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Theorem 3. The Closed Graph Theorem. A closed linear map from one Banach space into another is continuous. | Proof. Let \( L : X \rightarrow Y \) be closed and linear. In \( X \), define a new norm \( N\left( x\right) = \) \( \parallel x\parallel + \parallel {Lx}\parallel \) . Then \( \left( {X, N}\right) \) is complete. Indeed, if \( \left\lbrack {x}_{n}\right\rbrack \) is a Cauchy sequence with the norm \( N \), then \( \le... | Yes |
Theorem 4. A normed linear space that is the image of a Banach space by a bounded, linear, interior map is also a Banach space. | Proof. Let \( L : X \rightarrow Y \) be the bounded, linear, interior map. Assume that \( X \) is a Banach space. By Problem 1.2.38 (page 14), it suffices to prove that each absolutely convergent series in \( Y \) is convergent. Let \( {y}_{n} \in Y \) and \( \sum \begin{Vmatrix}{y}_{n}\end{Vmatrix} < \infty \) . By Pr... | Yes |
Theorem 5. Let \( L \) be a continuous linear transformation from one normed linear space to another. The range of \( L \) is dense if and only if \( {L}^{ * } \) is injective. | Proof. Let \( L : X \rightarrow Y \) . By Theorem 3 in Section 1.6 (page 37), applied to \( L\left( X\right) \), we have these equivalent assertions: (1) \( L\left( X\right) \) is dense in \( Y \) . (2) \( L{\left( X\right) }^{ \bot } = 0 \) . (3) If \( \phi \in L{\left( X\right) }^{ \bot } \), then \( \phi = 0 \) . (4... | Yes |
Theorem 6. The Closed Range Theorem. Let \( L \) be a bounded linear transformation defined on a normed linear space and taking values in another normed linear space. The range of \( L \) and the null space of \( {L}^{ * } \), denoted by \( \mathcal{N}\left( {L}^{ * }\right) \), are related by the fact that \( {\left\l... | Proof. Recall the notation \( {U}_{ \bot } \) for the set \( \{ x \in X : \phi \left( x\right) = 0 \) for all \( \phi \in U\} \) , where \( X \) is a normed linear space and \( U \) is a subset of \( {X}^{ * } \) . (See Problems 1.6.20 and 1.6.21, on page 38, as well as Problem 13 in this section, page 52.) We denote b... | No |
Theorem 7 Let \( L \) be a continuous, linear, injective map from one Banach space into another. The range of \( L \) is closed if and only if \( L \) is bounded below: \( \mathop{\inf }\limits_{{\parallel x\parallel = 1}}\parallel {Lx}\parallel > 0 \) . | Proof. Assume first that \( \parallel {Lx}\parallel \geq c > 0 \) when \( \parallel x\parallel = 1 \) . By homogeneity, \( \parallel {Lx}\parallel \geq c\parallel x\parallel \) for all \( x \) . To prove that the range, \( \mathcal{R}\left( L\right) \), is closed, let \( {y}_{n} \in \mathcal{R}\left( L\right) \) and \(... | Yes |
Theorem 1. In a finite-dimensional normed linear space, weak and strong convergence coincide. | Proof. Let \( X \) be a \( k \) -dimensional space. Select a base \( \left\{ {{b}_{1},\ldots ,{b}_{k}}\right\} \) for \( X \) and let \( {\phi }_{1},\ldots ,{\phi }_{k} \) be the linear functionals such that for each \( x \) ,\n\n\[ x = \mathop{\sum }\limits_{{i = 1}}^{k}{\phi }_{i}\left( x\right) {b}_{i} \]\n\nBy Coro... | Yes |
Theorem 2. If a sequence \( \left\lbrack {x}_{n}\right\rbrack \) in a normed linear space converges weakly to an element \( x \), then a sequence of linear combinations of the elements \( {x}_{n} \) converges strongly to \( x \) . | Proof. Another way of stating the conclusion is that \( x \) belongs to the closed subspace\n\n\[ Y = \operatorname{closure}\left( {\operatorname{span}\left\{ {{x}_{1},{x}_{2},\ldots }\right\} }\right) \]\n\nIf \( x \notin Y \), then by Corollary 2 of the Hahn-Banach Theorem (page 34), there is a continuous linear func... | Yes |
Theorem 3. If the sequence \( \left\lbrack {{x}_{0},{x}_{1},{x}_{2},\ldots }\right\rbrack \) is bounded in a normed linear space \( X \) and if \( \phi \left( {x}_{n}\right) \rightarrow \phi \left( {x}_{0}\right) \) for all \( \phi \) in a fundamental subset of \( {X}^{ * } \), then \( {x}_{n} \rightharpoonup x \) . | Proof. (The term \ | No |
Minkowski Inequality. If \( x \) and \( y \) are two members of \( {\ell }_{p} \), then\n\n\[ \parallel x + y{\parallel }_{p} \leq \parallel x{\parallel }_{p} + \parallel y{\parallel }_{p} \] | Proof. For \( p = 1 \) an elementary proof goes as follows:\n\n\[ \parallel x + y{\parallel }_{1} = \sum \left| {x\left( n\right) + y\left( n\right) }\right| \leq \sum \left| {x\left( n\right) }\right| + \sum \left| {y\left( n\right) }\right| = \parallel x{\parallel }_{1} + \parallel y{\parallel }_{1} \]\n\nNow assume ... | Yes |
Theorem 8. A subspace of a normed linear space is closed if and only if it is weakly sequentially closed. | Proof. Let \( Y \) be a weakly sequentially closed subspace in the normed space \( X \) . If \( {y}_{n} \in Y \) and \( {y}_{n} \rightarrow y \), then \( {y}_{n} \rightharpoonup y \) and \( y \in Y \) . Hence \( Y \) is norm-closed.\n\nFor the converse, suppose that \( Y \) is norm-closed, and let \( {y}_{n} \in Y,{y}_... | Yes |
Theorem 9. A linear continuous mapping between normed spaces is weakly sequentially continuous. | Proof. Let \( A : X \rightarrow Y \) be linear and norm-continuous. In order to prove that \( A \) is weakly continuous, let \( {x}_{n} \rightharpoonup x \) . For all \( \phi \in {Y}^{ * },\phi \circ A \in {X}^{ * } \) . Hence \( \phi \left( {A{x}_{n} - {Ax}}\right) \rightarrow 0 \) for all \( \phi \in {Y}^{ * } \) . | Yes |
Theorem 10. Let \( X \) be a separable normed linear space, and \( \left\lbrack {\phi }_{n}\right\rbrack \) a bounded sequence in \( {X}^{ * } \) . Then there is a subsequence \( \left\lbrack {\phi }_{{n}_{i}}\right\rbrack \) that converges in the weak* sense to an element of \( {X}^{ * } \) . | Proof. Since \( X \) is separable, it contains a countable dense set, \( \left\{ {{x}_{1},{x}_{2},\ldots }\right\} \) . Since \( \left\lbrack {\phi }_{n}\right\rbrack \) is bounded, so is the sequence \( \left\lbrack {{\phi }_{n}\left( {x}_{1}\right) }\right\rbrack \) . We can therefore find an increasing sequence \( {... | Yes |
Theorem 1. Each space \( {\ell }_{p} \), where \( 1 < p < \infty \), is reflexive. | Proof. If \( {p}^{-1} + {q}^{-1} = 1 \), then \( {\ell }_{p}^{ * } = {\ell }_{q} \) and \( {\ell }_{q}^{ * } = {\ell }_{p} \) by Theorem 4 of Section 1.9, page 56. Hence \( {\ell }_{p}^{* * } = {\ell }_{p} \) . But we must be sure that the isometry involved in this statement is the natural one, \( J \) . Let \( A : {\e... | Yes |
Theorem 2. A closed linear subspace in a reflexive Banach space is reflexive. | Proof. Let \( Y \) be a closed subspace in a reflexive Banach space \( X \) . Let \( J : X \rightarrow \) \( {X}^{* * } \) be the natural map. Define \( R : {X}^{ * } \rightarrow {Y}^{ * } \) by the equation \( {R\phi } = \phi \mid Y \) . (This is the restriction map.) Let \( f \in {Y}^{* * } \) . Define \( y = {J}^{-1... | Yes |
