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Theorem 12.7. The following statements are equivalent for a bilinear form:\n\n(i) \( a\left( {u, v}\right) \) is symmetric;\n\n(ii) \( \Im {ma}\left( u\right) = 0,\;u \in D\left( a\right) \) ;\n\n\[ \text{(iii)}\Re e\;a\left( {u, v}\right) = \Re e\;a\left( {v, u}\right) ,\;u, v \in D\left( a\right) \text{.} \] | Proof. That (i) implies (ii) is trivial. To show that (ii) implies (iii), note that\n\n\[ a\left( {{iu} + v}\right) = a\left( u\right) + {ia}\left( {u, v}\right) - {ia}\left( {v, u}\right) + a\left( v\right) .\n\nTaking the imaginary parts of both sides and using (ii), we get (iii). To prove that (iii) implies (i), obs... | Yes |
Theorem 12.14. Let \( A \) be a densely defined linear operator on \( H \) such that \( \overline{W\left( A\right) } \) is not the whole plane, a half-plane, a strip, or a line. Then \( A \) has a closed extension \( \widehat{A} \) such that\n\n(12.44)\n\n\[ \sigma \left( \widehat{A}\right) \subset \overline{W\left( A\... | The proof of Theorem 12.14 will be based on the following two theorems.\n\nTheorem 12.15. If \( A | No |
Theorem 12.16. Let \( a\left( {u, v}\right) \) be a densely defined bilinear form such that \( \overline{W\left( a\right) } \) is not the whole plane, a half-plane, a strip, or a line. Suppose that \( \left\{ {u}_{k}\right\} \subset D\left( a\right) ,{u}_{n} \rightarrow 0, a\left( {{u}_{n} - {u}_{m}}\right) \rightarrow... | Proof. Let \( a\left( {u, v}\right) \) be the bilinear form defined by\n\n(12.45)\n\n\[ a\left( {u, v}\right) = \left( {{Au}, v}\right) ,\;u, v \in D\left( A\right) ,\]\n\nwith \( D\left( a\right) = D\left( A\right) \) . Then \( W\left( a\right) = W\left( A\right) \) . By Theorem 12.13, there is a symmetric bilinear fo... | Yes |
Corollary 12.20. If \( a\left( {u, v}\right) \) is a bilinear form such that \( W\left( a\right) \) is not the whole plane, then there are constants \( \gamma ,{k}_{0} \) with \( \left| \gamma \right| = 1 \) such that\n\n(12.55)\n\n\[ \Re e\left\lbrack {{\gamma a}\left( u\right) + {k}_{0}\parallel u{\parallel }^{2}}\ri... | Corollary 12.20 follows easily from the lemma. In fact, we know that \( W\left( a\right) \) is convex from Theorem 12.9. Hence, it must be contained in a half-plane by Lemma 12.19. But every half-plane is of the form\n\n\[ \Re e\left\lbrack {{\gamma z} + {k}_{0}}\right\rbrack \geq 0,\;\left| \gamma \right| = 1. \]\n\nT... | Yes |
Theorem 12.21. Let \( a\left( {u, v}\right) \) be a densely defined bilinear form such that \( W\left( a\right) \) is not the whole plane. Let \( A \) be the operator associated with \( a\left( {u, v}\right) \) . If \( D\left( A\right) \) is dense in \( H \), then \( A \) is closable. | Proof. By Corollary 12.20, there are constants \( \gamma ,{k}_{0} \) such that (12.55) holds. Set\n\n\[ b\left( {u, v}\right) = {\gamma a}\left( {u, v}\right) + {k}_{0}\left( {u, v}\right) \]\n\nand\n\n\[ B = {\gamma A} + {k}_{0} \]\n\nThen \( B \) is the operator associated with \( b\left( {u, v}\right) \) . Moreover,... | Yes |
Theorem 12.22. Let \( b\left( {u, v}\right) \) be a closed, symmetric bilinear form on a Hilbert space \( H \) satisfying\n\n(12.70)\n\n\[ \parallel u{\parallel }^{2} \leq {Cb}\left( u\right) ,\;u \in D\left( b\right) . \]\n\nSuppose that \( a\left( {u, v}\right) \) is a bilinear form with \( D\left( b\right) \subset D... | Proof. Let \( X \) be \( D\left( b\right) \) with scalar product \( b\left( {u, v}\right) \) . Since \( b\left( {u, v}\right) \) is a closed bilinear form,(12.70) and (12.71) imply that \( X \) is a Hilbert space (see the proof of Theorem 12.11). Now \( a\left( {u, v}\right) \) is a bilinear form on \( X \) and\n\n(12.... | Yes |
Corollary 12.23. Let \( a\left( {u, v}\right) \) be a bilinear form defined on the whole of \( H \) such that\n\n(12.76)\n\n\[ \nm\parallel u{\parallel }^{2} \leq \left| {a\left( u\right) }\right| \leq M\parallel u{\parallel }^{2},\;u \in H \]\n\nholds for positive \( m, M \) . Then for each bounded linear functional \... | Proof. We merely take \( b\left( {u, v}\right) \) to be the scalar product of \( H \) in Theorem 12.22. | No |
Theorem 12.27. Let \( B \) be a densely defined linear operator on \( H \) such that \( W\left( B\right) \) is the line \( \Re {e\lambda } = 0 \) . Then a necessary and sufficient condition that \( B \) have a closed extension \( \widehat{B} \) such that\n\n\[ \sigma \left( \widehat{B}\right) \subset W\left( \widehat{B... | The last statement follows from the fact that \( R{\left( I + B\right) }^{ \bot } \) and \( R{\left( I - B\right) }^{ \bot } \) have complete orthonormal sequences \( \left\{ {\varphi }_{k}\right\} \) and \( \left\{ {\psi }_{k}\right\} \), respectively (Lemma 11.9). Moreover, these sequences are either both infinite or... | Yes |
Theorem 12.28. Let \( B \) be a densely defined linear operator on \( H \) such that \( \overline{W\left( B\right) } \) is the strip \( 1 - a \leq \Re {ez} \leq 0, a > 1 \) . If \( \overline{R\left( {a - B}\right) } = \overline{R\left( {a + B}\right) } \), then \( B \) has a closed extension \( \widehat{B} \) satisfyin... | Proof. On \( R{\left( a - B\right) }^{ \bot } = R{\left( a + B\right) }^{ \bot } \) we define \( \widehat{T} \) to be \( - I \) . Then \( \widehat{T} \) is isometric on this set. Thus,(12.104) [and hence,(12.105)] holds for \( u \in \overline{R\left( {a - B}\right) } \) and for \( u \in R{\left( a - B\right) }^{ \bot }... | Yes |
Lemma 12.29. Let \( A \) be a closed linear operator on a Banach space \( X \) . If \( \lambda \) is a boundary point of \( \rho \left( A\right) \) and \( \left\{ {\lambda }_{n}\right\} \) is a sequence of points in \( \rho \left( A\right) \) converging to \( \lambda \), then \( \begin{Vmatrix}{\left( A - {\lambda }_{n... | Proof. If the lemma were not true, there would be a sequence \( \left\{ {\lambda }_{n}\right\} \subset \rho \left( A\right) \) such that \( {\lambda }_{n} \rightarrow \lambda \) as \( n \rightarrow \infty \) while\n\n(12.114)\n\n\[ \begin{Vmatrix}{\left( A - {\lambda }_{n}\right) }^{-1}\end{Vmatrix} \leq C. \]\n\nSince... | Yes |
