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Theorem 18.4.1 Let \( \\left( {\\Omega ,\\mathcal{S},\\mu }\\right) \) be a finite measure space and let \( T : \\Omega \\rightarrow \\Omega \) satisfy \( {T}^{-1}\\left( E\\right) \\in \\mathcal{S}, T\\left( E\\right) \\in \\mathcal{S} \) for all \( E \\in \\mathcal{S} \). Also suppose for all positive integers, \( n ... | Proof: To begin with, it follows from simple considerations that\n\n\[ \n\\int {\\left| {\\mathcal{X}}_{A}\\left( {T}^{n}\\left( \\omega \\right) \\right) \\right| }^{p}{d\\mu } = \\int {\\left| {\\mathcal{X}}_{{T}^{-n}\\left( A\\right) }\\left( \\omega \\right) \\right| }^{p}{d\\mu } = \\mu \\left( {{T}^{-n}\\left( A\... | Yes |
For each \( r > 0 \) there exists a finite set of points\n\n\[ \left\{ {{y}_{1},\cdots ,{y}_{n}}\right\} \subseteq \overline{f\left( K\right) } \]\n\nand continuous functions \( {\psi }_{i} \) defined on \( \overline{f\left( K\right) } \) such that for \( x \in \overline{f\left( K\right) } \) ,\n\n\[ \mathop{\sum }\lim... | Proof: Using the compactness of \( \overline{f\left( K\right) } \), there exists\n\n\[ \left\{ {{y}_{1},\cdots ,{y}_{n}}\right\} \subseteq \overline{f\left( K\right) } \subseteq K \]\n\nsuch that\n\n\[ {\left\{ B\left( {y}_{i}, r\right) \right\} }_{i = 1}^{n} \]\ncovers \( \overline{f\left( K\right) } \) . Let\n\n\[ {\... | Yes |
Lemma 18.5.3 For each \( r > 0 \), there exists \( {x}_{r} \in \) convex hull of \( \overline{f\left( K\right) } \subseteq K \) such that\n\n\[ \n{f}_{r}\left( {x}_{r}\right) = {x}_{r},\begin{Vmatrix}{{f}_{r}\left( x\right) - f\left( x\right) }\end{Vmatrix} < r\text{ for all }x \n\] | Proof: If \( {f}_{r}\left( {x}_{r}\right) = {x}_{r} \) and\n\n\[ \n{x}_{r} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{y}_{i} \n\]\n\nfor \( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} = 1 \) and the \( {y}_{i} \) described in the above lemma, we need\n\n\[ \n{f}_{r}\left( {x}_{r}\right) = \mathop{\sum }\limits_{{i = 1... | Yes |
Theorem 18.5.4 Let \( K \) be a closed and convex subset of \( X \), a normed linear space. Let \( f : K \rightarrow K \) be continuous and suppose \( \overline{f\left( K\right) } \) is compact. Then \( f \) has a fixed point. | Proof: Recall that \( f\left( {x}_{r}\right) - {f}_{r}\left( {x}_{r}\right) \in B\left( {0, r}\right) \) and \( {f}_{r}\left( {x}_{r}\right) = {x}_{r} \) with \( {x}_{r} \in \) convex hull of \( \overline{f\left( K\right) } \subseteq K \) . \n\nThere is a subsequence, still denoted with subscript \( r \) such that \( f... | Yes |
Theorem 18.5.5 Let \( f : X \rightarrow X \) be a compact map. Then either\n\n1. There is a fixed point for \( {tf} \) for all \( t \in \left\lbrack {0,1}\right\rbrack \) or\n\n2. For every \( r > 0 \), there exists a solution to \( x = {tf}\left( x\right) \) for some \( t \in \left( {0,1}\right) \) such that \( \paral... | Proof: Suppose there is \( {t}_{0} \in \left\lbrack {0,1}\right\rbrack \) such that \( {t}_{0}f \) has no fixed point. Then \( {t}_{0} \neq 0.{t}_{0}f \) obviously has a fixed point if \( {t}_{0} = 0 \) . Thus \( {t}_{0} \in (0,1\rbrack \) . Then let \( {r}_{M} \) be the radial retraction onto \( \overline{B\left( {0, ... | Yes |
Lemma 18.5.6 For each \( U \in {\mathcal{B}}_{0} \), there exists a finite set of points\n\n\[ \n\left\{ {{y}_{1}\cdots {y}_{n}}\right\} \subseteq \overline{f\left( K\right) }\n\]\n\nand continuous functions \( {\psi }_{i} \) defined on \( \overline{f\left( K\right) } \) such that for \( x \in \overline{f\left( K\right... | Proof: Let \( U = {B}_{A}\left( {0, r}\right) \). Using the compactness of \( \overline{f\left( K\right) } \), there exists\n\n\[ \n\left\{ {{y}_{1}\cdots {y}_{n}}\right\} \subseteq \overline{f\left( K\right) }\n\]\n\nsuch that\n\n\[ \n{\left\{ {y}_{i} + U\right\} }_{i = 1}^{n}\n\]\n\ncovers \( \overline{f\left( K\righ... | Yes |
Lemma 18.5.7 For each \( U \in {\mathcal{B}}_{0} \), there exists \( {x}_{U} \in \) convex hull of \( \overline{f\left( K\right) } \subseteq K \) such that\n\n\[ \n{f}_{U}\left( {x}_{U}\right) = {x}_{U} \n\] | Proof: If \( {f}_{U}\left( {x}_{U}\right) = {x}_{U} \) and\n\n\[ \n{x}_{U} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{y}_{i} \n\]\n\nfor \( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} = 1 \), we need\n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{n}{y}_{j}{\psi }_{j}\left( {f\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i... | Yes |
Theorem 18.5.8 Let \( K \) be a closed and convex subset of \( X \), a locally convex topological vector space in which every point is closed. Let \( f : K \rightarrow K \) be continuous and suppose \( \overline{f\left( K\right) } \) is compact. Then \( f \) has a fixed point. | Proof: First consider the following claim which will yield a candidate for the fixed point. Recall that \( f\left( {x}_{U}\right) - {f}_{U}\left( {x}_{U}\right) \in U \) and \( {f}_{U}\left( {x}_{U}\right) = {x}_{U} \) with \( {x}_{U} \in \) convex hull of \( \overline{f\left( K\right) } \subseteq K \) .\n\nClaim: Ther... | Yes |
Theorem 18.5.9 Let \( \mathbf{f} : \left\lbrack {0, T}\right\rbrack \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be continuous and suppose there exists \( L > 0 \) such that for all \( \lambda \in \left( {0,1}\right) \), if \( {\mathbf{x}}^{\prime } = \lambda \mathbf{f}\left( {t,\mathbf{x}}\right) ,\mathbf{x... | Proof: Let \( N\mathbf{x}\left( t\right) \equiv {\int }_{0}^{t}\mathbf{f}\left( {s,\mathbf{x}\left( s\right) }\right) {ds}. \) Thus a solution to the initial value problem exists if there exists a solution to \( {\mathbf{x}}_{0} + N\left( \mathbf{x}\right) = \mathbf{x} \). Let \( m \equiv \max \left\{ {\left| {\mathbf{... | Yes |
Theorem 18.5.10 Let \( \mathbf{f} : \left\lbrack {0, T}\right\rbrack \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be continuous and suppose there exists \( L > 0 \) such that for all \( \lambda \in \left( {0,1}\right) \), if\n\n\[{\mathbf{x}}^{\prime } = \lambda \mathbf{f}\left( {t,\mathbf{x}}\right) ,\mathb... | Proof: Let \( F : X \rightarrow X \) where \( X \) described above.\n\n\[F\mathbf{y}\left( t\right) \equiv {\int }_{0}^{t}\mathbf{f}\left( {s,\mathbf{y}\left( s\right) + {\mathbf{x}}_{0}}\right) {ds}\]\n\nLet \( B \) be a bounded set in \( X \) . Then \( \left| {\mathbf{f}\left( {s,\mathbf{y}\left( s\right) + {\mathbf{... | Yes |
Theorem 18.6.2 Let \( X \) be a complete metric space and let \( \phi : X \rightarrow ( - \infty ,\infty \rbrack \) be proper, lower semicontinuous and bounded below. Let \( {x}_{0} \) be such that\n\n\[ \phi \left( {x}_{0}\right) \leq \mathop{\inf }\limits_{{x \in X}}\phi \left( x\right) + \varepsilon \]\n\nThen for e... | Proof: Let \( {x}_{1} = {x}_{0} \) and define\n\n\[ {S}_{1} \equiv \left\{ {z \in X : \phi \left( z\right) \leq \phi \left( {x}_{1}\right) - \frac{\varepsilon }{\lambda }d\left( {z,{x}_{1}}\right) }\right\} \]\n\nThen \( {S}_{1} \) contains \( {x}_{1} \) so it is nonempty. It is also clear that \( {S}_{1} \) is a close... | Yes |
Theorem 18.6.3 Let \( \phi \) be lower semicontinuous, proper, and bounded below on a complete metric space \( X \) and let \( F : X \rightarrow \mathcal{P}\left( X\right) \) be set valued such that \( F\left( x\right) \neq \varnothing \) for all \( x \) . Also suppose that for each \( x \in X \), there exists \( y \in... | Proof: In the above Ekeland variational principle, let \( \varepsilon = 1 = \lambda \) . Then there exists \( {x}_{0} \) such that for all \( y \neq {x}_{0} \)\n\n\[ \phi \left( {x}_{0}\right) - d\left( {y,{x}_{0}}\right) < \phi \left( y\right) \text{, so}\phi \left( {x}_{0}\right) < \phi \left( y\right) + d\left( {y,{... | Yes |
