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Theorem 1.33. If \( {\left\{ {e}_{n}\right\} }_{1}^{\infty } \) is an orthonormal sequence in a Hilbert space \( \mathcal{H} \), then the following conditions are equivalent:\n\n(a) \( {\left\{ {e}_{n}\right\} }_{1}^{\infty } \) is an orthonormal basis.\n\n(b) If \( h \in \mathcal{H} \) and \( h \bot {e}_{n} \) for all... | Proof. The equivalence of (a) and (b) follows almost immediately from the definition, since if \( 0 \neq h \) and \( h \bot {e}_{n} \) for all \( n \), then \( {\left\{ {e}_{n}\right\} }_{1}^{\infty } \cup \{ h/\parallel h\parallel \} \) is an orthonormal set.\n\nNow assume (b) and suppose \( h \in \mathcal{H} \) and l... | Yes |
Is \( {\left\{ {e}_{n}\right\} }_{n = 0}^{\infty } \) an orthonormal basis for \( {L}_{a}^{2}\left( \mathbb{D}\right) \)? | To that end, we will show that \( {\left\{ {e}_{n}\right\} }_{n = 0}^{\infty } \) is an orthonormal basis for \( {L}_{a}^{2}\left( \mathbb{D}\right) \) by showing that if \( f \in {L}_{a}^{2}\left( \mathbb{D}\right) \) and \( f \bot {e}_{n} \) for all \( n = 0,1,2\ldots \), then \( f = 0 \) . The assumption that \( f \... | Yes |
Let \( \left( {X,\mathfrak{M},\mu }\right) \) be any \( \sigma \) -finite measure space and choose \( \varphi \in \) \( {L}^{\infty }\left( {X,\mu }\right) \) . Define the multiplication operator \( {M}_{\varphi } : {L}^{2}\left( {X,\mu }\right) \rightarrow {L}^{2}\left( {X,\mu }\right) \) by \( {M}_{\varphi }\left( f\... | To see this suppose \( \alpha < \parallel \varphi {\parallel }_{\infty } \) . If \( E \) is defined to be \( \{ x : \left| {\varphi \left( x\right) }\right| > \alpha \} \), then \( \mu \left( E\right) > 0 \) . The idea is to consider something like \( {\chi }_{E} \) to show that \( \begin{Vmatrix}{M}_{\varphi }\end{Vma... | Yes |
Consider the Bergman space \( {L}_{a}^{2}\left( \mathbb{D}\right) \) and let \( \varphi \in {L}^{\infty }\left( {\mathbb{D},{dA}/\pi }\right) \) ; note that \( \varphi \) is not assumed to be analytic. Define the Toeplitz operator with symbol \( \varphi \) on \( {L}_{a}^{2}\left( \mathbb{D}\right) \) by \( {T}_{\varphi... | Clearly \( {T}_{\varphi } \) is linear, and since \( P : {L}^{2}\left( {\mathbb{D},{dA}/\pi }\right) \rightarrow {L}_{a}^{2}\left( \mathbb{D}\right) \) is bounded with norm 1, and \( {M}_{\varphi } : {L}^{2}\left( {\mathbb{D},{dA}/\pi }\right) \rightarrow {L}^{2}\left( {\mathbb{D},{dA}/\pi }\right) \) is bounded of nor... | Yes |
The next pair of operators are simple but important ones. They act from \( {\ell }^{2} \) to itself. The first, called the forward shift, is defined by \[ S\left( {{x}_{1},{x}_{2},\ldots }\right) = \left( {0,{x}_{1},{x}_{2},\ldots }\right) . \] | It is easy to see that it is a bounded linear operator of norm one; in fact it is an isometry, meaning \( \parallel {Sx}\parallel = \parallel x\parallel \) for every \( x = \left( {{x}_{1},{x}_{2},\ldots }\right) \in {\ell }^{2} \) . | Yes |
Suppose that \( \mathcal{H} \) is a Hilbert space with orthonormal basis \( {\left\{ {e}_{n}\right\} }_{1}^{\infty } \). Choose any bounded sequence of complex numbers \( {\left\{ {\alpha }_{n}\right\} }_{1}^{\infty } \) and set \( A{e}_{n} = {\alpha }_{n}{e}_{n} \). Extend \( A \) by linearity to any finite linear com... | Since\n\n\[ \parallel {Ah}{\parallel }^{2} = \mathop{\sum }\limits_{1}^{\infty }{\left| \left\langle h,{e}_{n}\right\rangle \right| }^{2}{\left| {\alpha }_{n}\right| }^{2} \leq \left( {\mathop{\sup }\limits_{n}{\left| {\alpha }_{n}\right| }^{2}}\right) \mathop{\sum }\limits_{1}^{\infty }{\left| \left\langle h,{e}_{n}\r... | Yes |
We describe a class of operators called integral operators. Start with a \( \sigma \) -finite measure space \( \left( {X,\mathfrak{M},\mu }\right) \) and a measurable function \( k : X \times X \rightarrow \mathbb{C} \) with \( k \in {L}^{2}\left( {X \times X,\mu \times \mu }\right) \) . Define \( K : {L}^{2}\left( {X,... | For \( f \in {L}^{2}\left( {X,\mu }\right) \) we have\n\n\[ \parallel K\left( f\right) {\parallel }^{2} = {\int }_{X}{\left| g\left( x\right) \right| }^{2}{d\mu }\left( x\right) = {\int }_{X}{\left| {\int }_{X}k\left( x, y\right) f\left( y\right) d\mu \left( y\right) \right| }^{2}{d\mu }\left( x\right) \]\n\n\[ \leq {\... | Yes |
Theorem 2.11. Let \( \mathcal{H} \) and \( \mathcal{K} \) be Hilbert spaces and suppose that \( u : \mathcal{H} \times \mathcal{K} \rightarrow \mathbb{C} \) is a bounded sesquilinear form. There exists a unique \( A \in \mathcal{B}\left( {\mathcal{H},\mathcal{K}}\right) \) such that\n\n\[ u\left( {h, k}\right) = \langl... | Proof. For fixed \( h \in \mathcal{H} \) we define a mapping \( {\Lambda }_{h} : \mathcal{K} \rightarrow \mathbb{C} \) by\n\n\[ {\Lambda }_{h}\left( k\right) = \overline{u\left( {h, k}\right) }.\]\n\nOne can easily check that \( {\Lambda }_{h} \) is linear. Moreover, since \( u \) is bounded by hypothesis,\n\n\[ \left|... | Yes |
Proposition 2.13. For A and B in \( \mathcal{B}\left( \mathcal{H}\right) \) we have\n\n(a) \( \;{A}^{* * } = A \) where \( {A}^{* * } = {\left( {A}^{ * }\right) }^{ * } \) . | Proof. For (a) we first note that by the definition of the adjoint we have \( \left\langle {{A}^{ * }x, y}\right\rangle = \) \( \left\langle {x,{A}^{* * }y}\right\rangle \) for all \( x \) and \( y \) in \( \mathcal{H} \) . Since also \( \langle {Ay}, x\rangle = \left\langle {y,{A}^{ * }x}\right\rangle \), taking conju... | Yes |
Proposition 2.14. If \( A \in \mathcal{B}\left( \mathcal{H}\right) \), then \( \parallel A\parallel = \begin{Vmatrix}{A}^{ * }\end{Vmatrix} \) and \( \begin{Vmatrix}{{A}^{ * }A}\end{Vmatrix} = \parallel A{\parallel }^{2} \) . | Proof. Take any vector \( h \in \mathcal{H} \) with \( \parallel h\parallel = 1 \) . We have\n\n\[ \parallel {Ah}{\parallel }^{2} = \langle {Ah},{Ah}\rangle = \left\langle {h,{A}^{ * }{Ah}}\right\rangle \leq \begin{Vmatrix}{{A}^{ * }{Ah}}\end{Vmatrix}\parallel h\parallel \leq \begin{Vmatrix}{{A}^{ * }A}\end{Vmatrix} \l... | Yes |
Proposition 2.16. If \( U : \mathcal{H} \rightarrow \mathcal{K} \) is an isomorphism, then \( {U}^{ * }U = {I}_{\mathcal{H}} \) (the identity on \( \mathcal{H} \) ) and \( U{U}^{ * } = {I}_{\mathcal{K}} \) . | Proof. Let \( h \) and \( g \) be in \( \mathcal{H} \) . We have\n\n\[ \left\langle {{U}^{ * }{Uh}, g}\right\rangle = \langle {Uh},{Ug}\rangle = \langle h, g\rangle \]\n\nso that \( \left\langle {{U}^{ * }{Uh} - h, g}\right\rangle = 0 \) . This says that for a fixed \( h,{U}^{ * }{Uh} - h \) is orthogonal to every vect... | Yes |
Proposition 2.19. If \( A \) is invertible, then so is \( {A}^{ * } \), and \( {\left( {A}^{ * }\right) }^{-1} = {\left( {A}^{-1}\right) }^{ * } \) . | Proof. When \( A \) is invertible we have \( A{A}^{-1} = I = {A}^{-1}A \) . Applying the \( {}^{ * } \) operation we have \( {\left( A{A}^{-1}\right) }^{ * } = {I}^{ * } = {\left( {A}^{-1}A\right) }^{ * } \) ; clearly \( {I}^{ * } = I \) so that by property (d) of Proposition 2.13 we have \( {\left( {A}^{-1}\right) }^{... | Yes |