Lemma 13.1. If \( F \) is a bounded linear selfadjoint projection on \( H \), then \( R\left( F\right) \) is closed and \( F \) is the orthogonal projection onto \( R\left( F\right) \) . | Proof. Set \( M = R\left( F\right) \) . Then \( M \) is closed. This follows from the fact that if \( F{u}_{n} \rightarrow v \), then \( {F}^{2}{u}_{n} \rightarrow {Fv} \) . Thus, \( v = {Fv} \), showing that \( v \in M \) . Now, if \( u \) is any element of \( H \), then\n\n\[ u = {Fu} + \left( {I - F}\right) u. \]\n\... | Yes |
Lemma 13.2. If \( A \) is in \( B\left( H\right) \), then\n\n(i) \( M \) is invariant under \( A \) if and only if \( {AE} = {EAE} \) ;\n\n(ii) \( M \) reduces \( A \) if and only if \( {AE} = {EA} \) . | Proof. (i) If \( M \) is invariant under \( A \), then \( {AEu} \in M \) for all \( u \in H \) . Hence, \( {EAEu} = {AEu} \) . Conversely, if \( {AE} = {EAE} \) and \( u \in M \), then \( {Au} = {AEu} = \) \( {EAEu} = {EAu} \) . Hence \( {Au} \in M \) .\n\n(ii) Suppose \( M \) reduces \( A \) . We have\n\n\[ \n{EA} = {... | Yes |
Theorem 13.3. Let \( A \) be any real number satisfying \( 0 \leq A \leq 1 \) . Then there exists a real number \( B \geq 0 \) such that \( {B}^{2} = A \) . | Proof. Suppose \( B \) exists. Set \( R = 1 - A, S = 1 - B \) . Then \( {\left( 1 - S\right) }^{2} = 1 - R \) , or\n\n(13.7)\n\n\[ S = \frac{1}{2}\left( {R + {S}^{2}}\right) \]\n\nConversely, if we can find a solution \( S \) of (13.7), then an easy calculation shows that \( B = 1 - S \) satisfies \( {B}^{2} = A \) . S... | Yes |
Lemma 13.4. If\n\n(13.12)\n\n\[ \n- {MI} \leq A \leq {MI},\;M \geq 0, \]\n\nthen\n\n(13.13)\n\n\[ \n\parallel A\parallel \leq M\text{.} \]\n | Proof. By (13.12),\n\n(13.14)\n\n\[ \n\left| \left( {{Au}, u}\right) \right| \leq M\parallel u{\parallel }^{2},\;u \in H. \]\n\nHence, by Lemma 12.4,\n\n\[ \n\left| \left( {{Au}, v}\right) \right| \leq M\parallel u\parallel \cdot \parallel v\parallel ,\;u, v \in H. \]\n\nIf we take \( v = {Au} \), we obtain\n\n\[ \n\pa... | Yes |
Theorem 13.5. If \( A \) is a positive operator in \( B\left( H\right) \), then there is a unique \( B \geq 0 \) such that \( {B}^{2} = A \) . Moreover, \( B \) commutes with any \( C \in B\left( H\right) \) which commutes with \( A \) . | Proof. It suffices to consider the case\n\n(13.15)\n\n\[ 0 \leq A \leq I \]\n\nTo see this, note that the operator \( A/\parallel A\parallel \) always satisfies (13.15), and if we can find an operator \( G \) such that \( {G}^{2} = A/\parallel A\parallel \), then \( B = \parallel A{\parallel }^{1/2}G \) satisfies \( {B... | Yes |
Lemma 13.6. If \( \\left\\{ {S}_{n}\\right\\} \) is a sequence of operators in \( B\\left( H\\right) \) satisfying (13.21), then there is an operator \( S \) in \( B\\left( H\\right) \) such that\n\n(13.22)\n\n\[ \n{S}_{n}u \\rightarrow {Su},\\;u \\in H.\n\] | Proof. Set\n\n\[ \n{S}_{mn} = {S}_{n} - {S}_{m},\\;m \\leq n.\n\]\n\nThen by (13.21),\n\n(13.23)\n\n\[ \n0 \\leq {S}_{mn} \\leq I\n\]\n\nHence,\n\n(13.24)\n\n\[ \n{\\begin{Vmatrix}{S}_{mn}u\\end{Vmatrix}}^{4} = {\\left( {S}_{mn}u,{S}_{mn}u\\right) }^{2}\n\]\n\n\[ \n\\leq \\left( {{S}_{mn}u, u}\\right) \\left( {{S}_{mn}... | Yes |
Corollary 13.7. If \( A \geq 0, B \geq 0 \) and \( {AB} = {BA} \), then \( {BA} \geq 0 \) . | Proof. By Theorem 13.5, \( A \) and \( B \) have square roots \( {A}^{1/2} \geq 0 \) and \( {B}^{1/2} \geq 0 \) which commute. Hence,\n\n\[\n\left( {{ABu}, u}\right) = {\begin{Vmatrix}{A}^{1/2}{B}^{1/2}u\end{Vmatrix}}^{2} \geq 0.\n\] | Yes |
Lemma 13.8. Let \( {M}_{1} \) and \( {M}_{2} \) be closed subspaces of \( H \), and let \( {E}_{1} \) and \( {E}_{2} \) be the orthogonal projections onto them, respectively. Then the following statements are equivalent:\n\n(a) \( {E}_{1} \leq {E}_{2} \) ;\n\n(b) \( \begin{Vmatrix}{{E}_{1}u}\end{Vmatrix} \leq \begin{Vm... | Proof. (a) implies (b). \( {\begin{Vmatrix}{E}_{1}u\end{Vmatrix}}^{2} = \left( {{E}_{1}u,{E}_{1}u}\right) = \left( {{E}_{1}^{2}u, u}\right) = \left( {{E}_{1}u, u}\right) \leq \) \( \left( {{E}_{2}u, u}\right) = {\begin{Vmatrix}{E}_{2}u\end{Vmatrix}}^{2} \).\n\n(b) implies (c). If \( u \in {M}_{1} \), then \( \parallel ... | Yes |
Lemma 13.9. If \( B \in B\left( H\right) \) commutes with \( A \), then it commutes with \( E \) . | Proof. As we mentioned above, \( B \) commutes with \( {A}^{ + } \) . Thus, \( B{A}^{ + } = {A}^{ + }B \) . This implies that \( N\left( {A}^{ + }\right) \) is invariant under \( B \) (see Section 13.1). Thus \( {BE} = {EBE} \) (Lemma 13.2). Taking adjoints, we get \( {EB} = {EBE} \), which implies \( {BE} = {EB} \) . | Yes |
Lemma 13.11. Let \( B \geq 0 \) be an operator in \( B\left( H\right) \) which commutes with \( A \) and satisfies \( B \geq A \) . Then \( B \geq {A}^{ + } \) . | Proof. By (13.39),\n\n\[ \nB \geq {A}^{ + } + {EB} \geq {A}^{ + } \n\]\n\nsince \( {BE} \geq 0 \) . | No |
Lemma 13.12. Let \( B \) be a positive operator in \( B\left( H\right) \) which commutes with \( A \) and satisfies \( B \geq - A \) . Then \( B \geq {A}^{ - } \) . | Proof. By (13.40), \( {EB} \geq {A}^{ - } \) . But \( \left( {I - E}\right) B \geq 0 \) . Hence, \( B \geq {A}^{ - } \) . | No |
Theorem 13.13. Let \( A \) be a selfadjoint operator in \( B\left( H\right) \). Set\n\n\[ m = \mathop{\inf }\limits_{{\parallel u\parallel = 1}}\left( {{Au}, u}\right) ,\;M = \mathop{\sup }\limits_{{\parallel u\parallel = 1}}\left( {{Au}, u}\right) .\n\]\n\nThen there is a family \( \{ E\left( \lambda \right) \} \) of ... | Proof. Set \( A\left( \lambda \right) = A - \lambda \). Then\n\n(13.46)\n\n\[ A\left( {\lambda }_{1}\right) \geq A\left( {\lambda }_{2}\right) \text{ for }{\lambda }_{1} \leq {\lambda }_{2} \]\n\nLet the operators \( \left| {A\left( \lambda \right) }\right| ,{A}^{ + }\left( \lambda \right) ,{A}^{ - }\left( \lambda \rig... | Yes |
