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Theorem 7.10. If \( A \in \Phi \left( {X, Y}\right) \) and \( B \) is \( A \) -compact, then \( A + B \in \Phi \left( {X, Y}\right) \) and \( i\left( {A + B}\right) = i\left( A\right) \) . | Proof. By Corollary 7.6, \( A \in \Phi \left( {W, Y}\right) \) . Since \( B \in K\left( {W, Y}\right) \), we see that \( A + B \in \Phi \left( {W, Y}\right) \) (Theorem 7.8). We now apply Lemma 7.7 to conclude that \( A + B \in \Phi \left( {X, Y}\right) \) . | No |
Theorem 7.11. If \( A \in \Phi \left( {X, Y}\right) \), then there is an \( \eta > 0 \) such that for every linear operator \( B \) from \( X \) to \( Y \) satisfying \( D\left( B\right) \supset D\left( A\right) \) and \[ \parallel {Bx}\parallel \leq \eta \left( {\parallel x\parallel + \parallel {Ax}\parallel }\right) ... | Proof. We introduce \( W \) as before and apply the same proof as in Theorem 7.9. | No |
Theorem 7.12. If \( A \in \Phi \left( {X, Y}\right) \) and \( B \) is a densely defined closed linear operator from \( Y \) to \( Z \) such that \( {BA} \in \Phi \left( {X, Z}\right) \), then \( B \in \Phi \left( {Y, Z}\right) \) . | Proof. Here we must exercise a bit of care. By Lemma 7.4, there is a finite-dimensional subspace \( {Y}_{0} \subset D\left( B\right) \) such that (7.5) holds. Let \( {A}_{0} \) be an operator given by Theorem 7.1. Because \( {Y}_{0} \subset D\left( B\right), A{A}_{0} \) maps \( D\left( B\right) \) into itself. Hence, \... | Yes |
Theorem 7.13. If \( B \in \Phi \left( {Y, Z}\right) \) and \( A \) is a densely defined closed linear operator from \( X \) to \( Y \) such that \( {BA} \in \Phi \left( {X, Z}\right) \), then the restriction of \( A \) to \( D\left( {BA}\right) \) is in \( \Phi \left( {X, V}\right) \), where \( V \) is \( D\left( B\rig... | Proof. By Theorem 7.1, there is an operator \( {B}_{0} \in B\left( {Z, Y}\right) \) such that\n\n\[ \n{B}_{0}B = I - {F}_{3}\text{ on }D\left( B\right) , \n\]\n\nwhere \( {F}_{3} \in B\left( Y\right) \) and \( R\left( {F}_{3}\right) = N\left( B\right) \) . Thus,\n\n\[ \n{B}_{0}{BA} = A - {F}_{3}A\text{ on }D\left( {BA}... | Yes |
Theorem 7.14. Let \( A \) be a densely defined closed linear operator from \( X \) to \( Y \) . Suppose \( B \) is in \( B\left( {Y, Z}\right) \) with \( \alpha \left( B\right) < \infty \) and \( {BA} \in \Phi \left( {X, Z}\right) \) . Then \( A \in \Phi \left( {X, Y}\right) \) . | Proof. We have \( R\left( B\right) \supset R\left( {BA}\right) \), which is closed and such that \( R{\left( BA\right) }^{ \circ } \) is finite-dimensional. Hence, \( R\left( B\right) \) is closed (Lemmas 5.3 and 5.6), and \( R{\left( B\right) }^{ \circ } \subset R{\left( BA\right) }^{ \circ } \) is finite-dimensional.... | Yes |
Theorem 7.16. If \( A \) is a densely defined closed linear operator from \( X \) to \( Y \) and \( R\left( {A}^{\prime }\right) \) is closed in \( {X}^{\prime } \), then \( R\left( A\right) = {}^{ \circ }N\left( {A}^{\prime }\right) \), and hence, is closed in \( Y \) . | The proof of Theorem 7.16 is a bit involved and requires a few steps. We begin by first showing that an adjoint operator is always closed. If \( {y}_{n}^{\prime } \in D\left( {A}^{\prime }\right) \) and \( {y}_{n}^{\prime } \rightarrow {y}^{\prime } \) in \( {Y}^{\prime } \) with \( {A}^{\prime }{y}_{n}^{\prime } \righ... | Yes |
Lemma 7.18. \( p\left( x\right) \) has the following properties:\n\n\[ \n\text{(a)}p\left( {x + y}\right) \leq p\left( x\right) + p\left( y\right) ,\;x, y \in X\text{;} \n\]\n\n\[ \n\text{(b)}p\left( {\alpha x}\right) = {\alpha p}\left( x\right) ,\;x \in X,\alpha > 0\text{;} \n\]\n\n(c) \( p\left( x\right) < 1 \) impli... | Proof. (a) Suppose \( {\alpha x} \in U \) and \( {\beta y} \in U \), where \( \alpha > 0 \) and \( \beta > 0 \) . Since \( U \) is convex,\n\n\[ \n\frac{{\alpha }^{-1}{\alpha x} + {\beta }^{-1}{\beta y}}{{\alpha }^{-1} + {\beta }^{-1}} = \frac{x + y}{{\alpha }^{-1} + {\beta }^{-1}} \n\]\n\nis in \( U \) . Hence,\n\n\[ ... | Yes |
Theorem 7.19. If \( A \) is in \( B\left( {X, Y}\right) \), then \( A \in \Phi \left( {X, Y}\right) \) if and only if \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^{\prime }}\right) \) | Proof. That \( A \in \Phi \left( {X, Y}\right) \) implies that \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^{\prime }}\right) \) is Theorem 5.15. If \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^{\prime }}\right) \), then \( \beta \left( A\right) = \alpha \left( {A}^{\prime }\right) < \infty \) and \( \alp... | Yes |
Theorem 7.20. Let \( A \) be a closed linear operator from \( X \) to \( Y \) with \( D\left( A\right) \) dense in \( X \) . Then \( D\left( {A}^{\prime }\right) \) is total in \( {Y}^{\prime } \) . | We shall give the simple proof of Theorem 7.20 at the end of the section. | No |
Theorem 7.23. Let \( A \) be a closed linear operator from \( X \) to \( Y \) with \( D\left( A\right) \) dense in \( X \) . If \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^{\prime }}\right) \), then \( A \in \Phi \left( {X, Y}\right) \) with \( i\left( A\right) = - i\left( {A}^{\prime }\right) \) . | Proof. Clearly, \( \beta \left( A\right) = \alpha \left( {A}^{\prime }\right) < \infty \), and by (5.24), we have\n\n\[ \alpha \left( A\right) \leq \beta \left( {A}^{\prime }\right) < \infty \]\n\nWe have just shown that \( R\left( A\right) \) is closed (Theorem 7.16). Thus \( A \in \Phi \left( {X, Y}\right) \) . Now w... | Yes |
Lemma 7.24. If \( X \) is a Banach space and \( A \) is closed, then \( \lambda \in \rho \left( A\right) \) if and only if\n\n(7.28)\n\n\[ \alpha \left( {A - \lambda }\right) = 0,\;R\left( {A - \lambda }\right) = X. \] | Proof. The \ | No |