Theorem 18.6.4 Let \( X \) be a Banach space and \( \phi : X \rightarrow \mathbb{R} \) be Gateaux differentiable, bounded from below, and lower semicontinuous. Then for every \( \varepsilon > 0 \) there exists \( x \in X \) such that\n\n\[ \phi \left( {x}_{\varepsilon }\right) \leq \mathop{\inf }\limits_{{x \in X}}\phi... | Proof: From the Ekeland variational principle with \( \lambda = 1 \), there exists \( {x}_{\varepsilon } \) such that\n\n\[ \phi \left( {x}_{\varepsilon }\right) \leq \phi \left( {x}_{0}\right) \leq \mathop{\inf }\limits_{{x \in X}}\phi \left( x\right) + \varepsilon \]\n\nand for all \( x \) ,\n\n\[ \phi \left( {x}_{\v... | Yes |
Theorem 18.6.5 Let \( X \) be a Banach space and \( \phi : X \rightarrow \mathbb{R} \) be Gateaux differentiable, bounded from below, and lower semicontinuous. Also suppose there exists \( a, c > 0 \) such that\n\n\[ a\parallel x\parallel - c \leq \phi \left( x\right) \text{ for all }x \in X \]\n\nThen \( \left\{ {{\ph... | Proof: Let \( {x}^{ * } \in {X}^{\prime },\begin{Vmatrix}{x}^{ * }\end{Vmatrix} \leq a \) . Let\n\n\[ \psi \left( x\right) = \phi \left( x\right) - \left\langle {{x}^{ * }, x}\right\rangle \]\n\nThis is lower semicontinuous. It is also bounded from below because\n\n\[ \psi \left( x\right) \geq \phi \left( x\right) - a\... | Yes |
Lemma 18.6.6 Let \( X \) be a Banach space and \( \phi : X \rightarrow \mathbb{R} \) be Gateaux differentiable, bounded from below, and lower semicontinuous. Suppose for all \( a > 0 \) there exists \( {ac} > 0 \) such that\n\n\[ \phi \left( x\right) \geq a\parallel x\parallel - c\text{for all}x \]\n\nThen \( \left\{ {... | If the above holds, then\n\n\[ \frac{\phi \left( x\right) }{\parallel x\parallel } \geq a - \frac{c}{\parallel x\parallel } \]\n\nand so, since \( a \) is arbitrary, it must be the case that\n\n\[ \mathop{\lim }\limits_{{\parallel x\parallel \rightarrow \infty }}\frac{\phi \left( x\right) }{\parallel x\parallel } = \in... | Yes |
Theorem 19.1.2 (Cauchy Schwarz) In any inner product space\n\n\\[ \n\\left| \\left( {x, y}\\right) \\right| \\leq \\parallel x\\parallel \\parallel y\\parallel \n\\] | Proof: Let \\( \\omega \\in \\mathbb{C},\\left| \\omega \\right| = 1 \\), and \\( \\bar{\\omega }\\left( {x, y}\\right) = \\left| \\left( {x, y}\\right) \\right| = \\operatorname{Re}\\left( {x,{y\\omega }}\\right) \\) . Let\n\n\\[ \nF\\left( t\\right) = \\left( {x + {ty\\omega }, x + {t\\omega y}}\\right) .\n\\]\n\nIf ... | Yes |
Proposition 19.1.3 For an inner product space, \( \parallel x\parallel \equiv {\left( x, x\right) }^{1/2} \) does specify a norm. | Proof: All the axioms are obvious except the triangle inequality. To verify this,\n\n\[ \parallel x + y{\parallel }^{2} \equiv \left( {x + y, x + y}\right) \equiv \parallel x{\parallel }^{2} + \parallel y{\parallel }^{2} + 2\operatorname{Re}\left( {x, y}\right) \]\n\n\[ \leq \parallel x{\parallel }^{2} + \parallel y{\p... | Yes |
Lemma 19.1.5 For \( x \in H \), an inner product space,\n\n\[ \parallel x\parallel = \mathop{\sup }\limits_{{\parallel y\parallel \leq 1}}\left| \left( {x, y}\right) \right| \]\n\n\( \left( {19.1.4}\right) \) | Proof: By the Cauchy Schwarz inequality, if \( x \neq 0 \), \n\n\[ \parallel x\parallel \geq \mathop{\sup }\limits_{{\parallel y\parallel \leq 1}}\left| \left( {x, y}\right) \right| \geq \left( {x,\frac{x}{\parallel x\parallel }}\right) = \parallel x\parallel . \]\n\nIt is obvious that 19.1.4 holds in the case that \( ... | Yes |
Theorem 19.1.8 Let \( K \) be a closed convex nonempty subset of a Hilbert space, \( H \) , and let \( x \in H \) . Then there exists a unique point \( {Px} \in K \) such that \( \parallel {Px} - x\parallel \leq \) \( \parallel y - x\parallel \) for all \( y \in K \) . | Proof: Consider uniqueness. Suppose that \( {z}_{1} \) and \( {z}_{2} \) are two elements of \( K \) such that for \( i = 1,2 \) ,\n\n\[ \begin{Vmatrix}{{z}_{i} - x}\end{Vmatrix} \leq \parallel y - x\parallel \]\n\nfor all \( y \in K \) . Also, note that since \( K \) is convex,\n\n\[ \frac{{z}_{1} + {z}_{2}}{2} \in K ... | Yes |
Let \( K \) be a closed, convex, nonempty subset of a Hilbert space, \( H \), and let \( x \in H \) . Then for \( z \in K, z = {Px} \) if and only if\n\n\[ \operatorname{Re}\left( {x - z, y - z}\right) \leq 0 \]\n\nfor all \( y \in K \) . | Proof of Corollary: Let \( z \in K \) and let \( y \in K \) also. Since \( K \) is convex, it follows that if \( t \in \left\lbrack {0,1}\right\rbrack \) ,\n\n\[ z + t\left( {y - z}\right) = \left( {1 - t}\right) z + {ty} \in K. \]\n\nFurthermore, every point of \( K \) can be written in this way. (Let \( t = 1 \) and ... | Yes |
Corollary 19.1.10 Let \( K \) be a nonempty convex closed subset of a Hilbert space, H. Then the projection map, \( P \) is continuous. In fact, \[ \left| {{Px} - {Py}}\right| \leq \left| {x - y}\right| . \] | Proof: Let \( x,{x}^{\prime } \in H \) . Then by Corollary 19.1.9, \[ \operatorname{Re}\left( {{x}^{\prime } - P{x}^{\prime },{Px} - P{x}^{\prime }}\right) \leq 0,\operatorname{Re}\left( {x - {Px}, P{x}^{\prime } - {Px}}\right) \leq 0 \] Hence \[ 0 \leq \operatorname{Re}\left( {x - {Px},{Px} - P{x}^{\prime }}\right) - ... | Yes |
Corollary 19.1.11 Let \( \mathbf{f} : K \rightarrow K \) where \( K \) is a convex compact subset of \( {\mathbb{R}}^{n} \). Then \( \mathbf{f} \) has a fixed point. | Proof: Let \( K \subseteq \overline{B\left( {\mathbf{0}, R}\right) } \) and let \( P \) be the projection map onto \( K \). Then consider the map \( \mathbf{f} \circ P \) which maps \( \overline{B\left( {\mathbf{0}, R}\right) } \) to \( \overline{B\left( {\mathbf{0}, R}\right) } \) and is continuous. By the Brouwer fix... | Yes |
Corollary 19.1.13 Let \( K \) be a closed subspace of a Hilbert space, \( H \), and let \( x \in H \) . Then for \( z \in K, z = {Px} \) if and only if\n\n\[ \left( {x - z, y}\right) = 0 \]\n\n(19.1.8)\n\nfor all \( y \in K \) . Furthermore, \( H = K \oplus {K}^{ \bot } \) where\n\n\[ {K}^{ \bot } \equiv \{ x \in H : \... | Proof: Since \( K \) is a subspace, the condition 19.1.6 implies \( \operatorname{Re}\left( {x - z, y}\right) \leq 0 \) for all \( y \in K \) . Replacing \( y \) with \( - y \), it follows \( \operatorname{Re}\left( {x - z, - y}\right) \leq 0 \) which implies \( \operatorname{Re}\left( {x - z, y}\right) \geq 0 \) for a... | Yes |
Theorem 19.1.14 Let \( H \) be a Hilbert space and let \( f \in {H}^{\prime } \). Then there exists a unique \( z \in H \) such that\n\n\[ f\left( x\right) = \left( {x, z}\right) \]\n\nfor all \( x \in H \). | Proof: Letting \( y, w \in H \) the assumption that \( f \) is linear implies\n\n\[ f\left( {{yf}\left( w\right) - f\left( y\right) w}\right) = f\left( w\right) f\left( y\right) - f\left( y\right) f\left( w\right) = 0 \]\n\nwhich shows that \( {yf}\left( w\right) - f\left( y\right) w \in {f}^{-1}\left( 0\right) \), whi... | Yes |
Lemma 19.2.2 In the context of the above definition, \( {L}^{-1}\left( y\right) \) is characterized by\n\n\[ \n{\left( {L}^{-1}\left( y\right), x\right) }_{U} = 0\text{ for all }x \in \ker \left( L\right) \]\n\n\[ \nL\left( {{L}^{-1}\left( y\right) }\right) = y,\;\left( {{L}^{-1}\left( y\right) \in {M}_{y}}\right) \]\n... | Proof: The point \( {L}^{-1}\left( y\right) \) is well defined as noted above. I claim it is characterized by the following for \( y \in L\left( U\right) \)\n\n\[ \n{\left( {L}^{-1}\left( y\right), x\right) }_{U} = 0\text{ for all }x \in \ker \left( L\right) \]\n\n\[ \nL\left( {{L}^{-1}\left( y\right) }\right) = y,\;\l... | Yes |