Proposition 2.20. If \( U \) is in \( \mathcal{B}\left( {\mathcal{H},\mathcal{K}}\right) \) with \( U \) invertible and \( {U}^{-1} = {U}^{ * } \), then \( U \) is an isomorphism. | Proof. We have already observed that \( U \) must be surjective, so we only need to check that it preserves inner products:\n\n\[ \langle {Uh},{Ug}\rangle = \left\langle {h,{U}^{ * }{Ug}}\right\rangle = \left\langle {h,{U}^{-1}{Ug}}\right\rangle = \langle h, g\rangle \]\n\nfor all \( h \) and \( g \) in \( \mathcal{H} ... | Yes |
Let \( S \) be the forward shift on \( {\ell }^{2} \), so that \( {S}^{ * } \) is the backward shift. We have \( {S}^{ * }S = I \), but \( S \) is not unitary, since its not surjective. | This example points out an important distinction with the finite-dimensional situation. For a linear map \( T \) from \( {\mathbb{C}}^{n} \) into itself, \( T \) is necessarily bijective if it is either one-to-one or surjective. | Yes |
Consider a multiplication operator \( {M}_{\varphi } \) on \( {L}^{2}\left( {X,\mu }\right) \) for \( \varphi \) in \( {L}^{\infty } \) \( \left( {X,\mu }\right) \) . When is \( {M}_{\varphi } \) unitary? | We want \( {M}_{\varphi }{M}_{\varphi }^{-1} = {M}_{\varphi }{M}_{\varphi }^{ * } = I \) . We know that \( {M}_{\varphi }^{ * } = {M}_{\bar{\varphi }} \), so that \( {M}_{\varphi } \) is unitary if and only if \( {\left| \varphi \right| }^{2}f = f \) for all \( f \) in \( {L}^{2}\left( {X,\mu }\right) \) ; that is, if ... | Yes |
Let \( F \) map \( {L}^{2}\left( {\left\lbrack {0,{2\pi }}\right\rbrack ,{dt}/\left( {2\pi }\right) }\right) \) into \[ {\ell }^{2}\left( \mathbb{Z}\right) \equiv \left\{ {{\left\{ {a}_{n}\right\} }_{-\infty }^{\infty } : \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left| {a}_{n}\right| }^{2} < \infty }\right\} \... | Linearity of \( F \) follows from linearity of the integral. Since \( \left\{ {e}^{int}\right\} \) is an orthonormal basis for \( {L}^{2}\left( {\left\lbrack {0,{2\pi }}\right\rbrack ,{dt}/\left( {2\pi }\right) }\right) \), part (f) of Theorem 1.33 guarantees that \( F \) preserves inner products. Given a sequence \( {... | Yes |
Theorem 2.25. If \( A \) is a bounded linear operator from a Hilbert space \( \mathcal{H} \) to a Hilbert space \( \mathcal{K} \), then \( A \) is invertible if and only if \( A \) is bounded below and has dense range. | Proof. The \ | No |
Consider the Banach space \( X = C\left\lbrack {0,1}\right\rbrack \) in the supremum norm. As in Exercise 2.3, let \( \varphi \) be a continuous map of \( \left\lbrack {0,1}\right\rbrack \) into \( \left\lbrack {0,1}\right\rbrack \) and define the bounded linear operator \( {C}_{\varphi } \) on \( X \) by \( {C}_{\varp... | \[ {C}_{\varphi }^{ * }\left( {\Lambda }_{p}\right) \left( f\right) = {\Lambda }_{p}\left( {{C}_{\varphi }\left( f\right) }\right) = {\Lambda }_{p}\left( {f \circ \varphi }\right) = f\left( {\varphi \left( p\right) }\right) = {\Lambda }_{\varphi \left( p\right) }\left( f\right) \] for every \( f \) in \( X \) . Thus \(... | Yes |
Corollary 3.3. Let \( X \neq \{ 0\} \) be a normed linear space. Given \( {x}_{0} \neq 0 \) in \( X \), there is a bounded linear functional \( \varphi \) on \( X \) of norm 1 with \( \varphi \left( {x}_{0}\right) = \begin{Vmatrix}{x}_{0}\end{Vmatrix} \) . | Proof. Set \( M = \left\{ {\alpha {x}_{0} : \alpha \in \mathbb{F}}\right\} \), a subspace of \( X \) . Define \( \varphi \) on \( M \) by \( \varphi \left( {\alpha {x}_{0}}\right) = \alpha \begin{Vmatrix}{x}_{0}\end{Vmatrix} \) . It is easy to see that \( \varphi \) is a bounded linear functional on \( M \) with norm 1... | Yes |
Corollary 3.4. Suppose \( X \neq \{ 0\} \) is a normed linear space. Given \( {x}_{1} \neq {x}_{2} \) we may find a bounded linear functional \( \varphi \) on \( X \) with \( \varphi \left( {x}_{1}\right) \neq \varphi \left( {x}_{2}\right) \) . | Proof. Apply Corollary 3.3 to \( {x}_{0} = {x}_{1} - {x}_{2} \) . | Yes |
Corollary 3.5. Suppose \( {x}_{0} \) is an element of a normed linear space \( X \) . We have\n\n\[ \begin{Vmatrix}{x}_{0}\end{Vmatrix} = \sup \left\{ {\left| {\varphi \left( {x}_{0}\right) }\right| : \varphi \in {X}^{ * },\parallel \varphi \parallel = 1}\right\} \]\n\nand moreover this supremum is attained. | Proof. The result if trivially true if \( {x}_{0} = 0 \) . In general, we have \( \left| {\varphi \left( {x}_{0}\right) }\right| \leq \begin{Vmatrix}{x}_{0}\end{Vmatrix} \) if \( \parallel \varphi \parallel = 1 \), so the supremum is at most \( \begin{Vmatrix}{x}_{0}\end{Vmatrix} \) . On the other hand, by Corollary 3.... | Yes |
Proposition 3.8. Suppose \( X \) and \( Y \) are normed linear spaces and \( T : X \rightarrow Y \) is linear. Then \( T \) is bounded if and only if \( {T}^{-1}\left( {\{ y \in Y : \parallel y\parallel < 1\} }\right) \), the preimage of the open unit ball in \( Y \) under \( T \), has nonempty interior. | Proof. First suppose \( T \) is bounded with \( \parallel T\parallel = M \) . If \( x \in B\left( {0,1/M}\right) \) we have \( \parallel {Tx}\parallel < \) 1, and thus \( B\left( {0,1/M}\right) \) is contained in the preimage of the open unit ball of \( Y \) under \( T \) .\n\nThe more interesting direction is the \ | No |
Theorem 3.10 (Baire Category Theorem). A complete metric space is not the union of a countable number of nowhere dense sets. | Proof. Let \( M \) be a complete metric space. Suppose, for a contradiction, that \( M \) is the countable union of sets \( {A}_{n} \) that are nowhere dense. We will construct a Cauchy sequence in \( M \) with no limit point in \( M \) .\n\nSince \( {A}_{1} \) is nowhere dense, we may find an open ball \( {B}_{1} \) w... | Yes |
Theorem 3.11 (Principle of Uniform Boundedness). Suppose \( X \) is a Banach space and \( \mathcal{F} \) is a family of bounded linear operators from \( X \) to a normed linear space \( Y \) . If, for every \( x \in X \) ,\n\n\[ \sup \{ \parallel {Tx}\parallel : T \in \mathcal{F}\} < \infty \]\n\nthen\n\n\[ \sup \{ \pa... | Proof. Define \( {A}_{n} \equiv \{ x \in X : \parallel {Tx}\parallel \leq n \) for all \( T \in \mathcal{F}\} \) . The hypothesis says that each \( x \in X \) is in some \( {A}_{n} \), so that \( X = { \cup }_{n = 1}^{\infty }{A}_{n} \) . By the Baire category theorem, for some \( n,\overline{{A}_{n}} \) has nonempty i... | Yes |
Theorem 3.12 (Banach-Steinhaus Theorem). Suppose \( \left\{ {T}_{n}\right\} \) is a sequence of bounded linear operators from a Banach space \( X \) to a Banach space \( Y \) . Assume further that for all \( x \in X,\mathop{\lim }\limits_{{n \rightarrow ∞ }}{T}_{n}x \) exists. Define \( T : X \rightarrow Y \) by \( {Tx... | Proof. It is easy to check that \( T \) is linear, and we leave the details of this to the reader. We will use the principle of uniform boundedness to show that \( T \) is bounded. For each \( x \in X,\mathop{\sup }\limits_{n}\begin{Vmatrix}{{T}_{n}x}\end{Vmatrix} < ∞ \) since \( \left\{ {{T}_{n}x}\right\} \) is a conv... | No |