Theorem 7.25. The set \( {\Phi }_{A} \) is open, and \( i\left( {A - \lambda }\right) \) is constant on each of its components. | Proof. That \( {\Phi }_{A} \) is an open set follows as in the proof of Theorem 6.3 (we use Theorem 7.9 here). To show that the index is constant on each component, let \( {\lambda }_{1},{\lambda }_{2} \) be any two points in \( {\Phi }_{A} \) which are connected by a smooth curve \( C \) whose points are all in \( {\P... | Yes |
Theorem 7.26. \( {\Phi }_{A + K} = {\Phi }_{A} \) for all \( K \) which are \( A \) -compact, and \( i(A + K - \lambda ) = i\left( {A - \lambda }\right) \) for all \( \lambda \in {\Phi }_{A} \) . | Proof. The proof is an immediate consequence of Theorem 7.10. | No |
Theorem 7.27. \( \lambda \notin {\sigma }_{e}\left( A\right) \) if and only if \( \lambda \in {\Phi }_{A} \) and \( i\left( {A - \lambda }\right) = 0 \) . | Proof. If \( \lambda \notin {\sigma }_{e}\left( A\right) \), then there is a \( K \in K\left( X\right) \) such that \( \lambda \in \rho \left( {A + K}\right) \) . In particular, \( \lambda \in {\Phi }_{A + K} \) and \( i\left( {A + K - \lambda }\right) = 0 \) (Lemma 7.24). Adding the operator \( - K \) to \( A + K \), ... | Yes |
Theorem 7.29. Let \( A \) be a closed linear operator from \( X \) to \( Y \) with domain \( D\left( A\right) \) dense in \( X \) . Then \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) if and only if there is a seminorm \( \left| \cdot \right| \) defined on \( D\left( A\right) \), which is compact relative to the graph n... | Proof. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \), then we can write\n\n(7.33)\n\n\[ X = N\left( A\right) \oplus {X}_{0} \]\n\nwhere \( {X}_{0} \) is a closed subspace of \( X \) (Lemma 5.1). Let \( P \) be the projection of \( X \) onto \( N\left( A\right) \) along \( {X}_{0} \), i.e., the operator defined by\n\... | Yes |
Theorem 7.30. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( B \) is \( A \) -compact, then \( A + B \in \) \( {\Phi }_{ + }\left( {X, Y}\right) \) . | Proof. By Theorem 7.29,\n\n\[ \parallel x\parallel \leq C\parallel \left( {A + B}\right) x\parallel + \left| x\right| + C\parallel {Bx}\parallel ,\;x \in D\left( A\right) .\n\]\n\nSet\n\n\[ {\left| x\right| }_{0} = \left| x\right| + C\parallel {Bx}\parallel .\n\]\n\nThen \( {\left. \mid \cdot \right| }_{0} \) is a semi... | Yes |
Theorem 7.31. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \), then there is an \( \eta > 0 \) such that \( A + B \in \) \( {\Phi }_{ + }\left( {X, Y}\right) \) and\n\n\[ \alpha \left( {A + B}\right) \leq \alpha \left( A\right) \]\n\nfor each linear operator \( B \) from \( X \) to \( Y \) with \( D\left( B\right) \su... | Proof. By (7.35),\n\n\[ \parallel x\parallel + \parallel {Ax}\parallel \leq \left( {C + 1}\right) \parallel \left( {A + B}\right) x\parallel + \parallel {Px}\parallel + \left( {C + 1}\right) \parallel {Bx}\parallel \]\n\n\[ \leq \left( {C + 1}\right) \parallel \left( {A + B}\right) x\parallel + \parallel {Px}\parallel ... | Yes |
Theorem 7.32. If \( A \in {\Phi }_{ + }\left( {X, Y}\right), B \in {\Phi }_{ + }\left( {Y, Z}\right) \) and \( D\left( {BA}\right) \) is dense in \( X \), then \( {BA} \in {\Phi }_{ + }\left( {X, Z}\right) \) . | Proof. That \( {BA} \) is a closed operator follows as in the proof of Theorem 7.3 (in fact, there we used only the facts that \( \alpha \left( B\right) < \infty, R\left( B\right) \) is closed, and that \( A \) and \( B \) are closed). To prove the rest, we note that\n\n\[ \parallel x\parallel \leq {C}_{1}\parallel {Ax... | Yes |
Lemma 7.33. \( A \notin {\Phi }_{ + } \) if and only if there is a bounded sequence \( \left\{ {u}_{k}\right\} \subset \) \( D\left( A\right) \) having no convergent subsequence such that \( \left\{ {A{u}_{k}}\right\} \) converges. | Proof. Suppose \( A \notin {\Phi }_{ + } \) . If \( \alpha \left( A\right) = \infty \), then there is a bounded sequence in \( N\left( A\right) \) having no convergent subsequence. If \( \alpha \left( A\right) < \infty \), then \( R\left( A\right) \) is not closed. Let \( P \) be a bounded projection onto \( N\left( A\... | Yes |
Theorem 7.35. Let \( X, Y, Z \) be Banach spaces, and assume that \( A \) is a densely defined, closed linear operator from \( X \) to \( Y \) such that \( R\left( A\right) \) is closed in \( Y \) and \( \beta \left( A\right) < \infty \) (i.e., \( A \in {\Phi }_{ - }\left( {X, Y}\right) \) ). Let \( B \) be a densely d... | That \( D\left( {BA}\right) \) is dense in \( X \) follows from | No |
Lemma 7.36. If \( A, B \) satisfy the hypotheses of Theorem 7.35 and \( x \) is any element in \( D\left( A\right) \), then there is a sequence \( \left\{ {x}_{k}\right\} \subset D\left( {BA}\right) \) such that \( A{x}_{k} \rightarrow {Ax} \) in \( Y \) and \( {x}_{k} \rightarrow x \) in \( X \) . Consequently, \( D\l... | Proof. Since \( D\left( B\right) \) is dense in \( Y \), we see by Lemma 7.4 that\n\n(7.46)\n\n\[ Y = R\left( A\right) \oplus {Y}_{0} \]\n\nwhere \( {Y}_{0} \subset D\left( B\right) \) . Let \( x \) be any element in \( D\left( A\right) \) . Then there is a sequence \( \left\{ {y}_{k}\right\} \subset D\left( B\right) \... | Yes |
Theorem 8.2. Every closed subspace of a reflexive Banach space is reflexive. | Proof. Let \( Z \) be a closed subspace of a reflexive space \( X \) . Then \( Z \) is a Banach space. Let \( {z}^{\prime \prime } \) be any element of \( {Z}^{\prime \prime } \) . For any \( {x}^{\prime } \in {X}^{\prime } \), the restriction \( {x}_{r}^{\prime } \) of \( {x}^{\prime } \) to \( Z \) is an element of \... | Yes |
