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A similar example is the space \( {l}_{2} \). It consists of all sequences of the form (1.30) for which\n\n\[ \mathop{\sum }\limits_{1}^{\infty }{\alpha }_{i}^{2} < \infty \]\n\nHere we are not immediately sure that the sum of two elements of \( {l}_{2} \) is in \( {l}_{2} \). Looking ahead a moment, we intend to inves... | Thus, if we can verify the triangle inequality, we will also have shown that the sum of two elements of \( {l}_{2} \) is in \( {l}_{2} \). Now, since the triangle inequality holds in \( {\mathbb{R}}^{n} \), we have\n\n\[ \mathop{\sum }\limits_{1}^{n}{\left( {\alpha }_{i} + {\beta }_{i}\right) }^{2} \leq \sum {\alpha }_... | Yes |
To check that \( B \) is a complete normed vector space, assume that \( \left\{ {\varphi }_{j}\right\} \) is a sequence satisfying\n\n\[ \begin{Vmatrix}{{\varphi }_{j} - {\varphi }_{k}}\end{Vmatrix} \rightarrow 0\text{ as }j, k \rightarrow \infty . \]\n\nThen for each \( \varepsilon > 0 \), there is an \( N \) satisfyi... | Thus, for each \( x \) in the interval \( \left\lbrack {a, b}\right\rbrack \), the sequence \( \left\{ {{\varphi }_{j}\left( x\right) }\right\} \) has a limit \( {c}_{x} \) as \( j \rightarrow \infty \) . Define the function \( \varphi \left( x\right) \) to have the value \( {c}_{x} \) at the point \( x \) . By (1.37),... | Yes |
Theorem 1.4. 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\n\n(1.56)\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{0}^{n}{\alpha }_{j}{\varphi }_{j} - f}\end{Vmatrix} \rightarrow 0\text{ as ... | Note that this correspondence is linear and one-to-one in both directions. We shall have more to say about this in Chapter 3. As usual, we are not satisfied with merely proving statements about \( {L}^{2} \) . We want to know if similar statements hold in other Hilbert spaces. So we examine the assumptions we have made... | No |
Theorem 1.5. Let \( \\left( {{\\alpha }_{1},{\\alpha }_{2},\\cdots }\\right) \) be a sequence of real numbers, and let \( \\left\\{ {\\varphi }_{n}\\right\\} \) be an orthonormal sequence in \( H \) . Then\n\n\[ \n\\mathop{\\sum }\\limits_{1}^{n}{\\alpha }_{i}{\\varphi }_{i}\n\]\n\nconverges in \( H \) as \( n \\righta... | Proof. For \( m < n \) ,\n\n\[ \n{\\begin{Vmatrix}\\mathop{\\sum }\\limits_{m}^{n}{\\alpha }_{i}{\\varphi }_{i}\\end{Vmatrix}}^{2} = \\mathop{\\sum }\\limits_{m}^{n}{\\alpha }_{i}^{2}\n\] | Yes |
Theorem 1.6. If \( \left\{ {\varphi }_{n}\right\} \) is complete, then for each \( f \in H \)\n\n\[ f = \mathop{\sum }\limits_{1}^{\infty }\left( {f,{\varphi }_{i}}\right) {\varphi }_{i} \]\n\nand\n\n(1.62)\n\n\[ \parallel f{\parallel }^{2} = \mathop{\sum }\limits_{1}^{\infty }{\left( f,{\varphi }_{i}\right) }^{2} \] | Proof. Let \( f \) be any element of \( H \), and let\n\n\[ {f}_{n} = \sum {\alpha }_{j}^{\left( n\right) }{\varphi }_{j} \]\n\nbe a sequence of sums of the form (1.61) which converges to \( f \) in \( H \) . In particular,(1.53) holds. Thus, there is a sequence \( \left( {{\alpha }_{0},{\alpha }_{1},\cdots }\right) \)... | Yes |
Theorem 1.7. If \( \left\{ {\varphi }_{n}\right\} \) is complete, then\n\n\[ \left( {f, g}\right) = \mathop{\sum }\limits_{1}^{\infty }\left( {f,{\varphi }_{n}}\right) \left( {g,{\varphi }_{n}}\right) \] | Proof. Set \( {\alpha }_{j} = \left( {f,{\varphi }_{j}}\right) \) . Since\n\n\[ f = \mathop{\sum }\limits_{1}^{\infty }{\alpha }_{j}{\varphi }_{j} \]\n\nwe have\n\n\[ \left( {f, g}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sum }\limits_{1}^{n}{\alpha }_{j}{\varphi }_{j}, g}\right) = \math... | Yes |
For every bounded linear functional \( F \) on a Hilbert space \( H \) there is a unique element \( y \in H \) such that\n\n\[ F\left( x\right) = \left( {x, y}\right) \text{ for all }x \in H. \]\n\nMoreover,\n\n\[ \parallel y\parallel = \mathop{\sup }\limits_{{x \in H, x \neq 0}}\frac{\left| F\left( x\right) \right| }{... | In order to get an idea how to go about proving it, let us examine (2.3) a bit more closely. If \( F \) assigns to each element \( x \) the value zero, then we can take \( y = 0 \), and the theorem is trivial. Otherwise the \( y \) we are searching for cannot vanish. However, it must be \ | No |
Theorem 2.2. Let \( N \) be a closed subspace of a Hilbert space \( H \), and let \( x \) be an element of \( H \) which is not in \( N \). Set\n\n(2.5)\n\n\[ d = \mathop{\inf }\limits_{{z \in N}}\parallel x - z\parallel \]\n\nThen there is an element \( z \in N \) such that \( \parallel x - z\parallel = d \). | Proof. By the definition of \( d \), there is a sequence \( \left\{ {z}_{n}\right\} \) of elements of \( N \) such that \( \begin{Vmatrix}{x - {z}_{n}}\end{Vmatrix} \rightarrow d \). We apply the parallelogram law (cf.(1.38)) to \( x - {z}_{n} \) and \( x - {z}_{m} \). Thus\n\n\[ {\begin{Vmatrix}\left( x - {z}_{n}\righ... | Yes |
Theorem 2.3. Let \( N \) be a closed subspace of a Hilbert space \( H \) . Then for each \( x \in H \), there are a \( v \in N \) and a \( w \) orthogonal to \( N \) such that \( x = v + w \) . This decomposition is unique. | Proof. If \( x \in N \), put \( v = x, w = 0 \) . If \( x \notin N \), let \( z \in N \) be such that \( \parallel x - z\parallel = d \), where \( d \) is given by (2.5). We set \( v = z, w = x - z \) and must show that \( w \) is orthogonal to \( N \) . Let \( u \neq 0 \) be any element of \( N \) and \( \alpha \) any... | Yes |
Corollary 2.4. If \( N \) is a closed subspace of a Hilbert space \( H \) but is not the whole of \( H \), then there is an element \( y \neq 0 \) in \( H \) which is orthogonal to \( N \) . | Proof. Let \( x \) be any element of \( H \) which is not in \( N \) . By Theorem 2.3, \( x = v + w \), where \( v \in N \) and \( w \) is orthogonal to \( N \) . Clearly, \( w \neq 0 \), for otherwise \( x \) would be in \( N \) . We can take \( w \) as the element \( y \) sought. | Yes |
Theorem 2.6. Let \( M \) be a subspace of a normed vector space \( X \), and suppose that \( f\left( x\right) \) is a bounded linear functional on \( M \). Set\n\n\[ \parallel f\parallel = \mathop{\sup }\limits_{{x \in M, x \neq 0}}\frac{\left| f\left( x\right) \right| }{\parallel x\parallel }.\]\n\nThen there is a bou... | Proof. Set\n\n\[ p\left( x\right) = \parallel f\parallel \cdot \parallel x\parallel ,\;x \in X. \]\n\nThen \( p\left( x\right) \) is a sublinear functional and\n\n\[ f\left( x\right) \leq p\left( x\right) ,\;x \in M. \]\n\nThen by the Hahn-Banach theorem there is a functional \( F\left( x\right) \) defined on the whole... | Yes |
