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Theorem 14.4.3 For each finite set\n\n\\[ \nJ = \\left( {{t}_{1},\\cdots ,{t}_{n}}\\right) \\subseteq I \n\\]\n\n\nsuppose there exists a Borel probability measure, \\( {\\nu }_{J} = {\\nu }_{{t}_{1}\\cdots {t}_{n}} \\) defined on the Borel sets of \\( \\mathop{\\prod }\\limits_{{t \\in J}}{M}_{t} \\) such that the fol... | Proof: Let \\( \\mathcal{E} \\) be the algebra of sets defined in Definition 14.4.1. I want to define a measure on \\( \\mathcal{E} \\) . For \\( \\mathbf{F} \\in \\mathcal{E} \\), there exists \\( J \\) such that \\( \\mathbf{F} \\) is the finite disjoint unions of sets of \\( {\\mathcal{R}}_{J} \\) . Define\n\n\\[ \n... | Yes |
Lemma 14.4.4 Let \( J \) be a finite subset of \( I \) . Then \( \mathbf{U} \) is a Borel set in \( \mathop{\prod }\limits_{{t \in J}}{M}_{t} \) if and only if there exists a Borel set, \( {\mathbf{U}}^{\prime } \) in \( \mathop{\prod }\limits_{{t \in J}}{M}_{t}^{\prime } \) such that \( \mathbf{U} = {\mathbf{U}}^{\pri... | Proof: A subbasis for the topology for \( \left\lbrack {-\infty ,\infty }\right\rbrack \) is sets of the form \( \lbrack - \infty, a) \) and \( (a,\infty \rbrack \) . Also a subbasis for the topology of \( {\left\lbrack -\infty ,\infty \right\rbrack }^{n} \) is sets of the form \( \mathop{\prod }\limits_{{i = 1}}^{n}\l... | Yes |
Theorem 14.4.5 (Kolmogorov extension theorem) For each finite set\n\n\\[ \nJ = \\left( {{t}_{1},\\cdots ,{t}_{n}}\\right) \\subseteq I \n\\]\n\nsuppose there exists a Borel probability measure, \\( {\\nu }_{J} = {\\nu }_{{t}_{1}\\cdots {t}_{n}} \\) defined on the Borel sets of \\( \\mathop{\\prod }\\limits_{{t \\in J}}... | Proof: Using Lemma 14.4.4, extend each measure, \\( {\\nu }_{J} \\) to \\( {M}_{t}^{\\prime } \\), defined by adding in the points \\( \\pm \\infty \\) at the ends, by letting \\( {\\nu }_{J}\\left( \\mathbf{E}\\right) \\equiv {\\nu }_{J}\\left( {\\mathbf{E} \\cap \\mathop{\\prod }\\limits_{{t \\in I}}{M}_{t}}\\right) ... | Yes |
Theorem 15.1.2 (Holder’s inequality) If \( f \) and \( g \) are measurable functions, then if \( p > 1 \), \[ \int \left| f\right| \left| g\right| {d\mu } \leq {\left( \int {\left| f\right| }^{p}d\mu \right) }^{\frac{1}{p}}{\left( \int {\left| g\right| }^{q}d\mu \right) }^{\frac{1}{q}}. \] | Proof: First here is a proof of Young's inequality . | No |
Lemma 15.1.3 If \( p > 1 \), and \( 0 \leq a, b \) then \( {ab} \leq \frac{{a}^{p}}{p} + \frac{{b}^{q}}{q} \) . | Proof: Consider the following picture:\n\n\n\nFrom this picture, the sum of the area between the \( x \) axis and the curve added to the area between the \( t \) axis and the curve is at least as large as \( {ab} \) ... | Yes |
Lemma 15.1.4 For \( a, b \geq 0 \) ,\n\n\[ \n{ab} \leq \frac{{a}^{p}}{p} + \frac{{b}^{q}}{q} \n\]\n\nand equality occurs when if and only if \( {a}^{p} = {b}^{q} \) . | Proof: If \( b = 0 \), the inequality is obvious. Fix \( b > 0 \) and consider\n\n\[ \nf\left( a\right) \equiv \frac{{a}^{p}}{p} + \frac{{b}^{q}}{q} - {ab}. \n\]\n\nThen \( {f}^{\prime }\left( a\right) = {a}^{p - 1} - b \) . This is negative when \( a < {b}^{1/\left( {p - 1}\right) } \) and is positive when \( a > {b}^... | Yes |
Lemma 15.1.5 Suppose \( x, y \in \mathbb{C} \) . Then\n\n\[ \n{\left| x + y\right| }^{p} \leq {2}^{p - 1}\left( {{\left| x\right| }^{p} + {\left| y\right| }^{p}}\right)\n\] | Proof: The function \( f\left( t\right) = {t}^{p} \) is concave up for \( t \geq 0 \) because \( p > 1 \) . Therefore, the secant line joining two points on the graph of this function must lie above the graph of the function. This is illustrated in the following picture.\n\n![3f4063cc-9f64-45dc-a428-31c15f6604d5_423_0.... | Yes |
Corollary 15.1.6 (Minkowski inequality) Let \( 1 \leq p < \infty \) . Then \[ {\left( \int {\left| f + g\right| }^{p}d\mu \right) }^{1/p} \leq {\left( \int {\left| f\right| }^{p}d\mu \right) }^{1/p} + {\left( \int {\left| g\right| }^{p}d\mu \right) }^{1/p}. \] | Proof: If \( p = 1 \), this is obvious because it is just the triangle inequality. Let \( p > 1 \) . Without loss of generality, assume \[ {\left( \int {\left| f\right| }^{p}d\mu \right) }^{1/p} + {\left( \int {\left| g\right| }^{p}d\mu \right) }^{1/p} < \infty \] and \( {\left( \int {\left| f + g\right| }^{p}d\mu \rig... | Yes |
Corollary 15.1.7 Let \( {f}_{i} \in {L}^{p}\left( \Omega \right) \) for \( i = 1,2,\cdots, n \) . Then\n\n\[ \n{\left( \int {\left| \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{i}\right| }^{p}d\mu \right) }^{1/p} \leq \mathop{\sum }\limits_{{i = 1}}^{n}{\left( \int {\left| {f}_{i}\right| }^{p}\right) }^{1/p}.\n\] | This shows that if \( f, g \in {L}^{p} \), then \( f + g \in {L}^{p} \) . Also, it is clear that if \( a \) is a constant and \( f \in {L}^{p} \), then \( {af} \in {L}^{p} \) because\n\n\[ \n\int {\left| af\right| }^{p}{d\mu } = {\left| a\right| }^{p}\int {\left| f\right| }^{p}{d\mu } < \infty .\n\]\n\nThus \( {L}^{p} ... | No |