Theorem 8.3. A subspace \( W \) of \( {X}^{\prime } \) is saturated if and only if \( W = {M}^{ \circ } \) for some subset \( M \) of \( X \) . | Proof. If \( W \) is saturated, set \( M = {}^{ \circ }W \) . Then clearly, \( W \subset {M}^{ \circ } \) . Now suppose \( {x}^{\prime } \notin W \) . Then there is an \( x \in {}^{ \circ }W = M \) such that \( {x}^{\prime }\left( x\right) \neq 0 \) . Therefore, \( {x}^{\prime } \) is not in \( {M}^{ \circ } \) . This ... | Yes |
Theorem 8.5. A subspace \( W \) of \( {X}^{\prime } \) is saturated if and only if it is weak* closed. | Proof. Suppose \( W \) is saturated, and let \( {x}^{\prime } \in {X}^{\prime } \) be such that for each \( x \in X \), there is a \( \left\{ {x}_{k}^{\prime }\right\} \subset W \) satisfying (8.8). If \( {x}^{\prime } \notin W \), then there is an \( x \in {}^{ \circ }W \) such that \( {x}^{\prime }\left( x\right) \ne... | Yes |
Theorem 8.6. A finite-dimensional subspace of \( {X}^{\prime } \) is always saturated. | Proof. Suppose that \( {x}_{1}^{\prime },\cdots ,{x}_{n}^{\prime } \) form a basis for \( W \subset {X}^{\prime } \) and assume that \( {x}_{0}^{\prime } \notin W \) . Then the functionals \( {x}_{0}^{\prime },{x}_{1}^{\prime },\cdots ,{x}_{n}^{\prime } \) are linearly independent. By Lemma 4.14, there are elements\n\n... | Yes |
Lemma 8.8. Let \( X \) be a normed vector space, and let \( {x}^{\prime } \) be an element of \( {X}^{\prime } \) . Let \( M \) be the set of those \( x \) in \( X \) such that \( {x}^{\prime }\left( x\right) = 0 \) (i.e., \( M = {}^{ \circ }\left\lbrack {x}^{\prime }\right\rbrack \) ). Let \( y \) be any element not i... | Proof. Clearly, \( N \cap M = \{ 0\} \) . Moreover, for any \( x \in X \), set\n\n(8.11)\n\n\[ z = x - \frac{{x}^{\prime }\left( x\right) }{{x}^{\prime }\left( y\right) }y. \]\n\nThen \( {x}^{\prime }\left( z\right) = 0 \), showing that \( z \in M \) . Since \( x = z + {\alpha y} \), the proof is complete. | Yes |
Lemma 8.10. If \( \dim X = n < \infty \), then \( \dim {X}^{\prime } = n \) . | Proof. Let \( {x}_{1},\cdots ,{x}_{n} \) be a basis for \( X \) . Then there are functionals \( {x}_{1}^{\prime },\cdots \) , \( {x}_{n}^{\prime } \) in \( {X}^{\prime } \) such that\n\n(8.13)\n\n\[ \n{x}_{j}^{\prime }\left( {x}_{k}\right) = {\delta }_{jk},\;1 \leq j, k \leq n.\n\]\n\nIf \( x \in X \), then\n\n\[ \nx =... | Yes |
Theorem 8.11. If \( {X}^{\prime } \) is separable, so is \( X \) . | Proof. Let \( \left\{ {x}_{n}^{\prime }\right\} \) be a dense set in \( {X}^{\prime } \) . For each \( n \), there is an \( {x}_{n} \in X \) such that \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} = 1 \) and\n\n\[ \left| {{x}_{n}^{\prime }\left( {x}_{n}\right) }\right| \geq \begin{Vmatrix}{x}_{n}^{\prime }\end{Vmatrix}/2 \]\n... | Yes |
Corollary 8.12. If \( X \) is reflexive and separable, then so is \( {X}^{\prime } \) . | Proof. Let \( \left\{ {x}_{k}\right\} \) be a sequence which is dense in \( X \), and let \( {x}^{\prime \prime } \) be any element of \( {X}^{\prime \prime } \) . Then there is an \( x \in X \) such that \( {Jx} = {x}^{\prime \prime } \) . Moreover, for any \( \varepsilon > 0 \), there is an \( {x}_{k} \) such that \(... | Yes |
Theorem 8.13. If \( X \) is separable, then every bounded sequence in \( {X}^{\prime } \) has a weak* convergent subsequence. | Proof. Let \( \left\{ {x}_{n}^{\prime }\right\} \) be a bounded sequence in \( {X}^{\prime } \), and let \( \left\{ {x}_{k}\right\} \) be a sequence dense in \( X \) . Now \( {x}_{n}^{\prime }\left( {x}_{1}\right) \) is a bounded sequence of scalars, and hence, it contains a convergent subsequence. Thus, there is a sub... | Yes |
Theorem 8.14. Every subspace of a separable space is separable. | Proof. Let \( M \) be a subspace of a separable space \( X \), and let \( \left\{ {x}_{k}\right\} \) be a dense sequence in \( X \) . For each pair of integers \( j, k \), we pick an element \( {x}_{jk} \in M \) , if there is one, such that\n\n\[ \begin{Vmatrix}{{x}_{jk} - {x}_{k}}\end{Vmatrix} < 1/j \]\n\nIf there is ... | Yes |
Lemma 8.15. A weakly convergent sequence is necessarily bounded. | Proof. Consider the sequence \( \left\{ {J{x}_{k}}\right\} \) of elements of \( {X}^{\prime \prime } \) . For each \( {x}^{\prime } \) we have\n\n\[ \mathop{\sup }\limits_{k}\left| {J{x}_{k}\left( {x}^{\prime }\right) }\right| < \infty \]\n\nThen by the Banach-Steinhaus theorem (Theorem 3.17), there is a constant \( C ... | Yes |
Theorem 8.16. If \( X \) is reflexive, then every bounded sequence has a weakly convergent subsequence. | Proof. Suppose \( X \) is reflexive, and let \( \left\{ {x}_{n}\right\} \) be a bounded sequence in \( X \) . Set \( M = \overline{\left\lbrack \left\{ {x}_{n}\right\} \right\rbrack } \), the closure of the set of linear combinations of the \( {x}_{n} \) . As we observed before, \( M \) is separable (see the proof of T... | Yes |
Theorem 8.17. If \( X \) is finite-dimensional, then a sequence converges weakly if and only if it converges in norm. | Proof. Let \( \left\{ {x}_{k}\right\} \) be a sequence that is weakly convergent to \( x \) . Since \( X \) is finite-dimensional, so is \( {X}^{\prime } \) (Lemma 8.10). Let \( {x}_{1}^{\prime },\cdots ,{x}_{n}^{\prime } \) be a basis for \( {X}^{\prime } \) . Then every \( {x}^{\prime } \in {X}^{\prime } \) can be pu... | Yes |
Theorem 8.18. If \( X \) is a Banach space such that every total subspace of \( {X}^{\prime } \) is dense in \( {X}^{\prime } \), then \( X \) is reflexive. | Proof. If \( X \) were not reflexive, there would be an \( {x}_{0}^{\prime \prime } \in {X}^{\prime \prime } \) which is not in \( R\left( J\right) \) . Let \( W \) be the set of those \( {x}^{\prime } \in {X}^{\prime } \) which annihilate \( {x}_{0}^{\prime \prime } \) . Since \( {x}_{0}^{\prime \prime } \neq 0, W \) ... | Yes |