Lemma 14.1. If \( A \in B\left( {X, Y}\right) \) and \( B \in B\left( {Y, Z}\right) \), then\n\n(14.1)\n\n\[ \parallel {BA}{\parallel }_{m} \leq \parallel B\parallel \cdot \parallel A{\parallel }_{m} \] | Proof. Let \( \varepsilon > 0 \) be given. Then there is a subspace \( M \) of \( X \) having finite codimension such that\n\n\[ \parallel {Ax}\parallel \leq \left( {\parallel A{\parallel }_{m} + \varepsilon }\right) \parallel x\parallel ,\;x \in M. \]\n\nThus,\n\n\[ \parallel {BAx}\parallel \leq \parallel B\parallel \... | Yes |
Corollary 14.3. For each \( A \) in \( B\left( {X, Y}\right) \) ,\n\n(14.7)\n\n\[ \parallel A + K{\parallel }_{m} = \parallel A{\parallel }_{m},\;K \in K\left( {X, Y}\right) . \] | Proof. Apply Theorem 14.2. | No |
Theorem 14.4. An operator \( A \) is in \( {\Phi }_{ + }\left( {X, Y}\right) \) with \( i\left( A\right) \leq 0 \) if and only if there are an operator \( K \in K\left( {X, Y}\right) \) and a constant \( C \) such that\n\n\[ \parallel x\parallel \leq C\parallel \left( {A - K}\right) x\parallel ,\;x \in D\left( A\right)... | Proof. If (14.8) holds, then we know that \( R\left( {A - K}\right) \) is closed in \( Y \) and that \( A - K \) is one-to-one (Theorem 3.12). Thus \( A - K \in {\Phi }_{ + }\left( {X, Y}\right) \) with \( \alpha \left( {A - K}\right) = 0 \) . Since \( \beta \left( {A - K}\right) \geq 0 \), we see that \( i\left( {A - ... | Yes |
Theorem 14.5. An operator \( A \) in \( B\left( {X, Y}\right) \) is in \( {\Phi }_{ + }\left( {X, Y}\right) \) if and only if for each Banach space \( Z \) there is a constant \( C \) such that\n\n(14.9)\n\n\[ \parallel T{\parallel }_{m} \leq C\parallel {AT}{\parallel }_{m},\;T \in B\left( {Z, X}\right) . \] | Proof. If \( A \in \Phi \left( {X, Y}\right) \), let \( {A}_{0} \) be the operator satisfying Theorem 5.4. Then \( F = {A}_{0}A - I \) is in \( K\left( X\right) \) . Thus, for \( T \in B\left( {Z, X}\right) \), we have\n\n\[ T = {A}_{0}{AT} - {FT} \]\n\nIn view of Theorem 14.2, this implies\n\n\[ \parallel T{\parallel ... | Yes |
Lemma 14.6. \( P\left( S\right) \) is a linear subspace of \( X \) . If, in addition, \( S \) is an open subset of \( X \), then \( P\left( S\right) \) is closed. | Proof. Suppose \( s \in S, a, b \in P\left( S\right) ,\alpha \neq 0 \) . Then \( {\alpha a} + s = \alpha \left( {a + s/\alpha }\right) \in S \) and \( \left( {a + b}\right) + s = a + \left( {b + s}\right) \in S \) . Thus, \( P\left( S\right) \) is a subspace. Assume \( S \) open. Then for each \( s \in S \), there is a... | Yes |
Lemma 14.7. Let \( S, T \) be subsets of \( X \) which satisfy (14.10). Assume that \( S \) is open, that \( S \subset T \) and that \( T \) does not contain any boundary points of \( S \) . Then \( P\left( T\right) \subset P\left( S\right) \) . | Proof. Suppose \( s \in S, b \in P\left( T\right) \) . Then\n\n\[ \n{\alpha b} + s = \alpha \left( {b + s/\alpha }\right) \in T,\;\alpha \neq 0.\n\]\n\nSince \( S \) is open, \( {\alpha b} + s \in S \) for \( \left| \alpha \right| \) sufficiently small. It follows that \( {\alpha b} + s \in \) \( S \) for all scalars \... | Yes |
Lemma 14.8. If \( {GS} \subset S \), then \( P\left( S\right) \) is a left ideal. If \( {SG} \subset S \), then \( P\left( S\right) \) is a right ideal. | Proof. Suppose \( a \in G, b \in P\left( S\right), s \in S \) . Then\n\n\[ \n{ab} + s = a\left( {b + {a}^{-1}s}\right) \in S.\n\]\n\nConsequently, \( {ab} \in P\left( S\right) \) . Since every element of \( B \) is the sum of two elements of \( G \) and \( P\left( S\right) \) is a subspace of \( B \), the first stateme... | Yes |
Theorem 14.10. \( P\left( G\right) = R \) . | Proof. Suppose \( b \in P\left( G\right) \) and \( a \in G \) . Then \( {a}^{-1} + b \in G \) . Hence, \( e + {ab} = \) \( a\left( {{a}^{-1} + b}\right) \in G \) . This means that \( b \in R \) . Conversely, suppose \( b \in R \) . If \( a \in G \) , then \( e + {a}^{-1}b \in G \) . Hence, \( a + b = a\left( {e + {a}^{... | Yes |
Theorem 14.12. \( P\left( {G}_{\ell }\right) = P\left( {G}_{r}\right) = R \) . | Proof. First we note that \( G \subset {G}_{\ell } \) and that \( {G}_{\ell } \) does not contain any boundary points of \( G \) . For if \( a \in {G}_{\ell },{a}_{k} \in G \) and \( {a}_{k} \rightarrow a \), then \( b{a}_{k} - e = \) \( b{a}_{k} - {ba} = b\left( {{a}_{k} - a}\right) \rightarrow 0 \) as \( k \rightarro... | Yes |
Theorem 14.13. \( \left\lbrack A\right\rbrack \) is in \( {G}_{\ell } \) if and only if \( A \in {\Phi }_{\ell }\left( X\right) .\left\lbrack A\right\rbrack \) is in \( {G}_{r} \) if and only if \( A \in {\Phi }_{r}\left( X\right) \) . | Proof. \( \left\lbrack A\right\rbrack \in {G}_{\ell } \) if and only if there is a \( {A}_{0} \in B\left( X\right) \) such that \( \left\lbrack {A}_{0}\right\rbrack \left\lbrack A\right\rbrack = \left\lbrack I\right\rbrack \) . Thus, \( {A}_{0}A = I - {K}_{1} \), where \( {K}_{1} \in K\left( X\right) \) . This is true ... | Yes |
Theorem 14.15. If \( {\Phi }_{Z} \) is not empty, then\n\n(14.12)\n\n\[ P\left( {\Phi }_{Z}\right) = F\left( X\right) \] | Proof. Set \( \widetilde{\Phi } = {\Phi }_{\ell } \cup {\Phi }_{r} \) . Note that \( {\Phi }_{Z} \) is an open set in \( B\left( X\right) \) and satisfies (14.10). Since \( {\Phi }_{Z} \subset \widetilde{\Phi } \) and \( \widetilde{\Phi } \) does not contain any boundary points of \( {\Phi }_{Z} \) , we have, by Lemma ... | Yes |
Theorem 14.16. \( \Gamma \left( {A + B}\right) \leq \Delta \left( A\right) + \Gamma \left( B\right) \) . | Proof. First we note that\n\n(14.16)\n\n\[ \n{\Gamma }_{N}\left( {A + B}\right) \leq {\Gamma }_{N}\left( A\right) + \begin{Vmatrix}{\left. B\right| }_{N}\end{Vmatrix}.\n\]\n\nTo see this, let \( \varepsilon > 0 \) be given. By definition, there is an infinite dimensional subspace \( V \) of \( N \) such that \( \begin{... | Yes |
Theorem 14.17. \( \Delta \left( {A + B}\right) \leq \Delta \left( A\right) + \Delta \left( B\right) \) . | Proof. For any \( N \), we have, by Theorem 14.16,\n\n\[{\Gamma }_{N}\left( {A + B}\right) \leq {\Delta }_{N}\left( A\right) + {\Gamma }_{N}\left( B\right) \leq \Delta \left( A\right) + \Delta \left( B\right)\]\n\nThe theorem now follows from the definition. | Yes |