Theorem 15.1.10 The following hold for \( {L}^{p}\left( \Omega \right) \)\n\na.) \( {L}^{p}\left( \Omega \right) \) is complete.\n\nb.) If \( \left\{ {f}_{n}\right\} \) is a Cauchy sequence in \( {L}^{p}\left( \Omega \right) \), then there exists \( f \in {L}^{p}\left( \Omega \right) \) and a subsequence which converge... | Proof: Let \( \left\{ {f}_{n}\right\} \) be a Cauchy sequence in \( {L}^{p}\left( \Omega \right) \) . This means that for every \( \varepsilon > 0 \) there exists \( N \) such that if \( n, m \geq N \), then \( {\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{p} < \varepsilon \) . Now select a subsequence as follows. L... | Yes |
Lemma 15.1.11 Let \( \\left( {X,\\mathcal{S},\\mu }\\right) \) and \( \\left( {Y,\\mathcal{F},\\lambda }\\right) \) be finite complete measure spaces and let \( f \) be \( \\overline{\\mu \\times \\lambda } \) measurable and uniformly bounded. Then the following inequality is valid for \( p \\geq 1 \) .\n\n\[ \n{\\int ... | Proof: Since \( f \) is bounded and \( \\mu \\left( X\\right) ,\\lambda \\left( Y\\right) < \\infty \) ,\n\n\[ \n{\\left( {\\int }_{Y}{\\left( {\\int }_{X}\\left| f\\left( x, y\\right) \\right| d\\mu \\right) }^{p}d\\lambda \\right) }^{\\frac{1}{p}} < \\infty .\n\]\n\nLet\n\n\[ \nJ\\left( y\\right) = {\\int }_{X}\\left... | Yes |
Theorem 15.1.12 Let \( \\left( {X,\\mathcal{S},\\mu }\\right) \) and \( \\left( {Y,\\mathcal{F},\\lambda }\\right) \) be \( \\sigma \) -finite measure spaces and let \( f \) be product measurable. Then the following inequality is valid for \( p \\geq 1 \) .\n\n\[ \n{\\int }_{X}{\\left( {\\int }_{Y}{\\left| f\\left( x, ... | Proof: Since the two measure spaces are \( \\sigma \) finite, there exist measurable sets, \( {X}_{m} \) and \( {Y}_{k} \) such that \( {X}_{m} \\subseteq {X}_{m + 1} \) for all \( m,{Y}_{k} \\subseteq {Y}_{k + 1} \) for all \( k \), and \( \\mu \\left( {X}_{m}\\right) ,\\lambda \\left( {Y}_{k}\\right) < \) \( \\infty ... | Yes |
Theorem 15.2.1 Let \( p \geq 1 \) and let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a measure space. Then the simple functions are dense in \( {L}^{p}\left( \Omega \right) \) . | Proof: Recall that a function, \( f \), having values in \( \mathbb{R} \) can be written in the form \( f = {f}^{ + } - {f}^{ - } \) where\n\n\[ \n{f}^{ + } = \max \left( {0, f}\right) ,{f}^{ - } = \max \left( {0, - f}\right) .\n\]\n\nTherefore, an arbitrary complex valued function, \( f \) is of the form\n\n\[ \nf = \... | Yes |
Lemma 15.2.3 Let \( \Omega \) be a metric space in which the closed balls are compact and let \( K \) be a compact subset of \( V \), an open set. Then there exists a continuous function \( f : \Omega \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( f\left( x\right) = 1 \) for all \( x \in K \) and \( \opera... | Proof: Let \( K \subseteq W \subseteq \bar{W} \subseteq V \) and \( \bar{W} \) is compact. To obtain this list of inclusions consider a point in \( K, x \), and take \( B\left( {x,{r}_{x}}\right) \) a ball containing \( x \) such that \( \overline{B\left( {x,{r}_{x}}\right) } \) is a compact subset of \( V \) . Next us... | Yes |
Theorem 15.2.4 Let \( \left( {\Omega ,\mathcal{S},\mu }\right) \) be a regular measure space as in Definition 15.2.2 where the conclusion of Lemma 15.2.3 holds. Then \( {C}_{c}\left( \Omega \right) \) is dense in \( {L}^{p}\left( \Omega \right) \) . | Proof: First consider a measurable set, \( E \) where \( \mu \left( E\right) < \infty \) . Let \( K \subseteq E \subseteq V \) where \( \mu \left( {V \smallsetminus K}\right) < \varepsilon \) . Now let \( K \prec h \prec V \) . Then\n\n\[ \n\int {\left| h - {\mathcal{X}}_{E}\right| }^{p}{d\mu } \leq \int {\mathcal{X}}_... | Yes |
Corollary 15.3.3 Let \( \Omega \) be any \( \mu \) measurable subset of \( {\mathbb{R}}^{n} \) and let \( \mu \) be a Radon measure. Then \( {L}^{p}\left( {\Omega ,\mu }\right) \) is separable. Here the \( \sigma \) algebra of measurable sets will consist of all intersections of measurable sets with \( \Omega \) and th... | Proof: Let \( \widetilde{\mathcal{D}} \) be the restrictions of \( \mathcal{D} \) to \( \Omega \) . If \( f \in {L}^{p}\left( \Omega \right) \), let \( F \) be the zero extension of \( f \) to all of \( {\mathbb{R}}^{n} \) . Let \( \varepsilon > 0 \) be given. By Theorem 15.3.1 or 15.3.2 there exists \( s \in \mathcal{... | Yes |
Theorem 15.4.2 (Continuity of translation in \( {L}^{p} \) ) Let \( f \in {L}^{p}\left( {\mathbb{R}}^{n}\right) \) with the measure being Lebesgue measure. Then\n\n\[ \mathop{\lim }\limits_{{\parallel \mathbf{w}\parallel \rightarrow 0}}{\begin{Vmatrix}{f}_{\mathbf{w}} - f\end{Vmatrix}}_{p} = 0 \] | Proof: Let \( \varepsilon > 0 \) be given and let \( g \in {C}_{c}\left( {\mathbb{R}}^{n}\right) \) with \( \parallel g - f{\parallel }_{p} < \frac{\varepsilon }{3} \) . Since Lebesgue measure is translation invariant \( \left( {{m}_{n}\left( {\mathbf{w} + E}\right) = {m}_{n}\left( E\right) }\right) \),\n\n\[ {\begin{V... | Yes |
