \( {c}_{-1} \) is called the residue of \( f \) at \( \zeta \) . By Laurent’s theorem, the residue is also given by\n\n(12)\n\n\[ {c}_{-1} = \frac{1}{2\pi i}{\int }_{C}f\left( z\right) {dz} \] | Yes |
The integral \( {\int }_{C}{e}^{z}/{z}^{4}{dz} \), where \( C \) is the unit circle, can be computed with the principle in Equation (12). Indeed, the given integral is \( {2\pi i} \) times the residue of \( {e}^{z}/{z}^{4} \) at 0 . | Since\n\n\[ \n{e}^{z}/{z}^{4} = \left( {1 + z + \frac{{z}^{2}}{2!} + \frac{{z}^{3}}{3!} + \cdots }\right) /{z}^{4} \n\]\n\n\[ \n= {z}^{-4} + {z}^{-3} + \frac{1}{2}{z}^{-2} + \frac{1}{6}{z}^{-1} + \cdots \n\]\n\nwe see that the residue is \( \frac{1}{6} \) and the integral is \( \frac{1}{3}{\pi i} \) . | Yes |
Theorem 3 The Residue Theorem. Let \( C \) be a simple closed curve inside of which \( f \) is analytic with the exception of isolated singularities at the points \( {\zeta }_{1},\ldots ,{\zeta }_{m} \) . Then \( \frac{1}{2\pi i}{\int }_{C}f\left( z\right) {dz} \) is the sum of the residues of \( f \) at \( {\zeta }_{1... | Proof. Draw mutually disjoint circles \( {C}_{1},\ldots ,{C}_{m} \) around the singularities and contained within \( C \) . The integral around the path shown in the figure is zero, by Cauchy’s integral theorem. (Figure 6.1a depicts the case \( m = 2 \) .) Therefore,\n\n\[ 0 = {\int }_{C}f\left( z\right) {dz} - {\int }... | Yes |
Example 5. Let us compute \( {\int }_{C}\frac{dz}{{z}^{2} + 1} \), where \( C \) is the circle described by \( \left| {z - i}\right| = 1 \). | By the preceding theorem, the integral is \( {2\pi i} \) times the sum of the residues inside \( C \) . We have\n\n\[ f\left( z\right) = \frac{1}{{z}^{2} + 1} = \frac{1}{\left( {z + i}\right) \left( {z - i}\right) } = \frac{i/2}{z + i} - \frac{i/2}{z - i} \]\n\nThe residue at \( i \) is therefore \( - i/2 \), and the v... | Yes |
Theorem 4. If \( f \) is a proper rational function and if the curve \( C \) encloses all the poles of \( f \), then \( {\int }_{C}f\left( z\right) {dz} = 0 \) . | Proof. Write \( f = p/q \), where \( p \) and \( q \) are polynomials. Since \( f \) is proper, the degree of \( p \) is less than that of \( q \) . Hence the point at \( \infty \) is not a singularity of \( f \) . Now, \( C \) is the boundary of one region containing the poles, and it is also the boundary of the compl... | Yes |
Theorem 4. If \( f \) is a proper rational function and if the curve \( C \) encloses all the poles of \( f \), then \( {\int }_{C}f\left( z\right) {dz} = 0 \) . | Proof. Write \( f = p/q \), where \( p \) and \( q \) are polynomials. Since \( f \) is proper, the degree of \( p \) is less than that of \( q \) . Hence the point at \( \infty \) is not a singularity of \( f \) . Now, \( C \) is the boundary of one region containing the poles, and it is also the boundary of the compl... | Yes |
Theorem 5. Let \( f \) be analytic in the closed upper half-plane with the exception of a finite number of poles, none of which are on the real axis. Define\n\n\[ \n{M}_{r} \equiv \sup \{ \left| {{zf}\left( z\right) }\right| : \left| z\right| = r,\mathcal{I}\left( z\right) \geq 0\}\n\]\n\nIf \( {M}_{r} \) converges to ... | Proof. Consider the region shown in Figure 6.1b, where \( C \) is the semicircular arc and \( r \) is chosen so large that all the poles of \( f \) lying in the upper half-plane are contained in the semicircular region. On \( C \) we have \( z = r{e}^{i\theta } \) and \( {dz} = {ir}{e}^{i\theta }{d\theta } \) . Hence\n... | Yes |
The simplest case of the heat equation is\n\n\[ \n{u}_{xx} = {u}_{t} \n\]\nin which the subscripts denote partial derivatives. The distribution of heat in an infinite bar would obey this equation for \( \infty < x < \infty \) and \( t \geq 0 \) . A fully defined practical problem would consist of the differential equat... | We define \( \widehat{u}\left( {y, t}\right) \) to be the Fourier transform of \( u \) in the space variable. Thus\n\n\[ \widehat{u}\left( {y, t}\right) = {\int }_{-\infty }^{\infty }u\left( {x, t}\right) {e}^{-{2\pi ixy}}{dx} \]\n\nTaking the Fourier transform in Equations (1) and (2) with respect to the space variabl... | Yes |