Theorem 19.2.3 Let \( U, H \) be Hilbert spaces and let \( L \in \mathcal{L}\left( {U, H}\right) \) . Then Definition 19.2.1 makes \( L\left( U\right) \) into a Hilbert space. Also \( L : U \rightarrow L\left( U\right) \) is continuous and \( {L}^{-1} : L\left( U\right) \rightarrow U \) is continuous. Also, | Proof: First consider the claim that \( L : U \rightarrow L\left( U\right) \) is continuous and \( {L}^{-1} \) : \( L\left( U\right) \rightarrow U \) is also continuous. Why is \( L \) continuous? Say \( {u}_{n} \rightarrow 0 \) in \( U \) . Then\n\n\[{\begin{Vmatrix}L{u}_{n}\end{Vmatrix}}_{L\left( U\right) } \equiv {\... | Yes |
Lemma 19.2.4 Let \( L \in \mathcal{L}\left( {U, H}\right) \) . Then \( L\left( \overline{B\left( {0, r}\right) }\right) \) is closed and convex. | Proof: It is clear this is convex since \( L \) is linear. Why is it closed? \( \overline{B\left( {0, r}\right) } \) is compact in the weak topology by the Banach Alaoglu theorem, Theorem 17.5.4 on Page 489. Furthermore, \( L \) is continuous with respect to the weak topologies on \( U \) and \( H \) . Here is why this... | Yes |
Theorem 19.2.5 Let \( {U}_{i}, i = 1,2 \) and \( H \) be Hilbert spaces and let \( {T}_{i} \in \mathcal{L}\left( {{U}_{i}, H}\right) \) . If there exists \( c \geq 0 \) such that for all \( x \in H \n\n\[ \n{\begin{Vmatrix}{T}_{1}^{ * }x\end{Vmatrix}}_{1} \leq c{\begin{Vmatrix}{T}_{2}^{ * }x\end{Vmatrix}}_{2}\n\]\n\nth... | Proof: Consider the first claim. If it is not so, then there exists \( {u}_{0},{\begin{Vmatrix}{u}_{0}\end{Vmatrix}}_{1} \leq 1 \n\nbut\n\n\[ \n{T}_{1}\left( {u}_{0}\right) \notin {T}_{2}\left( \overline{B\left( {0, c}\right) }\right)\n\]\n\nthe latter set being a closed convex nonempty set thanks to Lemma 19.2.4. Then... | Yes |
Theorem 19.3.1 Let \( \\left\\{ {{x}_{1},\\cdots ,{x}_{n}}\\right\\} \) be a basis for \( M \) a subspace of \( H \) a Hilbert space. Then there exists an orthonormal basis for \( M,\\left\\{ {{u}_{1},\\cdots ,{u}_{n}}\\right\\} \) which has the property that for each \( k \\leq n,\\operatorname{span}\\left( {{x}_{1},\... | Proof: Let \( \\left\\{ {{x}_{1},\\cdots ,{x}_{n}}\\right\\} \) be a basis for \( M \) . Let \( {u}_{1} \\equiv {x}_{1}/\\left| {x}_{1}\\right| \) . Thus for \( k = 1 \) , \( \\operatorname{span}\\left( {u}_{1}\\right) = \\operatorname{span}\\left( {x}_{1}\\right) \) and \( \\left\\{ {u}_{1}\\right\\} \) is an orthonor... | Yes |
Theorem 19.3.2 Let \( M \) be the span of \( \\left\\{ {{u}_{1},\\cdots ,{u}_{n}}\\right\\} \) in a Hilbert space, \( H \) and let \( y \\in H \) . Then \( {Py} \) is given by\n\n\\[ \n{Py} = \\mathop{\\sum }\\limits_{{k = 1}}^{n}\\left( {y,{u}_{k}}\\right) {u}_{k} \n\\]\n\n(19.3.17)\n\nand the distance is given by\n\n... | Proof:\n\n\\[ \n\\left( {y - \\mathop{\\sum }\\limits_{{k = 1}}^{n}\\left( {y,{u}_{k}}\\right) {u}_{k},{u}_{p}}\\right) = \\left( {y,{u}_{p}}\\right) - \\mathop{\\sum }\\limits_{{k = 1}}^{n}\\left( {y,{u}_{k}}\\right) \\left( {{u}_{k},{u}_{p}}\\right) \n\\]\n\n\\[ \n= \\left( {y,{u}_{p}}\\right) - \\left( {y,{u}_{p}}\\... | Yes |
Theorem 19.3.4 Let \( M = \operatorname{span}\left( {{x}_{1},\cdots ,{x}_{n}}\right) \subseteq H \), a Real Hilbert space where \( \left\{ {{x}_{1},\cdots ,{x}_{n}}\right\} \) is a basis and let \( y \in H \) . Then letting \( d \) be the distance from \( y \) to \( M \) , \[ {d}^{2} = \frac{G\left( {{x}_{1},\cdots ,{x... | Proof: By Theorem 19.3.1 \( M \) is a closed subspace of \( H \) . Let \( \mathop{\sum }\limits_{{k = 1}}^{n}{\alpha }_{k}{x}_{k} \) be the element of \( M \) which is closest to \( y \) . Then by Corollary 19.1.13, \[ \left( {y - \mathop{\sum }\limits_{{k = 1}}^{n}{\alpha }_{k}{x}_{k},{x}_{p}}\right) = 0 \] for each \... | Yes |
Lemma 19.4.3 Let \( {a}_{n} \neq 1,{a}_{n} > 0 \), and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . Then\n\n\[ \mathop{\prod }\limits_{{k = 1}}^{\infty }\left( {1 - {a}_{n}}\right) \equiv \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\prod }\limits_{{k = 1}}^{n}\left( {1 - {a}_{n}}\right)... | Proof:Without loss of generality, you can assume \( {a}_{n} < 1/2 \) because the two conditions are determined by the values of \( {a}_{n} \) for \( n \) large. By the above sketch the following is obtained.\n\n\[ \ln \mathop{\prod }\limits_{{k = 1}}^{n}\left( {1 - {a}_{k}}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\... | Yes |
Theorem 19.4.4 Let \( \left\{ {p}_{n}\right\} \) be a sequence of real numbers larger than \( - 1/2 \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{p}_{n} = \infty \) . Let \( S \) denote the set of finite linear combinations of the functions, \( \left\{ {{x}^{{p}_{1}},{x}^{{p}_{2}},\cdots }\right\} \) ... | Proof: The polynomials are dense in \( {L}^{2}\left( {0,1}\right) \) and so \( S \) is dense in \( {L}^{2}\left( {0,1}\right) \) if and only if for every \( \varepsilon > 0 \) there exists a function \( f \) from \( S \) such that for each integer \( m \geq 0,{\left( {\int }_{0}^{1}{\left| f\left( x\right) - {x}^{m}\ri... | Yes |
Theorem 19.4.5 Let \( S \) be finite linear combinations of \( \left\{ {1,{x}^{{p}_{1}},{x}^{{p}_{2}},\cdots }\right\} \) where \( {p}_{j} \geq 1 \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{p}_{n} = \infty \) . Then \( S \) is dense in \( C\left( \left\lbrack {0,1}\right\rbrack \right) \) if and only if \... | Proof: If \( S \) is dense in \( C\left( \left\lbrack {0,1}\right\rbrack \right) \) then \( S \) must also be dense in \( {L}^{2}\left( {0,1}\right) \) and so by Theorem 19.4.4 \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{p}_{k}} = \infty \) .\n\nSuppose then that \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\f... | Yes |
Theorem 19.6.1 Let \( S = {\left\{ {\phi }_{n}\right\} }_{n \in \mathbb{Z}} \) . Then \( \operatorname{span}\left( S\right) \) is dense in \( {L}^{2}\left( {0,{2\pi }}\right) \) . | Proof: By regularity of Lebesgue measure, it follows from Theorem 15.2.4 that \( {C}_{c}\left( {0,{2\pi }}\right) \) is dense in \( {L}^{2}\left( {0,{2\pi }}\right) \) . Therefore, it suffices to show that for \( g \in {C}_{c}\left( {0,{2\pi }}\right) \) , then for every \( \varepsilon > 0 \) there exists \( h \in \ope... | Yes |
Corollary 19.6.2 For \( f \in {L}^{2}\left( {0,{2\pi }}\right) \) , \n\n\[ \n\mathop{\lim }\limits_{{m \rightarrow \infty }}{\begin{Vmatrix}f - \mathop{\sum }\limits_{{k = - m}}^{m}\left( f,{\phi }_{k}\right) {\phi }_{k}\end{Vmatrix}}_{{L}^{2}\left( {0,{2\pi }}\right) } \n\] | Proof: This follows from Theorem 19.5.2 on Page 566. | No |
Theorem 19.7.2 Let \( A \) be a compact self adjoint operator defined on a Hilbert space, \( H \) . Then there exists a countable set of eigenvalues, \( \left\{ {\lambda }_{i}\right\} \) and an orthonormal set of eigenvectors, \( {u}_{i} \) satisfying\n\n\[{\lambda }_{i}\text{is real,}\left| {\lambda }_{n}\right| \geq ... | Proof: If \( \parallel A\parallel = 0 \) then pick \( u \in H \) with \( \parallel u\parallel = 1 \) and let \( {\lambda }_{1} = 0 \) . Since \( A\left( H\right) = 0 \) it follows the span of \( u \) is dense in \( A\left( H\right) \) and this proves the theorem in this uninteresting case.\n\nAssume from now on \( A \n... | Yes |