Theorem 9.6. If \( B \) is nontrivial, then for each \( a \) in \( B,\sigma \left( a\right) \) is not empty. | Proof. Suppose \( a \neq 0 \) and \( \rho \left( a\right) \) is the whole complex plane. Let \( {a}^{\prime } \neq 0 \) be any element of \( {B}^{\prime } \) (the dual space of \( B \) considered as a Banach space). Since the series in (9.12) converges in norm for \( \left| {\lambda - \mu }\right| \cdot \begin{Vmatrix}... | Yes |
Theorem 9.8. \( \rho \left( \left\lbrack A\right\rbrack \right) = {\Phi }_{A} \) . | Proof. By considering \( A + \lambda \) in place of \( A \), it suffices to prove that \( A \in \Phi \left( X\right) \) if and only if \( \left\lbrack A\right\rbrack \) is a regular element of \( C \) . If \( \left\lbrack A\right\rbrack \) is a regular element of \( C \) , then there is an \( {A}_{0} \in B\left( X\righ... | Yes |
Theorem 9.9. If there is an operator \( A \) in \( B\left( X\right) \) such that \( {\Phi }_{A} \) consists of the whole complex plane, then \( X \) is finite dimensional. | Proof. If \( {\Phi }_{A} \) is the whole complex plane, then so is \( \rho \left( \left\lbrack A\right\rbrack \right) \) (Theorem 9.8). By Theorem 9.6, \( C \) must be a trivial Banach algebra, i.e., \( \left\lbrack I\right\rbrack = \left\lbrack 0\right\rbrack \) . This means that the identity \( I \) is a compact oper... | Yes |
A complex number \( \lambda \) is in \( \sigma \left( a\right) \) if and only if there is a multiplicative linear functional \( m \) on \( B \) such that \( m\left( a\right) = \lambda \) . | This theorem is easy to prove in one direction. In fact, if \( \lambda \in \rho \left( A\right) \), then there is a \( b \in B \) such that\n\n(9.28)\n\n\[ b\left( {a - {\lambda e}}\right) = e. \]\n\nThen for any \( m \in M \) ,\n\n(9.29)\n\n\[ m\left( b\right) \left( {m\left( a\right) - {\lambda m}\left( e\right) }\ri... | No |
Theorem 9.12. If \( H \neq B \) is an ideal in \( B \), then there is an \( m \) in \( M \) such that \( m \) vanishes on \( H \) . | Proof. By Theorem 9.11, there is a maximal ideal \( N \) containing \( H \) . Thus, if \( a \in B \), then \( a = {a}_{1} + {\lambda e} \), where \( {a}_{1} \in N \) . Define \( m\left( a\right) \) to be \( \lambda \) . Clearly, \( m \) is a linear functional on \( B \) . It is also multiplicative. This follows from th... | Yes |
Theorem 9.13. A scalar \( \mu \) is in \( \sigma \left\lbrack {P\left( {{a}_{1},\cdots ,{a}_{n}}\right) }\right\rbrack \) if and only if there is a vector \( \left( {{\lambda }_{1},\cdots ,{\lambda }_{n}}\right) \) in \( \sigma \left( {{a}_{1},\cdots ,{a}_{n}}\right) \) such that \( \mu = P\left( {{\lambda }_{1},\cdots... | Proof. By Theorem 9.10, \( \mu \in \sigma \left\lbrack {P\left( {{a}_{1},\cdots ,{a}_{n}}\right) }\right\rbrack \) if and only if there is an \( m \in M \) such that\n\n\[ m\left\lbrack {P\left( {{a}_{1},\cdots ,{a}_{n}}\right) }\right\rbrack = \mu . \]\n\nBut\n\n\[ m\left\lbrack {P\left( {{a}_{1},\cdots ,{a}_{n}}\righ... | Yes |
Lemma 9.14. Maximal ideals are closed. | Proof. Let \( N \) be a maximal ideal in \( B \) . Then\n\n(9.38)\n\n\[ B = N \oplus \{ e\} \]\n\nLet \( \left\{ {a}_{n}\right\} \) be a sequence of elements in \( N \) which approach an element \( a \) in B. Now\n\n(9.39)\n\n\[ a = {a}_{1} + {\lambda e} \]\n\nwhere \( {a}_{1} \in N \) . Suppose \( \lambda \neq 0 \) . ... | Yes |
Theorem 9.15. If \( m \) is in \( M \), then \( m \) is bounded and \( \parallel m\parallel \leq 1 \) . | Proof. Let \( N \) be the set of those \( x \in B \) such that \( m\left( x\right) = 0 \) . Then \( N \) is a maximal ideal (see Section 9.3). If the inequality\n\n(9.40)\n\n\[ \left| {m\left( a\right) }\right| \leq \parallel a\parallel ,\;a \in B \]\n\nwere not true, there would exist an \( a \in B \) such that \( \le... | Yes |
Theorem 9.16. An ideal \( N \) is maximal if and only if the only ideal \( L \) satisfying \( B \neq L \supset N \) is \( L = N \) . | Proof. Suppose \( N \) is maximal and that \( L \) is an ideal satisfying \( B \neq L \supset N \) . Let \( a \) be any element in \( L \) . By the definition of a maximal ideal, \( a = {a}_{1} + {\lambda e} \) , where \( {a}_{1} \in N \) . Since \( a \) and \( {a}_{1} \) are both in \( L \), so is \( {\lambda e} \) . ... | Yes |
Lemma 9.18. \( E \) is in \( R\left( X\right) \) if and only if \( I + {\lambda E} \in \Phi \left( X\right) \) for all scalars \( \lambda \) . | Proof. If \( E \in R\left( X\right) \), the statement is true for \( \lambda = 0 \) . Otherwise, \( E + I/\lambda \in \) \( \Phi \left( X\right) \) . Hence, \( I + {\lambda E} \in \Phi \left( X\right) \) . Conversely, if \( \mu \neq 0 \), then \( \mu \left( {I + E/\mu }\right) \in \Phi \left( X\right) \) , showing that... | No |
Lemma 9.20. \( E \) is in \( R\left( X\right) \) if and only if \( {\begin{Vmatrix}{\left\lbrack E\right\rbrack }^{n}\end{Vmatrix}}^{1/n} \rightarrow 0 \) as \( n \rightarrow \infty \) . | Proof. This follows from the fact that \( E \in R\left( X\right) \) if and only if \( \lambda \in {\Phi }_{E} \) for all scalars \( \lambda \neq 0 \) . By Theorem 9.8 this is true if and only if \( \lambda \in \rho \left( \left\lbrack E\right\rbrack \right) \) for all \( \rho \neq 0 \) . We now apply Theorem 6.13 to co... | No |
Lemma 9.21. If \( E \) is in \( R\left( X\right) \) and \( K \) is in \( K\left( X\right) \), then \( E + K \) is in \( R\left( X\right) \) . | Proof. We have \( \left\lbrack {E + K - \lambda }\right\rbrack = \left\lbrack {E - \lambda }\right\rbrack \) . | No |
Lemma 9.22. If \( E \) is in \( R\left( X\right), B \) is in \( B\left( X\right) \) and \( B \smile E \), then \( {EB} \) and \( {BE} \) are in \( R\left( X\right) \) . | Proof. We note that \( \parallel {\left\lbrack EB\right\rbrack }^{n}{\parallel }^{1/n} = {\begin{Vmatrix}{\left\lbrack B\right\rbrack }^{n}{\left\lbrack E\right\rbrack }^{n}\end{Vmatrix}}^{1/n} \leq \parallel \left\lbrack B\right\rbrack \parallel \cdot \parallel {\left\lbrack E\right\rbrack }^{n}{\parallel }^{1/n} \rig... | Yes |