\[ \left\{ \begin{array}{ll} {u}_{xx} = {u}_{t} & x \geq 0, t \geq 0 \\ u\left( {x,0}\right) = f\left( x\right), u\left( {0, t}\right) = 0 & x \geq 0, t \geq 0 \end{array}\right. \] | The easiest way to ensure that this will be zero (and thus satisfy the boundary condition in our problem) is to extend \( f \) to be an odd function. Then the integrand in Equation (8) is odd, and \( u\left( {0, t}\right) = 0 \) automatically. So we define \( f\left( {-x}\right) = - f\left( x\right) \) for \( x > 0 \),... | No |
Example 3. Again, we consider the heat equation with boundary conditions:\n\n\[ \left\{ \begin{array}{ll} {u}_{xx} = {u}_{t} & x \geq 0, t \geq 0 \\ u\left( {x,0}\right) = f\left( x\right) & u\left( {0, t}\right) = g\left( t\right) \end{array}\right. \] | Because the differential equation is linear and homogeneous, the method of superposition can be applied. We solve two related problems, viz.,\n\n\[ {v}_{xx} = {v}_{t}\;v\left( {x,0}\right) = f\left( x\right) \;v\left( {0, t}\right) = 0 \]\n\n\[ {w}_{xx} = {w}_{t}\;w\left( {x,0}\right) = 0\;w\left( {0, t}\right) = g\lef... | Yes |
The Helmholtz Equation is\n\n\[ \n{\Delta u} - {gu} = f \n\]\n\nin which \( \Delta \) is the Laplacian, \( \mathop{\sum }\limits_{{k = 1}}^{n}{\partial }^{2}/\partial {x}_{k}^{2} \) . The functions \( f \) and \( g \) are prescribed on \( {\mathbb{R}}^{n} \), and \( u \) is the unknown function of \( n \) variables. We... | Carrying out the differentiation under the integral that defines the convolution, we obtain\n\n\[ \nf * {\Delta h} - f * h = f \n\]\n\nIs there a way to cancel the three occurrences of \( f \) in this equation? After all, \( {L}^{1} \) is a Banach algebra, with multiplication defined by convolution. But there are pitfa... | Yes |
Theorem 1. Every distribution having compact support is tempered. | Proof. Let \( T \) be a distribution with compact support \( K \) . Select \( \psi \in \mathfrak{D} \) so that \( \psi \left( x\right) = 1 \) for all \( x \) in an open neighborhood of \( K \) . We extend \( T \) by defining \( \bar{T}\left( \phi \right) = T\left( {\phi \psi }\right) \) when \( \phi \in \mathcal{S} \) ... | Yes |
Theorem 2. Let \( f \) be a measurable function such that \( f/P \in {L}^{1}\left( {\mathbb{R}}^{n}\right) \) for some polynomial \( P \) . Then \( \widetilde{f} \) is a tempered distribution. | Proof. For \( \phi \in \mathcal{S} \), we have\n\n\[ \widetilde{f}\left( \phi \right) = {\int }_{{\mathbb{R}}^{n}}f\left( x\right) \phi \left( x\right) {dx} \]\n\nSuppose that \( P \) is a polynomial such that \( f/P \in {L}^{1} \) . Write\n\n\[ \widetilde{f}\left( \phi \right) = \int \left( {f/P}\right) \left( {P \cdo... | Yes |
Theorem 3. If \( T \) is a tempered distribution, then so is \( \widehat{T} \) . Moreover, the map \( T \mapsto \widehat{T} \) is linear, injective, surjective, and continuous from \( {\mathcal{S}}^{\prime } \) to \( {\mathcal{S}}^{\prime } \) . | Proof. The Fourier operator \( \mathcal{F} \) is a continuous linear bijection from \( \mathcal{S} \) onto \( \mathcal{S} \) by Theorem 3 in Section 6.3, page 303. Also, \( {\mathcal{F}}^{-1} = {\mathcal{F}}^{3} \) . Since \( \widehat{T} = T \circ \mathcal{F} \), we see that \( \widehat{T} \) is the composition of two ... | Yes |
Theorem 4. If \( T \) is a tempered distribution and \( P \) is a polynomial, then \[ \widehat{PT} = P\left( \frac{-\partial }{2\pi i}\right) \widehat{T}\;\text{ and }\;P \cdot \widehat{T} = \widehat{P\left( \frac{\partial }{2\pi i}\right) }T \] | Proof. For \( \phi \) in \( \mathcal{S} \) we have \[ \widehat{PT}\left( \phi \right) = \left( {PT}\right) \left( \widehat{\phi }\right) = T\left( {P\widehat{\phi }}\right) = T\left\lbrack {\left( P\left( \frac{D}{2\pi i}\right) \phi \right) }^{ \land }\right\rbrack = \widehat{T}\left( {P\left( \frac{D}{2\pi i}\right) ... | No |