Corollary 2.8. If \( {x}_{1} \) is an element of \( X \) such that \( f\left( {x}_{1}\right) = 0 \) for every bounded linear functional \( f \) on \( X \), then \( {x}_{1} = 0 \) . | Corollary 2.8 is an immediate consequence of Theorem 2.7. If \( {x}_{1} \neq 0 \) , there would be a bounded linear functional \( F \) on \( X \) such that \( F\left( {x}_{1}\right) = \begin{Vmatrix}{x}_{1}\end{Vmatrix} \) . Thus, \( {x}_{1} = 0 \) . | Yes |
Theorem 2.9. Let \( M \) be a subspace of a normed vector space \( X \), and suppose \( {x}_{0} \) is an element of \( X \) satisfying\n\n\( \left( {2.23}\right) \)\n\n\[ d = d\left( {{x}_{0}, M}\right) = \mathop{\inf }\limits_{{x \in M}}\begin{Vmatrix}{{x}_{0} - x}\end{Vmatrix} > 0. \]\n\nThen there is a bounded linea... | Proof. Let \( {M}_{1} \) be the set of all elements \( z \in X \) of the form\n\n(2.24)\n\n\[ z = \alpha {x}_{0} + x,\;\alpha \in \mathbb{R}, x \in M. \]\n\nDefine the functional \( f \) on \( {M}_{1} \) by \( f\left( z\right) = {\alpha d} \). Now the representation (2.24) is unique, for if \( z = {\alpha }_{1}{x}_{0} ... | Yes |
Theorem 2.10. \( {X}^{\prime } \) is a Banach space whether or not \( X \) is. | Proof. Let \( \left\{ {f}_{n}\right\} \) be a Cauchy sequence in \( {X}^{\prime } \) . Thus for any \( \varepsilon > 0 \), there is an \( N \) such that\n\n\[ \begin{Vmatrix}{{f}_{n} - {f}_{m}}\end{Vmatrix} < \varepsilon \text{ for }m, n > N \]\n\nor, equivalently,\n\n(2.26)\n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{... | Yes |
Theorem 2.11. \( {l}_{p}^{\prime } = {l}_{q} \), where \( 1/p + 1/q = 1 \) . | Proof. Suppose \( x = \left( {{x}_{1},{x}_{2},\cdots }\right) \in {l}_{p} \), and \( f \in {l}_{p}^{\prime } \) . Set \( {e}_{1} = \left( {1,0,\cdots }\right) ,{e}_{2} = \) \( \left( {0,1,0,\cdots }\right) \), and in general \( {e}_{j} \) the vector having the \( j \) -th entry equal to one and all other entries equal ... | Yes |
Lemma 2.13. A necessary and sufficient condition for (2.50) to hold is that \( g\left( c\right) = g\left( a\right) = g\left( b\right) \) at all points \( c \) of continuity of \( g \) . | This shows us how to \ | No |
Theorem 2.14. For each bounded linear functional \( f \) on \( C\left\lbrack {a, b}\right\rbrack \) there is a unique normalized function \( \widehat{g} \) of bounded variation such that\n\n(2.55)\n\n\[ f\left( x\right) = {\int }_{a}^{b}x\left( t\right) d\widehat{g}\left( t\right) ,\;x \in C\left\lbrack {a, b}\right\rb... | Proof. In case you have forgotten, it remains to prove (2.51). Let \( \varepsilon > 0 \) be given. Then there is a \( \delta > 0 \) so small and a partition\n\n\[ c + \delta = {\tau }_{0} < {\tau }_{1} < \cdots < {\tau }_{m} = b \]\n\nof the interval \( \left\lbrack {c + \delta, b}\right\rbrack \) such that\n\n\[ {V}_{... | Yes |
Theorem 3.1. If a linear operator \( A \) is continuous at one point \( {x}_{0} \in X \) , then it is bounded, and hence continuous at every point. | Proof. If \( A \) were not bounded, then for each \( n \) we could find an element \( {x}_{n} \in X \) such that\n\n\[ \begin{Vmatrix}{A{x}_{n}}\end{Vmatrix} > n\begin{Vmatrix}{x}_{n}\end{Vmatrix}\text{.}\]\n\nSet\n\n\[ {z}_{n} = \frac{{x}_{n}}{n\begin{Vmatrix}{x}_{n}\end{Vmatrix}} + {x}_{0} \]\n\nThen \( {z}_{n} \righ... | Yes |
Theorem 3.2. If \( Y \) is a Banach space, so is \( B\left( {X, Y}\right) \) . | Proof. Suppose \( \left\{ {A}_{n}\right\} \) is a Cauchy sequence of operators in \( B\left( {X, Y}\right) \) . Then for each \( \varepsilon > 0 \) there is an integer \( N \) such that\n\n\[ \begin{Vmatrix}{{A}_{n} - {A}_{m}}\end{Vmatrix} < \varepsilon \text{ for }m, n > N. \]\n\nThus for each \( x \neq 0 \) ,\n\n(3.3... | Yes |
Theorem 3.3. \( {A}^{\prime } \in B\left( {{Y}^{\prime },{X}^{\prime }}\right) \), and \( \begin{Vmatrix}{A}^{\prime }\end{Vmatrix} = \parallel A\parallel \) . | Proof. We have by (3.5),\n\n\[ \left| {{A}^{\prime }{y}^{\prime }\left( x\right) }\right| = \left| {{y}^{\prime }\left( {Ax}\right) }\right| \leq \begin{Vmatrix}{y}^{\prime }\end{Vmatrix} \cdot \parallel A\parallel \cdot \parallel x\parallel .\n\]\n\nHence,\n\n\[ \begin{Vmatrix}{{A}^{\prime }{y}^{\prime }}\end{Vmatrix}... | Yes |
Lemma 3.4. \( {S}^{ \circ } \) and \( {}^{ \circ }T \) are closed subspaces. | Proof. We consider \( {S}^{ \circ } \) ; the proof for \( {}^{ \circ }T \) is similar. Clearly, \( {S}^{ \circ } \) is a subspace, for if \( {x}_{1},{x}_{2} \) annihilate \( S \), so does \( {\alpha }_{1}{x}_{1} + {\alpha }_{2}{x}_{2} \) . Suppose \( {x}_{n}^{\prime } \in {S}^{ \circ } \) and \( {x}_{n}^{\prime } \righ... | Yes |
Lemma 3.5. If \( M \) is a closed subspace of \( X \), then \( {}^{ \circ }\left( {M}^{ \circ }\right) = M \) . | Proof. Clearly, \( x \in {}^{ \circ }\left( {M}^{ \circ }\right) \) if, and only if, \( {x}^{\prime }\left( x\right) = 0 \) for all \( {x}^{\prime } \in {M}^{ \circ } \) . But this is satisfied by all \( x \in M \) . Hence, \( M \subset {}^{ \circ }\left( {M}^{ \circ }\right) \) . Now suppose \( {x}_{1} \) is an elemen... | Yes |
Lemma 3.6. If \( S \) is a subset of \( X \), and \( M \) is the closed subspace spanned by \( S \), then \( {M}^{ \circ } = {S}^{ \circ } \) and \( M = {}^{ \circ }\left( {S}^{ \circ }\right) \) . | Proof. The second statement follows from the first, since by Lemma 3.5, \( M = {}^{ \circ }\left( {M}^{ \circ }\right) = {}^{ \circ }\left( {S}^{ \circ }\right) \) . As for the first, since \( S \subset M \), we have clearly \( {M}^{ \circ } \subset {S}^{ \circ } \) . Moreover, if \( {x}_{j} \in S \) and \( {x}^{\prime... | Yes |
Theorem 3.9. If \( X \) is complete, then it is of the second category. | Proof. Suppose \( X \) were of the first category. Then\n\n(3.16)\n\n\[ X = \mathop{\bigcup }\limits_{1}^{\infty }{W}_{k} \]\n\nwhere each \( {W}_{k} \) is nowhere dense. Thus there is a point \( {x}_{1} \) not in \( {\bar{W}}_{1} \) . Since \( {x}_{1} \) is not a limit point of \( {W}_{1} \), there is an \( {r}_{1} \)... | Yes |
Theorem 3.11. If \( X, Y \) are Banach spaces and \( A \) is a closed linear operator from \( X \) to \( Y \) with \( R\left( A\right) = Y, N\left( A\right) = \{ 0\} \), then \( {A}^{-1} \in B\left( {Y, X}\right) \) . | The reason we said \ | No |
Lemma 3.15. If a subsequence of a Cauchy sequence converges, then the whole sequence converges. | Proof. Let \( \left\{ {x}_{n}\right\} \) be a Cauchy sequence in a normed vector space \( X \), and let \( \varepsilon > 0 \) be given. Then there is an \( N \) so large that\n\n\[ \begin{Vmatrix}{{x}_{n} - {x}_{m}}\end{Vmatrix} < \varepsilon ,\;m, n > N. \]\n\nNow if \( \left\{ {x}_{n}\right\} \) has a subsequence con... | Yes |