Lemma 9.23. If \( A \in \Phi \left( X\right) \), then there is an \( {A}_{0} \in \Phi \left( X\right) \) such that\n\n(9.45)\n\n\[ \left\lbrack {{A}_{0}A}\right\rbrack = \left\lbrack {A{A}_{0}}\right\rbrack = \left\lbrack I\right\rbrack \] | Proof. This follows from Theorem 5.4. | No |
Lemma 9.24. If \( E \in R\left( X\right), A \in \Phi \left( X\right) \) and \( A \smile E \), then \( {A}_{0} + E \in \Phi \left( X\right) \) . | Proof. We have \( \left\lbrack {A\left( {E + {A}_{0}}\right) }\right\rbrack = \left\lbrack {\left( {E + {A}_{0}}\right) A}\right\rbrack = \left\lbrack {{EA} + I}\right\rbrack \) . Since \( {EA} \in \) \( R\left( X\right) \) (Lemma 9.22), \( {EA} + I \in \Phi \left( X\right) \), and \( \left\lbrack {{EA} + I}\right\rbra... | Yes |
Lemma 9.25. If \( A \in \Phi \left( X\right), E \in R\left( X\right) \), and \( A \smile E \), then \( {A}_{0} \smile E \) . | Proof. We have \( \left\lbrack {{A}_{0}E}\right\rbrack = \left\lbrack {{A}_{0}{EA}{A}_{0}}\right\rbrack = \left\lbrack {{A}_{0}{AE}{A}_{0}}\right\rbrack = \left\lbrack {E{A}_{0}}\right\rbrack \) . | Yes |
Theorem 9.26. If \( A \in \Phi \left( X\right), E \in R\left( X\right) \) and \( A \smile E \), then \( A + E \in \Phi \left( X\right) \) . | Proof. We have \( {A}_{0} \in \Phi \left( X\right) \) and \( {A}_{0} \smile E \) (Lemmas 9.23 and 9.25). Thus \( A + E \in \Phi \left( X\right) \) (Lemma 9.24). | Yes |
Lemma 9.27. Suppose \( A \in \Phi \left( X\right) \) and \( E \in B\left( X\right) \) . Then \( {\lambda E} + A \in \Phi \left( X\right) \) for all \( \lambda \) if and only if \( E{A}_{0} \in R\left( X\right) \) . | Proof. If \( {\lambda E} + A \in \Phi \left( X\right) \), then \( \left\lbrack {\left( {{\lambda E} + A}\right) {A}_{0}}\right\rbrack = \left\lbrack {{A}_{0}\left( {{\lambda E} + A}\right) }\right\rbrack = \left\lbrack {{\lambda E}{A}_{0} + I}\right\rbrack \) is invertible in \( C \) . Hence, \( E{A}_{0} \in R\left( X\... | Yes |
Lemma 9.28. Suppose \( A \in \Phi \left( X\right) \) and \( E \in B\left( X\right) \) . Then \( {EA} \in R\left( X\right) \Leftrightarrow \) \( {AE} \in R\left( X\right) \) . | Proof. If \( {EA} \in R\left( X\right) \), then \( {\lambda EA} + I \in \Phi \left( X\right) \) for all \( \lambda \) . Hence, so is \( {\lambda E} + {A}_{0} \) , and consequently, so is \( {\lambda AE} + I \) . Therefore, \( {AE} \in R\left( X\right) \) . | No |
Theorem 9.29. The operator \( E \) in \( B\left( X\right) \) is in \( R\left( X\right) \) if and only if \( A + E \in \) \( \Phi \left( X\right) \) for all \( A \in \Phi \left( X\right) \) such that \( A \smile E \) . | Proof. By Theorem 9.26 we need only show the \ | No |
Theorem 9.30. If \( {E}_{1},{E}_{2} \in R\left( X\right) \) and \( {E}_{1} \smile {E}_{2} \), then \( {E}_{1} + {E}_{2} \in R\left( X\right) \) . | Proof. If \( \lambda \neq 0 \), then \( \lambda + {E}_{1} \in \Phi \left( X\right) \) . By Theorem 9.26, so is \( \lambda + {E}_{1} + {E}_{2} \) . Thus, \( {E}_{1} + {E}_{2} \in R\left( X\right) \) . | No |
Lemma 9.31. The operator \( E \) is in \( F\left( X\right) \) if and only if \( I + {AE} \in \Phi \) for all \( A \in \Phi \) . | Proof. Use Lemma 9.18. | No |
Theorem 9.32. \( E \) is in \( F\left( X\right) \) if and only if \( A + E \in \Phi \) for all \( A \in \Phi \) . Thus \( F\left( X\right) \) coincides with the set of Fredholm perturbations. | Proof. If \( E \in F\left( X\right) \) and \( A \in \Phi \), then \( {A}_{0}E \in R\left( X\right) \) (Lemma 9.18). Thus, \( \left( {I + {A}_{0}E}\right) \in \Phi \) (Lemma 9.18). Consequently, \( A\left( {I + {A}_{0}E}\right) \in \Phi \), showing that \( \left\lbrack {A + E}\right\rbrack \) is invertible in \( C \) . ... | Yes |
Lemma 9.34. For each \( B \) in \( B\left( X\right) \), there are operators \( {A}_{1},{A}_{2} \in \Phi \) such that \( B = {A}_{1} + {A}_{2} \) . | Proof. For \( \lambda \) sufficiently large, \( {A}_{1} = \lambda + B \) is invertible (Lemma 6.5). Take \( {A}_{2} = - {\lambda I} \) | No |
Corollary 9.35. If \( E \) is in \( F\left( X\right) \), then \( {BE} \) is in \( F\left( X\right) \) for all \( B \) in \( B\left( X\right) \) . | Proof. By Lemma 9.34, each \( B \in B\left( X\right) \) can be written in the form \( B = \) \( {A}_{1} + {A}_{2} \), where \( {A}_{j} \in \Phi \) . If \( A \) is any operator in \( \Phi \), then \( A{A}_{j}E \in R\left( X\right) \) . Thus, \( {A}_{j}E \in F\left( X\right) \) . Consequently, \( {BE} = {A}_{1}E + {A}_{2... | Yes |
Corollary 9.36. If \( E \in F\left( X\right) \), then \( {EA} \in R\left( X\right) \) for all \( A \in \Phi \) . | Proof. Use Lemma 9.28. | No |
Corollary 9.37. If \( E \) is in \( F\left( X\right) \), then \( {EB} \) is in \( F\left( X\right) \) for all \( B \) in \( B\left( X\right) \) . | Proof. Use Corollary 9.35. | No |
Corollary 9.38. If \( {E}_{n} \in F\left( X\right) \) and \( {E}_{n} \rightarrow E \) in \( B\left( X\right) \), then \( E \in F\left( X\right) \) . | Proof. If \( A \in \Phi \), we can take \( n \) so large that \( A - \left( {{E}_{n} - E}\right) \in \Phi \) (Theorem 5.11). Hence, \( A - \left( {{E}_{n} - E}\right) + {E}_{n} \in \Phi \) (Theorem 9.32). This shows that \( E \in F\left( X\right) \) . | No |
Lemma 9.40. If \( P \) is a projection in \( B\left( X\right) \) such that \( \dim R\left( P\right) < \infty \), then there is a constant \( C \) such that\n\n\[ \parallel x\parallel \leq {Cd}\left( {x, R\left( P\right) }\right) ,\;x \in N\left( P\right) . \]\n | Proof. Otherwise, there would be a sequence \( \left\{ {x}_{k}\right\} \in N\left( P\right) \) such that\n\n\[ \begin{Vmatrix}{x}_{k}\end{Vmatrix} = 1,\;d\left( {{x}_{k}, R\left( P\right) }\right) \rightarrow 0\text{ as }k \rightarrow \infty . \]\n\nHence, there is a sequence \( \left\{ {z}_{k}\right\} \in R\left( P\ri... | Yes |