Lemma 1. If \( f \in {L}^{p}\left( {\mathbb{R}}^{n}\right) \), and if \( {\psi }_{j} \) is as described above, then \( f * {\psi }_{j} \rightarrow f \) in \( {L}^{p}\left( {\mathbb{R}}^{n}\right) \), as \( j \rightarrow \infty \) . | Proof. The case \( p = 1 \) is contained in the proof of Theorem 1 in Section 6.4, page 306. Let \( {B}_{j} \) be the support of \( {\psi }_{j} \) (i.e., the ball at 0 of radius \( 1/j \) ). By familiar calculations and Hölder's inequality (Section 8.7, page 409) we have\n\n\[ \left| {\left( {f * {\psi }_{j}}\right) \l... | Yes |
Theorem 3. The set of functions in \( {W}^{k, p}\left( \Omega \right) \) that are of class \( {C}^{\infty } \) is dense in \( {W}^{k, p}\left( \Omega \right) \) . | Proof. Let \( {B}_{1},{B}_{2},\ldots \) be a sequence of open balls such that \( \overline{{B}_{i}} \subset \Omega \) for all \( i \) and \( \bigcup {B}_{i} = \Omega \) . The center and radius of \( {B}_{i} \) are indicated by writing \( {B}_{i} = B\left( {{x}_{i},{r}_{i}}\right) \) . Appealing to Theorem 1 in Section ... | Yes |
Every continuous function on the interval \( \left\lbrack {a, b}\right\rbrack \) is integrable. Hence, this simple containment relation is valid: \( C\left\lbrack {a, b}\right\rbrack \subset {L}^{1}\left\lbrack {a, b}\right\rbrack \) . Is this an embedding? We seek a constant \( c \) such that\n\n\[ \parallel f{\parall... | The constant \( c = b - a \) obviously serves:\n\n\[ \parallel f{\parallel }_{1} = {\int }_{a}^{b}\left| {f\left( x\right) }\right| {dx} \leq {\int }_{a}^{b}\parallel f{\parallel }_{\infty } = \left( {b - a}\right) \parallel f{\parallel }_{\infty } \] | Yes |
If \( 1 \leq s < r < \infty \) and if the domain \( \Omega \) has finite Lebesgue measure, then \( {L}^{r}\left( \Omega \right) \hookrightarrow {L}^{s}\left( \Omega \right) \). | To prove this, start with an \( f \) in \( {L}^{r}\left( \Omega \right) \) and write \( r = {ps} \). We may assume that \( f \geq 0 \). Then \( {f}^{s} \) is in \( {L}^{p}\left( \Omega \right) \) because \( \int {f}^{sp} = \int {f}^{r} \). Use the Hölder Inequality (page 409) with conjugate indices \( p \) and \( q = p... | Yes |
Theorem 5. \( \;{W}^{1,2}\left( \mathbb{R}\right) \hookrightarrow {W}^{0,\infty }\left( \mathbb{R}\right) \) . | Proof. (In outline. For details, see [LL], Chapter 8.) Let \( f \) be an element of \( {W}^{1,2}\left( \mathbb{R}\right) \) . Since \( \mathcal{D}\left( \mathbb{R}\right) \) is dense in \( {W}^{1,2}\left( \mathbb{R}\right) \), there exists a sequence \( \left\lbrack {f}_{i}\right\rbrack \) in \( \mathcal{D}\left( \math... | No |
Theorem 2. If a topological space has the fixed-point property, then the same is true of every space homeomorphic to it. | Proof. Let spaces \( X \) and \( Y \) be homeomorphic. This means that there is a homeomorphism \( h : X \rightarrow Y \) (a continuous map having a continuous inverse). Suppose that \( X \) has the fixed-point property. To prove that \( Y \) has the fixed-point property, let \( f \) be a continuous map of \( Y \) into... | Yes |
Theorem 3. The Schauder-Tychonoff Fixed-Point Theorem. Every compact convex set in a locally convex linear topological Hausdorff space has the fixed-point property. | Proof. ([Day],[Sma]) Let \( K \) be such a set, and let \( f \) be a continuous map of \( K \) into \( K \) . We denote the family of all convex, symmetric, open neighborhoods of 0 by \( \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) . The set \( A \) is simply an index set, which we partially order by writing \( \a... | Yes |
Theorem 6. Let \( D \) be a convex set in a locally convex linear topological Hausdorff space. If \( f \) maps \( D \) continuously into a compact subset of \( D \), then \( f \) has a fixed point. | Proof. As in the proof of Theorem 3, we use the family of neighborhoods \( {U}_{\alpha } \) . Let \( K \) be a compact subset of \( D \) that contains \( f\left( D\right) \) . Proceed as in the proof of Theorem 3, using the same set of neighborhoods \( {U}_{\alpha } \) . By the lemma, for each \( \alpha \) there is a f... | Yes |