Example 15.5.2 Let \( U = B\left( {\mathbf{z},{2r}}\right) \)\n\n\[ \psi \left( \mathbf{x}\right) = \left\{ \begin{array}{ll} \exp \left\lbrack {\left( {\left| \mathbf{x} - \mathbf{z}\right| }^{2} - {r}^{2}\right) }^{-1}\right\rbrack & \text{ if }\left| {\mathbf{x} - \mathbf{z}}\right| < r, \\ 0 & \text{ if }\left| {\m... | The following also is easily obtained. | No |
Lemma 15.5.3 Let \( U \) be any open set. Then \( {C}_{c}^{\infty }\left( U\right) \neq \varnothing \) . | Proof: Pick \( \mathbf{z} \in U \) and let \( r \) be small enough that \( B\left( {\mathbf{z},{2r}}\right) \subseteq U \) . Then let \( \psi \in {C}_{c}^{\infty }\left( {B\left( {\mathbf{z},{2r}}\right) }\right) \subseteq {C}_{c}^{\infty }\left( U\right) \) be the function of the above example. | Yes |
Let \( \psi \in {C}_{c}^{\infty }\left( {B\left( {0,1}\right) }\right) \;\left( {B\left( {0,1}\right) = \{ \mathbf{x} : \left| \mathbf{x}\right| < 1\} }\right) \) with \( \psi \left( \mathbf{x}\right) \geq 0 \) and \( \int {\psi dm} = 1 \) . Let \( {\psi }_{m}\left( \mathbf{x}\right) = {c}_{m}\psi \left( {m\mathbf{x}}\... | By the change of variables theorem \( {c}_{m} = {m}^{n} \) . | Yes |
Lemma 15.5.7 Let \( f \in {L}_{loc}^{1}\left( {{\mathbb{R}}^{n},\mu }\right) \), and \( g \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . Then \( f * g \) is an infinitely differentiable function. Here \( \mu \) is a Radon measure on \( {\mathbb{R}}^{n} \) . | Proof: Consider the difference quotient for calculating a partial derivative of \( f * g \) .
\[
\frac{f * g\left( {\mathbf{x} + t{\mathbf{e}}_{j}}\right) - f * g\left( \mathbf{x}\right) }{t} = \int f\left( \mathbf{y}\right) \frac{g\left( {\mathbf{x} + t{\mathbf{e}}_{j} - \mathbf{y}}\right) - g\left( {\mathbf{x} - \ma... | Yes |
Theorem 15.5.8 Let \( K \) be a compact subset of an open set, \( U \) . Then there exists a function, \( h \in {C}_{c}^{\infty }\left( U\right) \), such that \( h\left( \mathbf{x}\right) = 1 \) for all \( \mathbf{x} \in K \) and \( h\left( \mathbf{x}\right) \in \left\lbrack {0,1}\right\rbrack \) for all X. | Proof: Let \( r > 0 \) be small enough that \( K + B\left( {\mathbf{0},{3r}}\right) \subseteq U \) . The symbol, \( K + B\left( {\mathbf{0},{3r}}\right) \) means\n\n\[ \n\{ \mathbf{k} + \mathbf{x} : \mathbf{k} \in K\text{ and }\mathbf{x} \in B\left( {\mathbf{0},{3r}}\right) \} .\n\]\n\nThus this is simply a way to writ... | Yes |
Corollary 15.5.9 Let \( K \) be a compact set in \( {\mathbb{R}}^{n} \) and let \( {\left\{ {U}_{i}\right\} }_{i = 1}^{\infty } \) be an open cover of \( K \) . Then there exist functions, \( {\psi }_{k} \in {C}_{c}^{\infty }\left( {U}_{i}\right) \) such that \( {\psi }_{i} \prec {U}_{i} \) and for all \( \mathbf{x} \i... | Proof: This follows from a repeat of the proof of Theorem 12.2.11 on Page 304, replacing the lemma used in that proof with Theorem 15.5.8. | No |
Theorem 15.5.10 For each \( p \geq 1,{C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) is dense in \( {L}^{p}\left( {\mathbb{R}}^{n}\right) \) . Here the measure is Lebesgue measure. | Proof: Let \( f \in {L}^{p}\left( {\mathbb{R}}^{n}\right) \) and let \( \varepsilon > 0 \) be given. Choose \( g \in {C}_{c}\left( {\mathbb{R}}^{n}\right) \) such that \( \parallel f - g{\parallel }_{p} < \frac{\varepsilon }{2} \) . This can be done by using Theorem 15.2.4. Now let\n\n\[ \n{g}_{m}\left( \mathbf{x}\righ... | Yes |
Corollary 15.5.11 Let \( U \) be an open set. For each \( p \geq 1,{C}_{c}^{\infty }\left( U\right) \) is dense in \( {L}^{p}\left( U\right) \) . Here the measure is Lebesgue measure. | Proof: Let \( f \in {L}^{p}\left( U\right) \) and let \( \varepsilon > 0 \) be given. Choose \( g \in {C}_{c}\left( U\right) \) such that \( \parallel f - g{\parallel }_{p} < \frac{\varepsilon }{2} \) . This is possible because Lebesgue measure restricted to the open set, \( U \) is regular. Thus the existence of such ... | Yes |
Theorem 16.0.4 If \( \mathfrak{D} \) is locally finite then\n\n\[ \cup \{ \bar{D} : D \in \mathfrak{D}\} = \overline{\cup \{ D : D \in \mathfrak{D}\} }.\] | Proof: It is clear the left side is a subset of the right. Let \( p \) be a limit point of\n\n\[ \cup \{ D : D \in \mathfrak{D}\} \]\n\nand let \( p \in V \), an open set intersecting only finitely many sets of \( \mathfrak{D},{D}_{1}\ldots {D}_{n} \) . If \( p \) is not in any of \( \overline{{D}_{i}} \) then \( p \in... | Yes |