Lemma 14.18. If \( M, N \) are subspaces of \( X \) such that \( \dim {M}^{ \circ } < \infty \) and \( \dim N = \infty \), then \( \dim M \cap N = \infty \) . | Proof. We know that \( X = M \oplus L \), where \( \dim L < \infty \) (Lemma 5.3). If \( \dim M \cap N < \infty \), then \( M = M \cap N \oplus {M}_{1} \) and \( X = {M}_{1} \oplus {N}_{1} \), where \( {N}_{1} = M \cap N \oplus L \) and \( \dim {N}_{1} < \infty \) . Moreover, \( {M}_{1} \cap N = \{ 0\} \) since every e... | Yes |
Lemma 14.19. Let \( M \) be a subspace of \( X \) having infinite codimension. Then there is an infinite dimensional subspace \( W \) of \( X \) such that \( W \cap M = \{ 0\} \) . | Proof. Let \( \left\{ {x}_{k}^{\prime }\right\} \) be a sequence of linearly independent functionals in \( {M}^{ \circ } \) . Then there is a sequence \( \left\{ {x}_{k}\right\} \) of elements of \( X \) such that\n\n\[ \n{x}_{j}^{\prime }\left( {x}_{k}\right) = {\delta }_{jk},\;j, k = 1,2,\cdots \n\]\n\n(Lemma 5.12). ... | Yes |
Lemma 14.20. If \( M \) and \( N \) are subspaces of \( X \) having finite codimension in \( X \), then \( M \cap N \) has finite codimension in \( X \) . | Proof. By Lemma 5.3 there are finite dimensional subspaces \( {M}_{1},{N}_{1} \subset X \) such that\n\n\[ X = M \oplus {M}_{1},\;X = N \oplus {N}_{1} \]\n\nIf \( M \cap N \) did not have finite codimension in \( M \), then there would be an infinite dimensional subspace \( W \subset M \) such that \( W \cap \left( {M ... | Yes |
Theorem 14.21. If \( T \in B\left( {X, Y}\right) \) and \( A \in B\left( {Y, Z}\right) \), then\n\n(14.17)\n\n\[{\Gamma }_{M}\left( {AT}\right) \leq {\Gamma }_{M}\left( T\right) {\Delta }_{TM}\left( A\right)\]\n\nwhen \( \dim {TM} = \infty \) . If \( \dim {TM} < \infty \), then \( {\Gamma }_{M}\left( {AT}\right) = 0 \)... | Proof. Let \( \varepsilon > 0 \) be given. Then there is an \( N \subset M \) such that \( \begin{Vmatrix}{\left. T\right| }_{N}\end{Vmatrix} < \) \( {\Gamma }_{M}\left( T\right) + \varepsilon \) . Thus \n\nAssume th... | Yes |
Corollary 14.22. \( {\Delta }_{M}\left( {AT}\right) \leq {\Delta }_{M}\left( T\right) \Delta \left( A\right) \) . | Proof. For \( N \subset M \), we have\n\n\[ \n{\Gamma }_{N}\left( {AT}\right) \leq {\Gamma }_{N}\left( T\right) \Delta \left( A\right) \leq {\Delta }_{M}\left( T\right) \Delta \left( A\right) \n\]\n\nThe corollary now follows from the definition. | No |
Lemma 14.25. \( \Gamma \left( A\right) \tau \left( T\right) \leq \parallel {AT}\parallel \) . | Proof. By definition, for each \( \varepsilon > 0 \) there is a \( W \) such that\n\n\[ \parallel {Tz}\parallel \geq \left\lbrack {r\left( T\right) - \varepsilon }\right\rbrack \parallel z\parallel ,\;z \in W. \]\n\nIf \( \tau \left( T\right) = 0 \), the lemma is trivial. Otherwise, pick \( \varepsilon < \tau \left( T\... | Yes |
\[ \Gamma \left( A\right) = \mathop{\inf }\limits_{Z}\mathop{\inf }\limits_{{T \in B\left( {Z, X}\right) }}\frac{\parallel {AT}\parallel }{\tau \left( T\right) } \] | Proof. By Lemma 14.25, \( \Gamma \left( A\right) \leq \) the right hand side of (14.20). To show that it is equal, let \( \varepsilon > 0 \) be given. Then there is an \( M \subset X \) such that \[ \begin{Vmatrix}{\left. A\right| }_{M}\end{Vmatrix} < \Gamma \left( A\right) + \varepsilon \] We may assume that \( M \) i... | Yes |
\[ \Gamma \left( A\right) \geq \mathop{\inf }\limits_{Z}\mathop{\inf }\limits_{{T \in B\left( {Z, X}\right) }}\frac{\parallel {AT}\parallel }{\Delta \left( T\right) }. \] | Proof. Apply Theorems 14.24 and 14.26. | No |
Theorem 14.28. \( \Delta \left( A\right) \leq \parallel A{\parallel }_{m} \) . | Proof. Let \( \varepsilon > 0 \) be given, and let \( V \) be a subspace having finite codimen-sion such that\n\n\[ \begin{Vmatrix}{\left. A\right| }_{V}\end{Vmatrix} < \parallel A{\parallel }_{m} + \varepsilon \]\n\nLet \( M \) be given, and set \( N = M \cap V \) . Then \( \dim N = \infty \) and \( \begin{Vmatrix}{\l... | Yes |
Theorem 14.29. \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) if and only if \( \Gamma \left( A\right) \neq 0 \) . | Proof. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \), then there is a constant \( c > 0 \) such that\n\n\[ \parallel {AT}{\parallel }_{m} \geq c\parallel T{\parallel }_{m},\;T \in B\left( {Z, X}\right) ,\]\n\nwhere \( c \) does not depend on \( Z \) (Theorem 14.5). Thus,\n\n\[ \parallel {AT}\parallel \geq {c\Delta }... | Yes |
Theorem 14.30. If \( \Delta \left( B\right) < \Gamma \left( A\right) \), then \( A + B \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( i\left( {A + B}\right) = \) \( i\left( A\right) \) . | Proof. By Theorem 14.16,\n\n\[ \Gamma \left( {A + {\lambda B}}\right) \geq \Gamma \left( A\right) - {\lambda \Gamma }\left( B\right) > 0 \]\n\nfor \( 0 \leq \lambda \leq 1 \) . Thus \( A + {\lambda B} \in {\Phi }_{ + }\left( {X, Y}\right) \) for such \( \lambda \) (Theorem 14.29). The rest follows from the constancy of... | Yes |
Theorem 14.32. If \( \tau \left( B\right) < \nu \left( A\right) \), then \( A + B \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( i\left( {A + B}\right) = \) \( i\left( A\right) \) . | Proof. If \( A + B \notin {\Phi }_{ + }\left( {X, Y}\right) \), then there are an infinite dimensional subspace \( M \) and a \( K \in K\left( {X, Y}\right) \) such that \( {\left. \left( A + B - K\right) \right| }_{M} = 0 \) (Theorem 9.43). Let \( \varepsilon > 0 \) be given. Then there is a subspace \( W \) having fi... | Yes |
Corollary 14.33. If \( X = Y \) and \( \tau \left( A\right) < 1 \), then \( I - A \in \Phi \left( {X, Y}\right) \) with \( i\left( {I - A}\right) = 0 \) . | Proof. Clearly \( \nu \left( I\right) = 1 \) . Apply Theorem 14.32 and the fact that \( i\left( I\right) = \) 0. | No |