Theorem 3. A Banach space is reflexive if and only if its conjugate space is reflexive. | Proof. Let \( X \) be reflexive. Then the natural embedding \( J : X \rightarrow {X}^{* * } \) is surjective. Let \( \Phi \in {X}^{* * * } \), and define \( \phi \in {X}^{ * } \) by the equation \( \phi = \Phi \circ J \) . Then for arbitrary \( f \in {X}^{* * } \) we have \( f = {Jx} \) for some \( x \), and consequent... | Yes |
Theorem 1. The norm has these properties\na. \( \parallel x\parallel > 0 \) if \( x \neq 0 \)\nb. \( \parallel {\alpha x}\parallel = \left| \alpha \right| \parallel x\parallel \;\left( {\alpha \in \mathbb{C}}\right) \)\nc. \( \left| {\langle x, y\rangle }\right| \leq \parallel x\parallel \parallel y\parallel \; \) Cauc... | Proof. Only \( \mathbf{c} \) and \( \mathbf{d} \) offer any difficulty. For \( \mathbf{c} \), let \( \parallel y\parallel = 1 \) and write\n\n\[ 0 \leq \langle x - {\lambda y}, x - {\lambda y}\rangle = \langle x, x\rangle - \bar{\lambda }\langle x, y\rangle - \lambda \langle y, x\rangle + {\left| \lambda \right| }^{2}\... | Yes |
Theorem 1. The norm has these properties\na. \( \parallel x\parallel > 0 \) if \( x \neq 0 \)\nb. \( \parallel {\alpha x}\parallel = \left| \alpha \right| \parallel x\parallel \;\left( {\alpha \in \mathbb{C}}\right) \)\nc. \( \left| {\langle x, y\rangle }\right| \leq \parallel x\parallel \parallel y\parallel \; \) Cauc... | Proof. Only \( \mathbf{c} \) and \( \mathbf{d} \) offer any difficulty. For \( \mathbf{c} \), let \( \parallel y\parallel = 1 \) and write\n\n\[ 0 \leq \langle x - {\lambda y}, x - {\lambda y}\rangle = \langle x, x\rangle - \bar{\lambda }\langle x, y\rangle - \lambda \langle y, x\rangle + {\left| \lambda \right| }^{2}\... | Yes |
Theorem 1. The norm has these properties\na. \( \parallel x\parallel > 0 \) if \( x \neq 0 \)\nb. \( \parallel {\alpha x}\parallel = \left| \alpha \right| \parallel x\parallel \;\left( {\alpha \in \mathbb{C}}\right) \)\nc. \( \left| {\langle x, y\rangle }\right| \leq \parallel x\parallel \parallel y\parallel \; \) Cauc... | Proof. Only \( \mathbf{c} \) and \( \mathbf{d} \) offer any difficulty. For \( \mathbf{c} \), let \( \parallel y\parallel = 1 \) and write\n\n\[ 0 \leq \langle x - {\lambda y}, x - {\lambda y}\rangle = \langle x, x\rangle - \bar{\lambda }\langle x, y\rangle - \lambda \langle y, x\rangle + {\left| \lambda \right| }^{2}\... | Yes |
We write \( {L}^{2}\left\lbrack {a, b}\right\rbrack \) for the set of all complex-valued Lebesgue measurable functions on \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ \n{\int }_{a}^{b}{\left| x\left( t\right) \right| }^{2}{dt} < \infty \n\]\n\n(The concept of measurability is explained in Chapter 8, Section 4... | See Chapter 8, Section 7, page 411 for the proof. | No |
Example 5. Let \( \\left( {S,\\mathcal{A},\\mu }\\right) \) be any measure space. The notation \( {L}^{2}\\left( S\\right) \) then denotes the space of measurable complex functions on \( S \) such that \( \\int {\\left| f\\left( s\\right) \\right| }^{2}{d\\mu } < \\) \( \\infty \) . In \( {L}^{2}\\left( S\\right) \\), ... | See Theorem 3 in Section 8.7, page 411. | No |
Theorem 2. If \( K \) is a closed, convex, nonvoid set in a Hilbert space \( X \), then to each \( x \) in \( X \) there corresponds a unique point \( y \) in \( K \) closest to \( x \) ; that is,\n\n\[ \parallel x - y\parallel = \operatorname{dist}\left( {x, K}\right) \mathrel{\text{:=}} \inf \{ \parallel x - v\parall... | Proof. Put \( \alpha = \operatorname{dist}\left( {x, K}\right) \), and select \( {y}_{n} \in K \) so that \( \begin{Vmatrix}{x - {y}_{n}}\end{Vmatrix} \rightarrow \alpha \) . Notice that \( \frac{1}{2}\left( {{y}_{n} + {y}_{m}}\right) \in K \) by the convexity of \( K \) . Hence \( \begin{Vmatrix}{\frac{1}{2}\left( {{y... | Yes |
Theorem 2. If \( K \) is a closed, convex, nonvoid set in a Hilbert space \( X \), then to each \( x \) in \( X \) there corresponds a unique point \( y \) in \( K \) closest to \( x \) ; that is,\n\n\[ \parallel x - y\parallel = \operatorname{dist}\left( {x, K}\right) \mathrel{\text{:=}} \inf \{ \parallel x - v\parall... | Proof. Put \( \alpha = \operatorname{dist}\left( {x, K}\right) \), and select \( {y}_{n} \in K \) so that \( \begin{Vmatrix}{x - {y}_{n}}\end{Vmatrix} \rightarrow \alpha \) . Notice that \( \frac{1}{2}\left( {{y}_{n} + {y}_{m}}\right) \in K \) by the convexity of \( K \) . Hence \( \begin{Vmatrix}{\frac{1}{2}\left( {{y... | Yes |
Theorem 3. Let \( Y \) be a subspace in an inner-product space \( X \). Let\n\n\( x \in X \) and \( y \in Y \). These are equivalent assertions:\n\na. \( x - y \bot Y \), i.e., \( \langle x - y, v\rangle = 0 \) for all \( v \in Y \).\n\nb. \( y \) is the unique point of \( Y \) closest to \( x \). | Proof. If a is true, then for any \( u \in Y \) we have\n\n\[{\begin{Vmatrix}x - u\end{Vmatrix}}^{2} = {\begin{Vmatrix}\left( x - y\right) + \left( y - u\right) \end{Vmatrix}}^{2} = {\begin{Vmatrix}x - y\end{Vmatrix}}^{2} + {\begin{Vmatrix}y - u\end{Vmatrix}}^{2} \geq {\begin{Vmatrix}x - y\end{Vmatrix}}^{2}\]\n\nHere w... | Yes |
Theorem 4. If \( Y \) is a closed subspace of a Hilbert space \( X \), then\n\n\( X = Y \oplus {Y}^{ \bot } \) . | Proof. We have to prove that \( {Y}^{ \bot } \) is a subspace, that \( Y \cap {Y}^{ \bot } = 0 \), and that \( X \subset Y + {Y}^{ \bot } \) . If \( {v}_{1} \) and \( {v}_{2} \) belong to \( {Y}^{ \bot } \), then so does \( {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} \), since for \( y \in Y \)\n\n\[ \langle y,{\alpha ... | Yes |
Theorem 5. If the Parallelogram Law is valid in a normed linear space, then that space is an inner-product space. In other words, an inner product can be defined in such a way that \( \langle x, x\rangle = \parallel x{\parallel }^{2} \) . | Proof. We define the inner product by the equation\n\n\[ 4\langle x, y\rangle = \parallel x + y{\parallel }^{2} - \parallel x - y{\parallel }^{2} + i\parallel x + {iy}{\parallel }^{2} - i\parallel x - {iy}{\parallel }^{2} \]\n\nFrom the definition, it follows that\n\n\[ 4\mathcal{R}\langle x, y\rangle = \parallel x + y... | Yes |
Theorem 1. Pythagorean Law. If \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \) is a finite orthogonal set of \( n \) distinct elements in an inner-product space, then\n\n\[ \n{\begin{Vmatrix}\mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{j = 1}}^{n}{\begin{Vmatrix}{x}_{j}... | Proof. By our assumptions, \( {x}_{i} \neq {x}_{j} \) if \( i \neq j \), and consequently,\n\n\[ \n{\begin{Vmatrix}\mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}\end{Vmatrix}}^{2} = \left\langle {\mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j},\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}}\right\rangle = \mathop{\sum }\limits_{{j ... | Yes |