Theorem 9.41. \( A \in B\left( X\right) \) is in \( {\Phi }_{ + } \) if and only if there are a projection \( P \) in \( B\left( X\right) \) with \( \dim R\left( P\right) < \infty \) and a constant \( C \) such that\n\n\[ d\left( {x, R\left( P\right) }\right) \leq C\parallel {Ax}\parallel ,\;x \in N\left( P\right) . \] | Proof. Assume \( A \in {\Phi }_{ + } \) . By Lemmas 5.1 and 5.2, there is a projection \( P \in B\left( X\right) \) such that \( R\left( P\right) = N\left( A\right) \) . Since \( R\left( A\right) \) is closed, there is a constant \( C \) such that\n\n\[ d\left( {x, N\left( A\right) }\right) \leq C\parallel {Ax}\paralle... | Yes |
Theorem 9.42. If \( A \) is not in \( {\Phi }_{ + } \), then there are sequences \( \left\{ {x}_{k}\right\} \subset X \) , \( \left\{ {x}_{k}^{\prime }\right\} \subset {X}^{\prime } \) such that\n\n(9.47)\n\n\[ \n{x}_{j}^{\prime }\left( {x}_{k}\right) = {\delta }_{jk},\;\begin{Vmatrix}{x}_{k}^{\prime }\end{Vmatrix} \cd... | Proof. For \( k \geq 1 \), assume that \( {x}_{1},\cdots ,{x}_{k - 1},{x}_{1}^{\prime },\cdots ,{x}_{k - 1}^{\prime } \) have been found, and set\n\n\[ \n{Px} = \mathop{\sum }\limits_{{j = 1}}^{{k - 1}}{x}_{j}^{\prime }\left( x\right) {x}_{j},\;x \in X\n\]\n\nwhen \( k > 1 \), and \( P = 0 \), otherwise. Then \( P \) i... | Yes |
Theorem 9.43. \( A \) in \( B\left( X\right) \) is in \( {\Phi }_{ + } \) if and only if \( \alpha \left( {A - K}\right) < \infty \) for all \( K \in K\left( X\right) \) . | Proof. If \( A \in {\Phi }_{ + } \) and \( K \in K\left( X\right) \), then \( A - K \in {\Phi }_{ + } \) by Theorem 5.22. In particular, \( \alpha \left( {A - K}\right) < \infty \) . Conversely, suppose \( A \notin {\Phi }_{ + } \) . Then by Theorem 9.42 there are sequences \( \left\{ {x}_{k}\right\} ,\left\{ {x}_{k}^{... | Yes |
Theorem 9.44. \( A \in \Phi \) if and only if \( \alpha \left( {A - K}\right) < \infty \) and \( \beta \left( {A - K}\right) < \) \( \infty \) for all \( K \in K\left( X\right) \) . | Proof. If \( A \in \Phi \), then \( A - K \in \Phi \;\forall K \in K\left( X\right) \) . Consequently, \( \alpha (A - \) \( K) < \infty ,\beta \left( {A - K}\right) < \infty \;\forall K \in K\left( X\right) \) (Theorem 5.10). Conversely, if \( \alpha \left( {A - K}\right) < \infty \;\forall K \in K\left( X\right) \), t... | Yes |
Corollary 9.45. If \( {E}_{1},{E}_{2} \in {F}_{ + }\left( X\right) \), then \( {E}_{1} + {E}_{2} \in {F}_{ + }\left( X\right) \) . | Proof. Just use the definition. | No |
Theorem 9.46. \( E \in {F}_{ + }\left( X\right) \) if and only if \( \alpha \left( {A - E}\right) < \infty \) for all \( A \in {\Phi }_{ + } \) . | Proof. If \( E \in {F}_{ + }\left( X\right) \) and \( A \in {\Phi }_{ + } \), then \( A - E \in {\Phi }_{ + } \) by definition. Hence, \( \alpha \left( {A - E}\right) < \infty \) . If \( A \in {\Phi }_{ + } \) and \( A - E \notin {\Phi }_{ + } \), then there is a \( K \in K\left( X\right) \) such that \( \alpha \left( ... | Yes |
Theorem 9.47. \( E \) is in \( F\left( X\right) \) if and only if \( \alpha \left( {A - E}\right) < \infty \) for all \( A \in \Phi \) . | Proof. If \( E \in F\left( X\right) \) and \( A \in \Phi \), then \( A - E \in \Phi \) (Theorem 9.32). Thus, \( \alpha \left( {A - E}\right) < \infty \) . Conversely, suppose \( \alpha \left( {A - E}\right) < \infty \forall A \in \Phi \) . Let \( A \) be any particular operator in \( \Phi \) . Then \( \left( {A - K}\ri... | Yes |
Lemma 9.49. If \( {E}_{k} \in {F}_{ + }\left( X\right) \) and \( {E}_{k} \rightarrow E \), then \( E \in {F}_{ + }\left( X\right) \) . | Proof. Use the same reasoning as in the proof of Corollary 9.38. Use Theorem 5.23 in place of Theorem 5.11. | No |
Lemma 9.50. If \( E \in {F}_{ + }\left( X\right) \), then \( {AE} \) and \( {EA} \) are in \( {F}_{ + }\left( X\right) \) for all \( A \in \) \( \Phi \) . | Proof. If \( A \in \Phi \) and \( C \in {\Phi }_{ + } \), then \( E + {A}_{0}C \in {\Phi }_{ + } \) (Theorem 5.26). Thus, \( A\left( {E + {A}_{0}C}\right) \in {\Phi }_{ + } \) together with \( {AE} + C \) . This means that \( {AE} \in {F}_{ + }\left( X\right) \) . A similar argument works for \( {EA} \) . | No |
Lemma 9.51. If \( E \in {F}_{ + }\left( X\right) \), then \( {BE} \) and \( {EB} \) are in \( {F}_{ + }\left( X\right) \) for all \( B \in \) \( B\left( X\right) \) . | Proof. See the proof of Corollary 9.35. | No |
Theorem 9.52. \( {F}_{ + }\left( X\right) \) is a closed two-sided ideal. | Proof. See the proofs of Lemmas 9.49 and 9.51. | No |
If \( A \) is not in \( {\Phi }_{ - } \), then there are sequences \( \left\{ {x}_{k}\right\} \subset X \) , \( \left\{ {x}_{k}^{\prime }\right\} \subset {X}^{\prime } \) such that\n\n\[ \n{x}_{j}^{\prime }\left( {x}_{k}\right) = {\delta }_{jk},\begin{Vmatrix}{x}_{k}^{\prime }\end{Vmatrix} = 1,\begin{Vmatrix}{x}_{k}\en... | Proof. For \( n > 0 \), assume that \( {x}_{1},\cdots ,{x}_{n - 1},{x}_{1}^{\prime },\cdots ,{x}_{n - 1}^{\prime } \) have been found, and set\n\n\[ \nP{x}^{\prime } = \mathop{\sum }\limits_{1}^{{n - 1}}{x}^{\prime }\left( {x}_{k}\right) {x}_{k}^{\prime }\n\]\n\nNow by Theorem 9.41 there is an \( {x}_{n}^{\prime } \in ... | Yes |
Theorem 9.54. \( A \) in \( B\left( X\right) \) is in \( {\Phi }_{ - } \) if and only if \( \beta \left( {A - K}\right) < \infty \) for all \( K \) in \( K\left( X\right) \) . | Proof. If \( A \in {\Phi }_{ - } \) and \( K \in K\left( X\right) \), then \( A - K \in {\Phi }_{ - } \) by Theorem 5.28. In particular, \( \beta \left( {A - K}\right) < \infty \) . Conversely, if \( A \notin {\Phi }_{ - }\left( X\right) \), then there are sequences \( \left\{ {x}_{k}\right\} ,\left\{ {x}_{k}^{\prime }... | Yes |