Theorem 3.16. Let \( X, Y \) be Banach spaces, and assume that \( A \in B\left( {X, Y}\right) \) . If \( R\left( A\right) \) is closed in \( Y \), then\n\n(3.25)\n\n\[ R\left( {A}^{\prime }\right) = N{\left( A\right) }^{ \circ }, \]\n\nand hence, \( R\left( {A}^{\prime }\right) \) is closed in \( {X}^{\prime } \) (Lemm... | Proof. If \( {x}^{\prime } \in R\left( {A}^{\prime }\right) \), then there is a \( {y}^{\prime } \in {Y}^{\prime } \) such that \( {A}^{\prime }{y}^{\prime } = {x}^{\prime } \) . For \( x \in N\left( A\right) , \)\n\n\[ {x}^{\prime }\left( x\right) = {A}^{\prime }{y}^{\prime }\left( x\right) = {y}^{\prime }\left( {Ax}\... | Yes |
Theorem 3.17. Let \( X \) be a Banach space, and let \( Y \) be a normed vector space. Let \( W \) be any subset of \( B\left( {X, Y}\right) \) such that for each \( x \in X \) ,\n\n\[ \mathop{\sup }\limits_{{A \in W}}\parallel {Ax}\parallel < \infty \]\n\nThen there is a finite constant \( M \) such that \( \parallel ... | Proof. For each positive integer \( n \), let \( {S}_{n} \) denote the set of all \( x \in X \) such that \( \parallel {Ax}\parallel \leq n \) for all \( A \in W \) . Clearly, \( {S}_{n} \) is closed. For if \( \left\{ {x}_{k}\right\} \) is a sequence of elements in \( {S}_{n} \), and \( {x}_{k} \rightarrow x \), then ... | Yes |
Theorem 3.18. Let \( A \) be a closed operator from a Banach space \( X \) to a Banach space \( Y \) such that \( R\left( A\right) = Y \) . If \( Q \) is any open subset of \( D\left( A\right) \), then the image \( A\left( Q\right) \) of \( Q \) is open in \( Y \) . | Proof. Let \( D = D\left( A\right) /N\left( A\right) \), and define the operator \( \widehat{A} \) from \( D \) to \( Y \) by \( \widehat{A}\left\lbrack x\right\rbrack = {Ax} \) . Then \( \widehat{A} \) is a one-to-one operator from the Banach space \( D \) onto the Banach space \( Y \) . Consequently, it has a bounded... | Yes |
Corollary 3.19. Let \( A \) be a closed operator from a Banach space \( X \) to a Banach space \( Y \) such that \( R\left( A\right) = Y \) . If \( Q \) is any open subset of \( X \), then the image \( A\left( {Q \cap D\left( A\right) }\right) \) of \( Q \cap D\left( A\right) \) is open in \( Y \) . | This follows from the fact that \( Q \cap D\left( A\right) \) is open in \( D\left( A\right) \) . | No |
Theorem 4.1. Let \( X \) be a normed vector space, and let \( A = I - K \), where \( K \) is of the form (4.4). If \( N\left( A\right) = \{ 0\} \), then \( R\left( A\right) = X \) . Otherwise \( R\left( A\right) \) is closed in \( X \), and \( N\left( A\right) \) is finite dimensional, having the same dimension as \( N... | Yes, I shall explain everything. When \( {x}_{1}^{\prime }\left( {x}_{1}\right) = 1 \), then \( R\left( A\right) \) consists of the annihilators of \( {x}_{1}^{\prime } \), and hence, is closed (Lemma 3.4). Moreover, \( N\left( A\right) \) is the subspace spanned by \( {x}_{1} \), while \( N\left( {A}^{\prime }\right) ... | Yes |
Corollary 4.3. A finite dimensional normed vector space is always complete. | Proof. Suppose \( \dim X = n \), and let \( {x}_{1},\cdots ,{x}_{n} \) be a basis for \( X \) (i.e., a set of \( n \) linearly independent elements). Then each \( x \in X \) can be written in the form\n\n\[ x = {\alpha }_{1}{x}_{1} + \cdots + {\alpha }_{n}{x}_{n} \]\n\nSet\n\n\[ \parallel x{\parallel }_{1} = {\left( \m... | Yes |
Corollary 4.4. If \( M \) is a finite dimensional subspace of a normed vector space, then \( M \) is closed. | Proof. If \( \left\{ {x}^{\left( k\right) }\right\} \) is a sequence of elements of \( M \) such that \( {x}^{\left( k\right) } \rightarrow x \) in \( X \), it is a Cauchy sequence in \( M \) . Since \( M \) is complete with respect to any norm, \( {x}^{\left( k\right) } \) has a limit in \( M \) which must coincide wi... | "No" |
Corollary 4.5. If \( X \) is a finite dimensional normed vector space, then every bounded closed set \( T \) in \( X \) is compact. | Proof. Let \( {x}_{1},\cdots ,{x}_{n} \) be a basis for \( X \) . Then every element \( x \in X \) can be expressed in the form (4.15). Set\n\n(4.16)\n\n\[ \parallel x{\parallel }_{0} = \mathop{\sum }\limits_{1}^{n}\left| {\alpha }_{i}\right| \]\n\nThis is a norm on \( X \), and consequently it is equivalent to all oth... | Yes |
Lemma 4.7. Let \( M \) be a closed subspace of a normed vector space \( X \) . If \( M \) is not the whole of \( X \), then for each number \( \theta \) satisfying \( 0 < \theta < 1 \) there is an element \( {x}_{\theta } \in X \) such that\n\n\[ \begin{Vmatrix}{x}_{\theta }\end{Vmatrix} = 1\text{ and }d\left( {{x}_{\t... | Proof. Since \( M \neq X \), there is an \( {x}_{1} \in X \smallsetminus M \) (i.e., in \( X \) but not in \( M \) ). Since \( M \) is closed, \( d = d\left( {{x}_{1}, M}\right) > 0 \) . For any \( \varepsilon > 0 \) there is an \( {x}_{0} \in M \) such that\n\n\[ \begin{Vmatrix}{{x}_{1} - {x}_{0}}\end{Vmatrix} < d + \... | Yes |
Lemma 4.8. If \( V \) is an \( n \) -dimensional vector space, then every subspace of \( V \) is of dimension \( \leq n \) . | Proof. Let \( W \) be a subspace of \( V \) . If \( W \) consists only of the element 0, then \( \dim W = 0 \) . Otherwise, there is an element \( {w}_{1} \neq 0 \) in \( W \) . If there does not exist a \( w \in W \) such that \( {w}_{1} \) and \( w \) are linearly independent, then \( \dim W = 1 \) . Otherwise, let \... | Yes |
Theorem 4.10. Let \( X \) be a Banach space, and assume that \( K \in B\left( X\right) \) is the limit in norm of a sequence of operators of finite rank. If \( A = I - K \) , then \( R\left( A\right) \) is closed in \( X \), and \( \dim N\left( A\right) = \dim N\left( {A}^{\prime }\right) < \infty \) . | Proof. From (4.31) we see that \( N\left( A\right) = N\left( {I - {B}_{n}^{-1}{K}_{n}}\right) \) . Since \( {A}^{\prime } = (I - \) \( {\left. {B}_{n}^{-1}{K}_{n}\right) }^{\prime }{B}_{n}^{\prime } \), it follows that \( \dim N\left( {A}^{\prime }\right) = \dim N\left\lbrack {\left( I - {B}_{n}^{-1}{K}_{n}\right) }^{\... | Yes |
Theorem 4.11. Let \( X \) be a normed vector space and \( Y \) a Banach space. If \( L \) is in \( B\left( {X, Y}\right) \) and there is a sequence \( \left\{ {K}_{n}\right\} \subset K\left( {X, Y}\right) \) such that \[ \begin{Vmatrix}{L - {K}_{n}}\end{Vmatrix} \rightarrow 0\text{ as }n \rightarrow 0, \] then \( L \) ... | To illustrate Theorem 4.11, consider the operator \( K \) on \( {l}_{p} \) given by \[ K\left( {{x}_{1},{x}_{2},\cdots ,{x}_{k},\cdots }\right) = \left( {{x}_{1},{x}_{2}/2,\cdots ,{x}_{k}/k,\cdots }\right) . \] We shall see that \( K \) is a compact operator. To that end, set \[ {F}_{n}\left( {{x}_{1},{x}_{2},\cdots }\... | No |