Theorem 2. The General Pythagorean Law. Let \( \left\lbrack {x}_{j}\right\rbrack \) be an orthogonal sequence in a Hilbert space. The series \( \sum {x}_{j} \) converges if and only if \( \sum {\begin{Vmatrix}{x}_{j}\end{Vmatrix}}^{2} < \infty \) . If \( \sum {\begin{Vmatrix}{x}_{j}\end{Vmatrix}}^{2} = \lambda < \infty... | Proof. Put \( {S}_{n} = \mathop{\sum }\limits_{1}^{n}{x}_{j} \) and \( {s}_{n} = \mathop{\sum }\limits_{1}^{n}{\begin{Vmatrix}{x}_{j}\end{Vmatrix}}^{2} \) . \n\nBy the finite version of the Pythagorean Law, we have (for \( m > n \) ) \n\n\[ \n{\begin{Vmatrix}{S}_{m} - {S}_{n}\end{Vmatrix}}^{2} = {\begin{Vmatrix}\mathop... | Yes |
Theorem 3. If \( \left\lbrack {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right\rbrack \) is an orthonormal set in an inner-product space, and if \( Y \) is the linear span of \( \left\{ {{y}_{i} : 1 \leq i \leq n}\right\} \), then for any \( x \), the point in \( Y \) closest to \( x \) is \( \mathop{\sum }\limits_{{i = 1}}^{n}... | Proof. Let \( y = \mathop{\sum }\limits_{{i = 1}}^{n}\left\langle {x,{y}_{i}}\right\rangle {y}_{i} \) . By Theorem 3 in Section 2.1, page 65, it suffices to verify that \( x - y \bot Y \) . For this it is enough to verify that \( x - y \) is orthogonal to each basis vector \( {y}_{k} \) . We have\n\n\[ \langle x - y,{y... | Yes |
Theorem 4. Bessel’s Inequality. If \( \left\lbrack {{u}_{i} : i \in I}\right\rbrack \) is an orthonormal system in an inner-product space, then for every \( x \) ,\n\n\[ \sum {\left| \left\langle x,{u}_{i}\right\rangle \right| }^{2} \leq \parallel x{\parallel }^{2} \] | Proof. For \( j \) ranging over a finite subset \( J \) of \( I \), let \( y = \sum \left\langle {x,{u}_{j}}\right\rangle {u}_{j} \) . This vector \( y \) is the orthogonal projection of \( x \) onto the subspace \( U = \operatorname{span}\left\lbrack {{u}_{j} : j \in J}\right\rbrack \) . By Theorem \( 3, x - y \bot U ... | Yes |
Corollary 3. If \( \left\lbrack {{u}_{i} : i \in I}\right\rbrack \) is an orthonormal system, then for each \( x \) at most a countable number of the Fourier coefficients \( \left\langle {x,{u}_{i}}\right\rangle \) are nonzero. | Proof. Fixing \( x \), put \( {J}_{n} = \left\{ {i \in I : \left| \left\langle {x,{u}_{i}}\right\rangle \right| > 1/n}\right\} \) . By the Bessel Inequality,\n\n\[ \parallel x{\parallel }^{2} \geq \mathop{\sum }\limits_{{j \in {J}_{n}}}{\left| \left\langle x,{u}_{j}\right\rangle \right| }^{2} \geq \mathop{\sum }\limits... | Yes |
Theorem 5. Every nontrivial inner-product space has an orthonormal basis. | Proof. Call the space \( X \) . Since it is not 0, it contains a nonzero vector \( x \) . The set consisting solely of \( x/\parallel x\parallel \) is orthonormal. Now order the family of all orthonormal subsets of \( X \) in the natural way (by inclusion). In order to use Zorn's Lemma, one must verify that each chain ... | Yes |
Theorem 6. The Orthonormal Basis Theorem. For an orthonormal family \( \left\lbrack {u}_{i}\right\rbrack \) (not necessarily finite or countable) in a Hilbert space \( X \), the following properties are equivalent:\n\na. \( \left\lbrack {u}_{i}\right\rbrack \) is an orthonormal basis for \( X \).\n\nb. If \( x \in X \)... | Proof. To prove that a implies \( \mathbf{b} \), suppose that \( \mathbf{b} \) is false. Let \( x \neq 0 \) and \( x \bot {u}_{i} \) for all \( i \). Adjoin \( x/\parallel x\parallel \) to the family \( \left\lbrack {u}_{i}\right\rbrack \) to get a larger orthonormal family. Thus the original family is not maximal and ... | Yes |
One orthonormal basis in \( {\ell }^{2} \) is obtained by defining \( {u}_{n}\left( j\right) = {\delta }_{nj} \) . Thus\n\n\[ \n{u}_{1} = \left\lbrack {1,0,0,\ldots }\right\rbrack ,\;{u}_{2} = \left\lbrack {0,1,0,\ldots }\right\rbrack ,\text{ etc. } \n\] | To see that this is actually an orthonormal base, use the preceding theorem, in particular the equivalence of \( \mathbf{a} \) and \( \mathbf{b} \) . Suppose \( x \in {\ell }^{2} \) and \( \left\langle {x,{u}_{n}}\right\rangle = 0 \) for all \( n \) . Then \( x\left( n\right) = 0 \) for all \( n \), and \( x = 0 \) . | No |
An orthonormal basis for \( {L}^{2}\left\lbrack {0,1}\right\rbrack \) is provided by the functions \( {u}_{n}\left( t\right) = {e}^{2\pi int} \), where \( n \in \mathbb{Z} \). One verifies the orthonormality by computing the appropriate integrals. To show that \( \left\lbrack {u}_{n}\right\rbrack \) is a base, we use P... | \[ \left| {\langle x, p\rangle }\right| \geq \left| {\langle p, p\rangle }\right| - \left| {\langle y - p, p\rangle }\right| - \left| {\langle x - y, p\rangle }\right| \] \[ \geq \parallel p{\parallel }^{2} - \parallel y - p\parallel \parallel p\parallel - \parallel x - y\parallel \parallel p\parallel > 0 \] Thus it is... | Yes |
Theorem 7. The Orthogonal Projection Theorem. The orthogonal projection \( P \) of a Hilbert space \( X \) onto a closed subspace \( Y \) has these properties:\na. It is well-defined; i.e., \( {Px} \) exists and is unique in \( Y \).\nb. It is surjective, i.e., \( P\left( X\right) = Y \).\nc. It is linear.\nd. If \( Y ... | Proof. This is left to the problems. | No |
Theorem 8. The Gram-Schmidt Construction.\n\n\( \\left\\lbrack {{v}_{1},{v}_{2},{v}_{3},\\ldots }\\right\\rbrack \) be a linearly independent sequence in an inner product\n\nspace. Having set \( {u}_{1} = {v}_{1}/\\begin{Vmatrix}{v}_{1}\\end{Vmatrix} \), define recursively\n\n\[ \n{u}_{n} = \\frac{{v}_{n} - \\mathop{\\... | Notice that in the equation describing this algorithm there is a normalization process: the dividing of a vector by its norm to produce a new vector pointing in the same direction but having unit length. The other action being carried out is the subtraction from the vector \( {v}_{n} \) of its projection on the linear ... | Yes |
A nonseparable inner-product space cannot have a countable orthonormal base. | For an example, we consider the uncountable family of functions \( {u}_{\lambda }\left( t\right) = {e}^{i\lambda t} \), where \( t \in \mathbb{R} \) and \( \lambda \in \mathbb{R} \) . This family of functions is linearly independent (Problem 5), and is therefore a Hamel basis for a linear space \( X \) . We introduce a... | No |
An important example of an orthonormal basis is provided by the Legendre polynomials. We consider the space \( C\left\lbrack {-1,1}\right\rbrack \) and use the simple inner product\n\n\[ \langle f, g\rangle = {\int }_{-1}^{1}f\left( t\right) g\left( t\right) {dt} \]\n\nNow apply the Gram-Schmidt process to the monomial... | The orthonormal system is, of course, \( {p}_{n} = {P}_{n}/\begin{Vmatrix}{P}_{n}\end{Vmatrix} \). The completion of the space \( C\left\lbrack {-1,1}\right\rbrack \) with respect to the norm induced by the inner product is the space \( {L}^{2}\left\lbrack {-1,1}\right\rbrack \). Every function \( f \) in this space is... | Yes |