Theorem 14.34. \( \nu \left( A\right) \geq \Gamma \left( A\right) \) . | Proof. Let \( \varepsilon > 0 \) be given. Then there is an \( M \) such that\n\n\[ \begin{Vmatrix}{\left. A\right| }_{M}\end{Vmatrix} < \Gamma \left( A\right) + \varepsilon \]\n\nLet \( W \) having finite codimension be given, and set \( N = W \cap M \) . Then \( \dim N = \infty \), and\n\n\[ \mathop{\inf }\limits_{{x... | Yes |
An operator \( A \) in \( B\left( {X, Y}\right) \) is in \( {\Phi }_{ + }\left( {X, Y}\right) \) if and only if there is a constant \( C \) such that for any Banach space \( Z \), one has\n\n\[ \Delta \left( T\right) \leq {C\Delta }\left( {AT}\right) ,\;T \in B\left( {Z, X}\right) \] | Proof. If (14.23) holds, then\n\n\[ \tau \left( T\right) \leq \Delta \left( T\right) \leq {C\Delta }\left( {AT}\right) \leq C\parallel {AT}\parallel \]\n\nby Theorem 14.24. Thus, \( \Gamma \left( A\right) \neq 0 \) (Theorem 14.26). Consequently, \( A \in \) \( {\Phi }_{ + }\left( {X, Y}\right) \) (Theorem 14.29). Conve... | Yes |
Theorem 14.37. If \( Y \) has the compact approximation property with constant \( C \), then\n\n\[ \parallel A{\parallel }_{K} \leq C\parallel A{\parallel }_{q},\;A \in B\left( {X, Y}\right) . \] | Proof. Let \( \varepsilon > 0 \) be given. Then there exist elements \( {y}_{1},\cdots ,{y}_{n} \in Y \) such that (14.31) holds. By hypothesis, there exists an operator \( K \in K\left( Y\right) \) such that \( \parallel I - K\parallel \leq C \) and\n\n\[ \begin{Vmatrix}{{y}_{k} - K{y}_{k}}\end{Vmatrix} < \varepsilon ... | Yes |
Theorem 14.39. Compact operators are strictly singular. | Proof. Let \( K \) be an operator in \( K\left( {X, Y}\right) \), and suppose it has a bounded inverse \( B \) on a subspace \( M \subset X \) . If \( \left\{ {x}_{n}\right\} \) is a bounded sequence in \( M \), then \( \left\{ {K{x}_{n}}\right\} \) has a convergent subsequence (cf. Section 4.3). Hence, \( {x}_{n} = {B... | Yes |
Lemma 14.40. \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) if and only if it has a bounded inverse on some subspace having finite codimension. | Proof. Lemma 9.40 and Theorem 9.41. | No |
For \( A \) in \( B\left( {X, Y}\right) \), if \( A \notin {\Phi }_{ + }\left( {X, Y}\right) \), then for every \( \varepsilon > 0 \) there is a \( K \in K\left( {X, Y}\right) \) such that \( \parallel K\parallel \leq \varepsilon \) and \( \alpha \left( {A - K}\right) = \infty \) . Thus there is an infinite dimensional... | Proof. By Theorem 9.42 there are sequences \( \left\{ {x}_{k}\right\} \subset X,\left\{ {x}_{k}^{\prime }\right\} \subset {X}^{\prime } \) such that (9.47) holds. Let \( \varepsilon > 0 \) be given, and take \( n \) so large that \( {2}^{1 - n} < \varepsilon \) . Let \( M \) be the closed subspace spanned by \( \left\{... | Yes |
Corollary 14.42. If \( S \) is strictly singular, then for each \( \varepsilon > 0 \) there is an infinite dimensional subspace \( M \) such that the restriction of \( S \) to \( M \) is compact with norm \( \leq \varepsilon \) . | Proof. We merely note that \( S \) is not in \( {\Phi }_{ + }\left( {X, Y}\right) \) by Lemma 14.40 and apply Theorem 14.41. | No |
Strictly singular operators are in \( {F}_{ \pm }\left( {X, Y}\right) \) . | Proof. If the strictly singular operator \( S \) were not in \( {F}_{ + }\left( {X, Y}\right) \), then there would be an operator \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) such that \( \alpha \left( {A - S}\right) = \infty \) (Theorem 9.46). Let \( M = N\left( {A - S}\right) \) . Then \( {\left. A\right| }_{M} = {\... | Yes |
Theorem 14.44. An operator \( S \) is strictly singular if and only if \( {\Gamma }_{M}\left( S\right) = 0 \) for all \( M \subset X \) . | Proof. If \( S \) is strictly singular from \( X \) to \( Y \), then it is strictly singular from \( M \) to \( Y \) for each infinite dimensional subspace \( M \subset X \) . Thus by Corollary 14.42, for each \( \varepsilon > 0 \) there is an infinite dimensional subspace \( N \subset M \) such that \( {\left. S\right... | Yes |
Theorem 14.46. \( T \) in \( B\left( {X, Y}\right) \) is strictly singular if and only if\n\n(14.45)\n\n\[ \n{\Gamma }_{M}\left( {A + T}\right) = {\Gamma }_{M}\left( A\right) ,\;A \in B\left( {X, Y}\right) .\n\] | Proof. If \( T \) is strictly singular, then\n\n\[ \n{\Gamma }_{M}\left( {A + T}\right) \leq {\Gamma }_{M}\left( A\right) + {\Delta }_{M}\left( T\right) = {\Gamma }_{M}\left( A\right) \n\]\n\nfor any \( A \in B\left( {X, Y}\right) \) by Theorem 14.16 and Corollary 14.45. For the same reasons,\n\n\[ \n{\Gamma }_{M}\left... | Yes |
Corollary 14.47. \( T \) in \( B\left( {X, Y}\right) \) is strictly singular if and only if\n\n(14.46)\n\n\[ \Delta \left( {A + T}\right) = \Delta \left( A\right) ,\;A \in B\left( {X, Y}\right) . \] | Proof. Clearly,(14.45) implies (14.46). Thus, if \( T \) is strictly singular, (14.46) holds by Theorem 14.46. On the other hand, if (14.46) holds, then \( \Delta \left( T\right) = \Delta \left( {0 + T}\right) = \Delta \left( 0\right) = 0 \) . We now apply Corollary 14.45. | Yes |
Theorem 14.48. \( A \) is strictly singular if and only if \( \tau \left( A\right) = 0 \) . | Proof. If \( A \) is strictly singular, let \( \varepsilon > 0 \) be given. Then every \( M \) contains an \( N \) such that \( \begin{Vmatrix}{\left. A\right| }_{N}\end{Vmatrix} < \varepsilon \) (Corollary 14.42). Thus\n\n\[ \mathop{\inf }\limits_{{x \in M,\parallel x\parallel = 1}}\parallel {Ax}\parallel < \varepsilo... | Yes |
Theorem 14.49. The operator \( T \in B\left( {X, Y}\right) \) is strictly singular if and only if\n\n(14.47)\n\n\[ \tau \left( {A + T}\right) = \tau \left( A\right) ,\;A \in B\left( {X, Y}\right) \] | Proof. Suppose \( T \) is strictly singular, and let \( \varepsilon > 0 \) be given. For \( A \in \) \( B\left( {X, Y}\right) \) there is an \( M \) such that\n\n\[ \parallel {Ax}\parallel \geq \left\lbrack {\tau \left( A\right) - \varepsilon }\right\rbrack \parallel x\parallel ,\;x \in M. \]\n\nMoreover, there is an \... | Yes |