Theorem 4.12. Let \( X \) be a Banach space and let \( K \) be an operator in \( K\left( X\right) \). Set \( A = I - K \). Then, \( R\left( A\right) \) is closed in \( X \) and \( \dim N\left( A\right) = \dim N\left( {A}^{\prime }\right) \) is finite. In particular, either \( R\left( A\right) = X \) and \( N\left( A\ri... | The last statement of Theorem 4.12 is known as the Fredholm alternative. To show that \( R\left( A\right) \) is closed we make use of the trivial | No |
Lemma 4.13. Let \( X, Y \) be normed vector spaces, and let \( A \) be a linear operator from \( X \) to \( Y \) . Then for each \( x \) in \( D\left( A\right) \) and \( \varepsilon > 0 \) there is an element \( {x}_{0} \in D\left( A\right) \) such that\n\n\[ A{x}_{0} = {Ax},\;d\left( {{x}_{0}, N\left( A\right) }\right... | Proof. There is an \( {x}_{1} \in N\left( A\right) \) such that \( \begin{Vmatrix}{x - {x}_{1}}\end{Vmatrix} < d\left( {x, N\left( A\right) }\right) + \varepsilon \) . Set \( {x}_{0} = x - {x}_{1} \) | No |
Lemma 4.14. If \( {x}_{1}^{\prime },\cdots ,{x}_{m}^{\prime } \) are linearly independent vectors in \( {X}^{\prime } \), then there are vectors \( {x}_{1},\cdots ,{x}_{m} \) in \( X \) such that\n\n\[ \n{x}_{j}^{\prime }\left( {x}_{k}\right) = {\delta }_{jk} = \left\{ {\begin{array}{ll} 1 & j = k, \\ 0, & j \neq k, \e... | Proof. Assume that the lemma is true for \( m = l - 1 \geq 1 \) . Let \( {x}_{1}^{\prime },\cdots ,{x}_{l}^{\prime } \) be linearly independent vectors in \( {X}^{\prime } \) . Then, for any \( x \in X \), \n\n\[ \n{x}_{j}^{\prime }\left( {x - \mathop{\sum }\limits_{1}^{{l - 1}}{x}_{k}^{\prime }\left( x\right) {x}_{k}}... | Yes |
Lemma 4.16. Let \( X, Y \) be normed vector spaces. A linear operator \( K \) from \( X \) to \( Y \) is compact if, and only if, the image \( K\left( U\right) \) of a bounded set \( U \subset X \) is relatively compact in \( Y \). | Proof. Suppose \( K \in K\left( {X, Y}\right) \) . Since \( U \) is bounded, there is a constant \( C \) such that \( \parallel x\parallel \leq C \) for all \( x \in U \) . Let \( \left\{ {K{x}_{n}}\right\} \) be a sequence in \( K\left( U\right) \) . Then \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} \leq C \) . Since \( K \... | Yes |
Theorem 4.17. If a set \( U \subset X \) is relatively compact, then it is totally bounded. If \( X \) is complete and \( U \) is totally bounded, then \( U \) is relatively compact. | Proof. Assume that \( U \) is relatively compact, and let \( \varepsilon > 0 \) be given. We shall show that \( U \) has an \( \varepsilon \) -net consisting of a finite number of points of \( U \) . (We actually do not need to show that the points belong to \( U \) .) Let \( {x}_{1} \) be any point of \( U \) . If \( ... | No |
Theorem 5.4. If \( A \in \Phi \left( {X, Y}\right) \), there is a closed subspace \( {X}_{0} \) of \( X \) such that (5.2) holds and a subspace \( {Y}_{0} \) of \( Y \) of dimension \( \beta \left( A\right) \) such that (5.3) holds. Moreover, there is an operator \( {A}_{0} \in B\left( {Y, X}\right) \) such that\n\n(a)... | To prove statement \( \left( e\right) \), we note that the operator \( {F}_{1} = I - {A}_{0}A \) is equal to \( I \) on \( N\left( A\right) \) and vanishes on \( {X}_{0} \) . Hence, it is in \( B\left( X\right) \) by Lemma 5.2. Similar reasoning gives \( \left( f\right) \). | No |
Theorem 5.5. Let \( A \) be an operator in \( B\left( {X, Y}\right) \) and assume that there are operators \( {A}_{1},{A}_{2} \in B\left( {Y, X}\right) ,{K}_{1} \in K\left( X\right) ,{K}_{2} \in K\left( Y\right) \) such that\n\n(5.7)\n\n\[ \n{A}_{1}A = I - {K}_{1}\text{ on }X \n\]\n\nand\n\n(5.8)\n\n\[ \nA{A}_{2} = I -... | Proof. Since \( N\left( A\right) \subset N\left( {{A}_{1}A}\right) \), we have \( \alpha \left( A\right) \leq \alpha \left( {I - {K}_{1}}\right) < \infty \) (Theorem 4.12). Likewise, \( R\left( A\right) \supset R\left( {A{A}_{2}}\right) = R\left( {I - {K}_{2}}\right) \). Hence, \( N\left( {A}^{\prime }\right) \subset N... | No |
Lemma 5.6. Let \( X \) be a normed vector space, and suppose that \( X = N \) \( \oplus {X}_{0} \), where \( {X}_{0} \) is a closed subspace and \( N \) is finite dimensional. If \( {X}_{1} \) is a subspace of \( X \) containing \( {X}_{0} \), then \( {X}_{1} \) is closed. | Proof. Set \( M = N \cap {X}_{1} \) . Then \( {X}_{1} = {X}_{0} \oplus M \) . For if \( x \in {X}_{1}, x = {x}_{0} + z \) , where \( {x}_{0} \in {X}_{0} \) and \( z \in N \) . Since \( {x}_{0} \in {X}_{1} \), the same is true for \( z \) . Hence, \( z \in M \) . We now apply Lemma 5.2. | Yes |
Theorem 5.7. 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\n\n\[ i\left( {BA}\right) = i\left( B\right) + i\left( A\right) \] | Proof. By Theorem 5.4, there are operators\n\n\[ {A}_{0} \in B\left( {Y, X}\right) ,{B}_{0} \in B\left( {Z, Y}\right) ,{F}_{1} \in K\left( X\right) ,{F}_{2},{F}_{3} \in K\left( Y\right) ,{F}_{4} \in K\left( Z\right) \]\n\nsuch that\n\n\[ {A}_{0}A = I - {F}_{1}\text{ on }X,\;A{A}_{0} = I - {F}_{2}\text{ on }Y \]\n\nand\... | Yes |
Lemma 5.8. Let \( V, W \) be finite-dimensional vector spaces, and let \( L \) be a linear operator from \( V \) to \( W \) . Then \( \dim R\left( L\right) \leq \dim D\left( L\right) \) . If \( L \) is one-to-one, then \( \dim R\left( L\right) = \dim D\left( L\right) \) . | Proof. The second statement follows from the first, since \( {L}^{-1} \) exists and\n\n\[ D\left( {L}^{-1}\right) = R\left( L\right) ,\;R\left( {L}^{-1}\right) = D\left( L\right) . \]\n\nTo prove the first statement, assume that \( \dim D\left( L\right) < n \), and let \( {w}_{1},\cdots ,{w}_{n} \) be any \( n \) vecto... | Yes |
Lemma 5.9. Suppose \( A \in \Phi \left( {X, Y}\right) \), and let \( {A}_{0} \) be any operator satisfying (e) and (f) of Theorem 5.4. Then \( {A}_{0} \in \Phi \left( {Y, X}\right) \) and \( i\left( {A}_{0}\right) = - i\left( A\right) \) . | Proof. By hypothesis,(5.10) holds with \( {F}_{1} \in K\left( X\right) \) and \( {F}_{2} \in K\left( Y\right) \) . Thus, by Theorem 5.5 (with \( X \) and \( Y \) interchanged), we see that \( {A}_{0} \in \Phi \left( {Y, X}\right) \) . Moreover, by Theorem 5.7, \[ i\left( {A}_{0}\right) + i\left( A\right) = i\left( {I -... | Yes |
Theorem 5.10. If \( A \in \Phi \left( {X, Y}\right) \) and \( K \in K\left( {X, Y}\right) \), then \( A + K \in \Phi \left( {X, Y}\right) \) and\n\n\[ i\left( {A + K}\right) = i\left( A\right) \] | Proof. By Theorem 5.4, there are \( {A}_{0} \in B\left( {Y, X}\right) ,{F}_{1} \in K\left( X\right) ,{F}_{2} \in K\left( Y\right) \) such that\n\n\[ {A}_{0}A = I - {F}_{1}\text{ on }X,\;A{A}_{0} = I - {F}_{2}\text{ on }Y. \]\n\nHence,\n\n\[ {A}_{0}\left( {A + K}\right) = I - {F}_{1} + {A}_{0}K = I - {K}_{1}\text{ on }X... | Yes |