Suppose that \( {\left\{ {a}_{n}\right\} }_{1}^{\infty } \) is a sequence of complex numbers such that \( \mathop{\sum }\limits_{1}^{\infty }{a}_{n}{b}_{n} \) converges whenever \( {\left\{ {b}_{n}\right\} }_{1}^{\infty } \) is in \( {c}_{0} \). We will show that \( \mathop{\sum }\limits_{1}^{\infty }\left| {a}_{n}\rig... | To see this, define \( {T}_{k} : {c}_{0} \rightarrow \mathbb{C} \) by\n\n\[ \n{T}_{k}\left( \left\{ {b}_{n}\right\} \right) = \mathop{\sum }\limits_{{j = 1}}^{k}{a}_{j}{b}_{j} \n\]\n\nEach \( {T}_{k} \) is a bounded linear functional on \( {c}_{0} \) with \( \begin{Vmatrix}{T}_{k}\end{Vmatrix} \leq \mathop{\sum }\limit... | Yes |
What we will do in this example is use the principle of uniform boundedness to show there exists an \( f \in C\left( T\right) \) such that \( {s}_{n}\left( {f,0}\right) \) does not converge to \( f\left( 0\right) \), where \( {s}_{n}\left( {f,0}\right) \) denotes the symmetric partial sum \( \mathop{\sum }\limits_{{k =... | We begin with a calculation.\n\n\[ {s}_{N}\left( {f, t}\right) \equiv \mathop{\sum }\limits_{{k = - N}}^{N}\widehat{f}\left( k\right) {e}^{ikt} = \mathop{\sum }\limits_{{k = - N}}^{N}\left( {{\int }_{-\pi }^{\pi }f\left( x\right) {e}^{-{ikx}}\frac{dx}{2\pi }}\right) {e}^{ikt} \]\n\n\[ = {\int }_{-\pi }^{\pi }f\left( x\... | Yes |
Theorem 3.15. Let \( X \) be a normed linear space, and suppose \( A \) is a subset of \( X \) . If \( \sup \{ \left| {\varphi \left( x\right) }\right| : x \in A\} \) is finite for each fixed \( \varphi \) in \( {X}^{ * } \), then \( A \) is bounded. | Proof. Consider the natural map \( \Phi : X \rightarrow {X}^{* * } \) taking \( x \) to \( {x}^{* * } \) . Note that \( \Phi \left( A\right) \) is thus a collection of bounded linear functionals on \( {X}^{ * } \) . Since \( {X}^{ * } \) is a Banach space\n\n\[ \sup \{ \left| {\Phi \left( x\right) \left( \varphi \right... | Yes |
Corollary 3.17 (Inverse Mapping Theorem). Suppose \( X \) and \( Y \) are Banach spaces and \( T \in \mathcal{B}\left( {X, Y}\right) \) is bijective. Its set-theoretic inverse \( {T}^{-1} \) is then a bounded linear operator from \( Y \) to \( X \) . | Proof. We have already observed that \( {T}^{-1} \) exists as a linear map, so only boundedness of \( {T}^{-1} \) remains to be shown. By the open mapping theorem, \( T \) carries open sets to open sets. Now \( {T}^{-1} \) is bounded if and only if \( {T}^{-1} \) is continuous, and \( {T}^{-1} : Y \rightarrow X \) is c... | Yes |
Lemma 3.19. Suppose that \( X \) and \( Y \) are Banach spaces, and that \( A \) is a bounded linear operator mapping \( X \) onto \( Y \) . There is a positive number \( d \) with the following property: Given \( \varepsilon > 0 \) and \( y \in Y \) there exists \( x \in X \) such that \( \parallel {Ax} - y\parallel <... | Proof. Given \( y \in Y \) there exists \( \widetilde{x} \) in \( X \) with \( A\widetilde{x} = y \), since \( A \) is surjective. This means\n\n\[ Y = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }A\left( {k{B}_{X}}\right) \]\n\nwhere \( {B}_{X} \) is the open unit ball in \( X \) . Since \( Y \) is a complete metric sp... | Yes |
Theorem 3.21 (Closed Graph Theorem). If \( X \) and \( Y \) are Banach spaces and \( T \) : \( X \rightarrow Y \) is linear, then \( T \) is bounded if and only if graph \( \left( T\right) \) is closed in \( X \times Y \) . | Proof (Theorem 3.21). Only the \ | No |
Theorem 3.22 (Two-Norm Theorem). Suppose \( X \) is a normed linear space with two norms, \( \parallel \cdot {\parallel }_{1} \) and \( \parallel \cdot {\parallel }_{2} \), each of which make \( X \) into a Banach space. If there exists a finite constant \( M \) such that\n\n\[ \parallel x{\parallel }_{1} \leq M\parall... | Proof. Let \( I : \left( {X,\parallel \cdot {\parallel }_{1}}\right) \rightarrow \left( {X,\parallel \cdot {\parallel }_{2}}\right) \) be the identity map. Clearly \( I \) is linear, and we want to show that it is bounded. To do this we will apply the closed graph theorem. Suppose that \( {x}_{n} \rightarrow x \) in \(... | Yes |
Proposition 4.3. Any finite-dimensional normed linear space is a Banach space, and any finite-dimensional subspace of a normed linear space is necessarily a closed subspace. | Proof. Let \( \left( {X,\parallel \cdot \parallel }\right) \) be the given normed linear space, and suppose that \( X \) is finite-dimensional. Fix a basis \( \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right\} \) and let \( \parallel \cdot {\parallel }_{\infty } \) be a second norm defined on \( X \) as in the proof of ... | No |
Proposition 4.4. Every linear map from a finite-dimensional normed linear space into a normed linear space is continuous. | Proof. Suppose \( T : X \rightarrow Y \) is as in the statement, and fix a basis \( \left\{ {{b}_{1},\ldots ,{b}_{n}}\right\} \) in \( X \) . Define a second norm \( \parallel \cdot {\parallel }_{\infty } \) on \( X \) as in the proof of Theorem 4.2. The map \( T \) is continuous with respect to the original norm on \(... | Yes |
Proposition 4.6. If \( T \) is compact, then \( T \) is bounded. | Proof. If \( T \) is not bounded, we may find unit vectors \( {v}_{n} \) in \( X \) with \( \begin{Vmatrix}{T{v}_{n}}\end{Vmatrix} \uparrow \infty \) . This implies that \( \left\{ {T{v}_{n}}\right\} \) cannot have a convergence subsequence, since if \( T{v}_{{n}_{k}} \rightarrow y \), then \( \begin{Vmatrix}{T{v}_{{n}... | Yes |
Any linear operator \( T : {\mathbb{C}}^{n} \rightarrow {\mathbb{C}}^{n} \) is compact. | To see this, let \( \left\{ {x}_{n}\right\} \) be a bounded sequence of vectors in \( {\mathbb{C}}^{n} \) ; say \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} \leq M \) . Since \( T \) is bounded by Proposition 4.4, \( \left\{ {T{x}_{n}}\right\} \) is a set of vectors in the closed ball \( \overline{B\left( {0, R}\right) } \) ... | Yes |
Theorem 4.10. Suppose \( X \) is a Banach space. If \( \left\{ {T}_{n}\right\} \) is a sequence of compact operators in \( \mathcal{B}\left( X\right) \) and \( \begin{Vmatrix}{{T}_{n} - T}\end{Vmatrix} \rightarrow 0 \) for some \( T \in \mathcal{B}\left( X\right) \), then \( T \) is compact. | Proof. The argument we will use is sometimes referred to as the \ | No |
Proposition 4.12. If \( T \) is in \( \mathcal{B}\left( \mathcal{H}\right) \) for a separable Hilbert space \( \mathcal{H} \), then \( T \) is compact if and only if \( {T}^{ * } \) is compact. | Proof. Since \( {T}^{* * } = T \), it suffices to prove that \( T \) compact implies \( {T}^{ * } \) is compact. By Theorem 4.11, if \( T \) is compact, there are finite rank operators \( {T}_{n} \) that converge to \( T \) . Now \( \begin{Vmatrix}{{T}_{n} - T}\end{Vmatrix} = \begin{Vmatrix}{{T}_{n}^{ * } - {T}^{ * }}\... | Yes |