Theorem 14.50. If \( T \) is in \( B\left( {X, Y}\right) \) and\n\n(14.50)\n\n\[ \parallel T\parallel < \mu \left( A\right) \]\n\nthen \( A - T \in {\Phi }_{ + }\left( {X, Y}\right) \) and\n\n(14.51)\n\n\[ i\left( {A - T}\right) = i\left( A\right) \]\n\nIf\n\n(14.52)\n\n\[ \parallel T\parallel < {\mu }_{0}\left( A\righ... | Proof. Let \( \varepsilon > 0 \) be such that \( \parallel T\parallel + \varepsilon < \mu \left( A\right) \) . If \( A - T \notin {\Phi }_{ + }\left( {X, Y}\right) \), then there is an operator \( K \in K\left( {X, Y}\right) \) such that \( \parallel K\parallel < \varepsilon \) and \( \alpha \left( {A - T - K}\right) =... | Yes |
Theorem 14.51. If \( \alpha \left( A\right) < \infty \), then | \[ \gamma \left( A\right) \leq {\mu }_{0}\left( A\right) \leq \mu \left( A\right) \] | No |
Lemma 14.54. Under the hypotheses of Lemma 14.53 there is an element \( u \in M \) such that\n\n(14.56)\n\n\[ \parallel u\parallel = d\left( {u, N}\right) = 1 \] | Proof. Suppose \( T \in B\left( {X, Y}\right) \) is such that \( \alpha \left( {A - T}\right) > \alpha \left( A\right) \) . Then by Lemma 14.54 there is an element \( u \in N\left( {A - T}\right) \) such that\n\n\[ \parallel u\parallel = d\left( {u, N\left( A\right) }\right) = 1. \]\n\nConsequently,\n\n\[ \gamma \left(... | No |
If \( m\left( {T, A}\right) \) is a \( {\Phi }_{ + } \) perturbation function and \( A \in {\Phi }_{ + } \), then \( m\left( {T, A}\right) < 1 \) implies that\n\n(14.58)\n\n\[ i\left( {A - T}\right) = i\left( A\right) \] | Proof. If \( m\left( {T, A}\right) < 1 \), then (14.57) shows that \( A - {\theta T} \in {\Phi }_{ + } \) for \( 0 \leq \theta \leq 1 \) . By the constancy of the index (Theorem 5.11), we see that (14.58) holds. | Yes |
Theorem 14.57. There is a smallest perturbation function. | Proof. It is immediate from(14.64),(14.65) and (14.66) that \( {m}_{1},{m}_{2} \) and \( {m}_{3} \) are perturbation functions. If \( m\left( {T, A}\right) \) is a \( {\Phi }_{ + } \) perturbation function and \( m\left( {T, A}\right) < {m}_{1}\left( {T, A}\right) \) for some \( T \) and some \( A \in {\Phi }_{ + } \),... | Yes |
Theorem 14.58. Let \( A \) be a Fredholm operator on \( X \), and let \( {A}_{0} \) be as above. Then, setting \( C = {A}_{0}T \), the best Fredholm perturbation function \( {m}_{2} \) is given by\n\n(14.67)\n\n\[ \n{m}_{2}\left( {T, A}\right) = \lim {\left( \tau \left( {C}^{n}\right) \right) }^{1/n} \n\]\n\n(14.68)\n\... | Proof. Since \( {A}_{0} \) is also Fredholm, \( {\lambda A} - T \in \Phi \) if and only if \( {A}_{0}\left( {{\lambda A} - T}\right) = \) \( \lambda - {A}_{0}T + \lambda {K}_{1} \in \Phi \) if and only if \( \lambda + {A}_{0}T \in \Phi \) . Thus,(14.69) follows from the equation\n\n\[ \n{r}_{\sigma }\left( \left\lbrack... | Yes |
Lemma 14.59. Let \( H \) be a Hilbert space, and suppose that \( A \in B\left( H\right), M \) is a subspace and \( \nu \left( {\left. A\right| }_{M}\right) > 0 \) . Then for each \( \varepsilon > 0 \), there is an \( N \subset M \) such that \( \begin{Vmatrix}{\left. A\right| }_{N}\end{Vmatrix} < \nu \left( {\left. A\r... | Proof. To simplify the notation, assume that \( M = H \) . Then by definition, for each \( \varepsilon > 0 \) there is a subspace \( {M}^{\prime } \) of finite codimension with \( \nu \left( {\left. A\right| }_{{M}^{\prime }}\right) > 0 \) , and therefore \( A \in {\Phi }_{ + } \), and there is a norm one element \( {x... | Yes |
Theorem 14.60. On a Hilbert space, \( \tau \left( T\right) = \Delta \left( T\right) \) and \( \nu \left( A\right) = \Gamma \left( A\right) \) . Consequently, the two perturbation functions \( \Delta \left( T\right) /\Gamma \left( A\right) \) and \( \tau \left( T\right) /\nu \left( A\right) \) are equal. | Proof. To show that \( \Delta \left( T\right) \leq \tau \left( A\right) \), choose an \( M \) with \( \Gamma \left( {\left. T\right| }_{M}\right) > 0 \) . Then \( {\left. T\right| }_{M} \in {\Phi }_{ + } \), and consequently, \( \nu \left( {\left. T\right| }_{M}\right) > 0 \) (Lemma 14.31). Given an \( \varepsilon > 0 ... | Yes |
Theorem 14.61. Let the factored perturbation function \( m \) be given by (14.70). For each numerator \( {m}_{1} \) there is a best (i.e., largest) denominator \( d\left( {A,{m}_{1}}\right) \) for which \( {m}_{1}\left( T\right) /d\left( {A,{m}_{1}}\right) \) is a perturbation function of the same type as \( m\left( {\... | Proof. Suppose that \( m \) is a \( {\Phi }_{ + } \) perturbation function. Set\n\n(14.71)\n\n\[ d\left( A\right) = \inf \left\{ {{m}_{1}\left( T\right) : A - T \notin {\Phi }_{ + }}\right\} \]\n\nWe claim that \( {m}_{1}\left( T\right) /d\left( A\right) \) is the minimal \( {\Phi }_{ + } \) perturbation function with ... | Yes |
Lemma 14.62. Suppose that \( {m}_{1}\left( T\right) /{m}_{2}\left( A\right) \) is a \( {\Phi }_{ + } \) perturbation function. Let \( n\left( T\right) = n\left( {T,{m}_{2}}\right) \) be the best numerator for a given denominator \( {m}_{2} \), and let \( d\left( A\right) = d\left( {A,{m}_{1}}\right) \) be the best deno... | Proof. Property (i) is immediate. To see that (ii) holds, note that\n\n\[ d\left( {\lambda A}\right) = \inf \left\{ {{m}_{1}\left( T\right) : {\lambda A} + T \notin {\Phi }_{ + }}\right\} \]\n\n\[ = \inf \left\{ {m\left( {{\lambda T}/\lambda }\right) : A + T/\lambda \notin {\Phi }_{ + }}\right\} \]\n\n\[ = \left| \lamb... | Yes |
Corollary 14.63. Given the numerator \( \parallel T\parallel \) :\n\n(i): \( {\mu }_{0}\left( A\right) \) is the best denominator for a \( {\Phi }_{\alpha } \) perturbation function;\n\n(ii): \( \mu \left( A\right) \) is the best denominator for a \( {\Phi }_{ + } \) perturbation function;\n\n(iii): \( \mu \left( A\rig... | Proof. Statements (i) and (ii) follow from Theorem 14.50 and Theorem 14.61. By Lemma 14.55, \( \parallel T\parallel /\mu \left( A\right) \) is also a \( \Phi \) perturbation function, and since \( \mu \left( A\right) \) is the best demoninator for a \( {\Phi }_{ + } \) perturbation function, it is also the best denomin... | Yes |
Lemma 14.64. If \( {m}_{1}\left( T\right) /{m}_{2}\left( A\right) \) is an \( A \) -exact perturbation function, then the best denominator \( d\left( {A,{m}_{1}}\right) \) for \( {m}_{1} \) is \( {m}_{2}\left( A\right) \) . | Proof. See the proof of Theorem 14.61. | No |