Theorem 16.0.5 Let \( S \) be a regular topological space. (If \( p \in U \) open, then there exists an open set \( V \) such that \( p \in \bar{V} \subseteq U \) . ) The following are equivalent\n\n1.) Every open covering of \( S \) has a refinement that is open, covers \( S \) and is countably locally finite.\n\n2.) ... | Proof:\n\n1.) \( \Rightarrow \) 2.)\n\nLet \( \mathfrak{S} \) be an open cover of \( S \) and let \( \mathfrak{B} \) be an open countably locally finite refinement\n\n\[ \mathfrak{B} = { \cup }_{n = 1}^{\infty }{\mathfrak{B}}_{n} \]\n\nwhere \( {\mathfrak{B}}_{n} \) is an open refinement of \( \mathfrak{S} \) and \( {\... | Yes |
Theorem 16.0.6 If \( S \) is a metric space then \( S \) is paracompact (Every open cover has a locally finite open refinement also an open cover.) | Proof: Let \( \mathfrak{S} \) be an open cover. Well order \( \mathfrak{S} \) . For \( B \in \mathfrak{S} \) , \[ {B}_{n} \equiv \left\{ {x \in B : \operatorname{dist}\left( {x,{B}^{C}}\right) < \frac{1}{{2}^{n}}}\right\}, n = 1,2,\cdots . \] Thus \( {B}_{n} \) is contained in \( B \) but approximates it up to \( {2}^{... | Yes |
Corollary 16.1.2 Let \( S \) be a metric space and let \( \mathfrak{S} \) be any open cover of \( S \) . Then there exists a set \( \mathfrak{F} \), an open refinement of \( \mathfrak{S} \), and functions \( \left\{ {{\phi }_{F} : F \in \mathfrak{F}}\right\} \) such that\n\n\[{\phi }_{F} : S \rightarrow \left\lbrack {0... | Proof: Just change your open cover to consist of \( U \) and \( V \\smallsetminus H \) for each \( V \in \\mathfrak{S} \) . Then every function but one equals 0 on \( H \) and so exactly one of them equals 1 on\n\n\( H \) . \( \\blacksquare \) | Yes |
Lemma 16.2.1 Let \( A \) be a closed set in a metric space and let \( {x}_{n} \notin A,{x}_{n} \rightarrow {a}_{0} \in A \) and \( {a}_{n} \in A \) such that \( d\left( {{a}_{n},{x}_{n}}\right) < 6\operatorname{dist}\left( {{x}_{n}, A}\right) \) . Then \( {a}_{n} \rightarrow {a}_{0} \) . | Proof: By assumption,\n\n\[ d\left( {{a}_{n},{a}_{0}}\right) \leq d\left( {{a}_{n},{x}_{n}}\right) + d\left( {{x}_{n},{a}_{0}}\right) < 6\operatorname{dist}\left( {{x}_{n}, A}\right) + d\left( {{x}_{n},{a}_{0}}\right) \]\n\n\[ \leq {6d}\left( {{x}_{n},{a}_{0}}\right) + d\left( {{x}_{n},{a}_{0}}\right) = {7d}\left( {{x}... | Yes |
Theorem 16.2.5 Let \( K \) be closed and convex subset of \( X \) a Banach space. Then \( K \) is a retract. | Proof: By Theorem 16.2.3, there is a continuous function \( \widehat{I} \) extending \( I \) to all of \( X \) . Then also \( \widehat{I} \) has values in \( \operatorname{conv}\left( {IK}\right) = \operatorname{conv}\left( K\right) = K \) . Hence \( \widehat{I} \) is a continuous function which does what is needed. It... | No |
Lemma 16.3.1 Let \( \mathbf{g} : U \rightarrow {\mathbb{R}}^{n} \) be \( {C}^{2} \) where \( U \) is an open subset of \( {\mathbb{R}}^{n} \) . Then\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}\operatorname{cof}{\left( D\mathbf{g}\right) }_{{ij}, j} = 0 \]\n\nwhere here \( {\left( D\mathbf{g}\right) }_{ij} \equiv {g}_{i, ... | Proof: From the cofactor expansion theorem,\n\n\[ \det \left( {D\mathbf{g}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{g}_{i, j}\operatorname{cof}{\left( D\mathbf{g}\right) }_{ij} \]\n\nand so\n\n\[ \frac{\partial \det \left( {D\mathbf{g}}\right) }{\partial {g}_{i, j}} = \operatorname{cof}{\left( D\mathbf{g}\right) }... | Yes |
Lemma 16.3.3 There does not exist \( \mathbf{h} \in {C}^{2}\left( \overline{B\left( {\mathbf{0}, R}\right) }\right) \) such that \( \mathbf{h} : \overline{B\left( {\mathbf{0}, R}\right) } \rightarrow \) \( \partial B\left( {\mathbf{0}, R}\right) \) which also has the property that \( \mathbf{h}\left( \mathbf{x}\right) ... | Proof: Suppose such an \( \mathbf{h} \) exists. Let \( \lambda \in \left\lbrack {0,1}\right\rbrack \) and let \( {\mathbf{p}}_{\lambda }\left( \mathbf{x}\right) \equiv \mathbf{x} + \lambda \left( {\mathbf{h}\left( \mathbf{x}\right) - \mathbf{x}}\right) \) . This function, \( {\mathbf{p}}_{\lambda } \) is a homotopy of ... | Yes |
Theorem 16.3.5 Let \( \mathbf{f} : \overline{B\left( {\mathbf{0}, R}\right) } \rightarrow \overline{B\left( {\mathbf{0}, R}\right) } \) be continuous, this being a ball in \( {\mathbb{R}}^{p} \) . Then it has a fixed point \( \mathbf{x} \in \overline{B\left( {\mathbf{0}, R}\right) } \) such that \( \mathbf{f}\left( \ma... | Proof: You can extend \( \mathbf{f} \) to assume it is defined on all of \( {\mathbb{R}}^{p},\mathbf{f}\left( {\mathbb{R}}^{p}\right) \subseteq \overline{B\left( {\mathbf{0}, R}\right) } \) , the convex hull of \( \overline{B\left( {\mathbf{0}, R}\right) } \) . Then letting \( \left\{ {\psi }_{n}\right\} \) be a mollif... | Yes |