Proposition 4.14. Suppose that \( T \) is a bounded linear operator on a separable Hilbert space \( \mathcal{H} \) and \( {\left\{ {e}_{n}\right\} }_{n = 1}^{\infty } \) is an orthonormal basis for \( \mathcal{H} \) such that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{\begin{Vmatrix}T{e}_{n}\end{Vmatrix}}^{2} < \i... | Proof. The proof relies on repeated applications of Parseval’s identity. For each \( n \) we have\n\n\[ T{f}_{n} = \mathop{\sum }\limits_{{j = 1}}^{\infty }\left\langle {T{f}_{n},{f}_{j}}\right\rangle {f}_{j}\text{ and }{\begin{Vmatrix}T{f}_{n}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{{j = 1}}^{\infty }{\left| \left\... | Yes |
Theorem 4.15. Every Hilbert-Schmidt operator on a separable Hilbert space is compact. | Proof. Let \( A \) be Hilbert-Schmidt on \( \mathcal{H} \) . We will exhibit \( A \) as a limit of finite rank operators and use Theorem 4.10 to conclude \( A \) is compact. Fix an orthonormal basis \( {\left\{ {e}_{k}\right\} }_{1}^{\infty } \) of \( \mathcal{H} \), so that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }... | Yes |
Theorem 4.16. Suppose that \( {L}^{2}\left( {X,\mu }\right) \) is a separable Hilbert space and \( K \) is an integral operator on \( {L}^{2}\left( {X,\mu }\right) \), with kernel \( k\left( {x, y}\right) \in {L}^{2}\left( {X \times X}\right) \) . The operator \( K \) is Hilbert-Schmidt. | Proof. Let \( {\left\{ {e}_{n}\right\} }_{n = 1}^{\infty } \) be a basis for \( {L}^{2}\left( {X,\mu }\right) \) . For fixed \( x \in X \), write \( {k}_{x}\left( y\right) = k\left( {x, y}\right) \) ; then \( {k}_{x} \) is in \( {L}^{2}\left( \mu \right) \) for almost every \( x \) and we have\n\n\[ \left( {K{e}_{n}}\r... | Yes |
Lemma 4.18. Suppose \( T \) is a self-adjoint operator in \( \mathcal{B}\left( \mathcal{H}\right) \) for some Hilbert space \( \mathcal{H} \) . We have \[ \parallel T\parallel = \mathop{\sup }\limits_{{\parallel x\parallel = 1}}\left| {\langle {Tx}, x\rangle }\right| \] | Proof. Set \( M = \mathop{\sup }\limits_{{\parallel x\parallel = 1}}\left| {\langle {Tx}, x\rangle }\right| \) . Our goal is to show \( M = \parallel T\parallel \) . We make three easy observations (a) For each \( h \neq 0 \in \mathcal{H},\left| {\langle {Th}, h\rangle }\right| = \left| {\langle T\left( {\parallel h\pa... | Yes |
Theorem 4.19. If \( T \) is a compact self-adjoint operator in \( \mathcal{B}\left( \mathcal{H}\right) \), then at least one of the numbers \( \parallel T\parallel \) and \( - \parallel T\parallel \) is an eigenvalue of \( T \) . | Proof. Without loss of generality we assume \( \parallel T\parallel \neq 0 \), else \( T = 0 \) and 0 is trivially an eigenvalue of \( T \) . By Lemma 4.18 we have \( \parallel T\parallel = \mathop{\sup }\limits_{{\parallel x\parallel = 1}}\left| {\langle {Tx}, x\rangle }\right| \) . Find unit vectors \( {x}_{n} \) wit... | Yes |
Theorem 4.20. Suppose that \( T \) is self-adjoint in \( \mathcal{B}\left( \mathcal{H}\right) \) . Every eigenvalue of \( T \) is real, and the eigenvectors for distinct eigenvalues are orthogonal. | Proof. Suppose that for some nonzero vector \( h \) and some scalar \( \lambda ,{Th} = {\lambda h} \) . Since \( \langle {Th}, h\rangle = \langle {\lambda h}, h\rangle = \lambda \parallel h{\parallel }^{2} \) and \( \langle {Th}, h\rangle = \langle h,{Th}\rangle = \langle h,{\lambda h}\rangle = \bar{\lambda }\parallel ... | Yes |
Theorem 4.21. Suppose that \( T \) is a compact self-adjoint operator in \( \mathcal{B}\left( \mathcal{H}\right) \). The set of eigenvalues of \( T \) is a finite or countably infinite set of real numbers; if infinite, the eigenvalues form a sequence that converges to zero. | Proof. By Theorem 4.20, all the eigenvalues are real. Also observe that if \( {Tx} = {\lambda x} \), then \( \left| \lambda \right| \leq \parallel T\parallel \), so that no eigenvalue has absolute value greater than \( \parallel T\parallel \). There is nothing further to do if the set of eigenvalues is finite, so suppo... | Yes |
Lemma 4.22. Suppose that \( T \) is a bounded operator on a Hilbert space \( \mathcal{H} \) and that \( M \) is a closed subspace of \( \mathcal{H} \) . If \( {TM} \subseteq M \), then \( {T}^{ * }{M}^{ \bot } \subseteq {M}^{ \bot } \) . Conversely, if \( {T}^{ * }{M}^{ \bot } \subseteq {M}^{ \bot } \), then \( {TM} \s... | Proof. Since \( {T}^{* * } = T \) and \( {\left( {M}^{ \bot }\right) }^{ \bot } = M \), only the first assertion needs to be verified. Let \( n \) be in \( {M}^{ \bot } \) and let \( m \) be in \( M \) . We must show that \( {T}^{ * }n \bot m \), or equivalently, \( \left\langle {{T}^{ * }n, m}\right\rangle = 0 \) . We... | Yes |
Corollary 4.25. If \( T \) is a compact self-adjoint operator on a separable Hilbert space \( \mathcal{H} \), then there is an orthonormal basis \( \left\{ {e}_{n}\right\} \) of \( \mathcal{H} \) consisting of eigenvectors for \( T \) such that\n\n\[ \n{Tx} = \mathop{\sum }\limits_{n}{\lambda }_{n}\left\langle {x,{e}_{... | Proof. By Theorem 4.24, there is a finite or infinite orthonormal sequence \( \left\{ {g}_{n}\right\} \) such that\n\n\[ \n{Tx} = \mathop{\sum }\limits_{n}{\lambda }_{n}\left\langle {x,{g}_{n}}\right\rangle {g}_{n} \n\]\n\n(4.9)\n\nand \( T{g}_{n} = {\lambda }_{n}{g}_{n} \) . By the construction in Theorem 4.24, the \(... | Yes |
Proposition 4.27. If \( T \) is an operator in \( \mathcal{B}\left( X\right) \) for a Banach space \( X \), and if \( S \) is an invertible operator in \( \mathcal{B}\left( X\right) \), then \( \sigma \left( T\right) = \sigma \left( {{S}^{-1}{TS}}\right) \) . | Proof. If \( T - {\lambda I} \) is invertible, with inverse \( V \), then \( {S}^{-1}{TS} - {\lambda I} = {S}^{-1}\left( {T - {\lambda I}}\right) S \) has inverse \( {S}^{-1}{VS} \) . Conversely, if \( {S}^{-1}\left( {T - {\lambda I}}\right) S \) is invertible, then applying the first part we see that \( S\left\lbrack ... | Yes |
Consider the operator \( {M}_{x} \) of multiplication by \( \varphi \left( x\right) = x \) on the Hilbert space \( \left( {{L}^{2}\left\lbrack {0,1}\right\rbrack ,{dx}}\right) \). We claim, however, that each \( 0 \leq \lambda \leq 1 \) is in \( \sigma \left( {M}_{x}\right) \). | To see this, it is helpful to recall that an invertible operator is bounded below (meaning \( \parallel {Ag}\parallel \geq \delta \parallel g\parallel \) for some positive \( \delta \) and all \( g \) ; see Definition 2.24), and to observe that\n\n\[ \n{M}_{x} - {\lambda I} = {M}_{x - \lambda }\n\]\n\nthe operator of m... | Yes |
Theorem 4.29. Suppose \( X \) is a Banach space and \( T \in \mathcal{B}\left( X\right) \) is compact. If \( \lambda \neq 0 \) , then \( T - {\lambda I} \) has closed range. | Proof. For \( \lambda \neq 0, T - {\lambda I} = \lambda \left( {\frac{1}{\lambda }T - I}\right) \) . Since \( \frac{1}{\lambda }T \) is compact if \( T \) is, it suffices to prove the theorem for \( \lambda = 1 \) . Suppose, for a contradiction, that the range of \( T - I \) is not closed. Define a map \( S \) from the... | Yes |