Theorem 5.11. Assume that \( A \in \Phi \left( {X, Y}\right) \) . Then there is an \( \eta > 0 \) such that for any \( T \in B\left( {X, Y}\right) \) satisfying \( \parallel T\parallel < \eta \), one has \( A + T \in \Phi \left( {X, Y}\right) \), (5.15) \[ i\left( {A + T}\right) = i\left( A\right) \] and (5.16) \[ \alp... | Proof. By (5.14), we have \[ {A}_{0}\left( {A + T}\right) = I - {F}_{1} + {A}_{0}T\text{ on }X \] and \[ \left( {A + T}\right) {A}_{0} = I - {F}_{2} + T{A}_{0}\text{ on }Y. \] We take \( \eta = {\begin{Vmatrix}{A}_{0}\end{Vmatrix}}^{-1} \) . Then \( \begin{Vmatrix}{{A}_{0}T}\end{Vmatrix} \leq \begin{Vmatrix}{A}_{0}\end... | Yes |
Lemma 5.12. Let \( X \) be a vector space, and assume that \( X = N \oplus {X}_{0} \), where \( N \) is finite dimensional. If \( M \) is a subspace of \( X \) such that \( M \cap {X}_{0} = \{ 0\} \) , then \( \dim M \leq \dim N \) . | Proof. Suppose \( \dim N < n \), and let \( {x}_{1},\cdots ,{x}_{n} \) be any \( n \) vectors in \( M \) . By hypothesis,\n\n\[ \n{x}_{k} = {x}_{k0} + {x}_{k1},\;{x}_{k0} \in {X}_{0},{x}_{k1} \in N,1 \leq k \leq n.\n\]\n\nSince \( \dim N < n \), there are scalars \( {\alpha }_{1},\cdots ,{\alpha }_{n} \) not all zero s... | Yes |
Theorem 5.13. Assume that \( A \in B\left( {X, Y}\right) \) and \( B \in B\left( {Y, Z}\right) \) are such that \( {BA} \in \Phi \left( {X, Z}\right) \) . Then \( A \in \Phi \left( {X, Y}\right) \) if, and only if, \( B \in \Phi \left( {Y, Z}\right) \) . | Proof. First assume that \( A \in \Phi \left( {X, Y}\right) \), and let \( {A}_{0} \) be an operator satisfying Theorem 5.4. Thus,\n\n\[ \n{BA}{A}_{0} = B - B{F}_{2}\text{ on }Y \n\]\n\nwhere \( {F}_{2} \in K\left( Y\right) \) . Now \( {A}_{0} \in \Phi \left( {Y, X}\right) \) by Lemma 5.9, while \( {BA} \in \Phi \left(... | Yes |
Theorem 5.14. Assume that \( A \in B\left( {X, Y}\right) \) and \( B \) is in \( B\left( {Y, Z}\right) \) are such that \( {BA} \in \Phi \left( {X, Z}\right) \) . If \( \alpha \left( B\right) < \infty \), then \( A \in \Phi \left( {X, Y}\right) \) and \( B \in \Phi \left( {Y, Z}\right) \) . | Proof. Since \( R\left( B\right) \supset R\left( {BA}\right) \), we see by Lemma 5.6 that \( R\left( B\right) \) is closed. Moreover, \( \beta \left( B\right) \leq \beta \left( {BA}\right) \), and, hence, \( B \in \Phi \left( {Y, Z}\right) \) . We now apply Theorem 5.13. | No |
Theorem 5.16. Assume that \( A \) in \( B\left( {X, Y}\right) \) and \( B \) in \( B\left( {Y, Z}\right) \) are such that \( {BA} \in \Phi \left( {X, Z}\right) \) . If \( \beta \left( A\right) < \infty \), then \( A \in \Phi \left( {X, Y}\right) \) and \( B \in \Phi \left( {Y, Z}\right) \) . | Proof. Taking adjoints, we have \( {A}^{\prime }{B}^{\prime } \in \Phi \left( {{Z}^{\prime },{X}^{\prime }}\right) \) by Theorem 5.15. Moreover, \( \alpha \left( {A}^{\prime }\right) = \beta \left( A\right) < \infty \) . We can now apply Theorem 5.14 to conclude that \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^... | Yes |
Theorem 5.17. If \( K \) is in \( K\left( X\right) \) and \( A = I - K \), then there is an integer \( n \geq 1 \) such that \( N\left( {A}^{n}\right) = N\left( {A}^{k}\right) \) for all \( k \geq n \) . | Proof. First we note that the theorem is true if there is an integer \( k \) such that \( N\left( {A}^{k}\right) = N\left( {A}^{k + 1}\right) \) . For if \( j > k \) and \( x \in N\left( {A}^{j + 1}\right) \), then \( {A}^{j - k}x \in \) \( N\left( {A}^{k + 1}\right) = N\left( {A}^{k}\right) \), showing that \( x \in N... | Yes |
Theorem 5.18. A necessary and sufficient condition that \( A \in \Phi \left( X\right) \) with \( r\left( A\right) < \infty \) and \( {r}^{\prime }\left( A\right) < \infty \) is that there exist an integer \( n \geq 1 \) and operators \( E \) in \( B\left( X\right) \) and \( K \) in \( K\left( X\right) \) such that\n\n\... | Proof. To prove the sufficiency of (5.33), set \( W = {A}^{n} \) . Then by Theorem \( {5.5}, W \in \Phi \left( X\right) \) . Now by Theorem 5.17, there is an integer \( m \) such that\n\n\[ N\left\lbrack {\left( I - K\right) }^{j}\right\rbrack = N\left\lbrack {\left( I - K\right) }^{m}\right\rbrack, R\left\lbrack {\lef... | Yes |
Lemma 5.19. Let \( {A}_{1},\cdots ,{A}_{n} \) be operators in \( B\left( X\right) \) which commute, and suppose that their product \( A = {A}_{1}\cdots {A}_{n} \) is in \( \Phi \left( X\right) \) . Then each \( {A}_{k} \) is in \( \Phi \left( X\right) \) . | Proof. Clearly, \( N\left( {A}_{k}\right) \subset N\left( A\right) \) and \( R\left( {A}_{k}\right) \supset R\left( A\right) \) . Thus, \( \alpha \left( {A}_{k}\right) \) and \( \beta \left( {A}_{k}\right) \) are both finite. Moreover, by (5.3), \( X = R\left( A\right) \oplus {Y}_{0} \), where \( {Y}_{0} \) is a finite... | Yes |
Lemma 5.20. Assume that \( A \) is in \( B\left( {X, Y}\right) \) and that \( \alpha \left( A\right) < \infty \) . Let \( P \) be defined by (5.34). Then \( R\left( A\right) \) is closed in \( Y \) if and only if\n\n(5.35)\n\n\[ \parallel \left( {I - P}\right) x\parallel \leq \parallel {Ax}\parallel ,\;x \in X. \] | Proof. If \( R\left( A\right) \) is closed, then the restriction of \( A \) to \( {X}_{0} \) is one-to-one and has closed range. Hence, by Theorem 3.12,\n\n(5.36)\n\n\[ \parallel x\parallel \leq C\parallel {Ax}\parallel ,\;x \in {X}_{0}. \]\n\nBut for any \( x \in X,\left( {I - P}\right) x \in {X}_{0} \) and \( A\left(... | Yes |
Theorem 5.21. Suppose \( A \) is in \( B\left( {X, Y}\right) \) . Then \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) if and only if there is a seminorm \( \left| \cdot \right| \) compact relative to the norm of \( X \) such that (5.40) holds. | Proof. We have proved the \ | No |
Theorem 5.22. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( K \) is in \( K\left( {X, Y}\right) \), then \( A + K \in \) \( {\Phi }_{ + }\left( {X, Y}\right) \) and (5.13) holds. | Proof. By Theorem 5.21,\n\n\[ \parallel x\parallel \leq C\parallel \left( {A + K}\right) x\parallel + \left| x\right| + C\parallel {Kx}\parallel ,\;x \in X. \]\n\nSet\n\n\[ {\left| x\right| }_{0} = \left| x\right| + C\parallel {Kx}\parallel \]\n\nThen, \( {\left| x\right| }_{0} \) is a seminorm which is compact relativ... | Yes |
Lemma 5.24. Assume \( A \in {\Phi }_{ + }\left( {X, Y}\right) ,\widetilde{X} = N \oplus X, n = \dim N < \infty ,\widetilde{A} \in \) \( B\left( {\widetilde{X}, Y}\right) \) satisfies\n\n\[ \widetilde{A}\left( {z + x}\right) = {Cz} + {Ax},\;z \in N, x \in X, \]\n\nwhere \( C \) is in \( B\left( {N, Y}\right) \) . Then \... | Proof. We may assume that \( n = 1 \), i.e., \( N = \left\{ {z}_{0}\right\} \) . There are three cases.\n\nCase 1. \( C{z}_{0} = 0 \) . In this case we have \( N\left( \widetilde{A}\right) = N\left( A\right) \oplus \left\{ {z}_{0}\right\} ,\alpha \left( \widetilde{A}\right) = \) \( \alpha \left( A\right) + 1, R\left( \... | Yes |