Theorem 17.1.2 Let \( \left( {X, d}\right) \) be a complete metric space and let \( {\left\{ {U}_{n}\right\} }_{n = 1}^{\infty } \) be a sequence of open subsets of \( X \) satisfying \( \overline{{U}_{n}} = X \) ( \( {U}_{n} \) is dense). Then \( D \equiv { \cap }_{n = 1}^{\infty }{U}_{n} \) is a dense subset of \( X ... | Proof: Let \( p \in X \) and let \( {r}_{0} > 0 \) . I need to show \( D \cap B\left( {p,{r}_{0}}\right) \neq \varnothing \) . Since \( {U}_{1} \) is dense, there exists \( {p}_{1} \in {U}_{1} \cap B\left( {p,{r}_{0}}\right) \), an open set. Let \( {p}_{1} \in B\left( {{p}_{1},{r}_{1}}\right) \subseteq \overline{B\left... | No |
Corollary 17.1.3 Let \( X \) be a complete metric space and suppose \( X = { \cup }_{i = 1}^{\infty }{F}_{i} \) where each \( {F}_{i} \) is a closed set. Then for some \( i \), interior \( {F}_{i} \neq \varnothing \) . | Proof: If all \( {F}_{i} \) has empty interior, then \( {F}_{i}^{C} \) would be a dense open set. Therefore, from Theorem 17.1.2, it would follow that\n\n\[ \varnothing = {\left( { \cup }_{i = 1}^{\infty }{F}_{i}\right) }^{C} = { \cap }_{i = 1}^{\infty }{F}_{i}^{C} \neq \varnothing . \] | Yes |
Theorem 17.1.4 Let \( X \) and \( Y \) be two normed linear spaces and let \( L : X \rightarrow Y \) be linear \( \left( {L\left( {{ax} + {by}}\right) = {aL}\left( x\right) + {bL}\left( y\right) \text{for}a, b\text{scalars and}x, y \in X}\right) \) . The following are equivalent\n\na.) \( L \) is continuous at 0\n\nb.)... | Proof: a.) \( \Rightarrow \) b.) Let \( {x}_{n} \rightarrow x \) . It is necessary to show that \( L{x}_{n} \rightarrow {Lx} \) . But \( \left( {{x}_{n} - x}\right) \rightarrow 0 \) and so from continuity at 0, it follows\n\n\[ L\left( {{x}_{n} - x}\right) = L{x}_{n} - {Lx} \rightarrow 0 \]\n\nso \( L{x}_{n} \rightarro... | Yes |
Lemma 17.1.6 With \( \parallel L\parallel \) defined in 17.1.1, \( \mathcal{L}\left( {X, Y}\right) \) is a normed linear space. Also \( \left| \right| {Lx}\left| \right| \leq \left| \right| L\left| \right| \left| \right| x\left| \right| \) . | Proof: Let \( x \neq 0 \) then \( x/\parallel x\parallel \) has norm equal to 1 and so\n\n\[ \left| \left| {L\left( \frac{x}{\parallel x\parallel }\right) }\right| \right| \leq \parallel L\parallel \]\n\nTherefore, multiplying both sides by \( \parallel x\parallel ,\parallel {Lx}\parallel \leq \parallel L\parallel \par... | Yes |
Theorem 17.1.7 If \( Y \) is a Banach space, then \( \mathcal{L}\left( {X, Y}\right) \) is also a Banach space. | Proof: Let \( \left\{ {L}_{n}\right\} \) be a Cauchy sequence in \( \mathcal{L}\left( {X, Y}\right) \) and let \( x \in X \). \[ \begin{Vmatrix}{{L}_{n}x - {L}_{m}x}\end{Vmatrix} \leq \parallel x\parallel \begin{Vmatrix}{{L}_{n} - {L}_{m}}\end{Vmatrix}. \] Thus \( \left\{ {{L}_{n}x}\right\} \) is a Cauchy sequence. Let... | Yes |
Theorem 17.1.8 Let \( X \) be a Banach space and let \( Y \) be a normed linear space. Let \( {\left\{ {L}_{\alpha }\right\} }_{\alpha \in \Lambda } \) be a collection of elements of \( \mathcal{L}\left( {X, Y}\right) \) . Then one of the following happens.\n\na.) \( \sup \left\{ {\begin{Vmatrix}{L}_{\alpha }\end{Vmatr... | Proof: For each \( n \in \mathbb{N} \), define\n\n\[ {U}_{n} = \left\{ {x \in X : \sup \left\{ {\begin{Vmatrix}{{L}_{\alpha }x}\end{Vmatrix} : \alpha \in \Lambda }\right\} > n}\right\} .\n\nThen \( {U}_{n} \) is an open set because if \( x \in {U}_{n} \), then there exists \( \alpha \in \Lambda \) such that\n\n\[ \begi... | Yes |
Lemma 17.1.10 Let \( a \) and \( b \) be positive constants and suppose\n\n\[ B\left( {0, a}\right) \subseteq \overline{L\left( {B\left( {0, b}\right) }\right) }.\]\n\nThen\n\n\[ \overline{L\left( {B\left( {0, b}\right) }\right) } \subseteq L\left( {B\left( {0,{2b}}\right) }\right) . \] | Proof of Lemma 17.1.10: Let \( y \in \overline{L\left( {B\left( {0, b}\right) }\right) } \) . There exists \( {x}_{1} \in B\left( {0, b}\right) \) such that \( \begin{Vmatrix}{y - L{x}_{1}}\end{Vmatrix} < \frac{a}{2} \) . Now this implies\n\n\[ {2y} - {2L}{x}_{1} \in B\left( {0, a}\right) \subseteq \overline{L\left( {B... | Yes |
Lemma 17.1.13 The norm defined in Definition 17.1.12 on \( X \times Y \) along with the definition of addition and scalar multiplication given there make \( X \times Y \) into a normed linear space. | Proof: The only axiom for a norm which is not obvious is the triangle inequality. Therefore, consider\n\n\[ \begin{Vmatrix}{\left( {{x}_{1},{y}_{1}}\right) + \left( {{x}_{2},{y}_{2}}\right) }\end{Vmatrix} = \begin{Vmatrix}\left( {{x}_{1} + {x}_{2},{y}_{1} + {y}_{2}}\right) \end{Vmatrix} \]\n\n\[ = \max \left( {\begin{V... | Yes |
Lemma 17.1.14 If \( X \) and \( Y \) are Banach spaces, then \( X \times Y \) with the norm and vector space operations defined in Definition 17.1.12 is also a Banach space. | Proof: The only thing left to check is that the space is complete. But this follows from the simple observation that \( \left\{ \left( {{x}_{n},{y}_{n}}\right) \right\} \) is a Cauchy sequence in \( X \times Y \) if and only if \( \left\{ {x}_{n}\right\} \) and \( \left\{ {y}_{n}\right\} \) are Cauchy sequences in \( X... | No |