Theorem 9.56. \( E \in {F}_{ - }\left( X\right) \) if and only if \( \beta \left( {A - E}\right) < \infty \) for all \( A \in {\Phi }_{ - } \). | Proof. If \( E \in {F}_{ - }\left( X\right) \) and \( A \in {\Phi }_{ - } \), then \( A - E \in {\Phi }_{ - } \) by definition. Hence, \( \beta \left( {A - E}\right) < \infty \) . If \( A \in {\Phi }_{ - } \) and \( A - E \notin {\Phi }_{ - } \), then there is a \( K \in K\left( X\right) \) such that \( \beta \left( {A... | Yes |
Theorem 9.57. \( E \) is in \( F\left( X\right) \) if and only if \( \beta \left( {A - E}\right) < \infty \) for all \( A \in \Phi \) . | Proof. If \( E \in F\left( X\right) \) and \( A \in \Phi \), then \( A - E \in \Phi \) (Theorem 9.32). Thus, \( \beta \left( {A - E}\right) < \infty \) . Conversely, suppose \( \beta \left( {A - E}\right) < \infty \forall A \in \Phi \) . Let \( A \) be any particular operator in \( \Phi \) . Then \( \left( {A - K}\righ... | Yes |
Lemma 9.59. If \( {E}_{k} \in {F}_{ - }\left( X\right) \) and \( {E}_{k} \rightarrow E \), then \( E \in {F}_{ - }\left( X\right) \) . | Proof. Use the same reasoning as in the proof of Corollary 9.38. Use Theorem 5.23 in place of Theorem 5.11. | No |
Lemma 9.60. If \( E \in {F}_{ - }\left( X\right) \), then \( {AE} \) and \( {EA} \) are in \( {F}_{ - }\left( X\right) \) for all \( A \in \) \( \Phi \) . | Proof. If \( A \in \Phi \) and \( C \in {\Phi }_{ - } \), then \( E + {A}_{0}C \in {\Phi }_{ - } \) . Thus, \( A\left( {E + {A}_{0}C}\right) \in {\Phi }_{ - } \) together with \( {AE} + C \) . This means that \( {AE} \in {F}_{ - }\left( X\right) \) . A similar argument works for \( {EA} \) . | No |
Lemma 9.61. If \( E \in {F}_{ - }\left( X\right) \), then \( {BE} \) and \( {EB} \) are in \( {F}_{ - }\left( X\right) \) for all \( B \in \) \( B\left( X\right) \) . | Proof. See the proof of Corollary 9.35. | No |
Theorem 9.62. \( {F}_{ - }\left( X\right) \) is a closed two-sided ideal. | Proof. See the proofs of Lemmas 9.49 and 9.51. | No |
Theorem 10.1. Let \( A \) be a closed linear operator with dense domain \( D\left( A\right) \) on \( X \) having the interval \( \lbrack b,\infty ) \) in its resolvent set \( \rho \left( A\right) \), where \( b \geq 0 \), and such that there is a constant a satisfying\n\n(10.21)\n\n\[ \begin{Vmatrix}{\left( \lambda - A... | Before proving the theorem, we show how it gives the solution to our problem provided \( {u}_{0} \in D\left( A\right) \) . In fact,\n\n(10.23)\n\n\[ u\left( t\right) = {E}_{t}{u}_{0},\;t \geq 0 \]\n\nis a solution of (10.4) and (10.5). To see this, note that \( {E}_{t} \) maps \( D\left( A\right) \) into itself. The re... | Yes |
Lemma 10.2. Let \( D \) be a dense set in \( X \), and let \( \left\{ {B}_{\lambda }\right\} \) be a family of operators in \( B\left( X\right) \) satisfying\n\n(10.24)\n\n\[ \begin{Vmatrix}{B}_{\lambda }\end{Vmatrix} \leq M,\;\lambda \geq K. \]\n\nIf \( {B}_{\lambda }x \) converges as \( \lambda \rightarrow \infty \) ... | Proof. Let \( \varepsilon > 0 \) be given, and let \( x \) be any element of \( X \) . Then we can find an element \( \widetilde{x} \in D \) such that\n\n(10.27)\n\n\[ \parallel x - \widetilde{x}\parallel < \frac{\varepsilon }{3M}. \]\n\nThus,\n\n\[ \begin{Vmatrix}{{B}_{\lambda }x - {B}_{\mu }x}\end{Vmatrix} \leq \begi... | Yes |
Theorem 11.1. A Banach space \( X \) can be converted into a Hilbert space with the same norm if and only if (11.1) holds. | Proof. The simple proof of the \ | No |
Lemma 11.7. If \( A \) is normal, then\n\n\[ \n{r}_{\sigma }\left( A\right) = \parallel A\parallel \n\] | Proof. Let us show that\n\n\[ \n\begin{Vmatrix}{A}^{n}\end{Vmatrix} = \parallel A{\parallel }^{n},\;n = 1,2,\cdots ,\n\]\n\nwhen \( A \) is normal. For this purpose we note that\n\n\[ \n{\begin{Vmatrix}{A}^{k}u\end{Vmatrix}}^{2} = \left( {{A}^{k}u,{A}^{k}u}\right)\n\]\n\n\[ \n= \left( {{A}^{ * }{A}^{k}u,{A}^{k - 1}u}\r... | Yes |
If \( A \) is a normal compact operator, then there is an orthonormal sequence \( \left\{ {\varphi }_{k}\right\} \) of eigenvectors of \( A \) such that every element \( u \) in \( H \) can be written in the form\n\n\[ u = h + \sum \left( {u,{\varphi }_{k}}\right) {\varphi }_{k} \]\n\nwhere \( h \in N\left( A\right) \)... | Proof. By Theorem 11.3, \( A \) has an orthonormal set \( \left\{ {\varphi }_{k}\right\} \) of eigenvectors such that (11.14) holds. If \( u \) is any element of \( H \), set\n\n\[ h = u - \sum \left( {u,{\varphi }_{k}}\right) {\varphi }_{k} \]\n\nThen, by (11.14), \( {Ah} = 0 \), showing that \( h \in N\left( A\right)... | No |
Lemma 11.9. Every separable Hilbert space has a complete orthonormal sequence. | Proof. Let \( H \) be a separable Hilbert space, and let \( \left\{ {x}_{n}\right\} \) be a dense sequence in \( H \) . Remove from this sequence any element which is a linear combination of the preceding \( {x}_{j} \) . Let \( {N}_{n} \) be the subspace spanned by \( {x}_{1},\cdots ,{x}_{n} \), and let \( {\varphi }_{... | Yes |
Theorem 11.11. If\n\n(11.44)\n\n\[ K\\left( {x, y}\\right) \\overline{K\\left( {x, z}\\right) } = K\\left( {z, x}\\right) \\overline{K\\left( {y, x}\\right) },\\;a \\leq x, y, z \\leq b, \]\n\nthen there exists an orthonormal sequence \( \\left\{ {\\varphi }_{k}\\right\} \) (finite or infinite) of functions in \( {L}^{... | Proof. In view of Lemma 11.10, we can prove (11.45) by verifying that the operator (11.42) is normal and applying Theorem 11.3. This is simple, since\n\n(11.49)\n\n\[ {K}^{ * }v\\left( y\\right) = \\int \\overline{K\\left( {x, y}\\right) }v\\left( x\\right) {dx}. \]\n\nHence,\n\n\[ K{K}^{ * }u\\left( z\\right) = \\int ... | Yes |