Proposition 5.25. \( A \in \Phi \left( {X, Y}\right) \) if and only if \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( {A}^{\prime } \in \) \( {\Phi }_{ + }\left( {{Y}^{\prime },{X}^{\prime }}\right) \) | Proof. By Theorem 5.15, \( A \in \Phi \left( {X, Y}\right) \) implies \( {A}^{\prime } \in \Phi \left( {{Y}^{\prime },{X}^{\prime }}\right) \), so that the \ | No |
Theorem 5.26. 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 (5.9) holds. | Proof. By Theorem 5.21,\n\n\[ \parallel x\parallel \leq {C}_{1}\parallel {Ax}\parallel + {\left| x\right| }_{1},\;x \in X, \]\n\n\[ \parallel y\parallel \leq {C}_{2}\parallel {By}\parallel + {\left| y\right| }_{2},\;y \in Y \]\n\nwhere the seminorms \( {\left| \cdot \right| }_{1} \) and \( {\left| \cdot \right| }_{2} \... | Yes |
Theorem 5.28. If \( A \in {\Phi }_{ - }\left( {X, Y}\right) \) and \( K \) is in \( K\left( {X, Y}\right) \), then \( A + K \in \) \( {\Phi }_{ - }\left( {X, Y}\right) \) and (5.13) holds. | Proof. Now, \( {A}^{\prime } \in {\Phi }_{ + }\left( {{Y}^{\prime },{X}^{\prime }}\right) \) by hypothesis, and \( {K}^{\prime } \in K\left( {{Y}^{\prime },{X}^{\prime }}\right) \) by Theorem 4.15. Thus, \( {A}^{\prime } + {K}^{\prime } \in {\Phi }_{ + }\left( {{Y}^{\prime },{X}^{\prime }}\right) \) and\n\n\[ i\left( {... | Yes |
Theorem 5.29. If \( A \in {\Phi }_{ - }\left( {X, Y}\right) \), then there is an \( \eta > 0 \) such that \( \parallel T\parallel < \eta \) for \( T \) in \( B\left( {X, Y}\right) \) implies that \( A + T \in {\Phi }_{ - }\left( {X, Y}\right) \), and (5.15),(5.41) hold. | Proof. We know that \( {T}^{\prime } \in B\left( {{Y}^{\prime },{X}^{\prime }}\right) \) and that \( \begin{Vmatrix}{T}^{\prime }\end{Vmatrix} < \eta \) (Theorem 3.3). Hence, \( {A}^{\prime } + {T}^{\prime } \in {\Phi }_{ + }\left( {{Y}^{\prime },{X}^{\prime }}\right) \) for \( \eta \) sufficiently small, and\n\n\[ i\l... | Yes |
Theorem 5.30. 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 (5.9) holds. | Proof. By definition, \( {A}^{\prime } \in {\Phi }_{ + }\left( {{Y}^{\prime },{X}^{\prime }}\right) \) and \( {B}^{\prime } \in {\Phi }_{ + }\left( {{Z}^{\prime },{Y}^{\prime }}\right) \) . Thus, \( {\left( BA\right) }^{\prime } = \) \( {A}^{\prime }{B}^{\prime } \in {\Phi }_{ + }\left( {{Z}^{\prime },{X}^{\prime }}\ri... | Yes |
Theorem 5.31. If \( A \) is in \( B\left( {X, Y}\right), B \) is in \( B\left( {Y, Z}\right) \) and \( {BA} \in {\Phi }_{ - }\left( {X, Z}\right) \) , then \( B \in {\Phi }_{ - }\left( {Y, Z}\right) \) . | Proof. Since \( \dim R{\left( BA\right) }^{ \circ } < \infty \), there is a subspace \( {Z}_{0} \) such that \( \dim {Z}_{0} < \) \( \infty \) and \( Z = R\left( {BA}\right) \oplus {Z}_{0} \) (Lemma 5.3). Since \( R\left( B\right) \supset R\left( {BA}\right) \), we know that \( R\left( B\right) \) is closed (Lemma 5.6)... | Yes |
Theorem 5.32. If \( A \) is in \( B\left( {X, Y}\right), B \) is in \( B\left( {Y, Z}\right) \) and \( {BA} \in {\Phi }_{ + }\left( {X, Z}\right) \) , then \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) . | Proof. We merely note that \( {A}^{\prime }{B}^{\prime } \in {\Phi }_{ - }\left( {{Z}^{\prime },{X}^{\prime }}\right) \) . Theorem 5.31 now implies that \( {A}^{\prime } \in {\Phi }_{ - }\left( {{Y}^{\prime },{X}^{\prime }}\right) \), which means that \( A \in {\Phi }_{ + }\left( {X, Y}\right) \). | Yes |
Lemma 5.33. If \( {X}_{1} \) is a closed complemented subspace of a Banach space \( X \), then there is a bounded projection \( P \) on \( X \) with \( R\left( P\right) = {X}_{1} \) . | Proof. Let \( {X}_{2} \) be a closed complement of \( {X}_{1} \) in \( X \) . Then each \( x \in X \) is of the form\n\n\[ x = {x}_{1} + {x}_{2},\;{x}_{k} \in {X}_{k} \]\n\nDefine \( {Px} = {x}_{1} \) . Then \( P \) is a projection, and it is easily verified that it is a closed operator. Hence, it is bounded by the clo... | Yes |
Theorem 5.34. If \( A \in {\Phi }_{ + }\left( {X, Y}\right) \) and \( R\left( A\right) \) is complemented in \( Y \), then there is an \( {A}_{0} \in B\left( {Y, X}\right) \) such that \( {A}_{0}A \in \Phi \left( X\right) \) . | Proof. Let \( P \) be a bounded projection from \( Y \) to \( R\left( A\right) \) . There is a closed subspace \( {X}_{0} \in X \) such that\n\n\[ X = {X}_{0} \oplus N\left( A\right) \]\n\nThen \( A \) has a bounded inverse \( \widehat{A} \) from \( R\left( A\right) \) to \( {X}_{0} \) . Let \( {A}_{0} = \widehat{A}P \... | Yes |
Theorem 5.35. If \( A \in {\Phi }_{ - }\left( {X, Y}\right) \) with \( N\left( A\right) \) complemented, then there is an \( {A}_{0} \in B\left( {Y, X}\right) \) such that \( A{A}_{0} \in \Phi \left( Y\right) \) . | Proof. There is a finite dimensional subspace \( {Y}_{0} \in Y \) such that\n\n\[ Y = R\left( A\right) \oplus {Y}_{0} \]\n\nLet \( P \) be the bounded projection onto \( R\left( A\right) \) which vanishes on \( {Y}_{0} \), and define \( {A}_{0} \) as above. Then\n\n\[ A{A}_{0} = \left\{ \begin{array}{ll} I & \text{ on ... | Yes |
Theorem 5.36. If \( A \) is in \( B\left( {X, Y}\right), B \) is in \( B\left( {Y, Z}\right) \) and \( {BA} \in \Phi \left( {X, Z}\right) \) , then \( A \in {\Phi }_{ + }\left( {X, Y}\right) \), and \( B \in {\Phi }_{ - }\left( {Y, Z}\right) \) . Moreover, \( R\left( A\right) \) and \( N\left( B\right) \) are complemen... | Proof. The first statement follows from Theorems 5.31 and 5.32. Consequently, there is a closed subspace \( {X}_{0} \subset X \) such that\n\n\[ X = {X}_{0} \oplus N\left( A\right) \]\n\nLet\n\n\[ {Y}_{1} = R\left( A\right) \cap N\left( B\right) ,\;{X}_{1} = {A}^{-1}\left( {Y}_{1}\right) \cap {X}_{0}. \]\n\nSince \( {X... | Yes |
Theorem 6.3. \( {\Phi }_{A} \) and \( \rho \left( A\right) \) are open sets. Hence, \( \sigma \left( A\right) \) is a closed set. | Proof. If \( {\lambda }_{0} \in {\Phi }_{A} \), then by definition \( A - {\lambda }_{0} \in \Phi \left( X\right) \) . By Theorem 5.11, there is a constant \( \eta > 0 \) such that \( \left| \mu \right| < \eta \) implies that \( A - {\lambda }_{0} - \mu \in \Phi \left( X\right) \) and\n\n\[ i\left( {A - {\lambda }_{0} ... | Yes |
Theorem 6.4. For \( A \) in \( B\left( X\right) \), set\n\n(6.6)\n\n\[ \n{r}_{\sigma }\left( A\right) = \mathop{\inf }\limits_{n}{\begin{Vmatrix}{A}^{n}\end{Vmatrix}}^{1/n}. \]\n\nThen \( \rho \left( A\right) \) contains all scalars \( \lambda \) such that \( \left| \lambda \right| > {r}_{\sigma }\left( A\right) \) . | This theorem is an immediate consequence of the following two lemmas:\n\nLemma 6.5. If \( \left| | No |