Theorem 4.30. Suppose that \( T \) is a compact operator on a Hilbert space \( \mathcal{H} \) and let \( {M}_{j} \) be the range of the operator \( {\left( T - I\right) }^{j} \) for each \( j = 1,2,\ldots \) . There exists a positive integer \( j \) such that \( {M}_{j} = {M}_{j + 1} \) . | Proof. By the previous theorem, \( {M}_{1} \) is closed. For \( j > 1 \), we may expand \( {\left( I - T\right) }^{j} \) by the binomial theorem to write\n\n\[ \n{\left( I - T\right) }^{j} = I - {jT} + \frac{j\left( {j - 1}\right) }{2}{T}^{2} + \cdots + {\left( -1\right) }^{j}{T}^{j}, \]\n\nwhere \( A \equiv {jT} - j\l... | Yes |
Theorem 4.31. Suppose \( T \) is a compact operator on a Hilbert space \( \mathcal{H} \) and \( \lambda \neq 0 \) . If \( T - {\lambda I} \) is not invertible, then \( \lambda \) is an eigenvalue of \( T \) . | Proof. Since \( T - {\lambda I} = \lambda \left( {\frac{1}{\lambda }T - I}\right) \), there is no loss of generality in taking \( \lambda = 1 \) . Thus we are given that \( T - I \) is not invertible. Suppose that 1 is not an eigenvalue of \( T \) . This means \( \ker \left( {T - I}\right) = \{ 0\} \), and \( T - I \) ... | Yes |
Theorem 4.32. Let \( T \) be a compact operator on a Hilbert space \( \mathcal{H} \). Suppose \( \lambda \) is a nonzero complex number.\n\n(a) If \( T - {\lambda I} \) is one-to-one, then \( T - {\lambda I} \) is invertible.\n\n(b) If \( T - {\lambda I} \) maps \( \mathcal{H} \) onto \( \mathcal{H} \), then \( T - {\l... | Proof. The first statement is Theorem 4.31. For the second, take adjoints. If \( T - {\lambda I} \) is onto, then \( {T}^{ * } - \bar{\lambda }I \) is one-to-one (by Exercise 2.16 in Chapter 2). From the first part of the theorem, \( {T}^{ * } - \bar{\lambda }I \) is invertible; the adjoint of its inverse provides the ... | No |
Theorem 5.7. Suppose \( \mathcal{A} \) is a unital Banach algebra and let \( \mathcal{G} \) denote the invertible elements of \( \mathcal{A} \) . Then \( \mathcal{G} \) is an open set in \( \mathcal{A} \) . | To prove this theorem, we begin with a lemma which says that the open ball of radius 1 about the identity is contained in \( \mathcal{G} \) . | No |
Lemma 5.8. If \( B \in \mathcal{A} \) and \( \parallel I - B\parallel < 1 \), then \( B \) is invertible, and its inverse is given by \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{\left( I - B\right) }^{k} \) . | Proof (Lemma 5.8). Let \( C = I - B \) so that \( \parallel C\parallel = r < 1 \) and \( \begin{Vmatrix}{C}^{n}\end{Vmatrix} \leq \parallel C{\parallel }^{n} = {r}^{n} \) . Since \( r < 1 \) we have \( \mathop{\sum }\limits_{0}^{\infty }\begin{Vmatrix}{C}^{n}\end{Vmatrix} < \infty \) . This says the partial sums of \( ... | Yes |
Theorem 5.11. If \( f : \mathbb{C} \rightarrow \mathcal{A} \) is weakly analytic, where \( \mathcal{A} \) is a Banach space, and bounded in \( \mathbb{C} \), then \( f \) is constant. | Proof. We are given that \( \parallel f\left( z\right) \parallel \leq M < \infty \) for all \( z \in \mathbb{C} \) . If \( \varphi \) is arbitrary in \( {\mathcal{A}}^{ * } \), then \( \varphi \circ f \) is entire and\n\n\[ \left| {\varphi \left( {f\left( z\right) }\right) }\right| \leq \parallel \varphi \parallel \par... | Yes |
Theorem 5.12. If \( \mathcal{A} \) is a unital Banach algebra in which each nonzero element is invertible, then \( \mathcal{A} \) is isometrically isomorphic to \( \mathbb{C} \) . | Proof. Let \( A \in \mathcal{A} \) and suppose \( {\lambda }_{1},{\lambda }_{2} \) are two distinct complex numbers. At least one of \( A - {\lambda }_{1}, A - {\lambda }_{2} \) is invertible (since both can’t be 0 ). On the other hand, \( \sigma \left( A\right) \) is nonempty, so \( \sigma \left( A\right) \) consists ... | No |
Lemma 5.13. The product \( {\Pi }_{1}^{n}\left( {A - {\lambda }_{j}I}\right) \) is invertible if and only if each of the factors \( A - {\lambda }_{j}I \) is invertible. | Proof. The \ | No |
Theorem 5.14. Suppose \( \mathcal{A} \) is a unital Banach algebra, \( A \) is an element of \( \mathcal{A} \), and \( p \) is a polynomial. We have\n\n\[ \sigma \left( {p\left( A\right) }\right) = p\left( {\sigma \left( A\right) }\right) . \] | Proof. The result is easy when \( p \) is a constant, so we assume \( p \) has degree at least one. We’ll show the two inclusions: \( p\left( {\sigma \left( A\right) }\right) \subseteq \sigma \left( {p\left( A\right) }\right) \) and the reverse. For the first, let \( \lambda \in \sigma \left( A\right) \) and factor \( ... | Yes |
If \( A \) is a self-adjoint element of a unital \( {C}^{ * } \) -algebra, then we have\n\n\[ \begin{Vmatrix}{A}^{2}\end{Vmatrix} = \begin{Vmatrix}{{A}^{ * }A}\end{Vmatrix} = \parallel A{\parallel }^{2} \]\n\nand \( \begin{Vmatrix}{A}^{2}\end{Vmatrix} = \parallel A{\parallel }^{2} \) . | Since \( {A}^{2} \) is also self-adjoint, we may replace \( A \) by \( {A}^{2} \) to get\n\n\[ \begin{Vmatrix}{A}^{4}\end{Vmatrix} = {\begin{Vmatrix}{A}^{2}\end{Vmatrix}}^{2} = \parallel A{\parallel }^{4}. \]\n\nContinuing, an induction argument will show\n\n\[ \begin{Vmatrix}{A}^{{2}^{n}}\end{Vmatrix} = \parallel A{\p... | No |
Consider a (bounded) diagonal operator \( T \) on \( {\ell }^{2} \) with diagonal \( \left( {{\alpha }_{1},{\alpha }_{2},\ldots }\right) \). Clearly each scalar \( {\alpha }_{j} \) is an eigenvalue of \( T \). We leave it to the reader to check that the closure of the set \( \left\{ {\alpha }_{j}\right\} \) is \( \sigm... | To see this, note that given any point \( x = \left( {{x}_{1},{x}_{2},\ldots }\right) \) in \( {\ell }^{2} \) and any positive integer \( N \), the sequence\n\n\[ \left\{ {\frac{{x}_{1}}{{\alpha }_{1} - \lambda },\frac{{x}_{2}}{{\alpha }_{2} - \lambda },\cdots ,\frac{{x}_{N}}{{\alpha }_{N} - \lambda },0,0,\cdots }\righ... | Yes |
Theorem 5.20. Every nontrivial complex homomorphism \( \varphi \) of a unital Banach algebra is continuous and satisfies \( \parallel \varphi \parallel = 1 \) . | Proof. If \( A \in \mathcal{A} \) with \( \varphi \left( A\right) \neq 0 \), then\n\n\[ \varphi \left( {I - \frac{A}{\varphi \left( A\right) }}\right) = 0 \]\n\nAs we have just observed, this shows that \( I - A/\varphi \left( A\right) \) is not invertible, and so by Lemma 5.8 we must have \( \parallel A/\varphi \left(... | Yes |
Theorem 5.21. Suppose \( \mathcal{A} \) is a unital \( {C}^{ * } \) -algebra and \( \varphi : \mathcal{A} \rightarrow \mathbb{C} \) is a homomorphism. For every \( A \in \mathcal{A},\varphi \left( {A}^{ * }\right) = \overline{\varphi \left( A\right) } \) . | Proof. First suppose that \( A \) is a self-adjoint element of the \( {C}^{ * } \) -algebra \( \mathcal{A} \) and let \( t \) be a real number. Set \( B = A + {itI} \) so that \( {B}^{ * } = A - {itI} \) and \( {B}^{ * }B = {A}^{2} + {t}^{2}I \) . Since \( \varphi \) has norm 1 , we have\n\n\[ \n{\left| \varphi \left( ... | No |