Theorem 2. Existence of Adjoints. If \( A \) is a bounded linear operator on a Hilbert space \( X \) (thus \( A : X \rightarrow X \) ), then there is a uniquely defined bounded linear operator \( {A}^{ * } \) such that\n\n\[ \langle {Ax}, y\rangle = \left\langle {x,{A}^{ * }y}\right\rangle \;\left( {x, y \in X}\right) ... | Proof. For each fixed \( y \), the mapping \( x \mapsto \langle {Ax}, y\rangle \) is a bounded linear functional on \( X \) :\n\n\[ \langle A\left( {{\lambda x} + {\mu z}}\right), y\rangle = \langle {\lambda Ax} + {\mu Az}, y\rangle = \lambda \langle {Ax}, y\rangle + \mu \langle {Ax}, y\rangle \]\n\n\[ \left| {\langle ... | No |
Theorem 2. Existence of Adjoints. If \( A \) is a bounded linear operator on a Hilbert space \( X \) (thus \( A : X \rightarrow X \) ), then there is a uniquely defined bounded linear operator \( {A}^{ * } \) such that\n\n\[ \langle {Ax}, y\rangle = \left\langle {x,{A}^{ * }y}\right\rangle \;\left( {x, y \in X}\right) ... | Proof. For each fixed \( y \), the mapping \( x \mapsto \langle {Ax}, y\rangle \) is a bounded linear functional on \( X \) :\n\n\[ \langle A\left( {{\lambda x} + {\mu z}}\right), y\rangle = \langle {\lambda Ax} + {\mu Az}, y\rangle = \lambda \langle {Ax}, y\rangle + \mu \langle {Ax}, y\rangle \]\n\n\[ \left| {\langle ... | No |
Theorem 2. Existence of Adjoints. If \( A \) is a bounded linear operator on a Hilbert space \( X \) (thus \( A : X \rightarrow X \) ), then there is a uniquely defined bounded linear operator \( {A}^{ * } \) such that\n\n\[ \langle {Ax}, y\rangle = \left\langle {x,{A}^{ * }y}\right\rangle \;\left( {x, y \in X}\right) ... | Proof. For each fixed \( y \), the mapping \( x \mapsto \langle {Ax}, y\rangle \) is a bounded linear functional on \( X \) :\n\n\[ \langle A\left( {{\lambda x} + {\mu z}}\right), y\rangle = \langle {\lambda Ax} + {\mu Az}, y\rangle = \lambda \langle {Ax}, y\rangle + \mu \langle {Ax}, y\rangle \]\n\n\[ \left| {\langle ... | No |
If \( A \) is a bounded linear operator on a Hilbert space \( X \) (thus \( A : X \rightarrow X \) ), then there is a uniquely defined bounded linear operator \( {A}^{ * } \) such that\n\n\[ \langle {Ax}, y\rangle = \left\langle {x,{A}^{ * }y}\right\rangle \;\left( {x, y \in X}\right) \]\n\nFurthermore, \( \begin{Vmatr... | Proof. For each fixed \( y \), the mapping \( x \mapsto \langle {Ax}, y\rangle \) is a bounded linear functional on \( X \) :\n\n\[ \langle A\left( {{\lambda x} + {\mu z}}\right), y\rangle = \langle {\lambda Ax} + {\mu Az}, y\rangle = \lambda \langle {Ax}, y\rangle + \mu \langle {Ax}, y\rangle \]\n\n\[ \left| {\langle ... | No |
Theorem 3. If a linear map \( A \) on a Hilbert space satisfies \( \langle {Ax}, y\rangle = \langle x,{Ay}\rangle \) for all \( x \) and \( y \), then \( A \) is bounded and self-adjoint. | Proof. For each \( y \) in the unit ball, define a functional \( {\phi }_{y} \) by writing \( {\phi }_{y}\left( x\right) = \langle {Ax}, y\rangle \). It is obvious that \( {\phi }_{y} \) is linear, and we see also that it is bounded, since by the Cauchy-Schwarz inequality\n\n\[ \left| {{\phi }_{y}\left( x\right) }\righ... | Yes |
Theorem 3. If a linear map \( A \) on a Hilbert space satisfies \( \langle {Ax}, y\rangle = \langle x,{Ay}\rangle \) for all \( x \) and \( y \), then \( A \) is bounded and self-adjoint. | Proof. For each \( y \) in the unit ball, define a functional \( {\phi }_{y} \) by writing \( {\phi }_{y}\left( x\right) = \langle {Ax}, y\rangle \). It is obvious that \( {\phi }_{y} \) is linear, and we see also that it is bounded, since by the Cauchy-Schwarz inequality\n\n\[ \left| {{\phi }_{y}\left( x\right) }\righ... | Yes |
Lemma 1. Generalized Cauchy-Schwarz Inequality. If \( A \) is a Hermitian operator, then \[ \left| {\langle {Ax}, y\rangle }\right| \leq \parallel A\parallel \parallel x\parallel \parallel y\parallel \] | Proof. Consider these two elementary equations: \[ \langle A\left( {x + y}\right), x + y\rangle = \langle {Ax}, x\rangle + \langle {Ax}, y\rangle + \langle {Ay}, x\rangle + \langle {Ay}, y\rangle \] \[ - \langle A\left( {x - y}\right), x - y\rangle = - \langle {Ax}, x\rangle + \langle {Ax}, y\rangle + \langle {Ay}, x\r... | Yes |
Lemma 2. If \( A \) is Hermitian, then \( \parallel A\parallel = \parallel A\parallel \) . | Proof. By the Cauchy-Schwarz inequality,\n\n\[ \parallel A\parallel = \mathop{\sup }\limits_{{\parallel u\parallel = 1}}\left| {\langle {Au}, u\rangle }\right| \leq \mathop{\sup }\limits_{{\parallel u\parallel = 1}}\parallel {Au}\parallel \parallel u\parallel = \mathop{\sup }\limits_{{\parallel u\parallel = 1}}\paralle... | Yes |
Lemma 3. Every continuous linear operator (from one normed linear space into another) having finite-dimensional range is compact. | Proof. Let \( A \) be such an operator, and let \( \sum \) be the unit ball. Since \( A \) is continuous, \( A\left( \sum \right) \) is a bounded set in a finite-dimensional subspace, and its closure is compact, by Theorem 1 in Section 1.4, page 20. | Yes |
Theorem 4. If \( X \) and \( Y \) are Banach spaces, then the set of compact operators in \( \mathcal{L}\left( {X, Y}\right) \) is closed. | Proof. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a sequence of compact operators from \( X \) to \( Y \) . Suppose that \( \begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} \rightarrow 0 \) . To prove that \( A \) is compact, let \( \left\lbrack {x}_{i}\right\rbrack \) be a sequence in the unit ball of \( X \) . We wish t... | Yes |
Theorem 5. Let \( S \) be any measure space. In the space \( {L}^{2}\left( S\right) \) , consider the integral operator \( T \) defined by the equation\n\n\[ \n\left( {Tx}\right) \left( s\right) = {\int }_{S}k\left( {s, t}\right) x\left( t\right) {dt} \n\]\n\nIf the kernel \( k \) belongs to the space \( {L}^{2}\left( ... | Proof. Select an orthonormal basis \( \left\lbrack {u}_{n}\right\rbrack \) for \( {L}^{2}\left( S\right) \), and define \( {a}_{nm} = \) \( \left\langle {T{u}_{m},{u}_{n}}\right\rangle \) . This is the \ | No |
If \( \left\lbrack {u}_{n}\right\rbrack \) is an orthonormal sequence, then \( {u}_{n} \rightharpoonup 0 \). | This follows from Bessel's inequality, which shows that \( \left\langle {{u}_{n}, y}\right\rangle \rightarrow 0 \) for all \( y \) . | Yes |
Lemma 4. A weakly Cauchy sequence in a Hilbert space is weakly convergent to a point in the Hilbert space. | Proof. Let \( \left\lbrack {x}_{n}\right\rbrack \) be such a sequence. For each \( y \), the sequence \( \left\lbrack \left\langle {y,{x}_{n}}\right\rangle \right\rbrack \) has the Cauchy property, and is therefore bounded in \( \mathbb{C} \) . The linear functionals \( {\phi }_{n} \) defined by \( {\phi }_{n}\left( y\... | Yes |
Theorem 7. Let \( A \) be a continuous linear operator on a Hilbert space. If the range of \( A \) is closed, then it is the orthogonal complement of the null space of \( {A}^{ * } \) ; in symbols,\n\n\[\n\mathcal{R}\left( A\right) = {\left\lbrack \mathcal{N}\left( {A}^{ * }\right) \right\rbrack }^{ \bot }\n\] | Proof. This is similar to Theorem 6, and is therefore left to the problems. (Half of the theorem does not require the closed range.) | No |