Lemma 6.6. If \( \lambda \in \sigma \left( A\right) \), then for each \( n \) we have \( {\lambda }^{n} \in \sigma \left( {A}^{n}\right) \) . | Proof. Suppose \( {\lambda }^{n} \in \rho \left( {A}^{n}\right) \) . Now,\n\n(6.9)\n\n\[ {A}^{n} - {\lambda }^{n} = \left( {A - \lambda }\right) B = B\left( {A - \lambda }\right) \]\n\nwhere\n\n\[ B = {A}^{n - 1} + \lambda {A}^{n - 2} + \cdots + {\lambda }^{n - 2}A + {\lambda }^{n - 1}. \]\n\nThus, \( \alpha \left( {A ... | Yes |
Theorem 6.7. If \( \lambda \in \sigma \left( A\right) \), then \( p\left( \lambda \right) \in \sigma \left( {p\left( A\right) }\right) \) for any polynomial \( p\left( t\right) \) . | Proof. Since \( \lambda \) is a root of \( p\left( t\right) - p\left( \lambda \right) \), we have\n\n\[ p\left( t\right) - p\left( \lambda \right) = \left( {t - \lambda }\right) q\left( t\right) \]\n\nwhere \( q\left( t\right) \) is a polynomial with real coefficients. Hence,\n\n(6.10)\n\n\[ p\left( A\right) - p\left( ... | Yes |
Theorem 6.8. If \( X \) is a complex Banach space, then \( \mu \in \sigma \left( {p\left( A\right) }\right) \) if and only if \( \mu = p\left( \lambda \right) \) for some \( \lambda \in \sigma \left( A\right) \), i.e., if (6.13) holds. | Proof. We have proved it in one direction already (Theorem 6.7). To prove it in the other, let \( {\gamma }_{1},\cdots ,{\gamma }_{n} \) be the (complex) roots of \( p\left( t\right) - \mu \) . For a complex Banach space they are all scalars. Thus,\n\n\[ p\left( A\right) - \mu = c\left( {A - {\gamma }_{1}}\right) \cdot... | Yes |
Lemma 6.11. If \( \left| z\right| > \lim \sup {\begin{Vmatrix}{A}^{n}\end{Vmatrix}}^{1/n} \), then\n\n\[{\left( z - A\right) }^{-1} = \mathop{\sum }\limits_{1}^{\infty }{z}^{-n}{A}^{n - 1}\]\n\nwhere the convergence is in the norm of \( B\left( X\right) \) . | Proof. By hypothesis, there is a number \( \delta < 1 \) such that\n\n\[{\begin{Vmatrix}{A}^{n}\end{Vmatrix}}^{1/n} \leq \delta \left| z\right|\]\n\nfor \( n \) sufficiently large. Set \( B = {z}^{-1}A \). Then, by (6.15), we have\n\n\[\mathop{\sum }\limits_{0}^{\infty }\begin{Vmatrix}{B}^{n}\end{Vmatrix} < \infty\]\n\... | Yes |
Theorem 6.13. \( {r}_{\sigma }\left( A\right) = \mathop{\max }\limits_{{\lambda \in \sigma \left( A\right) }}\left| \lambda \right| \) and \( {\begin{Vmatrix}{A}^{n}\end{Vmatrix}}^{1/n} \rightarrow {r}_{\sigma }\left( A\right) \) as \( n \rightarrow \infty \) . | Proof. Set \( m = \mathop{\max }\limits_{{\lambda \in \sigma \left( A\right) }}\left| \lambda \right| \), and let \( \varepsilon > 0 \) be given. If \( C \) is a circle about the origin of radius \( a = m + \varepsilon \), we have by Theorem 6.12,\n\n\[ \begin{Vmatrix}{A}^{n}\end{Vmatrix} \leq \frac{1}{2\pi }{a}^{n}M\l... | Yes |
Theorem 6.14. If \( \left| z\right| > {r}_{\sigma }\left( A\right) \), then (6.14) holds with convergence in \( B\left( X\right) \) . | Now let \( b \) be any number greater than \( {r}_{\sigma }\left( A\right) \), and let \( f\left( z\right) \) be a complex valued function that is analytic in \( \left| z\right| < b \) . Thus,\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{0}^{\infty }{a}_{k}{z}^{k},\;\left| z\right| < b. \]\n\nWe can define \( f\left... | Yes |
Theorem 6.16. If \( A \) is in \( B\left( X\right) \) and \( f\left( z\right) \) is a function analytic in an open set \( \Omega \) containing \( \sigma \left( A\right) \) such that \( f\left( z\right) \neq 0 \) on \( \sigma \left( A\right) \), then \( f{\left( A\right) }^{-1} \) exists and is given by | \[ f{\left( A\right) }^{-1} = \frac{1}{2\pi i}{\oint }_{\partial \omega }\frac{1}{f\left( z\right) }{\left( z - A\right) }^{-1}{dz} \] where \( \omega \) is any open set such that (a) \( \sigma \left( A\right) \subset \omega ,\bar{\omega } \subset \Omega \) , (b) \( \partial \omega \) consists of a finite number of sim... | Yes |
Theorem 6.17. If \( f\left( z\right) \) is analytic in a neighborhood of \( \sigma \left( A\right) \), then\n\n(6.26)\n\n\[ \sigma \left( {f\left( A\right) }\right) = f\left( {\sigma \left( A\right) }\right) \]\n\ni.e., \( \mu \in \sigma \left( {f\left( A\right) }\right) \) if and only if \( \mu = f\left( \lambda \righ... | Proof. If \( f\left( \lambda \right) \neq \mu \) for all \( \lambda \in \sigma \left( A\right) \), then the function \( f\left( z\right) - \mu \) is analytic in a neighborhood of \( \sigma \left( A\right) \) and does not vanish there. Hence, \( f\left( A\right) - \mu \) has an inverse in \( B\left( X\right) \), i.e., \... | Yes |
Theorem 6.18. If \( \lambda ,\mu \) are in \( \rho \left( A\right) \), then\n\n\[{\left( \lambda - A\right) }^{-1} - {\left( \mu - A\right) }^{-1} = \left( {\mu - \lambda }\right) {\left( \lambda - A\right) }^{-1}{\left( \mu - A\right) }^{-1}.\]\n\nMoreover, if \( \left| {\lambda - \mu }\right| \cdot \begin{Vmatrix}{\l... | Proof. Let \( x \) be an arbitrary element of \( X \), and set \( u = {\left( \lambda - A\right) }^{-1}x \) . Thus, \( \left( {\lambda - A}\right) u = x \) and \( \left( {\mu - A}\right) u = x + \left( {\mu - \lambda }\right) u \) . Hence,\n\n\[u = {\left( \mu - A\right) }^{-1}x + \left( {\mu - \lambda }\right) {\left(... | Yes |
Theorem 6.19. \( \sigma \left( {A}_{i}\right) = {\sigma }_{i},\;i = 1,2 \) . | Proof. Let \( \mu \) be any point not in \( {\sigma }_{1} \) . Set \( g\left( z\right) = f\left( z\right) /\left( {\mu - z}\right) \), where \( f\left( z\right) \) is identically one on \( {\sigma }_{1} \) and vanishes on \( {\sigma }_{2} \) . Then \( g\left( z\right) \) is analytic in a neighborhood of \( \sigma \left... | Yes |
Lemma 6.21. If \( {X}_{1} \) and \( {X}_{2} \) are subspaces of a normed vector space \( X \), let \( {X}_{1} + {X}_{2} \) denote the set of all sums of the form \( {x}_{1} + {x}_{2} \), where \( {x}_{i} \in {X}_{i}, i = \) 1,2. If \( {X}_{1} \) is closed and \( {X}_{2} \) is finite-dimensional, then \( {X}_{1} + {X}_{... | Proof. Set \( {X}_{3} = {X}_{1} \cap {X}_{2} \) . Since \( {X}_{3} \) is finite-dimensional, there is a closed subspace \( {X}_{4} \) of \( {X}_{1} \) such that \( {X}_{1} = {X}_{3} \oplus {X}_{4} \) (Lemma 5.1). Clearly, \( {X}_{1} + {X}_{2} = \) \( {X}_{4} \oplus {X}_{2} \), and the latter is closed by Lemma 5.2. | Yes |
Corollary 6.22. Under the hypotheses of Theorem 6.20, there are operators \( E \) in \( B\left( X\right) \) and \( K \) in \( K\left( X\right) \) and an integer \( m > 0 \) such that \[ {\left( A - {\lambda }_{1}\right) }^{m}E = E{\left( A - {\lambda }_{1}\right) }^{m} = I - K. \] | Proof. By Theorem 6.20, \( {\Phi }_{A} \) contains the point \( {\lambda }_{1} \) . Since \( {\lambda }_{1} \) is an isolated point of \( \sigma \left( A\right), i\left( {A - \lambda }\right) = 0 \) in a neighborhood of \( {\lambda }_{1} \) (Theorem 5.11), and, hence, \( {r}^{\prime }\left( {A - {\lambda }_{1}}\right) ... | Yes |