Proposition 5.23. In a unital Banach algebra, the closure of a proper left, right, or two-sided ideal is a proper left, right, or two-sided ideal, respectively. | Proof. We give the proof for the case of a two-sided ideal and leave the other cases for the reader. Suppose \( \mathcal{J} \) is a two-sided proper ideal in \( \mathcal{A} \) and let \( \mathcal{G} \) be the set of invertible elements of \( \mathcal{A} \) . We know (Theorem 5.7) that \( \mathcal{G} \) is open and none... | No |
Proposition 5.24. In a unital Banach algebra, every maximal ideal is closed and every proper ideal is contained in a maximal ideal. | Proof. The first statement follows from the previous result: if \( \mathcal{J} \) is a nonclosed ideal, then \( \overline{\mathcal{J}} \) is a proper ideal strictly containing \( \mathcal{J} \), and hence \( \mathcal{J} \) cannot be maximal. For the second statement, we look at the collection \( \mathcal{P} \) of all p... | Yes |
Theorem 5.25. For a Banach algebra \( \mathcal{A} \) with proper closed ideal \( \mathcal{J} \), the quotient \( \mathcal{A}/\mathcal{J} \) is a Banach algebra. When \( \mathcal{A} \) is unital, so is \( \mathcal{A}/\mathcal{J} \) . The quotient map \( \Pi : \mathcal{A} \rightarrow \mathcal{A}/\mathcal{J} \) is a surje... | Proof. We already know that \( \mathcal{A}/\mathcal{J} \) is a Banach space. It is easy to check that the multiplication has the desired associative and distributive properties. We need to verify that\n\n\[ \parallel \left( {A + \mathcal{J}}\right) \left( {B + \mathcal{J}}\right) \parallel \leq \parallel A + \mathcal{J... | Yes |
Theorem 5.26. In a unital commutative Banach algebra \( \mathcal{A} \), for every \( \varphi \) in \( {\mathcal{M}}_{\mathcal{A}} \) , the kernel of \( \varphi \) is a maximal ideal of \( \mathcal{A} \), and conversely, every maximal ideal in \( \mathcal{A} \) is the kernel of some \( \varphi \in {\mathcal{M}}_{\mathca... | Proof. First suppose \( \mathcal{M} \) is a maximal ideal in \( \mathcal{A} \), so that \( \mathcal{M} \) is closed and \( \mathcal{A}/\mathcal{M} \) is a unital Banach algebra. We show that this quotient is isomorphic to \( \mathbb{C} \) by showing that every nonzero element is invertible and invoking Theorem 5.12, th... | Yes |
Suppose \( \mathcal{A} = C\left( X\right) \) for some compact Hausdorff space \( X \) . We claim that every nontrivial multiplicative linear functional on \( C\left( X\right) \) is an evaluation functional, i.e. has the form \( e{v}_{x}\left( f\right) = f\left( x\right) \) for some \( x \in X \) . This is equivalent to... | Since we already know that the evaluation functionals are multiplicative linear functionals with kernel \( {\mathcal{M}}_{x} \), it suffices to show that every proper ideal is contained in at least one \( {\mathcal{M}}_{x} \) . Assume, for a contradiction, that we have a proper ideal \( \mathcal{J} \) so that for each ... | Yes |
Theorem 5.28. Suppose \( \mathcal{A} \) is a commutative unital Banach algebra and let \( A \in \mathcal{A} \) . We have\n\n\[ \sigma \left( A\right) = \left\{ {\varphi \left( A\right) : \varphi \in {\mathcal{M}}_{\mathcal{A}}}\right\} \] | Proof. Fix \( A \in \mathcal{A} \) and suppose \( \lambda \in \sigma \left( A\right) \) . Then \( A - {\lambda I} \) is not invertible and\n\n\[ \{ \left( {A - {\lambda I}}\right) B : B \in \mathcal{A}\} \]\n\nis a proper ideal (it can’t contain \( I \) ), and hence is contained in a maximal ideal, which by Theorem 5.2... | Yes |
Corollary 5.29. An element A in a unital commutative Banach algebra \( \mathcal{A} \) is invertible if and only if \( \varphi \left( A\right) \neq 0 \) for all \( \varphi \in {\mathcal{M}}_{\mathcal{A}} \) . Furthermore, \( A \) is invertible if and only if \( A \) lies in no proper ideal of \( \mathcal{A} \) . | Proof. The first statement follows immediately from the theorem. For the second statement, notice that if \( A \) is not invertible, \( \{ {AB} : B \in \mathcal{A}\} \) is a proper ideal containing \( A \), and, as previously observed, no proper ideal contains an invertible element. | Yes |
Corollary 5.30. For every \( A \) in a unital commutative Banach algebra \( \mathcal{A} \), and every \( \varphi \in {\mathcal{M}}_{\mathcal{A}},\left| {\varphi \left( A\right) }\right| \leq r\left( A\right) \leq \parallel A\parallel . | Proof. Only the first inequality is new, and it is an immediate consequence of Theorem 5.28. | No |
Theorem 5.31. If \( \mathcal{A} \) is a commutative unital \( {C}^{ * } \) -algebra and \( A \in \mathcal{A} \) is self-adjoint, then \( \sigma \left( A\right) \) is contained in the real line. | Proof. If \( \lambda \in \sigma \left( A\right) \), then by Theorem \( {5.28\lambda } = \varphi \left( A\right) \) for some \( \varphi \in {\mathcal{M}}_{\mathcal{A}} \) . Using Theorem 5.21 we have\n\n\[ \bar{\lambda } = \overline{\varphi \left( A\right) } = \varphi \left( {A}^{ * }\right) = \varphi \left( A\right) = ... | Yes |
Proposition 5.32. Suppose \( \left\{ {\varphi }_{n}\right\} \) is a sequence in \( {\mathcal{A}}^{ * } \) . We have \( {\varphi }_{n} \rightarrow \varphi \) in the weak* topology if and only if \( {\varphi }_{n}\left( A\right) \rightarrow \varphi \left( A\right) \) for each \( A \) in \( \mathcal{A} \) . | Proof. If \( {\varphi }_{n} \rightarrow \varphi \left( {\text{weak*}}^{ * }\right) \) and \( A \) is in \( \mathcal{A} \), then by definition \( {A}^{* * } \) is continuous as a map of \( \left( {\mathcal{A},{\text{weak*}}^{ * }}\right) \) into \( \mathbb{C} \), so \( {A}^{* * }\left( {\varphi }_{n}\right) \rightarrow ... | Yes |
Consider the space \( {\ell }^{2} \), and recall that by Theorem 1.29, every bounded linear functional on \( {\ell }^{2} \) is given by \( \langle \cdot, y\rangle \) for some \( y \in {\ell }^{2} \). Let \( {e}_{n} \) be \( n \) th standard basis vector for \( {\ell }^{2} \), whose entries are all 0 except for a 1 in t... | By Proposition 5.32, this will follow if we can show that\n\n\[ \n{\varphi }_{n}\left( x\right) \rightarrow 0 \n\]\n\nfor each \( x \) in \( {\ell }^{2} \). By Parseval’s identity,\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left| \left\langle x,{e}_{n}\right\rangle \right| }^{2} < \infty \n\]\n\nand hence \( \l... | Yes |
Theorem 5.38. Suppose that \( A \) is a subset in a topological space \( X \) . We have \( x \in \bar{A} \) if and only if there is a net of points in \( A \) converging to \( x \) . | Proof. Suppose that \( x \) is in \( \bar{A} \) . Recall that \( \bar{A} \) can be described as the set of points \( y \) such that every open set containing \( y \) intersects \( A \) . Let \( I \) be the collection of all open sets containing our given \( x \), partially ordered by reverse inclusion; this is our dire... | Yes |