Lemma 17.1.15 Every closed subspace of a Banach space is a Banach space. | Proof: If \( F \subseteq X \) where \( X \) is a Banach space and \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence in \( F \), then since \( X \) is complete, there exists a unique \( x \in X \) such that \( {x}_{n} \rightarrow x \) . However this means \( x \in \bar{F} = F \) since \( F \) is closed. | Yes |
Theorem 17.1.17 Let \( X \) and \( Y \) be Banach spaces and suppose \( L : X \rightarrow Y \) is closed and linear. Then \( L \) is continuous. | Proof: Let \( G \) be the graph of \( L.G = \{ \left( {x,{Lx}}\right) : x \in X\} \) . By Lemma 17.1.15 it follows that \( G \) is a Banach space. Define \( P : G \rightarrow X \) by \( P\left( {x,{Lx}}\right) = x \) . \( P \) maps the Banach space \( G \) onto the Banach space \( X \) and is continuous and linear. By ... | Yes |
Corollary 17.1.18 Let \( L : D \subseteq X \rightarrow Y \) where \( X, Y \) are a Banach spaces, and \( L \) is a closed operator. Then define a new norm on \( D \) by\n\n\[ \parallel x{\parallel }_{D} \equiv \parallel x{\parallel }_{X} + \parallel {Lx}{\parallel }_{Y} \]\n\nThen \( D \) with this new norm is a Banach... | Proof: If \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence in \( D \) with this new norm, it follows both \( \left\{ {x}_{n}\right\} \) and \( \left\{ {L{x}_{n}}\right\} \) are Cauchy sequences and therefore, they converge. Since \( L \) is closed, \( {x}_{n} \rightarrow x \) and \( L{x}_{n} \rightarrow {Lx} \) for s... | Yes |
Theorem 17.2.5 (Hahn Banach theorem) Let \( X \) be a real vector space, let \( M \) be a subspace of \( X \), let \( f : M \rightarrow \mathbb{R} \) be linear, let \( \rho \) be a gauge function on \( X \), and suppose \( f\left( x\right) \leq \rho \left( x\right) \) for all \( x \in M \) . Then there exists a linear ... | Proof: Let \( \mathcal{F} = \{ \left( {V, g}\right) : V \supseteq M, V \) is a subspace of \( X, g : V \rightarrow \mathbb{R} \) is linear, \( g\left( x\right) = f\left( x\right) \) for all \( x \in M \), and \( g\left( x\right) \leq \rho \left( x\right) \) for \( x \in V\} \) . Then \( \left( {M, f}\right) \in \mathca... | Yes |
Lemma 17.2.9 Let \( X \) be a normed linear space and let \( x \in X \smallsetminus V \) where \( V \) is a closed subspace of \( X \). Then there exists \( {x}^{ * } \in {X}^{\prime } \) such that \( {x}^{ * }\left( x\right) = \parallel x\parallel ,{x}^{ * }\left( V\right) = \) \( \{ 0\} \), and | Proof: Let \( f : \mathbb{F}x + V \rightarrow \mathbb{F} \) be defined by \( f\left( {{\alpha x} + v}\right) = \alpha \parallel x\parallel \). First it is necessary to show \( f \) is well defined and continuous. If \( {\alpha }_{1}x + {v}_{1} = {\alpha }_{2}x + {v}_{2} \) then if \( {\alpha }_{1} \neq {\alpha }_{2} \)... | Yes |
Theorem 17.2.10 Let \( L \in \mathcal{L}\left( {X, Y}\right) \) where \( X \) and \( Y \) are Banach spaces. Then\n\na.) \( {L}^{ * } \in \mathcal{L}\left( {{Y}^{\prime },{X}^{\prime }}\right) \) as claimed and \( \begin{Vmatrix}{L}^{ * }\end{Vmatrix} = \parallel L\parallel \) . | Proof: It is routine to verify \( {L}^{ * }{y}^{ * } \) and \( {L}^{ * } \) are both linear. This follows immediately from the definition. As usual, the interesting thing concerns continuity.\n\n\[ \begin{Vmatrix}{{L}^{ * }{y}^{ * }}\end{Vmatrix} = \mathop{\sup }\limits_{{\parallel x\parallel \leq 1}}\left| {{L}^{ * }{... | Yes |
Lemma 17.3.2 Let \( 0 < p < 1 \) and let \( f, g \) be measurable functions. Also\n\n\[ \n{\int }_{\Omega }{\left| g\right| }^{p/\left( {p - 1}\right) }{d\mu } < \infty ,{\int }_{\Omega }{\left| f\right| }^{p}{d\mu } < \infty \n\]\n\nThen the following backwards Holder inequality holds.\n\n\[ \n{\int }_{\Omega }\left| ... | Proof: If \( \int \left| {fg}\right| {d\mu } = \infty \), there is nothing to prove. Hence assume this is finite.\n\nThen\n\[ \n\int {\left| f\right| }^{p}{d\mu } = \int {\left| g\right| }^{-p}{\left| fg\right| }^{p}{d\mu } \n\]\n\nThis makes sense because, due to the hypothesis on \( g \) it must be the case that \( g... | Yes |
Corollary 17.3.3 Let \( 0 < p < 1 \) and suppose \( \int {\left| h\right| }^{p}{d\mu } < \infty \) for \( h = f, g \) . Then\n\n\[{\left( \int {\left( \left| f\right| + \left| g\right| \right) }^{p}d\mu \right) }^{1/p} \geq {\left( \int {\left| f\right| }^{p}d\mu \right) }^{1/p} + {\left( \int {\left| g\right| }^{p}d\m... | Proof: If \( \int {\left( \left| f\right| + \left| g\right| \right) }^{p}{d\mu } = 0 \) then there is nothing to prove so assume this is\n\nnot zero.\n\[ \int {\left( \left| f\right| + \left| g\right| \right) }^{p}{d\mu } = \int {\left( \left| f\right| + \left| g\right| \right) }^{p - 1}\left( {\left| f\right| + \left|... | Yes |