Theorem 7. Rothe's Theorem. Let \( B \) denote the closed unit ball of a normed linear space \( X \) . If \( f \) maps \( B \) continuously into a compact subset of \( X \) and if \( f\left( {\partial B}\right) \subset B \), then \( f \) has a fixed point. | Proof. Let \( r \) denote the radial projection into \( B \) defined by \( r\left( x\right) = x \) if \( \parallel x\parallel \leq 1 \) and \( r\left( x\right) = x/\parallel x\parallel \) if \( \parallel x\parallel > 1 \) . This map is continuous (Problem 1). Hence \( r \circ f \) maps \( B \) into a compact subset of ... | Yes |
Theorem 8. Let \( B \) denote the closed unit ball in a normed space \( X \) . Let \( \left\{ {{f}_{t} : 0 \leq t \leq 1}\right\} \) be a family of continuous maps from \( B \) into one compact subset of \( X \) . Assume that\n\n(i) \( {f}_{0}\left( {\partial B}\right) \subset B \) .\n\n(ii) The map \( \left( {t, x}\ri... | Proof. (From [Sma]) If \( 0 < \varepsilon < 1 \), define\n\n\[ \n{g}_{\varepsilon }\left( x\right) = \left\{ \begin{array}{ll} {f}_{1}\left( \frac{x}{1 - \varepsilon }\right) & \parallel x\parallel \leq 1 - \varepsilon \\ {f}_{\left( {1 - \parallel x\parallel }\right) /\varepsilon }\left( \frac{x}{\parallel x\parallel ... | Yes |
Theorem 2 Let \( X \) be a paracompact space, \( Y \) a Banach space, and \( H \) a closed subspace in \( Y \) . Suppose that \( f : X \rightarrow Y \) is continuous and \( g : X \rightarrow H \) is bounded. Then for each \( \varepsilon > 0 \) there is a continuous map \( \bar{g} : X \rightarrow H \) that satisfies\n\n... | Proof. Let \( \lambda \) denote the number on the right in Inequality (1). For each \( x \in X \) , define\n\n\[ \Phi \left( x\right) = \{ h \in H : \parallel f\left( x\right) - h\parallel \leq \lambda \} \]\n\nThis set is nonempty because \( g\left( x\right) \in \Phi \left( x\right) \) . (Notice that \( g \) is a sele... | Yes |
Theorem 3. The Bartle-Graves Theorem. A continuous linear map of one Banach space onto another must have a continuous (but not necessarily linear) right inverse. | Proof. Let \( A : X \rightarrow Y \), as in the hypotheses. Since \( A \) is surjective, the equation \( {Ax} = y \) has solutions \( x \) for each \( y \in Y \) . At issue, then, is whether a continuous choice of \( x \) can be made. It is clear that we should set\n\n\[ \Phi \left( y\right) = \{ x \in X : {Ax} = y\} \... | Yes |
Theorem 1. Let \( X \) be a normed linear space and let \( K \) be a convex subset of \( X \) that contains 0 as an interior point. If \( z \in X \smallsetminus K \) , then there is a continuous linear functional \( \phi \) defined on \( X \) such that for all \( x \in K,\phi \left( x\right) \leq 1 \leq \phi \left( z\r... | Proof. Again, we need the Minkowski functional of \( K \) . It is\n\n\[ p\left( x\right) = \inf \{ \lambda : \lambda > 0\text{ and }x/\lambda \in K\} \]\n\nWe prove now that \( p\left( {x + y}\right) \leq p\left( x\right) + p\left( y\right) \) for all \( x \) and \( y \) . Select \( \lambda ,\mu > 0 \) so that \( x/\la... | Yes |
Theorem 2. Let \( {K}_{1},{K}_{2} \) be a disjoint pair of convex sets in a normed linear space \( X \) . If one of them has an interior point, then there is a nonzero functional \( \phi \in {X}^{ * } \) such that\n\n\[ \mathop{\sup }\limits_{{x \in {K}_{1}}}\phi \left( x\right) \leq \mathop{\inf }\limits_{{x \in {K}_{... | Proof. By performing a translation and by relabeling the two sets, we can assume that 0 is an interior point of \( {K}_{1} \) . Fix a point \( z \) in \( {K}_{2} \) and consider the set \( {K}_{1} - {K}_{2} + z \) . This set is convex and contains 0 as an interior point. Also, \( z \notin {K}_{1} - {K}_{2} + z \) becau... | Yes |