Theorem 5.39. Suppose that \( X \) and \( Y \) are topological spaces. A function \( f : X \rightarrow Y \) is continuous at \( {x}_{0} \) in \( X \) if and only if the net \( {\left\{ f\left( {x}_{\alpha }\right) \right\} }_{\alpha \in I} \) converges to \( f\left( {x}_{0}\right) \) in \( Y \) whenever \( {\left\{ {x}... | Proof. Assume that \( f \) is continuous at \( {x}_{0} \) and suppose \( \left\{ {x}_{\alpha }\right\} \) is a net in \( X \) converging to \( {x}_{0} \) . Let \( V \) be open in \( Y \) with \( f\left( {x}_{0}\right) \in V \) . By continuity, \( U \equiv {f}^{-1}\left( V\right) \) is open in \( X \) , and \( U \) cont... | No |
Theorem 5.41 (Banach-Alaoglu Theorem). Let \( {\mathcal{A}}^{ * } \) be the dual space of some Banach space \( \mathcal{A} \) . The norm-closed unit ball, \( \left\{ {\varphi \in {\mathcal{A}}^{ * } : \parallel \varphi \parallel \leq 1}\right\} \), is compact in the weak* topology. | Proof (Theorem 5.41). Suppose that \( {X}^{ * } \) is the dual of a Banach space \( X \) . Let \( B \) denote the norm-closed unit ball in \( {X}^{ * } \) :\n\n\[ B = \left\{ {\varphi \in {X}^{ * } : \parallel \varphi \parallel \leq 1}\right\} \]\n\nWe wish to show that \( B \) is compact in the weak* topology.\n\nFor ... | Yes |
Theorem 5.42. In the (relative) weak* topology, \( {\mathcal{M}}_{\mathcal{A}} \) is a compact Hausdorff space. | Proof. Since we already know that \( {\mathcal{A}}^{ * } \) with the weak* topology is Hausdorff, and a subspace of a Hausdorff space is Hausdorff, we need only check the compactness assertion. For this, it suffices to show \( {\mathcal{M}}_{\mathcal{A}} \) is closed in the weak* topology; then the Banach-Alaoglu theor... | Yes |
Theorem 5.43. Let \( \mathcal{A} \) be a commutative unital Banach algebra. For each \( A \) in \( \mathcal{A} \) , \( \widehat{A} \) is continuous on \( {\mathcal{M}}_{\mathcal{A}} \) . The map \( \Gamma \) is a continuous homomorphism of the commutative unital Banach algebra \( \mathcal{A} \) into \( C\left( {\mathca... | Proof. The continuity of \( \widehat{A} \) follows immediately from the definition of the weak* topology: If \( {\varphi }_{\alpha } \) is a net in \( {\mathcal{M}}_{\mathcal{A}} \) converging weak* to \( \varphi \), then \[ \widehat{A}\left( {\varphi }_{\alpha }\right) \equiv {\varphi }_{\alpha }\left( A\right) \right... | Yes |
What is the Gelfand map \( \Gamma : C\left( X\right) \rightarrow C\left( {\mathcal{M}}_{\mathcal{A}}\right) \) ? | By definition, we have \( \Gamma \left( f\right) = \widehat{f} \) where \( \widehat{f}\left( \varphi \right) = \varphi \left( f\right) \), for \( f \in \mathcal{A} = C\left( X\right) \) and \( \varphi \in {\mathcal{M}}_{\mathcal{A}} \) . But each \( \varphi \) has the form \( e{v}_{x} \) for some \( x \in X \), and \( ... | Yes |
Theorem 5.46. Suppose \( \mathcal{A} \) is a singly generated, commutative, unital \( {C}^{ * } \) -algebra, with \( \mathcal{A} = {C}^{ * }\left( A\right) \) for some \( A \) which is necessarily normal. There is a unique \( * \) - isomorphism of \( \mathcal{A} \) onto \( C\left( {\sigma \left( A\right) }\right) \) ma... | Proof. We know that \( \mathcal{A} \) is the closure of the polynomials in \( A \) and \( {A}^{ * } \), and that the Gelfand map \( \Gamma \) is an isometric \( * \) -isomorphism of \( {C}^{ * }\left( A\right) \) onto \( C\left( {\mathcal{M}}_{\mathcal{A}}\right) \) . Define \( \tau : {\mathcal{M}}_{\mathcal{A}} \mapst... | Yes |
Lemma 5.47. Suppose \( \mathcal{A} \) is a unital \( {C}^{ * } \) -algebra. An element \( A \in \mathcal{A} \) is invertible in \( \mathcal{A} \) if and only if \( A{A}^{ * } \) and \( {A}^{ * }A \) are both invertible. | Proof. The \ | No |
Theorem 5.48. Suppose \( \mathcal{A} \) and \( \mathcal{B} \) are \( {C}^{ * } \) -algebras with common identity \( I \), and assume \( \mathcal{B} \subseteq \mathcal{A} \) . If \( A \in \mathcal{B} \), then \( {\sigma }_{\mathcal{A}}\left( A\right) = {\sigma }_{\mathcal{B}}\left( A\right) \) . | Proof. First suppose \( A \) is self-adjoint, and let \( \mathcal{C} = {C}^{ * }\left( A\right) \), the \( {C}^{ * } \) -algebra generated by \( A = {A}^{ * } \) and \( I \) . Since \( \mathcal{C} \) is commutative and unital, the spectrum of the self-adjoint element \( A \) in \( \mathcal{C} \) is real (Theorem 5.31),... | Yes |
Theorem 5.49. If \( \mathcal{A} \) is a unital \( {C}^{ * } \) -algebra, and \( A \in \mathcal{A} \) is normal, then\n\n(a) \( A \) is self-adjoint if and only if \( \sigma \left( A\right) \subseteq \mathbb{R} \) . | Proof. We give the proof of (a) and leave the remaining parts to the reader as Exercise 5.35.\n\nLet \( \mathcal{B} \) be the commutative \( {C}^{ * } \) -algebra \( {C}^{ * }\left( A\right) \) ; Theorem 5.48 tells us that \( {\sigma }_{\mathcal{B}}\left( A\right) = \) \( {\sigma }_{\mathcal{A}}\left( A\right) \) . The... | No |
The space \( W \) consists of all complex-valued functions \( f \) on the unit circle \( T \) that can be expressed as an absolutely convergent Fourier series: | \[ f\left( {e}^{i\theta }\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{e}^{in\theta } \] (5.2) normed by \[ \parallel f\parallel = \parallel f{\parallel }_{W} \equiv \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\left| {a}_{n}\right| < \infty . \] The series defining \( f \) converges uniformly ... | Yes |
Theorem 5.51. Suppose \( f\left( {e}^{i\theta }\right) = \mathop{\sum }\limits_{{-\infty }}^{\infty }{a}_{n}{e}^{in\theta } \) lies in \( W \) . If \( f \) does not vanish on \( T \) , then \( 1/f \) is also in \( W \), that is, there exist \( \left\{ {b}_{n}\right\} \) with \( \mathop{\sum }\limits_{{-\infty }}^{\inft... | Proof (Theorem 5.51). The hypothesis on \( f \) says that \( {\varphi }_{\lambda }\left( f\right) = f\left( \lambda \right) \) does not vanish as \( \lambda \) ranges over \( T \) . Since we have shown that the functionals \( {\varphi }_{\lambda } \) exhaust \( {\mathcal{M}}_{W} \), we may apply Corollary 5.29 to concl... | Yes |
Theorem 5.52. Suppose \( N \) is a normal element in a unital \( {C}^{ * } \) -algebra \( \mathcal{A} \) and let \( f \in C\left( {\sigma \left( N\right) }\right) \) . We have \( \sigma \left( {f\left( N\right) }\right) = f\left( {\sigma \left( N\right) }\right) \) . | Proof. Since \( f \mapsto f\left( N\right) \equiv {\gamma }^{-1}\left( f\right) \) is a \( * \) -isomorphism of \( C\left( {\sigma \left( N\right) }\right) \) onto \( {C}^{ * }\left( N\right) \) we have \[ \sigma \left( {f\left( N\right) }\right) = {\sigma }_{C\left( {\sigma \left( N\right) }\right) }\left( f\right) = ... | Yes |
Theorem 5.55. If \( A \in {\mathcal{A}}_{ + } \) and \( n \in \mathbb{N} \), there is a unique \( B \in {\mathcal{A}}_{ + } \) satisfying \( {B}^{n} = A \) . | Proof. We establish existence first. Since by assumption, \( \sigma \left( A\right) \subseteq \lbrack 0,\infty ) \), the real-valued function \( f\left( t\right) = {t}^{1/n} \) is continuous on \( \sigma \left( A\right) \) . Thus \( f\left( A\right) \) is defined by the functional calculus; set \( B = f\left( A\right) ... | Yes |