Lemma 17.3.4 For any \( p \geq 2 \) the following inequality holds for any \( t \in \left\lbrack {0,1}\right\rbrack \) , \[ {\left| \frac{1 + t}{2}\right| }^{p} + {\left| \frac{1 - t}{2}\right| }^{p} \leq \frac{1}{2}\left( {{\left| t\right| }^{p} + 1}\right) \] | Proof: It is clear that, since \( p \geq 2 \), the inequality holds for \( t = 0 \) and \( t = 1 \) . Thus it suffices to consider only \( t \in \left( {0,1}\right) \) . Let \( x = 1/t \) . Then, dividing by \( 1/{t}^{p} \), the inequality holds if and only if \[ {\left( \frac{x + 1}{2}\right) }^{p} + {\left( \frac{x -... | Yes |
Corollary 17.3.5 If \( z, w \in \mathbb{C} \) and \( p \geq 2 \), then\n\n\[ \n{\left| \frac{z + w}{2}\right| }^{p} + {\left| \frac{z - w}{2}\right| }^{p} \leq \frac{1}{2}\left( {{\left| z\right| }^{p} + {\left| w\right| }^{p}}\right)\n\] | Proof: One of \( \left| w\right| ,\left| z\right| \) is larger. Say \( \left| z\right| \geq \left| w\right| \) . Then dividing both sides of the proposed inequality by \( {\left| z\right| }^{p} \) it suffices to verify that for all complex \( t \) having \( \left| t\right| \leq 1 \) ,\n\n\[ \n{\left| \frac{1 + t}{2}\ri... | Yes |
Theorem 17.3.6 Let \( p \geq 2 \) . Then\n\n\[ \n{\left| \left| \frac{f + g}{2}\right| \right| }_{{L}^{p}}^{p} + {\left| \left| \frac{f - g}{2}\right| \right| }_{{L}^{p}}^{p} \leq \frac{1}{2}\left( {\parallel f{\parallel }_{{L}^{p}}^{p} + \parallel g{\parallel }_{{L}^{p}}^{p}}\right)\n\] | Proof: This follows right away from the above corollary.\n\n\[ \n{\int }_{\Omega }{\left| \frac{f + g}{2}\right| }^{p}{d\mu } + {\int }_{\Omega }{\left| \frac{f - g}{2}\right| }^{p}{d\mu } \leq \frac{1}{2}{\int }_{\Omega }\left( {{\left| f\right| }^{p} + {\left| g\right| }^{p}}\right) {d\mu }\blacksquare\n\] | Yes |
Corollary 17.3.8 Let \( z, w \in \mathbb{C} \) . Then for \( p \in \left( {1,2}\right) \) , \[ {\left| \frac{z + w}{2}\right| }^{q} + {\left| \frac{z - w}{2}\right| }^{q} \leq {\left( \frac{1}{2}{\left| z\right| }^{p} + \frac{1}{2}{\left| w\right| }^{p}\right) }^{q/p} \] | Proof: One of \( \left| w\right| ,\left| z\right| \) is larger. Say \( \left| w\right| \geq \left| z\right| \) . Then dividing by \( {\left| w\right| }^{q} \), for \( t = z/w \), showing the above inequality is equivalent to showing that for all \( t \in \mathbb{C} \) , \( \left| t\right| \leq 1 \) \[ {\left| \frac{t +... | Yes |
Theorem 17.3.9 Let \( 2 \leq p \) . Then\n\n\[ \n{\left| \left| \frac{f + g}{2}\right| \right| }_{{L}^{p}}^{p} + {\left| \left| \frac{f - g}{2}\right| \right| }_{{L}^{p}}^{p} \leq \frac{1}{2}\left( {\parallel f{\parallel }_{{L}^{p}}^{p} + \parallel g{\parallel }_{{L}^{p}}^{p}}\right) \n\] | Proof: The first was established above. | No |
Theorem 17.3.10 The \( {L}^{p} \) spaces are uniformly convex. | Proof: First suppose \( p \geq 2 \) . Suppose \( {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{{L}^{p}},{\begin{Vmatrix}{g}_{n}\end{Vmatrix}}_{{L}^{p}} \leq 1 \) and \( {\begin{Vmatrix}\frac{{f}_{n} + {g}_{n}}{2}\end{Vmatrix}}_{{L}^{p}} \rightarrow 1 \) . Then from the first Clarkson inequality,\n\n\[{\begin{Vmatrix}\frac{{f}... | Yes |
Theorem 17.4.1 Let \( X \) be a Banach space and let \( V = \operatorname{span}\left( {{x}_{1},\cdots ,{x}_{n}}\right) \) . Then \( V \) is a closed subspace of \( X \) . | Proof: Without loss of generality, it can be assumed \( \left\{ {{x}_{1},\cdots ,{x}_{n}}\right\} \) is linearly independent. Otherwise, delete those vectors which are in the span of the others till a linearly independent set is obtained. Let\n\n\[ x = \mathop{\lim }\limits_{{p \rightarrow \infty }}\mathop{\sum }\limit... | Yes |
Lemma 17.5.1 The sets, \( {B}_{{A}^{\prime }}\left( {x, r}\right) \) where \( {A}^{\prime } \) is a finite subset of \( {X}^{\prime } \) and \( x \in X \) form a basis for a topology on \( X \) known as the weak topology. The sets \( {B}_{A}\left( {{x}^{ * }, r}\right) \) where \( A \) is a finite subset of \( X \) and... | Proof: The two assertions are very similar. I will verify the one for the weak topology. The union of these sets, \( {B}_{{A}^{\prime }}\left( {x, r}\right) \) for \( x \in X \) and \( r > 0 \) is all of \( X \) . Now suppose \( z \) is contained in the intersection of two of these sets. Say\n\n\[ z \in {B}_{{A}^{\prim... | Yes |
Theorem 17.5.4 Let \( {B}^{\prime } \) be the closed unit ball in \( {X}^{\prime } \) . Then \( {B}^{\prime } \) is compact in the weak \( * \) topology. | Proof: By the Tychonoff theorem, Theorem 17.5.3\n\n\[ P \equiv \mathop{\prod }\limits_{{x \in X}}\overline{B\left( {0,\parallel x\parallel }\right) } \]\n\nis compact in the product topology where the topology on \( \overline{B\left( {0,\parallel x\parallel }\right) } \) is the usual topology of \( \mathbb{F} \) . Reca... | Yes |