Theorem 3. Let \( {K}_{1},{K}_{2} \) be a disjoint pair of closed convex sets in a normed linear space \( X \) . Assume that at least one of the sets is compact. Then there is a \( \phi \in {X}^{ * } \) such that\n\n\[ \mathop{\sup }\limits_{{x \in {K}_{2}}}\phi \left( x\right) < \mathop{\inf }\limits_{{x \in {K}_{1}}}... | Proof. The set \( {K}_{1} - {K}_{2} \) is closed and convex. (See Problems 1.2.19 on page 12 and 1.4.17 on page 23.) Also, \( 0 \notin {K}_{1} - {K}_{2} \), and consequently there is a ball \( B\left( {0, r}\right) \) that is disjoint from \( {K}_{1} - {K}_{2} \) . By the preceding theorem, there is a nonzero continuou... | Yes |
Theorem 4. Let \( U \) be a compact set in a real Hilbert space. In order that the system of linear inequalities\n\n(1)\n\n\[ \langle u, x\rangle > 0\;\left( {u \in U}\right) \]\n\nbe consistent (i.e., have a solution, \( x \) ) it is necessary and sufficient that 0 not be in the closed convex hull of \( U \) . | Proof. For the sufficiency of the condition, assume the condition to be true. Thus, \( 0 \notin \overline{\operatorname{co}}\left( U\right) \) . By Theorem 3, there is a vector \( x \) and a real number \( \lambda \) such that \( \overline{\mathrm{{co}}}\left( U\right) \) and 0 are on opposite sides of the hyperplane\n... | Yes |
Theorem 5. For an \( m \times n \) matrix \( A \), either \( {Ax} \geq 0 \) for some \( x \in {S}_{n} \) , or \( {y}^{T}A < 0 \) for some \( y \in {S}_{m} \) . | Proof. Suppose that there is no \( x \) in the simplex \( {S}_{n} \) for which \( {Ax} \geq 0 \) . Then \( A\left( {S}_{n}\right) \) contains no point in the nonnegative orthant, \[ {P}_{m} = \left\{ {y \in {\mathbb{R}}^{m} : y \geq 0}\right\} \] Consequently, the convex sets \( A\left( {S}_{n}\right) \) and \( {P}_{m}... | Yes |
Theorem 2. Arzelà-Ascoli Theorem II. Let \( X \) be a compact metric space. A subset of \( C\left( X\right) \) is compact if and only if it is closed, bounded, and equicontinuous. | Proof. Suppose that \( K \) is a compact set in \( C\left( X\right) \) . Then it is closed. It is also totally bounded, and can be covered by a finite number of balls of radius 1 :\n\n\[ K \subset \mathop{\bigcup }\limits_{{i = 1}}^{n}B\left( {{f}_{i},1}\right) \]\n\nFor any \( g \in K \) there is an index \( i \) for ... | Yes |
Theorem 3. Dini’s Theorem. Let \( {f}_{1},{f}_{2},\ldots \) be continuous real-valued functions on a compact topological space. For each \( x \) assume that \( \left| {{f}_{n}\left( x\right) }\right| \downarrow 0 \) . Then this convergence is uniform. | Proof. Given \( \varepsilon > 0 \), put \( {S}_{k} = \left\{ {x : \left| {{f}_{k}\left( x\right) }\right| \geq \varepsilon }\right\} \) . Then each \( {S}_{k} \) is closed, and \( {S}_{k + 1} \subset {S}_{k} \) . For each \( x \) there is an index \( k \) such that \( x \notin {S}_{k} \) . Hence \( \mathop{\bigcap }\li... | Yes |
Lemma 1. Let \( A \) be a compact operator on a normed linear space.\n\nIf \( I + A \) is surjective, then it is injective. | Proof. Let \( B = I + A \) and \( {X}_{n} = \ker \left( {B}^{n}\right) \) . Suppose that \( B \) is surjective but not injective. We shall be looking for a contradiction. Note that \( 0 \subset {X}_{1} \subset {X}_{2} \subset \cdots \) It is now to be proved that these inclusions are proper. Select a nonzero element \(... | Yes |
Lemma 3. Let \( A \) be a compact operator on a Banach space. If \( I + A \) is injective, then it is surjective. | Proof. Let \( B = I + A \) and let \( {X}_{n} \) denote the range of \( {B}^{n} \) . We have \[ {B}^{n} = {\left( I + A\right) }^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {A}^{k} = I + \mathop{\sum }\limits_{{k = 1}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {... | Yes |
Theorem 1. The Fredholm Alternative. Let \( A \) be a compact linear operator on a Banach space. The operator \( I + A \) is surjective if and only if it is injective. | ## Proof. This is the result of putting Lemmas 1 and 3 together. | No |
Theorem 3. Let \( B \) be a bounded linear invertible operator, and let \( A \) be a compact operator, both defined on one Banach space and taking values in another. Then \( B + A \) is surjective if and only if it is injective. | Proof. Suppose that \( B + A \) is injective. Then so are \( {B}^{-1}\left( {B + A}\right) \) and \( I + {B}^{-1}A \) . Now, the product of a compact operator with a bounded operator is compact. (See Problem 7.) Thus, Theorem 1 is applicable, and \( I + {B}^{-1}A \) is surjective. Hence so are \( B\left( {I + {B}^{-1}A... | Yes |