Corollary 19.7.3 The main conclusion of the above theorem can be written as\n\n\[ A = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\lambda }_{k}{u}_{k} \otimes {u}_{k} \]\n\nwhere the convergence of the partial sums takes place in the operator norm. | Proof: Using 19.7.31\n\n\[ \left| \left( {\left( {A - \mathop{\sum }\limits_{{k = 1}}^{n}{\lambda }_{k}{u}_{k} \otimes {u}_{k}}\right) x, y}\right) \right| = \left| \left( {{Ax} - \mathop{\sum }\limits_{{k = 1}}^{n}{\lambda }_{k}\left( {x,{u}_{k}}\right) {u}_{k}, y}\right) \right| \]\n\n\[ = \left| \left( {\mathop{\sum... | Yes |
Lemma 19.7.4 If \( {V}_{\lambda } \) is the eigenspace for \( \lambda \neq 0 \) and \( B : {V}_{\lambda } \rightarrow {V}_{\lambda } \) is a compact self adjoint operator, then \( {V}_{\lambda } \) must be finite dimensional. | Proof: This follows from the above theorem because it gives a sequence of eigenvalues on restrictions of \( B \) to subspaces with \( {\lambda }_{k} \downarrow 0 \) . Hence, eventually \( {\lambda }_{n} = 0 \) because there is no other eigenvalue in \( {V}_{\lambda } \) than \( \lambda \) . Hence there can be no eigenv... | No |
Corollary 19.7.5 Let \( A \) be a compact self adjoint operator defined on a separable Hilbert space, \( H \) . Then there exists a countable set of eigenvalues, \( \left\{ {\lambda }_{i}\right\} \) and an orthonormal set of eigenvectors, \( {u}_{i} \) satisfying\n\n\[ A{v}_{i} = {\lambda }_{i}{v}_{i},\begin{Vmatrix}{u... | Proof: Let \( B \) be the restriction of \( A \) to \( {V}_{{\lambda }_{i}} \) . Thus \( B \) is a compact self adjoint operator which maps \( {V}_{\lambda } \) to \( {V}_{\lambda } \) and has only one eigenvalue \( {\lambda }_{i} \) on \( {V}_{{\lambda }_{i}} \) . By Lemma 19.7.4, \( {V}_{\lambda } \) is finite dimens... | Yes |
Corollary 19.7.6 Let \( A \) be a compact self adjoint operator and let \( \lambda \notin {\left\{ {\lambda }_{n}\right\} }_{n = 1}^{\infty } \) and \( \lambda \neq 0 \) where the \( {\lambda }_{n} \) are the eigenvalues of \( A \) . \( \left( {{Ax} = {\lambda x}, x \neq 0}\right) \) Then | \[ {\left( A - \lambda I\right) }^{-1}x = - \frac{1}{\lambda }x + \frac{1}{\lambda }\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\lambda }_{k}}{{\lambda }_{k} - \lambda }\left( {x,{u}_{k}}\right) {u}_{k}. \] Proof: Let \( m < n \) . Then since the \( \left\{ {u}_{k}\right\} \) form an orthonormal set, \[ \left| {\ma... | Yes |
Proposition 19.8.1 Suppose \( {y}_{i} \) solves the boundary conditions and the differential equation for \( \lambda = {\lambda }_{i} \) where \( {\lambda }_{1} \neq {\lambda }_{2} \) . Then we have the orthogonality relation\n\n\[ \n{\int }_{a}^{b}q\left( x\right) {y}_{1}\left( x\right) {y}_{2}\left( x\right) {dx} = 0... | Proof: The orthogonality relation, 19.8.44 follows from the fundamental assumption, 19.8.43 and 19.8.42. | No |
Lemma 19.8.6 Suppose \( \\left\\{ {A}_{n}\\right\\} \) is a sequence of compact operators in \( \\mathcal{L}\\left( {X, Y}\\right) \) for two Banach spaces, \( X \) and \( Y \) and suppose \( A \\in \\mathcal{L}\\left( {X, Y}\\right) \) and\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\begin{Vmatrix}{A ... | Proof: Let \( B \) be a bounded set in \( X \) such that \( \\parallel b\\parallel \\leq C \) for all \( b \\in B \) . I need to verify \( {AB} \) is totally bounded. Suppose then it is not. Then there exists \( \\varepsilon > 0 \) and a sequence, \( \\left\\{ {A{b}_{i}}\\right\\} \) where \( {b}_{i} \\in B \) and\n\n\... | Yes |
Theorem 19.8.9 Definition 19.8.8 is well defined and equals \( \mathop{\sum }\limits_{{j = 1}}^{\infty }{\lambda }_{j} \) where the \( {\lambda }_{j} \) are the eigenvalues of \( A \) . | Proof: Suppose \( \left\{ {u}_{k}\right\} \) is some other orthonormal basis. Then\n\n\[ \n{e}_{k} = \mathop{\sum }\limits_{{j = 1}}^{\infty }{u}_{j}\left( {{e}_{k},{u}_{j}}\right) \n\] \n\nBy Lemma 19.8.7 \( A \) is compact and so\n\n\[ \nA = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\lambda }_{k}{u}_{k} \otimes {u}_{... | Yes |
Lemma 19.10.2 Suppose \( A \in \mathcal{L}\left( {X, X}\right) \) is compact for \( X \) a Banach space. Then \( \left( {I - A}\right) \left( X\right) \) is a closed subspace of \( X \) . | Proof: Suppose \( \left( {I - A}\right) {x}_{n} \rightarrow y \) . Let\n\n\[ \n{\alpha }_{n} \equiv \operatorname{dist}\left( {{x}_{n},\ker \left( {I - A}\right) }\right) \n\]\n\nand let \( {z}_{n} \in \ker \left( {I - A}\right) \) be such that\n\n\[ \n{\alpha }_{n} \leq \begin{Vmatrix}{{x}_{n} - {z}_{n}}\end{Vmatrix} ... | Yes |
Theorem 19.10.3 Let \( A \in \mathcal{L}\left( {X, X}\right) \) be a compact operator and let \( f \in X \) . Then there exists a solution, \( x \), to\n\n\[ x - {Ax} = f \]\n\n(19.10.66)\n\nif and only if\n\n\[ {x}^{ * }\left( f\right) = 0 \]\n\n\( \left( {19.10.67}\right) \)\n\nfor all \( {x}^{ * } \in \ker \left( {I... | Proof: Suppose \( x \) is a solution to 19.10.66 and let \( {x}^{ * } \in \ker \left( {I - {A}^{ * }}\right) \) . Then\n\n\[ {x}^{ * }\left( f\right) = {x}^{ * }\left( {\left( {I - A}\right) \left( x\right) }\right) = \left( {\left( {I - {A}^{ * }}\right) {x}^{ * }}\right) \left( x\right) = 0.\]\n\nNext suppose \( {x}^... | Yes |
Let \( A \in \mathcal{L}\left( {X, X}\right) \) be a compact operator. Then there exists a solution to the equation\n\n\[ x - {Ax} = f \]\n\nfor all \( f \in X \) if and only if \( \left( {I - {A}^{ * }}\right) \) is one to one on \( {X}^{\prime } \) . | Proof: Suppose \( \left( {I - {A}^{ * }}\right) \) is one to one first. Then if \( {x}^{ * } - {A}^{ * }{x}^{ * } = 0 \) it follows \( {x}^{ * } = 0 \) and so for any \( f \in X,{x}^{ * }\left( f\right) = 0 \) for all \( {x}^{ * } \in \ker \left( {I - {A}^{ * }}\right) \) . By 19.10.3 there exists a solution to \( \lef... | Yes |
Lemma 19.12.2 \( \parallel \cdot {\parallel }_{\gamma } \) is a norm for \( {BC}\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \) and \( {BC}\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \) is a complete normed linear space. Also, a sequence is Cauchy in \( \parallel \cdot {\parallel }_{\gamma } \) if or onl... | Proof: First consider the claim about \( \parallel \cdot {\parallel }_{\gamma } \) being a norm. To simplify notation, let \( T = \left\lbrack {a, b}\right\rbrack \) . It is clear that \( \parallel f{\parallel }_{\gamma } = 0 \) if and only if \( f = 0 \) and \( \parallel f{\parallel }_{\gamma } \geq 0 \) . Also,\n\n\[... | Yes |
Theorem 19.12.3 Let 19.12.75 hold. Then there exists a unique solution to 19.12.74 in \( {BC}\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \) . | Proof: Use the norm of 19.12.73 where \( \gamma \neq 0 \) is described later. Let \( T \) : \( {BC}\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \rightarrow {BC}\left( {\left\lbrack {a, b}\right\rbrack ;X}\right) \) be defined by\n\n\[ \n{Tx}\left( t\right) \equiv {x}_{0} - {\int }_{a}^{t}F\left( {s, x\left( s\ri... | Yes |
Theorem 19.12.6 Suppose \( u \) is nonnegative, continuous, and real valued and that\n\n\[ u\left( t\right) \leq C + {\int }_{0}^{t}{ku}\left( s\right) {ds}, k \geq 0 \]\n\nThen \( u\left( t\right) \leq C{e}^{kt} \) . | Proof: Let \( w\left( t\right) \equiv {\int }_{0}^{t}{ku}\left( s\right) {ds} \) . Then\n\n\[ {w}^{\prime }\left( t\right) = {ku}\left( t\right) \leq {kC} + {kw}\left( t\right) \]\n\nand so \( {w}^{\prime }\left( t\right) - {kw}\left( t\right) \leq {kC} \) which implies \( \frac{d}{dt}\left( {{e}^{-{kt}}w\left( t\right... | Yes |