Theorem 6.23. Suppose \( A \) is in \( B\left( X\right) \) and \( {\lambda }_{1} \) is a point of \( \sigma \left( A\right) \) such that \( R\left( {A - {\lambda }_{1}}\right) \) is closed in \( X \) . Then any two of the following conditions imply the others.\n\n\[ \text{(a)}r\left( {A - {\lambda }_{1}}\right) < \inft... | Proof. We first note that Theorem 6.20 shows that (a) and (d) imply (b) and (c), because once it is known that \( {\lambda }_{1} \in {\Phi }_{A} \), then we know that \( i(A - \) \( \left. {\lambda }_{1}\right) = 0 \), and consequently, \( {r}^{\prime }\left( {A - {\lambda }_{1}}\right) = 0 \) .\n\nTo show that (b) and... | No |
Theorem 6.24. For \( A \) in \( B\left( X\right) ,\sigma \left( {A}^{\prime }\right) = \sigma \left( A\right) \) . | Proof. Suppose \( \lambda \in \rho \left( A\right) \) . Then there is an operator \( B \in B\left( X\right) \) such that\n\n\[ \left( {A - \lambda }\right) B = B\left( {A - \lambda }\right) = I \]\n\n(Theorem 3.8). Taking adjoints we get\n\n\[ {B}^{\prime }\left( {{A}^{\prime } - \lambda }\right) = \left( {{A}^{\prime ... | Yes |
Theorem 6.25. Equation (6.12) has a unique solution for each \( y \) in \( X \) if and only if \( p\left( \lambda \right) \neq 0 \) for all \( \lambda \in \sigma \left( \widehat{A}\right) \) . | In the example given at the end of Section 6.1, the operator \( \widehat{A} \) has eigenvalues \( i \) and \( - i \) . Hence, \( - 1 \) is in the spectrum of \( {\widehat{A}}^{2} \) and also in that of \( {A}^{2} \) . Thus, the equation\n\n\[ \left( {{A}^{2} + 1}\right) x = y \]\n\ncannot be solved uniquely for all \( ... | Yes |
Theorem 6.26. Let \( V \) be a complex vector space, and let \( p \) be a real valued functional on \( V \) such that\n\n\[ \n\text{(i)}p\left( {u + v}\right) \leq p\left( u\right) + p\left( v\right) ,\;u, v \in V\text{,}\n\]\n\n\[ \n\text{(ii)}p\left( {\alpha u}\right) = \left| \alpha \right| p\left( u\right) ,\;\alph... | Proof. Let us try to reduce the \ | No |
Corollary 6.27. Let \( M \) be a subspace of a complex normed vector space \( X \) . If \( f \) is a bounded linear functional on \( M \), then there is a bounded linear functional \( F \) on \( X \) such that\n\n\[ F\left( x\right) = f\left( x\right) ,\;x \in M, \]\n\n\[ \parallel F\parallel = \parallel f\parallel \] | This corollary follows from Theorem 6.26 as in the real case. | No |
Lemma 6.28. Let \( \Omega \) be an open set in \( {\mathbb{R}}^{2} \), and let \( K \) be a bounded closed set in \( \Omega \) . Then there exists a bounded open set \( \omega \) such that\n\n(1) \( \omega \supset K \) ,\n\n(2) \( \bar{\omega } \subset \Omega \) ,\n\n(3) \( \partial \omega \) consists of a finite numbe... | Proof. By considering the intersection of \( \Omega \) with a sufficiently large disk, we may assume that \( \Omega \) is bounded. Let \( \delta \) be the distance from \( K \) to \( \partial \Omega \) . Since both of these sets are compact and do not intersect, we must have \( \delta > 0 \) . Cover \( {\mathbb{R}}^{2}... | Yes |
Theorem 7.1. If \( A \in \Phi \left( {X, Y}\right) \), then there is an \( {A}_{0} \in B\left( {Y, X}\right) \) such that\n\n(a) \( N\left( {A}_{0}\right) = {Y}_{0} \),\n\n(b) \( R\left( {A}_{0}\right) = {X}_{0} \cap D\left( A\right) \),\n\n(c) \( {A}_{0}A = I \) on \( {X}_{0} \cap D\left( A\right) \),\n\n(d) \( A{A}_{... | The proof of Theorem 7.1 is the same as that of Theorem 5.4. | No |
Theorem 7.2. Let \( A \) be a densely defined closed linear operator from \( X \) to \( Y \) . Suppose there are operators \( {A}_{1},{A}_{2} \in B\left( {Y, X}\right) ,{K}_{1} \in K\left( X\right) ,{K}_{2} \in K\left( Y\right) \) such that\n\n(7.7)\n\n\[ \n{A}_{1}A = I - {K}_{1}\text{ on }D\left( A\right) \]\n\nand\n\... | The proof is identical to that of Theorem 5.5. Note that for any operators, \( A, B \), we define \( D\left( {BA}\right) \) to be the set of those \( x \in D\left( A\right) \) such that \( {Ax} \in D\left( B\right) \) . | No |
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 \( i\left( {BA}\right) = i\left( A\right) + i\left( B\right) \) | Proof. We must show that (a) \( D\left( {BA}\right) \) is dense in \( X \) , (b) \( {BA} \) is a closed operator, (c) \( R\left( {BA}\right) \) is closed in \( Z \) , (d) \( \alpha \left( {BA}\right) < \infty ,\beta \left( {BA}\right) < \infty \), and (7.9) holds. The only part that can be carried over from the bounded... | Yes |
Lemma 7.5. Suppose that \( A \in \Phi \left( {X, Y}\right) \) and \( P \) is in \( B\left( {W, X}\right) \) . If \( P \) is one-to-one, \( R\left( P\right) \supset D\left( A\right) \) and \( {P}^{-1}\left( {D\left( A\right) }\right) \) is dense in \( W \), then \( {AP} \in \) \( \Phi \left( {W, Y}\right) ,\alpha \left(... | Proof. Since \( D\left( {AP}\right) = {P}^{-1}\left( {D\left( A\right) }\right) \), it is dense in \( W \) by assumption. Moreover, \( {AP} \) is a closed operator. For if \( {w}_{n} \rightarrow w \) in \( W \) and \( {AP}{w}_{n} \rightarrow y \) in \( Y \), then \( P{w}_{n} \rightarrow {Pw} \) in \( X \) . Since \( A ... | No |
Lemma 7.7. Assume that \( W \) is continuously embedded and dense in \( X \) . If \( A \in \Phi \left( {W, Y}\right) \), then \( A \in \Phi \left( {X, Y}\right) \) with \( N\left( A\right) \) and \( R\left( A\right) \) unchanged. | Proof. Let \( P \) be the operator embedding \( W \) into \( X \) . Let \( Q \) be the linear operator from \( X \) to \( W \) with \( D\left( Q\right) = R\left( P\right) \) defined by \( {Qx} = w \) when \( x = {Pw} \) . Since \( P \) is one-to-one, \( Q \) is well defined. Moreover, one checks easily that \( Q \in \P... | Yes |
Theorem 7.8. If \( A \in \Phi \left( {X, Y}\right) \) and \( K \) is in \( K\left( {X, Y}\right) \), then \( A + K \in \Phi \left( {X, Y}\right) \) and\n\n(7.14)\n\n\[ i\left( {A + K}\right) = i\left( A\right) \] | Proof. The fact that \( A + K \in \Phi \left( {X, Y}\right) \) follows from Theorems 7.1 and 7.2 as before. To prove (7.14), we need a trick. Since \( A \) is closed, one can make \( D\left( A\right) \) into a Banach space \( W \) by equipping it with the graph norm\n\n\[ \parallel x{\parallel }_{D\left( A\right) } = \... | Yes |
Theorem 7.9. 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 \( A + T \in \Phi \left( {X, Y}\right) \) | The proof of Theorem 7.9 is almost identical to that of Theorem 5.11. | No |
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