Theorem 4. A compact linear transformation operating from one normed linear space to another maps weakly convergent sequences into strongly convergent sequences. | Proof. Let \( A \) be such an operator, \( A : X \rightarrow Y \) . Let \( {x}_{n} \rightharpoonup x \) (weak convergence) in \( X \) . It suffices to consider only the case when \( x = 0 \) . Thus we want to prove that \( A{x}_{n} \rightarrow 0 \) . By the weak convergence, \( \phi \left( {x}_{n}\right) \rightarrow 0 ... | Yes |
Lemma 4. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a bounded sequence of continuous linear transformations from one normed linear space to another. If \( {A}_{n}x \rightarrow 0 \) for each \( x \) in a compact set \( K \), then this convergence is uniform on \( K \) . | Proof. Suppose that the convergence in question is not uniform. Then there exist a positive \( \varepsilon \), a sequence of integers \( {n}_{i} \), and points \( {x}_{{n}_{i}} \in K \) such that \( \begin{Vmatrix}{{A}_{{n}_{i}}{x}_{{n}_{i}}}\end{Vmatrix} \geq \varepsilon \) . Since \( K \) is compact, we can assume at... | Yes |
Theorem 5. Let \( X \) and \( Y \) be Banach spaces. If \( Y \) has a (Schauder) basis, then every compact operator from \( X \) to \( Y \) is a limit of finite-rank operators. | Proof. If \( \left\lbrack {v}_{n}\right\rbrack \) is a basis for \( Y \), then each \( y \) in \( Y \) has a unique representation of the form\n\n\[ y = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\lambda }_{k}\left( y\right) {v}_{k} \]\n\n(See Problems 24-26 in Section 1.6, pages 38-39.) The functionals \( {\lambda }_{k... | No |
Theorem 6. Let \( A \) be a compact operator acting between two Banach spaces. If the range of \( A \) is closed, then it is finite dimensional. | Proof. Since \( A \) is compact, it is continuous and has a closed graph. Assume that \( A : X \rightarrow Y \) and that \( A\left( X\right) \) is closed in \( Y \) . Then \( A\left( X\right) \) is a Banach space. Let \( S \) denote the unit ball in \( X \) . By the Interior Mapping Theorem (Section 1.8, page 48), \( A... | Yes |
Lemma 5. In the definition of the degenerate kernel \( k = \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}{v}_{i}, \) there is no loss of generality in supposing that \( \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) and \( \left\{ {{v}_{1},\ldots ,{u}_{n}}\right\} \) are linearly independent sets. | Proof. Suppose that \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is linearly dependent. Then one vector is a linear combination of the others, say \( {v}_{n} = \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}{a}_{i}{v}_{i} \) . Then we can write the kernel with a sum of fewer terms as follows:\n\n\[ \n{Kx} = \mathop{\sum }... | Yes |
Theorem 7. Let \( A \) be a compact operator on a Banach space. Each nonzero element of the spectrum of \( A \) is an eigenvalue of \( A \) . | Proof. Let \( \lambda \neq 0 \) and suppose that \( \lambda \) is not an eigenvalue of \( A \) . We want to show that \( \lambda \) is not in the spectrum, or equivalently, that \( A - {\lambda I} \) is invertible. Since \( \lambda \) is not an eigenvalue, the equation \( \left( {A - {\lambda I}}\right) x = 0 \) has on... | Yes |
Let \( T = \left\lbrack {a, b}\right\rbrack \subset \mathbb{R} \). Let \( \mathcal{A} \) be the algebra of all polynomials in \( C\left( T\right) \). Then \( \mathcal{A} \) is dense in \( C\left( T\right) \), by the Stone-Weierstrass Theorem. This implies that for any continuous function \( f \) defined on \( \left\lbr... | \[ \parallel f - p\parallel \equiv \max \{ \left| {f\left( t\right) - p\left( t\right) }\right| : a \leq t \leq b\} < \epsilon \] | Yes |