Theorem 1.4. (p. 22) There is a one-to-one correspondence between \( {l}_{2} \) and \( {L}^{2} \) such that if \( \left( {{\alpha }_{0},{\alpha }_{1},\cdots }\right) \) corresponds to \( f \), then | \[ \begin{Vmatrix}{\mathop{\sum }\limits_{0}^{n}{\alpha }_{j}{\varphi }_{j} - f}\end{Vmatrix} \rightarrow 0\text{ as }n \rightarrow \infty \] and \[ \parallel f{\parallel }^{2} = \mathop{\sum }\limits_{0}^{\infty }{\alpha }_{j}^{2},\;{\alpha }_{j} = \left( {f,{\varphi }_{j}}\right) \] | No |
Theorem 1.6. (p. 24) If \( \left\{ {\varphi }_{n}\right\} \) is complete, then for each \( f \in H \)\n\n\[ f = \mathop{\sum }\limits_{1}^{\infty }\left( {f,{\varphi }_{i}}\right) {\varphi }_{i} \] | \[ \parallel f{\parallel }^{2} = \mathop{\sum }\limits_{1}^{\infty }{\left( f,{\varphi }_{i}\right) }^{2} \] | No |
Theorem 2.1. (Riesz Representation Theorem) (p. 29) For every bounded linear functional \( F \) on a Hilbert space \( H \) there is a unique element \( y \in H \) such that\n\n(B.5)\n\n\[ F\left( x\right) = \left( {x, y}\right) \text{ for all }x \in H. \] | Moreover,\n\n(B.6)\n\n\[ \parallel y\parallel = \mathop{\sup }\limits_{{x \in H, x \neq 0}}\frac{\left| F\left( x\right) \right| }{\parallel x\parallel }. \] | Yes |
Theorem 7.3. (p. 157) If \( A \in \Phi \left( {X, Y}\right) \) and \( B \in \Phi \left( {Y, Z}\right) \), then \( {BA} \in \Phi \left( {X, Z}\right) \) and | \[ i\left( {BA}\right) = i\left( A\right) + i\left( B\right) \] | Yes |
Theorem 7.9. (p. 161) For \( A \in \Phi \left( {X, Y}\right) \), there is an \( \eta > 0 \) such that for every \( T \) in \( B\left( {X, Y}\right) \) satisfying \( \parallel T\parallel < \eta \), one has \( A + T \in \Phi \left( {X, Y}\right) \) | \[ i\left( {A + T}\right) = i\left( A\right) \] and \[ \alpha \left( {A + T}\right) \leq \alpha \left( A\right) \] | Yes |
Theorem 7.29. (p. 173) 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 th... | \[ \parallel x\parallel \leq C\parallel {Ax}\parallel + \left| x\right| ,\;x \in D\left( A\right) . \] | Yes |
Theorem 7.35. (p. 178) 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 ... | \[ {\left( BA\right) }^{\prime } = {A}^{\prime }{B}^{\prime }\text{.} \] | No |
Theorem 12.8. (p. 270) Let \( a\left( {u, v}\right) \) be a densely defined closed bilinear form with associated operator \( A \) . If \( \overline{W\left( a\right) } \) is not the whole plane, a half-plane, a strip, or a line, then \( A \) is closed and | \[ \sigma \left( A\right) \subset \overline{W\left( a\right) } = \overline{W\left( A\right) } \] | Yes |
Theorem 12.14. (p. 274) Let \( A \) be a densely defined linear operator on \( H \) such that \( \overline{W\left( A\right) } \) is not the whole plane, a half-plane, a strip, or a line. Then \( A \) has a closed extension \( \widehat{A} \) such that | \[ \sigma \left( \widehat{A}\right) \subset \overline{W\left( A\right) } = \overline{W\left( \widehat{A}\right) } \] | Yes |
Theorem 14.37. (p. 341) If \( Y \) has the compact approximation property with constant \( C \), then | \[ \parallel A{\parallel }_{K} \leq C\parallel A{\parallel }_{q},\;A \in B\left( {X, Y}\right) . \] | Yes |
Theorem 14.51. (p. 346) If \( \alpha \left( A\right) < \infty \), then | \[ \gamma \left( A\right) \leq {\mu }_{0}\left( A\right) \leq \mu \left( A\right) \] | No |
If \( P \) dollars is deposited in a bank that pays an interest rate of 6 percent per year, compounded semiannually, then after \( t \) years the accumulated amount is\n\n\[ A = P{\left( 1 + {0.03}\right) }^{2t}. \]\n\nMore generally, if the interest rate is \( {100k} \) percent \( \left( {k = {0.06}\text{for 6 percent... | \[ {\left( 1 + \frac{k}{n}\right) }^{nt} = {\left\lbrack {\left( 1 + \frac{k}{n}\right) }^{n/k}\right\rbrack }^{kt} \rightarrow {e}^{kt} \]\n\nso\n\n\[ A = P{e}^{kt}. \] | Yes |
Suppose that \( {x}_{0} \) bacteria are placed in a nutrient solution at time \( t = 0 \), and that \( x = x\left( t\right) \) is the population of the colony at a later time \( t \) . If food and living space are unlimited, and if as a consequence the population at any moment is increasing at a rate proportional to th... | Since the rate of increase of \( x \) is proportional to \( x \) itself, we can write down the differential equation\n\n\[ \frac{dx}{dt} = {kx} \]\n\nBy separating the variables and integrating, we get\n\n\[ \frac{dx}{x} = {kdt},\;\log x = {kt} + c. \]\n\nSince \( x = {x}_{0} \) when \( t = 0 \), we have \( c = \log {x... | Yes |
Suppose, for instance, that \( {x}_{0} \) grams of matter are present initially, and decompose in a first order reaction. If \( x \) is the number of grams present at a later time \( t \), then the principle stated above yields the following differential equation:\n\n\[ - \frac{dx}{dt} = {kx},\;k > 0. \] | [Since \( {dx}/{dt} \) is the rate of growth of \( x, - {dx}/{dt} \) is its rate of decay, and (6) says that the rate of decay is proportional to \( x \) .] If we separate the variables in (6) and integrate, we obtain\n\n\[ \frac{dx}{x} = - {kdt},\;\log x = - {kt} + c. \]\n\n\( {}^{9} \) It is worth mentioning that the... | Yes |
A tank contains 50 gallons of brine in which 75 pounds of salt are dissolved. Beginning at time \( t = 0 \), brine containing 3 pounds of salt per gallon flows in at the rate of 2 gallons per minute, and the mixture (which is kept uniform by stirring) flows out at the same rate. When will there be 125 pounds of dissolv... | If \( x = x\left( t\right) \) is the number of pounds of dissolved salt in the tank at time \( t \geq 0 \), then the concentration at that time is \( x/{50} \) pounds per gallon. The rate of change of \( x \) is\n\n\[ \frac{dx}{dt} = \text{rate at which salt enters tank - rate at which salt leaves tank.} \]\n\nSince\n\... | Yes |