Lemma 19.13.2 Suppose \( A = {A}^{ * } \) and \( \left( {{Ax}, x}\right) \geq \varepsilon {\left| x\right| }^{2} \) . Then\n\n\[ \begin{Vmatrix}{e}^{-{At}}\end{Vmatrix} \leq {e}^{-{\varepsilon t}} \] | Proof: Let \( \widehat{x}\left( t\right) = x\left( t\right) {e}^{\varepsilon t} \) . Then the equation for \( {e}^{-{At}}{x}_{0} \equiv x\left( t\right) \) becomes\n\n\[ {\widehat{x}}^{\prime }\left( t\right) - \varepsilon \widehat{x}\left( t\right) + A\widehat{x}\left( t\right) = 0,\widehat{x}\left( 0\right) = {x}_{0}... | Yes |
Lemma 19.13.5 For \( \alpha ,\beta > 0,\Gamma \left( \alpha \right) \Gamma \left( \beta \right) = \Gamma \left( {\alpha + \beta }\right) {\int }_{0}^{1}{\left( 1 - v\right) }^{\alpha - 1}{v}^{\beta - 1}{dv} \) | \[ \Gamma \left( \alpha \right) \Gamma \left( \beta \right) \equiv {\int }_{0}^{\infty }{\int }_{0}^{\infty }{e}^{-\left( {t + s}\right) }{t}^{\alpha - 1}{s}^{\beta - 1}{dtds} = {\int }_{0}^{\infty }{\int }_{s}^{\infty }{e}^{-u}{\left( u - s\right) }^{\alpha - 1}{s}^{\beta - 1}{duds} \] \[ = {\int }_{0}^{\infty }{e}^{-... | Yes |
Lemma 19.13.6 For \( \alpha \in \left( {0,1}\right) ,{A}^{-\alpha }{A}^{-\left( {1 - \alpha }\right) } = {A}^{-1} \). More generally, if \( \alpha + \beta < 1,{A}^{-\alpha }{A}^{-\beta } = {A}^{-\left( {\alpha + \beta }\right) }.\) | Proof: The product is the following where \( \beta = 1 - \alpha \)\n\n\[ \frac{1}{\Gamma \left( \alpha \right) }{\int }_{0}^{\infty }{e}^{-{At}}{t}^{a - 1}{dt}\frac{1}{\Gamma \left( \beta \right) }{\int }_{0}^{\infty }{e}^{-{As}}{s}^{\beta - 1}{ds} \]\n\nThen this equals\n\n\[ \frac{1}{\Gamma \left( \alpha \right) \Gam... | Yes |
Lemma 19.13.7 If \( \alpha ,\beta \in \left( {0,1}\right) ,\alpha + \beta \leq 1 \), then \( {A}^{\alpha }{A}^{\beta } = {A}^{\alpha + \beta } \) . Also \( {A}^{\alpha } \) commutes with every operator in \( \mathcal{L}\left( {X, X}\right) \) which commutes with \( A \) . | Proof: The last assertion follows right away from the fact noted above that \( {A}^{-\left( {1 - \alpha }\right) } \) commutes with all operators which commute with \( A \) and that so does \( A \) . Thus if \( C \) is such a commuting operator,\n\n\[ C{A}^{\alpha } = {CA}{A}^{-\left( {1 - \alpha }\right) } = {AC}{A}^{... | Yes |
Theorem 19.13.11 In the situation of the definition, if \( \alpha + \beta \leq 1 \), for \( \alpha ,\beta \in \left( {0,1}\right) \), \[ {A}^{\alpha }{A}^{\beta } = {A}^{\alpha + \beta } \] and in particular, \[ {A}^{\alpha }{A}^{1 - \alpha } = A. \] Also, \( {A}^{\alpha } \) commutes with every operator which commutes... | Proof: \( {A}^{\alpha + \beta } \equiv \mathop{\lim }\limits_{{\varepsilon \rightarrow 0}}\left( {{\varepsilon I} + A}\right) {\left( \varepsilon I + A\right) }^{-\left( {1 - \left( {\alpha + \beta }\right) }\right) } \). Then since \( \left( {{\varepsilon I} + A}\right) \) commutes with \( {e}^{-\left( {{\varepsilon I... | Yes |
Lemma 19.14.2 Let \( M \equiv \sup \{ \parallel S\left( t\right) \parallel : t \in \left\lbrack {0, T}\right\rbrack \} \) . Then \( M < \infty \) . | Proof: If this is not true, then there exists \( {t}_{n} \in \left\lbrack {0, T}\right\rbrack \) such that \( \begin{Vmatrix}{S\left( {t}_{n}\right) }\end{Vmatrix} \geq n \) . That is the operators \( S\left( {t}_{n}\right) \) are not uniformly bounded. By the uniform boundedness principle, Theorem 17.1.8, there exists... | Yes |
Theorem 19.14.3 For \( M \) described in Lemma 19.14.2, there exists \( \alpha \) such that\n\n\[ \n\parallel S\left( t\right) \parallel \leq M{e}^{\alpha t}, t \geq 0 \n\]\n\nIn fact, \( \alpha \) can be chosen such that \( {M}^{1/T} = {e}^{\alpha } \) . | Proof: Let \( t \) be arbitrary. Then \( t = {mT} + r\left( t\right) \) where \( 0 \leq r\left( t\right) < T \) . Then by the semigroup property\n\n\[ \n\parallel S\left( t\right) \parallel = \parallel S\left( {{mT} + r\left( t\right) }\right) \parallel = \begin{Vmatrix}{S\left( {r\left( t\right) }\right) S{\left( T\ri... | Yes |
Proposition 19.14.5 Given a continuous semigroup \( S\left( t\right) \), its generator \( A \) exists and is a closed densely defined operator. Furthermore, for | Proof: First note \( D\left( A\right) \neq \varnothing \) . In fact \( 0 \in D\left( A\right) \) . It follows from Theorem 19.14.3 that for all \( \lambda \) larger than \( \alpha \), one can define a Laplace transform, \( R\left( \lambda \right) x \equiv \) \( {\int }_{0}^{\infty }{e}^{-{\lambda t}}S\left( t\right) {x... | No |
Corollary 19.14.7 Let \( S\left( t\right) \) be a continuous semigroup and let \( A \) be its generator. Then for \( 0 < a < b \) and \( x \in D\left( A\right) \)\n\n\[ S\left( b\right) x - S\left( a\right) x = {\int }_{a}^{b}S\left( t\right) {Axdt} \]\n\nand also for \( t > 0 \) you can take the derivative from the le... | Proof:Letting \( {y}^{ * } \in {X}^{\prime } \), \n\n\[ {y}^{ * }\left( {{\int }_{a}^{b}S\left( t\right) {Axdt}}\right) = {\int }_{a}^{b}{y}^{ * }\left( {S\left( t\right) \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{S\left( h\right) x - x}{h}}\right) {dt} \]\n\nThe difference quotients are bounded because they conver... | Yes |
Theorem 19.14.8 Suppose \( A \) is a densely defined linear operator which has the property that for all \( \lambda > 0 \) ,\n\n\[ \n{\left( \lambda I - A\right) }^{-1} \in \mathcal{L}\left( {X, X}\right) \n\]\n\nwhich means that \( {\lambda I} - A : D\left( A\right) \rightarrow X \) is one to one and onto with continu... | Proof: The condition 19.14.83 implies,\n\n\[ \n\begin{Vmatrix}{\left( \lambda I - A\right) }^{-1}\end{Vmatrix} \leq \frac{M}{\lambda} \n\]\n\nConsider, for \( \lambda > 0 \), the operator which is defined on \( D\left( A\right) \) ,\n\n\[ \n\lambda {\left( \lambda I - A\right) }^{-1}A \n\]\n\nOn \( D\left( A\right) \),... | Yes |
Lemma 19.14.9 There is a bounded linear operator given for \( \lambda > 0 \) by\n\n\[ \n- {\lambda I} + {\lambda }^{2}{\left( \lambda I - A\right) }^{-1} = {\lambda A}{\left( \lambda I - A\right) }^{-1} \equiv {A}_{\lambda }\n\]\n\nOn \( D\left( A\right) ,{A}_{\lambda } = \lambda {\left( \lambda I - A\right) }^{-1}A \)... | Now \( {L}_{\lambda }x \rightarrow x \) on a dense subset of \( X,{L}_{\lambda } \equiv \lambda {\left( \lambda I - A\right) }^{-1} \) . Also, from the hypothesis, \( \begin{Vmatrix}{L}_{\lambda }\end{Vmatrix} \leq M \) . Say \( x \) is arbitrary. Then does \( {L}_{\lambda }x \rightarrow x \) ? Let \( \widehat{x} \in D... | Yes |
Lemma 19.14.10 For all \( x \in D\left( A\right) ,\mathop{\lim }\limits_{{\lambda \rightarrow \infty }}\begin{Vmatrix}{{A}_{\lambda }x - {Ax}}\end{Vmatrix} = 0 \) . | Now from Corollary 19.12.5, there exists an approximate continuous semigroup \( {S}_{\lambda }\left( t\right) \) generated by \( {A}_{\lambda } \) which is the solution to\n\n\[ \n{S}_{\lambda }^{\prime }\left( t\right) = {A}_{\lambda }{S}_{\lambda }\left( t\right) ,{S}_{\lambda }\left( 0\right) = I \]\n\n(19.14.87)\n\... | Yes |
Lemma 19.14.11 For \( \lambda ,\mu > 0,{\left( \lambda I - A\right) }^{-1} \) and \( {\left( \mu I - A\right) }^{-1} \) commute. | Proof: Suppose\n\n\[ y = {\left( \mu I - A\right) }^{-1}{\left( \lambda I - A\right) }^{-1}x \]\n\n(19.14.90)\n\n\[ z = {\left( \lambda I - A\right) }^{-1}{\left( \mu I - A\right) }^{-1}x \]\n\n(19.14.91)\n\nI need to show \( y = z \) . First note \( z, y \in D\left( A\right) \) . Then also \( \left( {{\mu I} - A}\righ... | Yes |