Lemma 11.12. If \( \\left\\{ {\\varphi }_{k}\\right\\} \) is a complete orthonormal sequence in a Hilbert space \( H \), and \( K \) is an operator in \( B\\left( H\\right) \) satisfying\n\n(11.54)\n\n\[ \n\\mathop{\\sum }\\limits_{1}^{\\infty }{\\begin{Vmatrix}K{\\varphi }_{k}\\end{Vmatrix}}^{2} < \\infty \n\]\n\nthen... | Proof. First note that\n\n(11.56)\n\n\[ \n{Ku} = \\mathop{\\sum }\\limits_{1}^{\\infty }\\left( {u,{\\varphi }_{k}}\\right) K{\\varphi }_{k},\\;u \\in H.\n\]\n\nIn fact, since\n\n\[ \n\\mathop{\\sum }\\limits_{1}^{n}\\left( {u,{\\varphi }_{k}}\\right) {\\varphi }_{k}\n\]\n\nconverges to \( u \), and \( K \) is continuo... | Yes |
Theorem 11.13. If \( A \) is seminormal, then\n\n\[ \n{r}_{\sigma }\left( A\right) = \parallel A\parallel \n\] | Proof. First, assume that \( A \) is hyponormal. Then (11.31) holds. The proof is the same as that for normal operators with the exception that in this case we have\n\n\[ \n\begin{Vmatrix}{{A}^{ * }{A}^{k}u}\end{Vmatrix} \cdot \begin{Vmatrix}{{A}^{k - 1}u}\end{Vmatrix} \leq \begin{Vmatrix}{{A}^{k + 1}u}\end{Vmatrix} \c... | Yes |
Lemma 11.15. If \( A \) is hyponormal, then so is \( B = A - \lambda \) for any complex \( \lambda \) . | Proof.\n\n\[ \n{\begin{Vmatrix}{B}^{ * }u\end{Vmatrix}}^{2} = \left( {\left\lbrack {A{A}^{ * } - \lambda {A}^{ * } - \bar{\lambda }A + {\left| \lambda \right| }^{2}}\right\rbrack u, u}\right) \]\n\n\[ \leq \left( {\left\lbrack {{A}^{ * }A - \lambda {A}^{ * } - \bar{\lambda }A - {\left| \lambda \right| }^{2}}\right\rbra... | Yes |
Lemma 11.16. If \( A \) is hyponormal and maps a closed subspace \( M \) into itself, then the restriction of \( A \) to \( M \) is hyponormal. | Proof. Let \( {A}_{1} \) be the restriction of \( A \) to \( M \) . Then for \( u, v \in M \), we have\n\n(11.70)\n\n\[ \left( {u,{A}^{ * }v}\right) = \left( {{Au}, v}\right) = \left( {{A}_{1}u, v}\right) = \left( {u,{A}_{1}^{ * }v}\right) . \]\n\nIn particular,\n\n\[ {\begin{Vmatrix}{A}_{1}^{ * }u\end{Vmatrix}}^{2} = ... | Yes |
Lemma 11.17. If \( B \) is hyponormal with 0 an isolated point of \( \sigma \left( B\right) \) and either \( \alpha \left( B\right) \) or \( \beta \left( B\right) \) is finite, then \( B \in \Phi \left( H\right) \) and \( i\left( B\right) = 0 \) . | Proof. Set\n\n\[ P = \frac{1}{2\pi i}{\oint }_{\left| z\right| = \varepsilon }{\left( z - B\right) }^{-1}{dz} \]\n\nwhere \( \varepsilon > 0 \) is so small that the points \( 0 < \left| z\right| < \varepsilon \) are in \( \rho \left( B\right) \) . Then by (6.30), we have\n\n(11.73)\n\n\[ H = R\left( P\right) \oplus N\l... | Yes |
Corollary 11.18. If \( A \) is seminormal and \( \lambda \) is an isolated point of \( \sigma \left( A\right) \) , then \( \lambda \) is an eigenvalue of \( A \) . | Proof. Set \( B = A - \lambda \) . If \( A \) is hyponormal, so is \( B \) (Lemma 11.15). If \( \alpha \left( B\right) = 0 \), then by Lemma 11.17, we have \( B \in \Phi \left( H\right) \) and \( i\left( B\right) = 0 \) . But then \( R\left( B\right) = H \), showing that \( \lambda \in \rho \left( A\right) \) . Thus, w... | Yes |
Theorem 12.3. Let \( a\left( {u, v}\right) \) be a densely defined bilinear form with associated operator \( A \) . Then\n\n(a) If \( \lambda \notin \overline{W\left( a\right) } \), then \( A - \lambda \) is one-to-one and\n\n(12.11)\n\n\[ \parallel u\parallel \leq C\parallel \left( {A - \lambda }\right) u\parallel ,\;... | Proof. (a) Since \( \lambda \notin \overline{W\left( a\right) } \), there is a \( \delta > 0 \) such that\n\n(12.12)\n\n\[ \left| {a\left( u\right) - \lambda }\right| \geq \delta ,\;\parallel u\parallel = 1, u \in D\left( a\right) . \]\n\nThus\n\n(12.13)\n\n\[ \left| {a\left( u\right) - \lambda \parallel u{\parallel }^... | Yes |
Lemma 12.4. Let \( a\left( {u, v}\right) \) and \( b\left( {u, v}\right) \) be symmetric bilinear forms satisfying\n\n(12.17)\n\n\[ \left| {a\left( u\right) }\right| \leq {Mb}\left( u\right) ,\;u \in D\left( a\right) \cap D\left( b\right) . \]\n\nThen\n\n(12.18)\n\n\[ {\left| a\left( u, v\right) \right| }^{2} \leq {M}^... | Proof. Assume first that \( a\left( {u, v}\right) \) is real. Then\n\n(12.19)\n\n\[ a\left( {u \pm v}\right) = a\left( u\right) \pm {2a}\left( {u, v}\right) + a\left( v\right) ,\]\n\nand hence,\n\n\[ {4a}\left( {u, v}\right) = a\left( {u + v}\right) - a\left( {u - v}\right) \]\n\nThus\n\n\[ 4\left| {a\left( {u, v}\righ... | Yes |
Corollary 12.5. If \( b\left( {u, v}\right) \) is symmetric but \( a\left( {u, v}\right) \) is not and (12.17) holds, then\n\n\( \left( {12.22}\right) \)\n\n\[ \n{\left| a\left( u, v\right) \right| }^{2} \leq 4{M}^{2}b\left( u\right) b\left( v\right) ,\;u, v \in D\left( a\right) \cap D\left( b\right) .\n\] | Proof. Set\n\n(12.23)\n\n\[ \n{a}_{1}\left( {u, v}\right) = \frac{1}{2}\left\lbrack {a\left( {u, v}\right) + \overline{a\left( {v, u}\right) }}\right\rbrack \n\]\n\n(12.24)\n\n\[ \n{a}_{2}\left( {u, v}\right) = \frac{1}{2i}\left\lbrack {a\left( {u, v}\right) - \overline{a\left( {v, u}\right) }}\right\rbrack .\n\]\n\nTh... | Yes |
If \( b\left( {u, v}\right) \) is a symmetric bilinear form such that\n\n(12.26)\n\n\[ b\left( u\right) \geq 0,\;u \in D\left( b\right) \]\n\nthen\n\n(12.27)\n\n\[ {\left| b\left( u, v\right) \right| }^{2} \leq b\left( u\right) b\left( v\right) ,\;u, v \in D\left( b\right) ,\]\n\nand\n\n(12.28)\n\n\[ b{\left( u + v\rig... | Proof. By (12.26), we have \( \left| {b\left( u\right) }\right| = b\left( u\right) \) . Setting \( a\left( {u, v}\right) = b\left( {u, v}\right) \) in Lemma 12.4, we get (12.27). Inequality (12.28) follows from (12.27) in the usual fashion. In fact,\n\n\[ b\left( {u + v}\right) = b\left( u\right) + b\left( {u, v}\right... | Yes |
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