Lemma 1. If \( A \) is a Hermitian operator on an inner-product space, then:\n(1) All eigenvalues of \( A \) are real.\n(2) Any two eigenvectors of \( A \) belonging to different eigenvalues are orthogonal to each other.\n(3) The quadratic form \( x \mapsto \langle {Ax}, x\rangle \) is real-valued. | Proof. Let \( {Ax} = {\lambda x},{Ay} = {\mu y}, x \neq 0, y \neq 0,\lambda \neq \mu \) . Then\n\n\[ \lambda \langle x, x\rangle = \langle {\lambda x}, x\rangle = \langle {Ax}, x\rangle = \langle x,{Ax}\rangle = \langle x,{\lambda x}\rangle = \bar{\lambda }\langle x, x\rangle \]\n\nThus \( \lambda \) is real. To see th... | Yes |
Theorem 2. Let \( A \) be a compact operator (on an inner-product space) having spectral decomposition \( {Ax} = \sum {\lambda }_{n}\left\langle {x,{e}_{n}}\right\rangle {e}_{n} \) . (We allow \( {\lambda }_{n} \) to be complex.) If \( 0 \neq \lambda \notin \left\{ {\lambda }_{n}\right\} \), then \( A - {\lambda I} \) ... | Proof. If the series converges, then our formula is correct. Indeed, by the continuity of \( A - {\lambda I} \) we have by straightforward calculation\n\n\[left( {A - {\lambda I}}\right) {Bx} = B\left( {A - {\lambda I}}\right) x = x\]\n\nwhere \( {Bx} \) is defined by the right side of the equation in the statement of ... | Yes |
Theorem 3. Let \( A \) be an operator on an inner-product space having the form \( {Ax} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\lambda }_{n}\left\langle {x,{e}_{n}}\right\rangle {e}_{n} \), where \( \left\{ {e}_{n}\right\} \) is an orthonormal sequence and \( \left\lbrack {\lambda }_{n}\right\rbrack \) is a bounde... | Proof. The following are equivalent properties of a vector \( x \) :\n\n(a) \( x \in \ker \left( A\right) \)\n\n(b) \( \parallel {Ax}{\parallel }^{2} = 0 \)\n\n(c) \( \sum {\left| {\lambda }_{n}\left\langle x,{e}_{n}\right\rangle \right| }^{2} = 0 \)\n\n(d) \( \left\langle {x,{e}_{n}}\right\rangle = 0 \) for all \( n \... | Yes |
Theorem 4. Adopt the hypotheses of Theorem 3. The orthonormal set \( \left\{ {e}_{n}\right\} \) is maximal if and only if \( \ker \left( A\right) = 0 \) . | Proof. By Theorem 3, \( \ker \left( A\right) = 0 \) if and only if \( {M}^{ \bot } = 0 \) . (In these equations,0 denotes the 0 subspace.) The condition \( {M}^{ \bot } = 0 \) is equivalent to the maximality of \( \left\{ {e}_{n}\right\} \) . Here refer to Theorem 6 in Section 2.2, page 73, and observe that the equival... | Yes |
Theorem 5. Let \( A \) be an operator on a Hilbert space such that \( {Ax} = \) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{\lambda }_{n}\left\langle {x,{e}_{n}}\right\rangle {e}_{n} \), where \( \left\lbrack {e}_{n}\right\rbrack \) is an orthonormal sequence and \( \left\lbrack {\lambda }_{n}\right\rbrack \) is a bou... | Proof. Since \( v \) is in the range of \( A, v = {Az} \) for some \( z \) . Hence\n\n\[ \left\langle {v,{e}_{m}}\right\rangle = \left\langle {{Az},{e}_{m}}\right\rangle = \left\langle {\mathop{\sum }\limits_{{n = 1}}^{\infty }{\lambda }_{n}\left\langle {z,{e}_{n}}\right\rangle {e}_{n},{e}_{m}}\right\rangle = {\lambda ... | Yes |
Theorem 6. Singular-Value Decomposition for Compact Operators. Every compact operator on a separable Hilbert space is expressible in the form\n\n\[ \n{Ax} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\langle {x,{u}_{n}}\right\rangle {v}_{n} \]\n\nin which \( \left\lbrack {u}_{n}\right\rbrack \) is an orthonormal bas... | Proof. The operator \( {A}^{ * }A \) is compact and Hermitian. Its eigenvalues are nonnegative, because if \( {A}^{ * }{Ax} = {\beta x} \), then\n\n\[ \n0 \leq \langle {Ax},{Ax}\rangle = \left\langle {x,{A}^{ * }{Ax}}\right\rangle = \langle x,{\beta x}\rangle = \beta \langle x, x\rangle \]\n\nNow apply the spectral the... | Yes |
Theorem 7. Let \( \left\lbrack {u}_{\alpha }\right\rbrack \) and \( \left\lbrack {v}_{\beta }\right\rbrack \) be two orthonormal bases for a Hilbert space. Every linear operator \( A \) on the space satisfies\n\n\[ \mathop{\sum }\limits_{\alpha }{\begin{Vmatrix}A{u}_{\alpha }\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{... | Proof. By the Orthonormal Basis Theorem, Section 2.2 (page 73), we have\n\n\[ \mathop{\sum }\limits_{\alpha }{\begin{Vmatrix}A{u}_{\alpha }\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{\alpha }\mathop{\sum }\limits_{\beta }{\left| \left\langle A{u}_{\alpha },{v}_{\beta }\right\rangle \right| }^{2} = \mathop{\sum }\limits... | Yes |
Theorem 1. Under the preceding hypotheses, \( A \) is a Hermitian operator on \( X \) . | Proof. Let \( x, y \in X \) . We want to prove that \( \langle {Ax}, y\rangle = \langle x,{Ay}\rangle \) . We compute\n\n\[ \langle {Ax}, y\rangle - \langle x,{Ay}\rangle = {\int }_{a}^{b}\left\lbrack {\bar{y}{Ax} - {xA}\bar{y}}\right\rbrack = {\int }_{a}^{b}\left\lbrack {\bar{y}{\left( p{x}^{\prime }\right) }^{\prime ... | Yes |
Example 1. If \( {Ax} = - {x}^{\prime \prime } \) (i.e., \( p\left( t\right) = - 1 \) and \( q\left( t\right) = 0 \) ), what are the eigenvalues and eigenfunctions? | The solutions to \( - {x}^{\prime \prime } = {\lambda x} \) are of the form \( {c}_{1}\sin \sqrt{\lambda }t + {c}_{2}\cos \sqrt{\lambda }t \) . Hence every complex number \( \lambda \) is an eigenvalue, and each eigenspace is of dimension 2. | Yes |
Theorem 2. A right inverse of \( A \) in Equation (4) is the operator \( B \) defined by (5) \[ \left( {By}\right) \left( s\right) = {\int }_{a}^{b}g\left( {s, t}\right) y\left( t\right) {dt} \] | Proof. It is to be proved that \( {AB} = I \) . Let \( y \in C\left\lbrack {a, b}\right\rbrack \) and put \( x = {By} \) . We show first that \( {Ax} = y \) . From the equation \[ x\left( s\right) = {\int }_{a}^{b}g\left( {s, t}\right) y\left( t\right) {dt} \] \[ = {\int }_{a}^{s}u\left( s\right) v\left( t\right) y\lef... | Yes |
Consider the boundary-value problem\n\n\\[ \n{Ax} \\equiv {x}^{\\prime \\prime } + x = y\\;{x}^{\\prime }\\left( 0\\right) = x\\left( \\pi \\right) = 0 \n\\] | We shall solve it by means of a Green’s function. For the functions \\( u \\) and \\( v \\) we can take \\( u\\left( t\\right) = \\sin t \\) and \\( v\\left( t\\right) = \\cos t \\) . In this case the Green’s function is\n\n\\[ \ng\\left( {s, t}\\right) = \\left\\{ \\begin{array}{ll} \\sin s\\cos t & 0 \\leq t \\leq s ... | No |
Let us solve the problem in Example 3 by using the Spectral Theorem. The eigenvalues and eigenvectors of the differential operator \( A \) are obtained by solving \( {x}^{\prime \prime } + x = {\mu x} \) . | The general solution of the differential equation is\n\n\[ x\left( t\right) = {c}_{1}\sin \sqrt{1 - \mu }t + {c}_{2}\cos \sqrt{1 - \mu }t \]\n\nImposing the conditions \( {x}^{\prime }\left( 0\right) = x\left( \pi \right) = 0 \), we find that the eigenvalues are \( {\mu }_{n} = 1 - {\left( n - \frac{1}{2}\right) }^{2} ... | Yes |