Theorem 20.1.2 (Radon Nikodym) Let \( \lambda \) and \( \mu \) be finite measures defined on a \( \sigma \) - algebra, \( \mathcal{S} \), of subsets of \( \Omega \) . Suppose \( \lambda \ll \mu \) . Then there exists a unique \( f \in {L}^{1}\left( {\Omega ,\mu }\right) \) such that \( f\left( x\right) \geq 0 \) and\n\... | Proof: Let \( \Lambda : {L}^{2}\left( {\Omega ,\mu + \lambda }\right) \rightarrow \mathbb{C} \) be defined by\n\n\[ {\Lambda g} = {\int }_{\Omega }{gd\lambda } \]\n\nBy Holder's inequality,\n\n\[ \left| {\Lambda g}\right| \leq {\left( {\int }_{\Omega }{1}^{2}d\lambda \right) }^{1/2}{\left( {\int }_{\Omega }{\left| g\ri... | Yes |
Corollary 20.2.4 Suppose \( \left( {\Omega ,\mathcal{F}}\right) \) is a set with a \( \sigma \) algebra of subsets \( \mathcal{F} \) and suppose \( \mu : \mathcal{F} \rightarrow \mathbb{C} \) is only finitely additive. That is, \( \mu \left( {{ \cup }_{i = 1}^{n}{E}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mu ... | Proof: Say \( E \cap F = \varnothing \) for \( E, F \in \mathcal{F} \) . Let \( \pi \left( E\right) ,\pi \left( F\right) \) suitable partitions for which the following holds.\n\n\[ \left| \mu \right| \left( {E \cup F}\right) \geq \mathop{\sum }\limits_{{A \in \pi \left( E\right) }}\left| {\mu \left( A\right) }\right| +... | Yes |
Theorem 20.2.6 Let \( \\left( {\\Omega ,\\mathcal{S}}\\right) \) be a measure space and let \( \\lambda : \\mathcal{S} \\rightarrow \\mathbb{C} \) be a complex vector measure. Thus \( \\left| \\lambda \\right| \\left( \\Omega \\right) < \\infty \) . Let \( \\mu : \\mathcal{S} \\rightarrow \\left\\lbrack {0,\\mu \\left(... | Proof: It is clear that \( \\operatorname{Re}\\lambda \) and \( \\operatorname{Im}\\lambda \) are real-valued vector measures on \( \\mathcal{S} \) . Since \( \\left| \\lambda \\right| \\left( \\Omega \\right) < \\infty \), it follows easily that \( \\left| {\\operatorname{Re}\\lambda }\\right| \\left( \\Omega \\right)... | Yes |
Lemma 20.2.7 Suppose \( \left( {\Omega ,\mathcal{S},\mu }\right) \) is a measure space and \( f \) is a function in \( {L}^{1}\left( {\Omega ,\mu }\right) \) with the property that\n\n\[ \left| {{\int }_{E}{fd\mu }}\right| \leq \mu \left( E\right) \]\n\nfor all \( E \in \mathcal{S} \) . Then \( \left| f\right| \leq 1 \... | Proof of the lemma: Consider the following picture.\n\n\n\nwhere \( B\left( {p, r}\right) \cap B\left( {0,1}\right) = \varnothing \) . Let \( E = {f}^{-1}\left( {B\left( {p, r}\right) }\right) \) . In fact \( \mu \le... | Yes |
Corollary 20.2.8 Let \( \lambda \) be a complex vector measure with \( \left| \lambda \right| \left( \Omega \right) < {\infty }^{1} \) Then there exists a unique \( f \in {L}^{1}\left( \Omega \right) \) such that \( \lambda \left( E\right) = {\int }_{E}{fd}\left| \lambda \right| \) . Furthermore, \( \left| f\right| = 1... | Proof: First note that \( \lambda \ll \left| \lambda \right| \) and so such an \( {L}^{1} \) function exists and is unique. It is required to show \( \left| f\right| = 1 \) a.e. If \( \left| \lambda \right| \left( E\right) \neq 0 \) ,\n\n\[ \left| \frac{\lambda \left( E\right) }{\left| \lambda \right| \left( E\right) }... | Yes |
Corollary 20.2.9 Let \( \lambda \) be a complex vector measure such that \( \lambda \ll \mu \) where \( \mu \) is \( \sigma \) finite. Then there exists a unique \( g \in {L}^{1}\left( {\Omega ,\mu }\right) \) such that \( \lambda \left( E\right) = {\int }_{E}{gd\mu } \) . | Proof: By Corollary 20.2.8 and Theorem 20.2.5 which says that \( \left| \lambda \right| \) is finite, there exists a unique \( f \) such that \( \left| f\right| = 1\left| \lambda \right| \) a.e. and\n\n\[ \lambda \left( E\right) = {\int }_{E}{fd}\left| \lambda \right| \]\n\nNow \( \left| \lambda \right| \ll \mu \) and ... | Yes |
Corollary 20.2.10 Suppose \( \left( {\Omega ,\mathcal{S}}\right) \) is a measure space and \( \mu \) is a finite nonnegative measure on \( \mathcal{S} \) . Then for \( h \in {L}^{1}\left( \mu \right) \), define a complex measure, \( \lambda \) by\n\n\[ \lambda \left( E\right) \equiv {\int }_{E}{hd\mu } \]\n\nThen\n\n\[... | Proof: From Corollary 20.2.8 there exists \( g \) such that \( \left| g\right| = 1,\left| \lambda \right| a.e. and for all \( E \in \mathcal{S} \)\n\n\[ \lambda \left( E\right) = {\int }_{E}{gd}\left| \lambda \right| = {\int }_{E}{hd\mu }.\]\n\nLet \( {s}_{n} \) be a sequence of simple functions converging pointwise to... | Yes |
Theorem 20.3.1 (Riesz representation theorem) Let \( p > 1 \) and let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a finite measure space. If \( \Lambda \in {\left( {L}^{p}\left( \Omega \right) \right) }^{\prime } \), then there exists a unique \( h \in {L}^{q}\left( \Omega \right) \left( {\frac{1}{p} + \frac{1}{q... | Proof: (Uniqueness) If \( {h}_{1} \) and \( {h}_{2} \) both represent \( \Lambda \), consider\n\n\[f = {\left| {h}_{1} - {h}_{2}\right| }^{q - 2}\left( {\overline{{h}_{1}} - \overline{{h}_{2}}}\right),\]\n\nwhere \( \bar{h} \) denotes complex conjugation. By Holder’s inequality, it is easy to see that \( f \in {L}^{p}\... | No |
Theorem 20.3.4 (Riesz representation theorem) Let \( \\left( {\\Omega ,\\mathcal{S},\\mu }\\right) \) be a finite measure space. If \( \\Lambda \\in {\\left( {L}^{1}\\left( \\Omega \\right) \\right) }^{\\prime } \), then there exists a unique \( h \\in {L}^{\\infty }\\left( \\Omega \\right) \) such that\n\n\[ \n\\Lambd... | Proof: Just as in the proof of Theorem 20.3.1, there exists a unique \( h \\in {L}^{1}\\left( \\Omega \\right) \) such that for all simple functions, \( s \) ,\n\n\[ \n\\Lambda \\left( s\\right) = \\int {hsd\\mu } \n\]\n\n(20.3.11)\n\nTo show \( h \\in {L}^{\\infty }\\left( \\Omega \\right) \), let \( \\varepsilon > 0 ... | Yes |
Theorem 20.3.6 (Riesz representation theorem) Let \( \\left( {\\Omega ,\\mathcal{S},\\mu }\\right) \) be \( \\sigma \) finite and let\n\n\[ \n\\Lambda \\in {\\left( {L}^{p}\\left( \\Omega ,\\mu \\right) \\right) }^{\\prime }, p \\geq 1.\n\]\n\nThen there exists a unique \( h \\in {L}^{q}\\left( {\\Omega ,\\mu }\\right)... | Proof: Let \( \\left\\{ {\\Omega }_{n}\\right\\} \) be a sequence of disjoint elements of \( \\mathcal{S} \) having the property that\n\n\[ \n0 < \\mu \\left( {\\Omega }_{n}\\right) < \\infty ,{ \\cup }_{n = 1}^{\\infty }{\\Omega }_{n} = \\Omega .\n\]\n\nDefine\n\n\[ \nr\\left( x\\right) = \\mathop{\\sum }\\limits_{{n ... | No |
Theorem 20.3.7 For \( \left( {\Omega ,\mathcal{S},\mu }\right) \) a \( \sigma \) finite measure space and \( p > 1,{L}^{p}\left( \Omega \right) \) is reflexive. | Proof: Let \( {\delta }_{r} : {\left( {L}^{r}\left( \Omega \right) \right) }^{\prime } \rightarrow {L}^{{r}^{\prime }}\left( \Omega \right) \) be defined for \( \frac{1}{r} + \frac{1}{{r}^{\prime }} = 1 \) by\n\n\[ \n\int \left( {{\delta }_{r}\Lambda }\right) {gd\mu } = {\Lambda g} \n\] \n\nfor all \( g \in {L}^{r}\lef... | Yes |
Corollary 20.4.2 Suppose \( \left( {\Omega ,\mathcal{F}}\right) \) is a measure space as above and suppose \( \mu \) is a measure defined on \( \mathcal{F} \) . Denote by \( {BV}\left( {\Omega ;\mu }\right) \) those finitely additive measures of \( {BV}\left( \Omega \right) \nu \) such that \( \nu \ll \mu \) in the usu... | Proof: It is clear that it is a subspace. Is it closed? Suppose \( {\nu }_{n} \rightarrow \nu \) and each \( {\nu }_{n} \) is in \( {BV}\left( {\Omega ;\mu }\right) \) . Then if \( \mu \left( E\right) = 0 \), it follows that \( {\nu }_{n}\left( E\right) = 0 \) and so \( \nu \left( E\right) = 0 \) also, being the limit ... | Yes |