Theorem 9. Let \( {A}_{0},{A}_{1},\ldots \) be compact operators on a Banach space, and suppose \( \mathop{\lim }\limits_{n}{A}_{n} = {A}_{0} \) . If \( \lambda \) is not an eigenvalue of \( {A}_{0} \) and if for each \( n \) there is a point \( {x}_{n} \) such that \( {A}_{n}{x}_{n} - \lambda {x}_{n} = b \), then for ... | Proof. Since \( \lambda \) is not an eigenvalue of \( {A}_{0} \), it is not in the spectrum of \( {A}_{0} \), by Theorem 7. Hence \( {A}_{0} - {\lambda I} \) is invertible. Select \( m \) such that for \( n \geq m \)\n\n\[ \begin{Vmatrix}{\left( {{A}_{n} - {\lambda I}}\right) - \left( {{A}_{0} - {\lambda I}}\right) }\e... | Yes |
Theorem 1. A point \( y \) is in the closure of a set \( S \) in a topological space if and only if some net in \( S \) converges to \( y \) . | Proof. If the net \( \left\lbrack {x}_{\alpha }\right\rbrack \) is in \( S \) and converges to \( y \), then to each neighborhood \( U \) of \( y \) there corresponds an index \( \beta \) such that \( {x}_{\alpha } \in U \) whenever \( \beta \prec \alpha \) . In particular, \( {x}_{\beta } \in U \) . Thus each neighbor... | Yes |
Lemma 1 In a linear topological space, a set \( V \) is a neighborhood of a point \( z \) if and only if \( - z + V \) is a neighborhood of 0 . | Proof. Hold \( z \) fixed, and define \( f\left( x\right) = x + z \) . This mapping sends 0 to \( z \) . Let \( V \) be a neighborhood of \( z \) . Since \( f \) is continuous, \( {f}^{-1}\left( V\right) \) is a neighborhood of 0. Observe, now, that \( {f}^{-1}\left( V\right) = \{ x : f\left( x\right) \in V\} = \{ x : ... | Yes |
Theorem 1. A linear topological space is a Hausdorff space if and only if 0 is the only element common to all neighborhoods of 0. | Proof. The Hausdorff property is that for any pair of points \( x \neq y \) there must exist neighborhoods \( U \) and \( V \) of \( x \) and \( y \) respectively such that the pair \( U, V \) is disjoint. Select a neighborhood \( W \) of 0 such that \( x - y \notin W \) . Then (using the continuity of subtraction) sel... | Yes |
Theorem 2. Let \( X \) be a linear topological space, and \( U \) a neighborhood of 0 . Then the polar set\n\n\[ {U}^{ \circ } = \left\{ {\phi \in {X}^{ * } : \left| {\phi \left( x\right) }\right| \leq 1\text{ for all }x \in U}\right\} \]\n\nis compact in the weak* topology of \( {X}^{ * } \) . | Proof. The linear space \( {X}^{ * } \) (whose elements are continuous linear functionals) is a subspace of \( {X}^{\prime } \) (whose elements are linear functionals). The weak* topology in \( {X}^{ * } \) is the relative topology in \( {X}^{ * } \) derived from the weak* topology on \( {X}^{\prime } \) . By the prece... | Yes |
Theorem 3. (The Banach-Alaoglu Theorem) The unit ball in the conjugate space of a normed linear space is compact in the weak* topology. | Proof. In the preceding theorem, take \( U \) to be the unit ball of \( X \) . The polar of \( U \) will then be the unit ball in \( {X}^{ * } \) . | No |
Theorem 4. For any locally convex linear topological space there is a family of continuous seminorms that induces the topology. | Proof. Let \( P \) be the family of all continuous seminorms defined on the given space. Let \( U \) be a neighborhood of 0 in the original topology. First we must prove that \( U \) contains one of the sets \( V\left( {\varepsilon ;{p}_{1},\ldots ,{p}_{n}}\right) \) . Since the space is locally convex, \( U \) contain... | Yes |
Lemma 2. In a locally convex linear topological space, the convex hull of a totally bounded set is totally bounded. | Proof. Let \( Y \) be such a set and let \( U \) be any neighborhood of 0 . Select a convex neighborhood \( V \) of 0 such that \( V + V \subset U \) . Since \( Y \) is totally bounded, there is a finite set \( F \) such that \( Y \subset F + V \) . Let \( Z = \operatorname{co}\left( F\right) \) . The set \( Z \) is co... | Yes |