Example 5. Find the Green's function for this Sturm-Liouville problem:\n\n\[ \n{x}^{\prime \prime } = y\;x\left( 0\right) = {x}^{\prime }\left( 0\right) = 0\;x \in {C}^{2}\left\lbrack {0,1}\right\rbrack \n\] | The preceding theorem asserts that \( {g}^{t} \) should solve the homogeneous differential equation in the intervals \( 0 < s < t < 1 \) and \( 0 < t < s < 1 \) . Furthermore, \( {g}^{t} \) should be continuous, and it should satisfy the boundary conditions. Lastly, \( {g}^{\prime }\left( {s, t}\right) \) should have a... | Yes |
Example 6. Find the Green's function for the problem\n\n\[ \n{x}^{\prime \prime } - {x}^{\prime } - {2x} = y\;x\left( 0\right) = 0 = x\left( 1\right) \n\] | We tentatively set\n\n(7)\n\n\[ \ng\left( {s, t}\right) = \left\{ \begin{array}{ll} u\left( s\right) v\left( t\right) & 0 \leq s \leq t \leq 1 \\ v\left( s\right) u\left( t\right) & 0 \leq t \leq s \leq 1 \end{array}\right. \n\]\n\nand try to determine the functions \( u \) and \( v \) . The homogeneous differential eq... | Yes |
Example 7. Find the Green's function for this Sturm-Liouville problem:\n\n\[ \n{x}^{\prime \prime } + {9x} = y\;x\left( 0\right) = x\left( {\pi /2}\right) = 0 \n\] | According to the preceding theorem, \( g \) should be a continuous function on the square \( 0 \leq s, t \leq \pi /2 \), and \( {g}^{t} \) should solve the homogeneous problem in the intervals \( 0 \leq s \leq t \) and \( t \leq s \leq \pi /2 \) . Finally, \( \partial g/\partial s \) should have a jump of magnitude -1 ... | Yes |
Theorem 1. If \( f \) is differentiable at \( x \), then the mapping \( A \) in the definition is uniquely defined. (It depends on \( x \) as well as \( f \) .) | Proof. Suppose that \( {A}_{1} \) and \( {A}_{2} \) are two linear maps having the required property, expressed in Equation (1). Then to each \( \varepsilon > 0 \) there corresponds a \( \delta > 0 \) such that \[ \parallel f\left( {x + h}\right) - f\left( x\right) - {A}_{i}h\parallel < \varepsilon \parallel h\parallel... | Yes |
Theorem 2. If \( f \) is bounded in a neighborhood of \( x \) and if a linear map \( A \) has the property in Equation (1), then \( A \) is a bounded linear map; in other words, \( A \) is the Fréchet derivative of \( f \) at \( x \) . | Proof. Choose \( \delta > 0 \) so that whenever \( \parallel h\parallel \leq \delta \) we will have\n\n\[ \parallel f\left( {x + h}\right) \parallel \leq M\;\text{ and }\;\parallel f\left( {x + h}\right) - f\left( x\right) - {Ah}\parallel \leq \parallel h\parallel \]\n\nThen for \( \parallel h\parallel \leq \delta \) w... | Yes |
Let \( X = Y = \mathbb{R} \) . Let \( f \) be a function whose derivative (in the elementary sense) at \( x \) is \( \lambda \) . Then the Fréchet derivative of \( f \) at \( x \) is the linear map \( h \mapsto {\lambda h} \) | \[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\left| f\left( x + h\right) - f\left( x\right) - \lambda h\right| }{\left| h\right| } = \mathop{\lim }\limits_{{h \rightarrow 0}}\left| {\frac{f\left( {x + h}\right) - f\left( x\right) }{h} - \lambda }\right| = 0 \] | Yes |
Theorem 3. If \( f \) is differentiable at \( x \), then it is continuous at \( x \) . | Proof. Let \( A = {f}^{\prime }\left( x\right) \) . Then \( A \in \mathcal{L}\left( {X, Y}\right) \) . Given \( \varepsilon > 0 \), select \( \delta > 0 \) so that \( \delta < \varepsilon /\left( {1 + \parallel A\parallel }\right) \) and so that the following implication is valid:\n\n\[ \parallel h\parallel < \delta \;... | Yes |
Example 4. Let \( X = Y = C\left\lbrack {0,1}\right\rbrack \) and let \( \phi : \mathbb{R} \rightarrow \mathbb{R} \) be continuously differentiable. Define \( f : X \rightarrow Y \) by the equation \( f\left( x\right) = \phi \circ x \), where \( x \) is any element of \( C\left\lbrack {0,1}\right\rbrack \) . What is \(... | To answer this, we undertake a calculation of \( f\left( {x + h}\right) - f\left( x\right) \), using the classical mean value theorem:\n\n\[ \left\lbrack {f\left( {x + h}\right) - f\left( x\right) }\right\rbrack \left( t\right) = \phi \left( {x\left( t\right) + h\left( t\right) }\right) - \phi \left( {x\left( t\right) ... | Yes |
Theorem 4. Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) . If each of the partial derivatives \( {D}_{i}f\left( { = \partial f/\partial {x}_{i}}\right) \) exists in a neighborhood of \( x \) and is continuous at \( x \) then \( {f}^{\prime }\left( x\right) \) exists, and a formula for it is\n\n\[ \n{f}^{\prime ... | Proof. We must prove that\n\n\[ \n\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{\parallel h\parallel }\left\lbrack {f\left( {x + h}\right) - f\left( x\right) - \mathop{\sum }\limits_{{i = 1}}^{n}{h}_{i}{D}_{i}f\left( x\right) }\right\rbrack = 0 \n\]\nWe begin by writing\n\n\[ \nf\left( {x + h}\right) - f\left( x\ri... | Yes |
Theorem 5. Let \( f : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \), and let \( {f}_{1},\ldots ,{f}_{m} \) be the component functions of \( f \) . If all partial derivatives \( {D}_{j}{f}_{i} \) exist in a neighborhood of \( x \) and are continuous at \( x \), then \( {f}^{\prime }\left( x\right) \) exists, and\n\n\... | Proof. By the definition of the Euclidean norm,\n\n\[ \n\frac{1}{{\left\| h\right\| }^{2}}{\left\| f\left( x + h\right) - f\left( x\right) - Jh\right\| }^{2} = \frac{1}{{\left\| h\right\| }^{2}}\mathop{\sum }\limits_{{i = 1}}^{m}{\left\lbrack {f}_{i}\left( x + h\right) - {f}_{i}\left( x\right) - \mathop{\sum }\limits_{... | No |
Example 5. Let \( f\left( x\right) = \sqrt{\left| {x}_{1}{x}_{2}\right| } \) . Then the two partial derivatives of \( f \) exist at \( \left( {0,0}\right) \), but \( {f}^{\prime }\left( {0,0}\right) \) does not exist. | Details are left to Problem 16. | No |
Let \( L \) be a bounded linear operator on a real Hilbert space \( X \) . Define \( F : X \rightarrow \mathbb{R} \) by the equation \( F\left( x\right) = \langle x,{Lx}\rangle \) . In order to discover whether \( F \) is differentiable at \( x \), we write | \[ F\left( {x + h}\right) - F\left( x\right) = \langle x + h,{Lx} + {Lh}\rangle - \langle x,{Lx}\rangle \] \[ = \langle x,{Lh}\rangle + \langle h,{Lx}\rangle + \langle h,{Lh}\rangle \] Since the derivative is a linear map, we guess that \( A \) should be \( {Ah} = \langle x,{Lh}\rangle + \langle h,{Lx}\rangle \) . With... | Yes |
Theorem 1. The Chain Rule. If \( f \) is differentiable at \( x \) and if \( g \) is differentiable at \( f\left( x\right) \), then \( g \circ f \) is differentiable at \( x \), and\n\n\[{\left( g \circ f\right) }^{\prime }\left( x\right) = {g}^{\prime }\left( {f\left( x\right) }\right) \circ {f}^{\prime }\left( x\righ... | Proof. Define \( F = g \circ f, A = {f}^{\prime }\left( x\right), y = f\left( x\right), B = {g}^{\prime }\left( y\right) \), and\n\n\[{o}_{1}\left( h\right) = f\left( {x + h}\right) - f\left( x\right) - {Ah}\;\left( {h \in X}\right)\]\n\n\[{o}_{2}\left( k\right) = g\left( {y + k}\right) - g\left( y\right) - {Bk}\;\left... | Yes |