Lemma 20.4.4 The above definition of the integral with respect to a finitely additive measure in \( {BV}\left( {\Omega ;\mu }\right) \) is well defined. | Proof: First consider the claim about the integral being well defined on the simple functions. This is clearly true if it is required that the \( {c}_{k} \) are disjoint and the \( {E}_{k} \) also disjoint having union equal to \( \Omega \) . Thus define the integral of a simple function in this manner. First write the... | Yes |
Lemma 20.4.5 For \( {T}_{\nu } \) just defined,\n\n\[ \left| {{T}_{\nu }f}\right| \leq \parallel f{\parallel }_{{L}^{\infty }}\parallel \nu \parallel \] | Proof: As noted above, the conclusion true if \( f \) is simple. Now if \( f \) is in \( {L}^{\infty } \) , then it is the uniform limit of simple functions off a set of \( \mu \) measure zero. Therefore, by the definition of the \( {T}_{\nu } \),\n\n\[ \left| {{T}_{\nu }f}\right| = \mathop{\lim }\limits_{{n \rightarro... | Yes |
Theorem 20.4.6 Let \( \theta : {BV}\left( {\Omega ;\mu }\right) \rightarrow {\left( {L}^{\infty }\left( \Omega ;\mu \right) \right) }^{\prime } \) be given by \( \theta \left( \nu \right) \equiv {T}_{\nu } \) . Then \( \theta \) is one to one, onto and preserves norms. | Proof: It was shown in the above lemma that \( \theta \) maps into \( {\left( {L}^{\infty }\left( \Omega ;\mu \right) \right) }^{\prime } \) . It is obvious that \( \theta \) is linear. Why does it preserve norms? From the above lemma,\n\n\[ \parallel {\theta \nu }\parallel \equiv \mathop{\sup }\limits_{{\parallel f{\p... | No |
Lemma 20.5.3 Let \( X \) be uniformly convex and let \( \phi \in {X}^{\prime } \) . Then there exists \( x \in X \) such that\n\n\[ \parallel x\parallel = 1,\phi \left( x\right) = \parallel \phi \parallel \] | Proof: Let \( \parallel \begin{Vmatrix}{\widetilde{x}}_{n}\end{Vmatrix} \leq 1 \) and \( \left| {\phi \left( {\widetilde{x}}_{n}\right) }\right| \rightarrow \parallel \phi \parallel \) . Let \( {x}_{n} = {w}_{n}{\widetilde{x}}_{n} \) where \( \left| {w}_{n}\right| = 1 \) and\n\n\[ {w}_{n}\phi {\widetilde{x}}_{n} = \lef... | Yes |
Theorem 20.5.5 (Riesz representation theorem \( p > 1 \) ) The map \( \eta \) is 1-1, onto, continuous, and \[ \parallel {\eta g}\parallel = \parallel g\parallel ,\parallel \eta \parallel = 1. \] | Proof: Obviously \( \eta \) is linear. Suppose \( {\eta g} = 0 \) . Then \( 0 = \int {gfd\mu } \) for all \( f \in {L}^{p} \) . Let \( f = {\left| g\right| }^{q - 2}\bar{g} \) . Then \( f \in {L}^{p} \) and so \( 0 = \int {\left| g\right| }^{q}{d\mu } \) . Hence \( g = 0 \) and \( \eta \) is one to one. That \( {\eta g... | Yes |
Lemma 20.6.4 Let \( L \in {C}_{0}{\left( X\right) }^{\prime } \) as above. Then letting \( \mu \) be the Radon measure just described, it follows \( \mu \) is finite and\n\n\[ \mu \left( X\right) = \parallel \Lambda \parallel = \parallel L\parallel \] | Proof: First of all, why is \( \parallel \Lambda \parallel = \parallel L\parallel \) ? From 20.6.20 it follows \( \parallel \Lambda \parallel \leq \parallel L\parallel \) . But also\n\n\[ \left| {Lg}\right| \leq \lambda \left( \left| g\right| \right) = \Lambda \left( \left| g\right| \right) \leq \parallel \Lambda \para... | Yes |
Corollary 20.7.1 Let \( L \in {\left( {C}_{0}\left( X\right) \right) }^{\prime } \) where \( X \) is a locally compact Hausdorff space. Then there exists \( \sigma \in {L}^{\infty }\left( {X,\mu }\right) \) for \( \mu \) a finite Radon measure such that for all \( f \in {C}_{0}\left( X\right) \)\n\n\[ L\left( f\right) ... | Proof: Let\n\n\[ \widetilde{D} \equiv \{ f \in C\left( \widetilde{X}\right) : f\left( \infty \right) = 0\} . \]\n\nThus \( \widetilde{D} \) is a closed subspace of the Banach space \( C\left( \widetilde{X}\right) \) . Let \( \theta : {C}_{0}\left( X\right) \rightarrow \widetilde{D} \) be defined by\n\n\[ {\theta f}\lef... | Yes |
Lemma 20.8.2 Suppose \( {\lambda }_{i}, i = 1,2 \) is a complex Borel measure with total variation finite \( {}^{2} \) defined on \( X \), a locally compact Hausdorf space. Then \( {\lambda }_{1} - {\lambda }_{2} \) is also a regular measure on the Borel sets. | Proof: Let \( E \) be a Borel set. That way it is in the \( \sigma \) algebras associated with both \( {\lambda }_{i} \). Then by regularity of \( {\lambda }_{i} \), there exist \( K \) and \( V \) compact and open respectively such that \( K \subseteq E \subseteq V \) and \( \left| {\lambda }_{i}\right| \left( {V \sma... | Yes |
Lemma 20.9.1 Let \( \mathcal{C} \equiv {\left\{ {E}_{i}\right\} }_{i = 1}^{\infty } \) be a countable collection of sets and let \( {\Omega }_{1} \equiv \) \( { \cup }_{i = 1}^{\infty }{E}_{i} \) . Then there exists an algebra of sets, \( \mathcal{A} \), such that \( \mathcal{A} \supseteq \mathcal{C} \) and \( \mathcal... | Proof: Let \( {\mathcal{C}}_{1} \) denote all finite unions of sets of \( \mathcal{C} \) and also include \( {\Omega }_{1} \) and \( \varnothing \) . Thus \( {\mathcal{C}}_{1} \) is countable. Next let \( {\mathcal{B}}_{1} \) denote all sets of the form \( {\Omega }_{1} \smallsetminus A \) such that \( A \in {\mathcal{... | Yes |
Lemma 20.9.2 Let \( \left\{ {f}_{n}\right\} \) be a sequence of functions in \( {L}^{1}\left( {\Omega ,\mathcal{S},\mu }\right) \). Then there exists a \( \sigma \) finite set of \( \mathcal{S},{\Omega }_{1} \), and a \( \sigma \) algebra of subsets of \( {\Omega }_{1},{\mathcal{S}}_{1} \), such that \( {\mathcal{S}}_{... | Proof: Let \( {\mathcal{E}}_{n} \) denote the sets which are of the form\n\n\[ \left\{ {{f}_{n}^{-1}\left( {B\left( {z, r}\right) }\right) : z \in \mathbb{Q} + i\mathbb{Q}, r > 0, r \in \mathbb{Q},\text{ and }0 \notin \overline{B\left( {z, r}\right) }}\right\} \]\n\nSince each \( {\mathcal{E}}_{n} \) is countable, so i... | Yes |
Lemma 20.9.5 Let \( K \) be an equi integrable set. Then there exists \( C > 0 \) such that for all \( f \in K \) ,\n\n\[ \parallel f{\parallel }_{{L}^{1}} \leq C \]\n\nand \( K \) also satisfies the property that if \( \left\{ {E}_{n}\right\} \) is a decreasing sequence of measurable sets such that \( { \cap }_{n = 1}... | Proof: Choose \( {\lambda }_{0} \) such that\n\n\[ \mathop{\sup }\limits_{{f \in K}}{\int }_{\left\lbrack \left| f\right| \geq {\lambda }_{0}\right\rbrack }\left| f\right| {d\mu } \leq 1 \]\n\nThen for \( f \in K \),\n\n\[ {\int }_{\Omega }\left| f\right| {d\mu } = {\int }_{\left\lbrack \left| f\right| \geq {\lambda }_... | Yes |
Corollary 20.9.6 Let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a measure space in which \( \mu \left( \Omega \right) < \infty \) and let \( K \subseteq {L}^{1}\left( {\Omega ,\mathcal{S},\mu }\right) \) be equi integrable. Then every sequence from \( K \) has a weakly convergent subsequence. | Proof: From Lemma 20.9.5 the hypotheses of Theorem 20.9.3 are satisfied. | No |
Proposition 20.9.7 Let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a measure space in which \( \mu \left( \Omega \right) < \infty \) . Then \( K \subseteq {L}^{1}\left( {\Omega ,\mathcal{S},\mu }\right) \) is equi integrable if and only if \( K \) is uniformly integrable and there exists a constant, \( M \) such ... | Proof: First suppose \( K \) is equi integrable. Then pick \( \lambda \) such that for all \( f \in K \) ,\n\n\[ \n{\int }_{\left\lbrack \left| f\right| \geq \lambda \right\rbrack }\left| f\right| {d\mu } < 1 \n\]\n\nThen for \( f \in K \)\n\n\[ \n{\int }_{\Omega }\left| f\right| {d\mu } = {\int }_{\left\lbrack \left| ... | Yes |