Theorem 7. Mazur's Theorem. The closed convex hull of a totally bounded set in a complete locally convex linear topological space is compact. | Proof. Let \( K \) be such a set in such a space. By the preceding lemma, \( \operatorname{co}\left( K\right) \) is totally bounded. Hence \( \overline{\mathrm{{co}}}\left( K\right) \) is closed and totally bounded. Since the ambient space is complete, \( \overline{\mathrm{{co}}}\left( K\right) \) is complete and total... | Yes |
Theorem 3. Let \( f : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{R} \) . Assume that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {n, m}\right) \) exists for each \( m \) and that \( \mathop{\lim }\limits_{{m \rightarrow \infty }}f\left( {n, m}\right) \) exists for each \( n \), uniformly in \( n \) ... | Proof. Define \( g\left( m\right) = \mathop{\lim }\limits_{n}f\left( {n, m}\right) \) and \( h\left( n\right) = \mathop{\lim }\limits_{m}f\left( {n, m}\right) \) . Let \( \varepsilon > 0 \) . Find a positive integer \( M \) such that\n\n\[ m \geq M \Rightarrow \left| {f\left( {n, m}\right) - h\left( n\right) }\right| <... | Yes |
Theorem 4. The Kharshiladze-Lozinski Theorem. For each \( n = 0,1,2,\ldots \) let \( {P}_{n} \) be a projection of the space \( C\left\lbrack {-1,1}\right\rbrack \) onto the subspace \( {\Pi }_{n} \) of polynomials of degree at most \( n \) . Then \( \begin{Vmatrix}{P}_{n}\end{Vmatrix} \rightarrow \infty \) . | It is readily seen that the equation\n\n\[ \n{P}_{n}\left( f\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}\left( f\right) {p}_{n} \n\]\n\nwhere the coefficients \( {a}_{k} \) are as above, defines a projection of the type appearing in Theorem 4. That is, \( {P}_{n} \) is a continuous linear idempotent map from \(... | Yes |
Theorem 7. Under the hypotheses given above, Equation (3) is true for the point \( x = {x}_{0} \) . | Proof. By Hypothesis (A) we are allowed to define\n\n\[ f\left( x\right) = {\int }_{T}g\left( {x, t}\right) {d\mu }\left( t\right) \]\n\nThe derivative \( {f}^{\prime }\left( {x}_{0}\right) \) exists if and only if for each sequence \( \left\lbrack {x}_{n}\right\rbrack \) converging to \( {x}_{0} \) we have\n\n\[ {f}^{... | Yes |
Theorem 8. Let \( \left( {T,\mathcal{A},\mu }\right) \) be a measure space such that \( \mu \left( T\right) < \infty \). Let \( g : \left( {a, b}\right) \times T \rightarrow \mathbb{R} \). Assume that for each \( n,\left( {{\partial }^{n}g/\partial {x}^{n}}\right) \left( {x, t}\right) \) exists, is measurable, and is b... | Proof. Since \( \mu \left( T\right) < \infty \), any bounded measurable function on \( T \) is integrable. To see that Hypothesis (B) of the preceding theorem is true, use the mean value theorem: \[ \left| \frac{g\left( {x, t}\right) - g\left( {{x}_{0}, t}\right) }{x - {x}_{0}}\right| = \left| {\frac{\partial g}{\parti... | Yes |
Let \( X \) be an arbitrary set, and \( \mathcal{C} \) a collection of subsets of \( X \), countably many of which cover \( X \) . Let \( \beta \) be a function from \( \mathcal{C} \) to \( {\mathbb{R}}^{ * } \) such that\n\n\[ \inf \{ \beta \left( C\right) : C \in \mathcal{C}\} = 0 \]\n\nThen the equation\n\n\[ \mu \l... | Assume all the hypotheses. There are now three postulates for an outer measure to be verified. Our assumption about \( \beta \) implies that \( \beta \left( C\right) \geq 0 \) for all \( C \in \mathcal{C} \) . Therefore, \( \mu \left( A\right) \geq 0 \) for all \( A \) . Since \( \varnothing \subset C \) for all \( C \... | Yes |
Theorem 1. Let \( X \) be an arbitrary set, and \( \mathcal{C} \) a collection of subsets of \( X \), countably many of which cover \( X \) . Let \( \beta \) be a function from \( \mathcal{C} \) to \( {\mathbb{R}}^{ * } \) such that\n\n\[ \inf \{ \beta \left( C\right) : C \in \mathcal{C}\} = 0 \]\n\nThen the equation\n... | Proof. Assume all the hypotheses. There are now three postulates for an outer measure to be verified. Our assumption about \( \beta \) implies that \( \beta \left( C\right) \geq 0 \) for all \( C \in \mathcal{C} \) . Therefore, \( \mu \left( A\right) \geq 0 \) for all \( A \) . Since \( \varnothing \subset C \) for all... | Yes |
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