The mean annual rainfall in New York City is 42 in. The annual rainfall over many years is closely approximated by the normal density function with \( m = {42} \) and standard deviation \( \sigma = 2 \), \[ f\left( x\right) = \frac{1}{2\sqrt{2\pi }}{e}^{-{\left( x - {42}\right) }^{2}/8}. \] A sketch of this normal curv... | Solution (a) The proportion of years with rainfall between 40 and 44 in is \[ \frac{1}{2\sqrt{2\pi }}{\int }_{40}^{44}{e}^{-{\left( x - {42}\right) }^{2}/8}{dx} \] With the change of variable \( t = \left( {x - {42}}\right) /2 \) -and access to table of values of \( \Phi \left( t\right) \) -this becomes \[ \frac{1}{\sq... | Yes |
Example 1. Solve \( \left( {x + y}\right) {dx} - \left( {x - y}\right) {dy} = 0 \) . | We begin by writing the equation in the form suggested by the above discussion:\n\n\[ \frac{dy}{dx} = \frac{x + y}{x - y} \]\n\nSince the function on the right is clearly homogeneous of degree 0 , we know that it can be expressed as a function of \( z = y/x \) . This is easily accomplished by dividing numerator and den... | Yes |
Test the equation \( {e}^{y}{dx} + \left( {x{e}^{y} + {2y}}\right) {dy} = 0 \) for exactness, and solve it if it is exact. | Here we have\n\n\[ M = {e}^{y}\;\text{ and }\;N = x{e}^{y} + {2y}, \]\n\nso\n\n\[ \frac{\partial M}{\partial y} = {e}^{y}\;\text{ and }\;\frac{\partial N}{\partial x} = {e}^{y}. \]\n\nThus condition (4) is satisfied, and the equation is exact. This tells us that there exists a function \( f\left( {x, y}\right) \) such ... | Yes |
In the case of equation (1) we have\n\n\\[ \n\\frac{\\partial M/\\partial y - \\partial N/\\partial x}{N} = \\frac{1 - \\left( {{2xy} - 1}\\right) }{{x}^{2}y - x} = \\frac{-2\\left( {{xy} - 1}\\right) }{x\\left( {{xy} - 1}\\right) } = - \\frac{2}{x}, \n\\]\n\nwhich is a function only of \\( x \\) . | Accordingly,\n\n\\[ \n\\mu = {e}^{\\int - \\left( {2/x}\\right) {dx}} = {e}^{-2\\log x} = {x}^{-2} \n\\]\n\nis an integrating factor for (1), as we have already seen. | Yes |
Find the shape of a curved mirror such that light from a source at the origin will be reflected in a beam of rays parallel to the \( x \) -axis. | By symmetry, the mirror will have the shape of the surface of revolution generated by revolving a curve \( {APB} \) (Figure 19) about the \( x \) -axis.\n\nIt follows from the law of reflection that \( \alpha = {2\beta } \) . By the geometry of the situation, \( \phi = \beta \) and \( \theta = \alpha + \phi = {2\beta }... | Yes |
Example 1. Solve \( \frac{dy}{dx} + \frac{1}{x}y = {3x} \) . | This equation is obviously linear with \( P = 1/x \), so we have\n\n\[ \int {Pdx} = \int \frac{1}{x}{dx} = \log x\;\text{ and }\;{e}^{\int {Pdx}} = {e}^{\log x} = x. \]\n\nOn multiplying through by \( x \) and remembering (3), we obtain\n\n\[ \frac{d}{dx}\left( {xy}\right) = 3{x}^{2} \]\n\nso\n\n\[ {xy} = {x}^{3} + \ma... | Yes |
Solve \( x{y}^{\prime \prime } - {y}^{\prime } = 3{x}^{2} \) . | The variable \( y \) is missing from this equation, so (2) reduces it to\n\n\[ x\frac{dp}{dx} - p = 3{x}^{2} \]\n\nor\n\n\[ \frac{dp}{dx} - \frac{1}{x}p = {3x} \]\n\nwhich is linear. On solving this by the method of Section 10, we obtain\n\n\[ p = \frac{dy}{dx} = 3{x}^{2} + {c}_{1}x \]\n\nso\n\n\[ y = {x}^{3} + \frac{1... | Yes |
Example 2. Solve \( {y}^{\prime \prime } + {k}^{2}y = 0 \) . | With the aid of (5), we can write this in the form\n\n\[ p\frac{dp}{dy} + {k}^{2}y = 0\;\text{ or }\;{pdp} + {k}^{2}{ydy} = 0. \]\n\nIntegration yields\n\n\[ {p}^{2} + {k}^{2}{y}^{2} = {k}^{2}{a}^{2} \]\n\nso\n\n\[ p = \frac{dy}{dx} = \pm k\sqrt{{a}^{2} - {y}^{2}} \]\n\nor\n\n\[ \frac{dy}{\sqrt{{a}^{2} - {y}^{2}}} = \p... | Yes |
A point \( P \) is dragged along the \( {xy} \) -plane by a string \( {PT} \) of length \( a \) . If \( T \) starts at the origin and moves along the positive \( y \) -axis, and if \( P \) starts at \( \left( {a,0}\right) \), what is the path of \( P \) ? This curve is called a tractrix (from the Latin tractum, meaning... | It is easy to see from Figure 21 that the differential equation of the path is\n\n\[ \frac{dy}{dx} = - \frac{\sqrt{{a}^{2} - {x}^{2}}}{x}. \]\n\nOn separating variables and integrating, and using the fact that \( y = 0 \) when \( x = a \), we find that\n\n\[ y = a\log \left( \frac{a + \sqrt{{a}^{2} - {x}^{2}}}{x}\right... | Yes |
A rabbit starts at the origin and runs up the \( y \) -axis with speed \( a \) . At the same time a dog, running with speed \( b \), starts at the point \( \left( {c,0}\right) \) and pursues the rabbit. What is the path of the dog? | At time \( t \), measured from the instant both start, the rabbit will be at the point \( R = \left( {0,{at}}\right) \) and the dog at \( D = \left( {x, y}\right) \) (Figure 22). Since the line \( {DR} \) is tangent to the path, we have\n\n\[ \n\frac{dy}{dx} = \frac{y - {at}}{x}\;\text{ or }\;x{y}^{\prime } - y = - {at... | No |
Example 4. The \( y \) -axis and the line \( x = c \) are the banks of a river whose current has uniform speed \( a \) in the negative \( y \) -direction. A boat enters the river at the point \( \left( {c,0}\right) \) and heads directly toward the origin with speed \( b \) relative to the water. What is the path of the... | The components of the boat's velocity (Figure 23) are\n\n\[ \frac{dx}{dt} = - b\cos \theta \;\text{ and }\;\frac{dy}{dt} = - a + b\sin \theta ,\]\n\nso\n\n\[ \frac{dy}{dx} = \frac{-a + b\sin \theta }{-b\cos \theta } = \frac{-a + b\left( {-y/\sqrt{{x}^{2} + {y}^{2}}}\right) }{-b\left( {x/\sqrt{{x}^{2} + {y}^{2}}}\right)... | No |
Solve equation (4) for the case in which an initial current \( {I}_{0} \) is flowing and a constant emf \( {E}_{0} \) is impressed on the circuit at time \( t = 0 \) . For \( t \geq 0 \), our equation is\n\n\[ L\frac{dI}{dt} + {RI} = {E}_{0} \] | The variables can be separated, yielding\n\n\[ \frac{dI}{{E}_{0} - {RI}} = \frac{1}{L}{dt} \]\n\nOn integrating and using the initial condition \( I = {I}_{0} \) when \( t = 0 \), we get\n\n\[ \log \left( {{E}_{0} - {RI}}\right) = - \frac{R}{L}t + \log \left( {{E}_{0} - R{I}_{0}}\right) \]\n\nso\n\n\[ I = \frac{{E}_{0}... | Yes |
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