Theorem 2. Mean Value Theorem I. Let \( f \) be a real-valued mapping defined on an open set \( D \) in a normed linear space. Let \( a, b \in D \) . Assume that the line segment\n\n\[ \left\lbrack {a, b}\right\rbrack = \{ a + t\left( {b - a}\right) : 0 \leq t \leq 1\} \]\n\nlies in \( D \) . If \( f \) is continuous o... | Proof. Put \( g\left( t\right) = f\left( {a + t\left( {b - a}\right) }\right) \) . Then \( g \) is continuous on the interval \( \left\lbrack {0,1}\right\rbrack \) and differentiable on \( \left( {0,1}\right) \) . By the chain rule,\n\n\[ {g}^{\prime }\left( t\right) = {f}^{\prime }\left( {a + t\left( {b - a}\right) }\... | Yes |
Theorem 3. Mean Value Theorem II. Let \( f \) be a continuous map of a compact interval \( \left\lbrack {a, b}\right\rbrack \) of the real line into a normed linear space \( Y \) . If, for each \( x \) in \( \left( {a, b}\right) ,{f}^{\prime }\left( x\right) \) exists and satisfies \( \begin{Vmatrix}{{f}^{\prime }\left... | Proof. It suffices to prove that if \( a < \alpha < \beta < b \), then \( \parallel f\left( \beta \right) - f\left( \alpha \right) \parallel \leq M\left( {b - a}\right) \) because, the desired result would follow from this by continuity. Also, it suffices to prove \( \parallel f\left( \beta \right) - f\left( \alpha \ri... | Yes |
Theorem 4. Mean Value Theorem III. Let \( f \) be a map from an open set \( D \) in one normed linear space into another normed linear space. If the line segment\n\n\[ S = \{ {ta} + \left( {1 - t}\right) b : 0 \leq t \leq 1\} \]\n\nlies in \( D \) and if \( {f}^{\prime }\left( x\right) \) exists at each point of \( S \... | Proof. Define \( g\left( t\right) = f\left( {{ta} + \left( {1 - t}\right) b}\right) \) for \( 0 \leq t \leq 1 \) . By the chain rule, \( {g}^{\prime } \) exists and \( {g}^{\prime }\left( t\right) = {f}^{\prime }\left( {{ta} + \left( {1 - t}\right) b}\right) \left( {a - b}\right) \) . By the second Mean Value Theorem\n... | Yes |
Theorem 5. Let \( X \) and \( Y \) be normed spaces, \( D \) a connected open set in \( X \), and \( f \) a differentiable map of \( D \) into \( Y \) . If \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in D \), then \( f \) is a constant function. | Proof. Since \( {f}^{\prime }\left( x\right) \) exists for all \( x \in D, f \) is continuous on \( D \) (by Theorem 3 of Section 3.1, page 117). Select \( {x}_{0} \in D \) and define \( A = \left\{ {x \in D : f\left( x\right) = f\left( {x}_{0}\right) }\right\} \) . This is a closed subset of \( D \) (i.e., the interse... | Yes |
Theorem 1. Let \( f \) be a function from \( \mathbb{R} \) to \( \mathbb{R} \). Assume that \( {f}^{\prime \prime } \) is bounded, that \( f\left( r\right) = 0 \), and that \( {f}^{\prime }\left( r\right) \neq 0 \). Let \( \delta \) be a positive number such that\n\n\[ \rho \equiv \frac{1}{2}\delta \mathop{\max }\limit... | Proof. Define \( {e}_{n} = {x}_{n} - r \). Then\n\n\[ 0 = f\left( r\right) = f\left( {{x}_{n} - {e}_{n}}\right) = f\left( {x}_{n}\right) - {e}_{n}{f}^{\prime }\left( {x}_{n}\right) + \frac{1}{2}{e}_{n}^{2}{f}^{\prime \prime }\left( {\xi }_{n}\right) \]\n\nIn this equation, the point \( {\xi }_{n} \) is between \( {x}_{... | Yes |
For finding the square root of a given positive number \( a \), one can solve the equation \( {x}^{2} - a = 0 \) by Newton’s method. The iteration formula turns out to be\n\n\[ \n{x}_{n + 1} = \frac{1}{2}\left( {{x}_{n} + \frac{a}{{x}_{n}}}\right)\n\]\n\nThis formula was known to the ancient Greeks and is called Heron'... | In order to see how well it performs, we can use a computer system such as Mathematica, Maple, or Matlab to obtain the Newton approximations to \( \sqrt{2} \) . The iteration function is \( g\left( x\right) = \left( {x + 2/x}\right) /2 \), and a reasonable starting point is \( {x}_{0} = 1 \) . Mathematica is capable of... | Yes |
We illustrate the mechanics of Newton's method in higher dimensions with the following problem:\n\n\[ \n\\left\\{ \\begin{array}{l} x - y + 1 = 0 \\\\ {x}^{2} + {y}^{2} - 4 = 0 \\end{array}\\right.\n\]\n\nwhere \( x \) and \( y \) are real variables. We have here a mapping \( f : {\\mathbb{R}}^{2} \\rightarrow {\\mathb... | Hence the iteration formula, in detail, is this:\n\n\[ \\left\\lbrack \\begin{array}{l} {x}_{n + 1} \\\\ {y}_{n + 1} \\end{array}\\right\\rbrack = \\left\\lbrack \\begin{array}{l} {x}_{n} \\\\ {y}_{n} \\end{array}\\right\\rbrack - \\frac{1}{2{x}_{n} + 2{y}_{n}}\\left\\lbrack \\begin{array}{rr} 2{y}_{n} & 1 \\\\ - 2{x}_... | Yes |
Theorem 4. There is a neighborhood of \( {x}^{ * } \) such that the iteration sequence defined in Equation (7) converges to \( {x}^{ * } \) for arbitrary starting points in that neighborhood. | Proof. Select \( \varepsilon > 0 \) such that\n\n(10)\n\n\[ \theta \equiv \lambda + {M\varepsilon } < 1 \]\n\nBy the definition of the Fréchet derivative \( {F}^{\prime }\left( {x}^{ * }\right) \), we can write\n\n(11)\n\n\[ F\left( x\right) = F\left( {x}^{ * }\right) + {F}^{\prime }\left( {x}^{ * }\right) \left( {x - ... | Yes |
Theorem 5. If \( \left| \lambda \right| k < 1 \), then the integral equation (13) above has a unique solution. | Proof. Apply the Contraction Mapping Theorem (Chapter 4, Section 2, page 177) to the mapping \( F \) defined on \( C\left\lbrack {0,1}\right\rbrack \) by \( \left( {Fx}\right) \left( s\right) = v\left( s\right) + \lambda {\int }_{0}^{1}g\left( {s, t, x\left( t\right) }\right) {dt} \) . We see easily that\n\n\[ \begin{V... | Yes |
Theorem 6. The operator \( B \), as just defined, is also an integral operator, having the form\n\n(15)\n\n\[ \left( {Bh}\right) \left( s\right) = {\int }_{0}^{1}r\left( {s, t}\right) h\left( t\right) {dt} \]\n\nThe kernel satisfies these two integral equations\n\n(16)\n\n\[ \left\{ \begin{array}{l} r\left( {s, t}\righ... | Proof. From the definition of \( B \) we have \( {\lambda B} = {\left( I - \lambda A\right) }^{-1} - I \) or \( I + {\lambda B} = \) \( {\left( I - \lambda A\right) }^{-1} \) . Consequently, we have\n\n\[ \left( {I + {\lambda B}}\right) \left( {I - {\lambda A}}\right) = \left( {I - {\lambda A}}\right) \left( {I + {\lam... | Yes |
Solve the integral equation\n\n\[ \nx\\left( s\\right) - {\\int }_{0}^{1}{st}\\operatorname{Arctan}x\\left( t\\right) {dt} = 1 + {s}^{2} - {0.485s} \n\] | This conforms to the general theory outlined above. We have as kernel \( g\\left( {s, t, u}\\right) = {st}\\operatorname{Arctan}u \), and \( {g}_{3}\\left( {s, t, u}\\right) = {st}/\\left( {1 + {u}^{2}}\\right) \) . We take as starting point for the Newton iteration the constant function \( {x}_{0}\\left( t\\right) = 3... | Yes |
Theorem 2. Implicit Function Theorem for Many Variables.\n\nLet \( F : {\mathbb{R}}^{n} \times \mathbb{R} \rightarrow \mathbb{R} \), and suppose that \( F\left( {{x}_{0},{y}_{0}}\right) = 0 \) for some \( {x}_{0} \in {\mathbb{R}}^{n} \) and \( {y}_{0} \in \mathbb{R} \) . If all \( n + 1 \) partial derivatives \( {D}_{i... | Proof. This is left as a problem (Problem 3.4.4). | No |