Proposition 5.57. Let \( \mathcal{H} \) be a Hilbert space. We have the following:\n\n(a) If \( T \in \mathcal{B}\left( \mathcal{H}\right) \) satisfies \( \langle {Th}, h\rangle \geq 0 \) for all \( h \in \mathcal{H} \), then \( T \) is positive, i.e., a positive element of the \( {C}^{ * } \) -algebra \( \mathcal{B}\l... | Proof. In part (a), the hypothesis that \( \langle {Th}, h\rangle \geq 0 \) for all \( h \in \mathcal{H} \) implies that \( T \) is self-adjoint (see Exercise 4.12 in Chapter 4), and so its spectrum is contained in \( \mathbb{R} \) . We want to show further that its spectrum is contained in \( \lbrack 0,\infty ) \) . L... | Yes |
In this example we look at the linear operator on \( {\mathbb{C}}^{3} \) given by the matrix \( A = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right\rbrack \). Note that for any polynomial \( p \) and any vector \( h \in {\mathbb{C}}^{3} \), the first two components of \( p\left( A\... | Fix a column vector \( h = {\left( {h}_{1},{h}_{2},{h}_{3}\right) }^{t} \). First observe that no \( {h}_{j}, j = 1,2,3 \) could be zero if \( h \) is a cyclic vector, since the corresponding component of \( p\left( A\right) h \) would be zero for all polynomials \( p \). Next, the necessarily nonzero vector \( {\left(... | Yes |
Theorem 6.5. Let \( A \) be a normal operator in \( \mathcal{B}\left( \mathcal{H}\right) \), where \( \mathcal{H} \) is a separable Hilbert space. If the \( {C}^{ * } \) -algebra \( {C}^{ * }\left( A\right) \) has a cyclic vector, then the operator \( A \) is unitarily equivalent to a multiplication. | Proof (Theorem 6.5). Fix a cyclic vector \( h \) in \( \mathcal{H} \) . Let \( X = \sigma \left( A\right) \), the spectrum of \( A \) , which is a compact subset of \( \mathbb{C} \) . We know by Theorem 5.46 that there is a unique isometric \( * \) -isomorphism \( \gamma \) between \( {C}^{ * }\left( A\right) \) and \(... | Yes |
Lemma 6.6. Let \( A \) be a normal operator on a separable Hilbert space \( \mathcal{H} \). There is a finite or countable collection of nonzero, pairwise orthogonal subspaces \( {\mathcal{H}}_{n} \) of \( \mathcal{H} \) satisfying\n\n(a) \( \mathcal{H} = {\mathcal{H}}_{1} \oplus {\mathcal{H}}_{2} \oplus \cdots \).\n\n... | Proof. We use a Zorn’s lemma argument. Consider the following family \( \mathcal{F} \). An element of \( \mathcal{F} \) is a collection of nonzero pairwise orthogonal, \( \mathcal{A} \)-invariant closed subspaces \( {\mathcal{H}}_{\alpha } \) of \( \mathcal{H} \), each containing a cyclic vector for \( {C}^{ * }\left( ... | Yes |
Lemma 6.7. If \( {A}_{1},{A}_{2},\ldots \) is a finite or countable collection of operators on, respectively, separable Hilbert spaces \( {\mathcal{H}}_{1},{\mathcal{H}}_{2},\ldots \), with each \( {A}_{n} \) unitarily equivalent to a multiplication and \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{A}_{n}\end{Vmatrix} < ... | Proof. We are given the existence of \( \sigma \) -finite measure spaces \( \left( {{X}_{n},{\mu }_{n}}\right) \), unitary operators \( {W}_{n} : {L}^{2}\left( {{X}_{n},{\mu }_{n}}\right) \rightarrow {\mathcal{H}}_{n} \), and functions \( {f}_{n} \) in \( {L}^{\infty }\left( {{X}_{n},{\mu }_{n}}\right) \) such that\n\n... | Yes |
To find an easy example of a spectral measure, let \( \left( {X,\mathcal{F},\mu }\right) \) be a measure space and set \( \mathcal{H} = {L}^{2}\left( {X,\mu }\right) \) . Define \( E : \mathcal{F} \rightarrow \mathcal{B}\left( \mathcal{H}\right) \) by \( E\left( S\right) = {M}_{{\chi }_{S}} \), the operator of multipli... | The reader is encouraged to check the details verifying properties (1)-(4) in Definition 6.8. | No |
Suppose that \( {M}_{\varphi } \) is a multiplication operator on \( {L}^{2}\left( {X,\mu }\right) \), where \( \left( {X,\mu }\right) \) is a measure space. The function \( E : \mathcal{F} \rightarrow \mathcal{B}\left( {{L}^{2}\left( {X,\mu }\right) }\right) \) given by\n\n\[ E\left( S\right) = {M}_{{\chi }_{{\varphi ... | Clearly conditions (1) and (2) of Definition 6.8 hold. For (3) observe that if \( {S}_{1} \cap {S}_{2} = \varnothing \), then for any \( {h}_{1},{h}_{2} \in {L}^{2}\left( {X,\mu }\right) \) we have\n\n\[ \left\langle {E\left( {S}_{1}\right) {h}_{1}, E\left( {S}_{2}\right) {h}_{2}}\right\rangle = \left\langle {{M}_{{\ch... | Yes |
Suppose \( \left( {X,\mathcal{F}}\right) \) is a measurable space and \( E : \mathcal{F} \rightarrow \mathcal{B}\left( \mathcal{H}\right) \) is a spectral measure. If \( \mathcal{K} \) is another Hilbert space and \( W : \mathcal{H} \rightarrow \mathcal{K} \) is unitary, then the formula\n\n\[ F\left( S\right) = {WE}\l... | The reader is asked to verify the details in Exercise 6.6. | No |
Lemma 6.13. Suppose that \( \mathcal{H} \) is a Hilbert space and that \( {\left\{ {E}_{k}\right\} }_{k = 1}^{\infty } \) is a sequence of orthogonal projections on \( \mathcal{H} \) with \( {E}_{j}\mathcal{H} \bot {E}_{k}\mathcal{H} \) for all \( j \neq k \) . We have\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{E}... | Proof. Let \( {P}_{n} = \mathop{\sum }\limits_{1}^{n}{E}_{k} \) . We want to show that \( {P}_{n}h \rightarrow {Eh} \) for each \( h \in \mathcal{H} \), where \( E \) is defined as in the statement of the lemma. First suppose that \( h \) is in the subspace \( \mathop{\sum }\limits_{1}^{\infty } \oplus {E}_{k}\mathcal{... | Yes |
Proposition 6.14. Let \( E : \mathcal{F} \rightarrow \mathcal{B}\left( \mathcal{H}\right) \) be a spectral measure, as in Definition 6.8. For each \( h \) and \( g \) in \( \mathcal{H} \), define\n\n\[ \n{\mu }_{h, g}\left( S\right) = \langle E\left( S\right) h, g\rangle \n\]\n\n(6.6)\n\nfor \( S \) in \( \mathcal{F} \... | Proof. We first show that \( {\mu }_{h, g} \) is countably additive. Suppose that \( \left\{ {S}_{k}\right\} \) is a sequence of pairwise disjoint sets in \( \mathcal{F} \), and let \( S \) denote their union. We have\n\n\[ \n{\mu }_{h, g}\left( S\right) = \langle E\left( S\right) h, g\rangle = \left\langle {\mathop{\s... | Yes |
For each \( f \in B\left( {X,\mathcal{F}}\right) ,\pi \left( f\right) \) is a normal operator in \( \mathcal{B}\left( \mathcal{H}\right) \) . | We have\n\n\[ \pi \left( f\right) \pi {\left( f\right) }^{ * } = \pi \left( f\right) \pi \left( \bar{f}\right) = \pi \left( {f\bar{f}}\right) = \pi {\left( f\right) }^{ * }\pi \left( f\right) . \] | Yes |
Proposition 6.19. Let \( \left( {X,\mathcal{F}}\right), E \), and \( f \) be as in Proposition 6.18. We have\n\n\[ \ker \left( {\int {fdE}}\right) = E\left( {S}_{0}\right) \mathcal{H} \]\n\nwhere\n\n\[ {S}_{0} = \{ x \in X : f\left( x\right) = 0\} . \] | Proof. Write \( \pi \left( f\right) = \int {fdE} \) . By Proposition 6.18,\n\n\[ \parallel \pi \left( f\right) h{\parallel }^{2} = \int {\left| f\right| }^{2}d{\mu }_{h, h} \]\n\nfor any \( h \in \mathcal{H} \) . Thus\n\n\[ h \in \ker \pi \left( f\right) \Leftrightarrow \int {\left| f\right| }^{2}d{\mu }_{h, h} = 0 \Le... | Yes |
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