Lemma 21.1.2 Let \( x \) be strongly measurable. Then \( \parallel x\parallel \) is a real valued measurable function. There exists a sequence of simple functions \( \left\{ {y}_{n}\right\} \) which converges to \( f\left( s\right) \) pointwise and also \( \begin{Vmatrix}{{y}_{n}\left( s\right) }\end{Vmatrix} \leq 2\pa... | Proof: Consider the first claim. Letting \( {x}_{n} \) be a sequence of simple functions converging to \( x \) pointwise, it follows that \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} \) is a real valued measurable function. Since \( \parallel x\parallel \) is a pointwise limit, so is \( \parallel x\parallel \) a real valued ... | Yes |
Lemma 21.1.3 Suppose \( S \) is a nonempty subset of a metric space \( \left( {X, d}\right) \) and \( S \subseteq T \) where \( T \) is separable. Then there exists a countable dense subset of \( S \) . | Proof: Let \( D \) be the countable dense subset of \( T \) . Now consider the countable set \( \mathcal{B} \) of balls having center at a point of \( D \) and radius a positive rational number such that also, each ball in \( \mathcal{B} \) has nonempty intersection with \( S \) . Let \( \mathcal{D} \) consist of a poi... | Yes |
Lemma 21.1.5 Let \( x \in X \) a normed linear space. Then there exists \( f \in {X}^{\prime } \) such that \( \parallel f\parallel = 1 \) and \( f\left( x\right) = \parallel x\parallel \) . | Proof: Consider the one dimensional subspace\n\n\[ M \equiv \left\{ {\alpha \frac{x}{\parallel x\parallel } : \alpha \in \mathbb{F}}\right\} \]\n\nand define a continuous linear functional on \( M \) by \( g\left( {\alpha \frac{x}{\parallel x\parallel }}\right) \equiv \alpha \) . Then the norm of \( \parallel g\paralle... | Yes |
Lemma 21.1.6 If \( X \) is a separable Banach space with \( {B}^{\prime } \) the closed unit ball in \( {X}^{\prime } \), then there exists a sequence \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \equiv {D}^{\prime } \subseteq {B}^{\prime } \) with the property that for every \( x \in X \)\n\n\[ \parallel x\parallel... | Proof: Let \( {\left\{ {a}_{k}\right\} }_{k = 1}^{\infty } \) be a countable dense set in \( X \) and consider the mapping\n\n\[ {\phi }_{n} : {B}^{\prime } \rightarrow {\mathbb{F}}^{n} \]\n\ngiven by\n\n\[ {\phi }_{n}\left( f\right) \equiv \left( {f\left( {a}_{1}\right) ,\cdots, f\left( {a}_{n}\right) }\right) . \]\n\... | Yes |
Theorem 21.1.7 If \( x \) has values in a separable Banach space \( X \), then \( x \) is weakly measurable if and only if \( x \) is strongly measurable. | Proof: \( \Rightarrow \) It is necessary to show \( {x}^{-1}\left( U\right) \) is measurable whenever \( U \) is open. Since every open set is a countable union of balls, it suffices to show \( {x}^{-1}\left( {B\left( {a, r}\right) }\right) \) is measurable for any ball, \( B\left( {a, r}\right) \) . Since every open b... | Yes |
Corollary 21.1.8 Let \( X \) be a separable Banach space and let \( \mathcal{B}\left( X\right) \) denote the \( \sigma \) algebra of Borel sets. Let \( H \) be a dense subset of \( {X}^{\prime } \) . Then \( \mathcal{B}\left( X\right) = \sigma \left( H\right) \equiv \mathcal{F} \) , the smallest \( \sigma \) algebra of... | Proof: First I need to show \( \mathcal{F} \) contains open balls because then \( \mathcal{F} \) will contain the open sets and hence the Borel sets. As noted above, it suffices to show \( \mathcal{F} \) contains closed balls. Let \( {D}^{\prime } \) be those functionals in \( {B}^{\prime } \) defined in Lemma 21.1.6 c... | Yes |
Lemma 21.1.9 Let \( X \) be a metric space and suppose \( V \) is an open set in \( V \) . Then there exists open sets \( {V}_{m} \) such that\n\n\[ \cdots {V}_{m} \subseteq {\\bar{V}}_{m} \subseteq {V}_{m + 1} \subseteq \cdots, V = \\mathop{\\bigcup }\\limits_{{m = 1}}^{\\infty }{V}_{m}. \]\n\n(21.1.1) | Proof: Recall that if \( S \) is a nonempty set, \( x \\rightarrow \\operatorname{dist}\\left( {x, S}\\right) \) is a continuous map from \( X \) to \( \\mathbb{R} \) . First assume \( V \\neq X \) . Let\n\n\[ {V}_{m} \\equiv \\left\\{ {x \\in V : \\operatorname{dist}\\left( {x,{V}^{C}}\\right) > \\frac{1}{m}}\\right\\... | Yes |
Theorem 21.1.10 Let \( {x}_{n} \) and \( x \) be functions mapping \( \Omega \) to \( X \) where \( \mathcal{F} \) is a \( \sigma \) algebra of measurable sets of \( \Omega \) and \( X \) is a Banach space. Thus \( X \) satisfies 21.1.1. Then if \( {x}_{n} \) is strongly measurable, and \( x\left( s\right) = \mathop{\l... | Proof: Let \( \left\{ {V}_{m}\right\} \) be the sequence of 21.1.1. Since \( x \) is the pointwise limit of \( {x}_{n} \) ,\n\n\[ \n{x}^{-1}\left( {V}_{m}\right) \subseteq \left\{ {s : {x}_{k}\left( s\right) \in {V}_{m}}\right. \text{for all}k\text{large enough}\} \subseteq {x}^{-1}\left( \overline{{V}_{m}}\right) \tex... | Yes |
Lemma 21.1.12 Let \( X \) be a Banach space and let \( x : \left( {\Omega ,\mathcal{F}}\right) \rightarrow K \subseteq X \) where \( K \) is weakly compact and \( {X}^{\prime } \) is separable. Then \( x \) is weakly measurable if and only if \( {x}^{-1}\left( U\right) \in \mathcal{F} \) whenever \( U \) is a weakly op... | Proof: By Corollary 17.5.9 on Page 491, there exists a metric \( d \), such that the metric space topology with respect to \( d \) coincides with the weak topology on \( K \) . Since \( K \) is compact, it follows that \( K \) is also separable. Hence it is completely separable and so there exists a countable basis of ... | Yes |
Lemma 21.1.13 Let \( B \) be the closed unit ball in \( X \) . If \( {X}^{\prime } \) is separable, there exists a sequence \( {\left\{ {x}_{m}\right\} }_{m = 1}^{\infty } \equiv D \subseteq B \) with the property that for all \( {y}^{ * } \in {X}^{\prime } \) , \[ \begin{Vmatrix}{y}^{ * }\end{Vmatrix} = \mathop{\sup }... | Proof: Let \( {\left\{ {x}_{k}^{ * }\right\} }_{k = 1}^{\infty } \) be the dense subset of \( {X}^{\prime } \) . Define \( {\phi }_{n} : B \rightarrow {\mathbb{F}}^{n} \) by \[ {\phi }_{n}\left( x\right) \equiv \left( {{x}_{1}^{ * }\left( x\right) ,\cdots ,{x}_{n}^{ * }\left( x\right) }\right) . \] Then \( \left| {{x}_... | Yes |
Theorem 21.1.15 If \( {X}^{\prime } \) is separable and \( y : \Omega \rightarrow {X}^{\prime } \) is weak* measurable meaning \( s \rightarrow y\left( s\right) \left( x\right) \) is a \( \mathbb{F} \) valued measurable function, then \( y \) is strongly measurable. | Proof: It is necessary to show \( {y}^{-1}\left( {B\left( {{a}^{ * }, r}\right) }\right) \) is measurable for \( {a}^{ * } \in {X}^{\prime } \) . This will suffice because the separability of \( {X}^{\prime } \) implies every open set is the countable union of such balls of the form \( B\left( {{a}^{ * }, r}\right) \) ... | Yes |
Theorem 21.1.16 If \( {X}^{\prime } \) is separable, then so is \( X \) . | Proof: Let \( D = \left\{ {x}_{m}\right\} \subseteq B \), the unit ball of \( X \), be the sequence promised by Lemma 21.1.13. Let \( V \) be all finite linear combinations of elements of \( \left\{ {x}_{m}\right\} \) with rational scalars. Thus \( \bar{V} \) is a separable subspace of \( X \) . The claim is that \( \b... | Yes |
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