Theorem 3. General Implicit Function Theorem. Let \( X, Y \) , and \( Z \) be normed linear spaces, \( Y \) being assumed complete. Let \( \Omega \) be an open set in \( X \times Y \) . Let \( F : \Omega \rightarrow Z \) . Let \( \left( {{x}_{0},{y}_{0}}\right) \in \dot{\Omega } \) . Assume that \( F \) is continuous a... | Proof. We can assume that \( \left( {{x}_{0},{y}_{0}}\right) = \left( {0,0}\right) \) . Select \( \delta > 0 \) so that\n\n\[ \{ \left( {x, y}\right) : \parallel x\parallel \leq \delta ,\parallel y\parallel \leq \delta \} \subset \Omega \]\n\nPut \( A = {D}_{2}F\left( {0,0}\right) \) . Then \( A \in \mathcal{L}\left( {... | Yes |
Theorem 5. Inverse Function Theorem I. Let \( f \) be a continuously differentiable map from an open set \( \Omega \) in a Banach space into a normed linear space. If \( {x}_{0} \in \Omega \) and if \( {f}^{\prime }\left( {x}_{0}\right) \) is invertible, then there is a continuously differentiable function \( g \) defi... | Proof. For \( x \) in \( \Omega \) and \( y \) in the second space, define \( F\left( {x, y}\right) = f\left( x\right) - y \) . Put \( {y}_{0} = f\left( {x}_{0}\right) \) so that \( F\left( {{x}_{0},{y}_{0}}\right) = 0 \) . Note that \( {D}_{1}F\left( {x, y}\right) = {f}^{\prime }\left( x\right) \), and thus \( {D}_{1}... | Yes |
Theorem 6. Surjective Mapping Theorem I. Let \( X \) and \( Y \) be Banach spaces, \( \Omega \) an open set in \( X \) . Let \( f : \Omega \rightarrow Y \) be a continuously differentiable map. Let \( {x}_{0} \in \Omega \) and \( {y}_{0} = f\left( {x}_{0}\right) \) . If \( {f}^{\prime }\left( {x}_{0}\right) \) is inver... | Proof. Define \( F : \Omega \times Y \rightarrow Y \) by putting \( F\left( {x, y}\right) = f\left( x\right) - y \) . Then \( F\left( {{x}_{0},{y}_{0}}\right) = \) 0 and \( {D}_{1}F\left( {{x}_{0},{y}_{0}}\right) = {f}^{\prime }\left( {x}_{0}\right) \) . ( \( {D}_{1} \) is a partial derivative, as defined previously.) ... | Yes |
Theorem 7. A Fixed Point Theorem. Let \( \Omega \) be an open set in a Banach space \( X \), and let \( G \) be a differentiable map from \( \widehat{\Omega } \) to \( X \) . Suppose that there is a closed ball \( B \equiv B\left( {{x}_{0}, r}\right) \) in \( \Omega \) such that\n\n(i) \( k \equiv \mathop{\sup }\limits... | Proof. First, we show that \( G \mid B \) is a contraction. If \( {x}_{1} \) and \( {x}_{2} \) are in \( B \), then by the Mean Value Theorem (Theorem 4 in Section 3.2, page 123)\n\n\[ \begin{Vmatrix}{G\left( {x}_{1}\right) - G\left( {x}_{2}\right) }\end{Vmatrix} \leq \mathop{\sup }\limits_{{0 \leq \lambda \leq 1}}\beg... | Yes |
Theorem 8. Inverse Function Theorem II. Let \( \Omega \) be an open set in a Banach space \( X \) . Let \( f \) be a differentiable map from \( \Omega \) to a normed space \( Y \) . Assume that \( \Omega \) contains a closed ball \( B \equiv B\left( {{x}_{0}, r}\right) \) such that\n\n(i) The linear transformation \( A... | Proof. Let \( y \) be as hypothesized, and define \( G\left( x\right) = x - {A}^{-1}\left\lbrack {f\left( x\right) - y}\right\rbrack \) . It is clear that \( f\left( x\right) = y \) if and only if \( x \) is a fixed point of \( G \) . The map \( G \) is differentiable in \( \Omega \), and \( {G}^{\prime }\left( x\right... | Yes |
Consider a nonlinear Volterra integral equation\n\n\\[ \nx\\left( t\\right) - {2x}\\left( 0\\right) + \\frac{1}{2}{\\int }_{0}^{t}\\cos \\left( {st}\\right) {\\left\\lbrack x\\left( s\\right) \\right\\rbrack }^{2}{ds} = y\\left( t\\right) \\;\\left( {0 \\leq t \\leq 1}\\right)\n\\]\n\nin which \\( y \\in C\\left\\lbrac... | Let \\( A = {f}^{\\prime }\\left( 0\\right) \\), so that \\( {Ah} = h - {2h}\\left( 0\\right) \\) . One verifies easily that \\( {A}^{2}h = h \\), from which it follows that \\( {A}^{-1} = A \\) . In order to use the preceding theorem, with \\( {x}_{0} = 0 \\), we must verify its hypotheses. We have just seen that \\( ... | Yes |
Let \( f : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) be given by\n\n\[ f\left( x\right) = y\;x = \left( {{\xi }_{1},{\xi }_{2},{\xi }_{3}}\right) \;y = \left( {{\eta }_{1},{\eta }_{2},{\eta }_{3}}\right) \]\n\n\[ {\eta }_{1} = 2{\xi }_{1}^{4} + {\xi }_{3}\cos {\xi }_{2} - {\xi }_{1}{\xi }_{3} \]\n\n\[ {\eta }_{2... | To answer this, one can use the Inverse Function Theorem. We compute the Fréchet derivative or Jacobian:\n\n\[ {f}^{\prime }\left( x\right) = \left\lbrack \begin{matrix} 8{\xi }_{1}^{3} - {\xi }_{3} & - {\xi }_{3}\sin {\xi }_{2} & \cos {\xi }_{2} - {\xi }_{1} \\ 3{\left( {\xi }_{1} + {\xi }_{3}\right) }^{2} & - 4\cos {... | Yes |
Theorem 10. Let \( f \) be defined on an open set \( \Omega \) in the direct-sum space \( X = \mathop{\sum }\limits_{{i = 1}}^{n} \oplus {X}_{i} \) and take values in a normed space \( Y \) . Assume that all the partial derivatives \( {D}_{i}f \) exist in \( \Omega \) and are continuous at a point \( x \) in \( \Omega ... | Proof. Equation (1) defines a linear transformation from \( X \) to \( Y \), and\n\n\[ \n\begin{Vmatrix}{{f}^{\prime }\left( x\right) h}\end{Vmatrix} \leq \mathop{\sum }\limits_{{i = 1}}^{n}\begin{Vmatrix}{{D}_{i}f\left( x\right) {h}_{i}}\end{Vmatrix} \leq \mathop{\sum }\limits_{{i = 1}}^{n}\begin{Vmatrix}{{D}_{i}f\lef... | Yes |
Theorem 1. Necessary Condition for Extremum. Let \( \Omega \) be an open set in a normed linear space, and let \( f : \Omega \rightarrow \mathbb{R} \) . If \( {x}_{0} \) is a minimum point of \( f \) and if \( {f}^{\prime }\left( {x}_{0}\right) \) exists, then \( {f}^{\prime }\left( {x}_{0}\right) = 0 \) . | Proof. Let \( X \) be the Banach space, and assume \( {f}^{\prime }\left( {x}_{0}\right) \neq 0 \) . Then there exists \( v \in X \) such that \( {f}^{\prime }\left( {x}_{0}\right) v = - 1 \) . By the definition of \( {f}^{\prime }\left( {x}_{0}\right) \) we can take \( \lambda > 0 \) and so small that \( {x}_{0} + {\l... | Yes |
Let \( f \) and \( g \) be functions from \( {\mathbb{R}}^{2} \) to \( \mathbb{R} \) defined by \( f\left( {x, y}\right) = \) \( {x}^{2} + {y}^{2}, g\left( {x, y}\right) = x - y + 1 \) . The set \( M = \{ \left( {x, y}\right) : g\left( {x, y}\right) = 0\} \) is the straight line shown in Figure 3.3. Also shown are some... | The solution is \( \left( {-\frac{1}{2},\frac{1}{2}}\right) \). | Yes |
Example 2. Let \( f\left( {x, y}\right) = {x}^{2} - {y}^{2} \) and \( g\left( {x, y}\right) = {x}^{2} + {y}^{2} - 1 \) . Again we show \( M \) and some level sets of \( f \), which are hyperbolas and straight lines. There are four extrema; some are maxima and some are minima. Which are which? The \( H \) -function is \... | The \( \left( {x, y,\lambda }\right) \) solutions are \( \left( {0,1,1}\right) ,\left( {0, - 1,1}\right) ,\left( {1,0, - 1}\right) ,\left( {-1,0, - 1}\right) \) . | Yes |
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