Theorem 17.5.5 If \( K \subseteq {X}^{\prime } \) is compact in the weak \( * \) topology and \( X \) is separable in the weak topology then there exists a metric, \( d \), on \( K \) such that if \( {\tau }_{d} \) is the topology on \( K \) induced by \( d \) and if \( \tau \) is the topology on \( K \) induced by the... | Proof: Let \( D = \left\{ {x}_{n}\right\} \) be the dense countable subset in \( X \) . The metric is\n\n\[ d\left( {f, g}\right) \equiv \mathop{\sum }\limits_{{n = 1}}^{\infty }{2}^{-n}\frac{{\rho }_{{x}_{n}}\left( {f - g}\right) }{1 + {\rho }_{{x}_{n}}\left( {f - g}\right) } \]\n\nwhere \( {\rho }_{{x}_{n}}\left( f\r... | Yes |
Corollary 17.5.6 If \( X \) is weakly separable and \( K \subseteq {X}^{\prime } \) is compact in the weak \( * \) topology, then \( K \) is sequentially compact. That is, if \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \subseteq K \), then there exists a subsequence \( {f}_{{n}_{k}} \) and \( f \in K \) such that f... | Proof: By Theorem 17.5.5, \( K \) is a metric space for the metric described there and it is compact. Therefore by the characterization of compact metric spaces, Proposition 7.6.5 on Page 151, \( K \) is sequentially compact. This proves the corollary. | Yes |
Lemma 17.5.7 Let \( J : X \rightarrow {X}^{\prime \prime } \) be the James map\n\n\[ \n{Jx}\left( f\right) \equiv f\left( x\right) \n\]\n\nand let \( X \) be reflexive so that \( J \) is onto. Then \( J \) is a homeomorphism of \( \left( {X\text{, weak topology}}\right) \) and \( \left( {{X}^{\prime \prime }\text{, wea... | Proof: Let \( f \in {X}^{\prime } \) and let\n\n\[ \n{B}_{f}\left( {x, r}\right) \equiv \{ y : \left| {f\left( x\right) - f\left( y\right) }\right| < r\} . \n\]\n\nThus \( {B}_{f}\left( {x, r}\right) \) is a subbasic set for the weak topology on \( X \) . I claim that\n\n\[ \nJ{B}_{f}\left( {x, r}\right) = {B}_{f}\left... | Yes |
Corollary 17.5.8 If \( X \) is a reflexive Banach space, then the closed unit ball is weakly compact. | Proof: Let \( B \) be the closed unit ball. Then \( B = {J}^{-1}\left( {B}^{* * }\right) \) where \( {B}^{* * } \) is the unit ball in \( {X}^{\prime \prime } \) which is compact in the weak \( * \) topology. Therefore \( B \) is weakly compact because \( {J}^{-1} \) is continuous. | Yes |
Corollary 17.5.9 Let \( X \) be a reflexive Banach space. If \( K \subseteq X \) is compact in the weak topology and \( {X}^{\prime } \) is separable in the weak \( * \) topology, then there exists a metric \( d \), on \( K \) such that if \( {\tau }_{d} \) is the topology on \( K \) induced by \( d \) and if \( \tau \... | Proof: This follows from Theorem 17.5.5 and Lemma 17.5.7. Lemma 17.5.7 implies \( J\left( K\right) \) is compact in \( {X}^{\prime \prime } \) . Then since \( {X}^{\prime } \) is separable in the weak \( * \) topology, \( X \) is separable in the weak topology and so there is a metric, \( {d}^{\prime \prime } \) on \( ... | Yes |
Lemma 17.5.10 Let \( Y \) be a closed subspace of a Banach space \( X \) and let \( y \in X \smallsetminus Y \) . Then there exists \( {x}^{ * } \in {X}^{\prime } \) such that \( {x}^{ * }\left( Y\right) = 0 \) but \( {x}^{ * }\left( y\right) \neq 0 \) . | Proof: Define \( f\left( {x + {\alpha y}}\right) \equiv \parallel y\parallel \alpha \) . Thus \( f \) is linear on \( Y \oplus \mathbb{F}y \) . I claim that \( f \) is also continuous on this subspace of \( X \) . If not, then there exists \( {x}_{n} + {\alpha }_{n}y \rightarrow 0 \) but \( \left| {f\left( {{x}_{n} + {... | Yes |
Lemma 17.5.11 A closed subspace of a reflexive Banach space is reflexive. | Proof: Let \( Y \) be the closed subspace of the reflexive space, \( X \) . Consider the following diagram\n\n\[ \n{Y}^{\prime \prime }\overset{{i}^{* * }1\text{-}1}{ \rightarrow }{X}^{\prime \prime }\n\]\n\n\[ \n{Y}^{\prime }\overset{{i}^{ * }\text{ onto }}{ \leftarrow }{X}^{\prime }\n\]\n\n\[ \nY\;\overset{i}{ \right... | Yes |
Theorem 17.5.12 (Eberlein Smulian) The closed unit ball in a reflexive Banach space \( X \), is weakly sequentially compact. By this is meant that if \( \left\{ {x}_{n}\right\} \) is contained in the closed unit ball, there exists a subsequence, \( \left\{ {x}_{{n}_{k}}\right\} \) and \( x \in X \) such that for all \(... | Proof: Let \( \left\{ {x}_{n}\right\} \subseteq B \equiv \overline{B\left( {0,1}\right) } \) . Let \( Y \) be the closure of the linear span of \( \left\{ {x}_{n}\right\} \) . Thus \( Y \) is a separable. It is reflexive because it is a closed subspace of a reflexive space so the above lemma applies. By the Banach Alao... | Yes |
Corollary 17.5.13 Let \( \\left\\{ {x}_{n}\\right\\} \) be any bounded sequence in a reflexive Banach space \( X \) . Then there exists \( x \\in X \) and a subsequence, \( \\left\\{ {x}_{{n}_{k}}\\right\\} \) such that for all \( {x}^{ * } \\in {X}^{\prime } \) , | Proof: If a subsequence, \( {x}_{{n}_{k}} \) has \( \\begin{Vmatrix}{x}_{{n}_{k}}\\end{Vmatrix} \\rightarrow 0 \), then the conclusion follows. Simply let \( x = 0 \) . Suppose then that \( \\begin{Vmatrix}{x}_{n}\\end{Vmatrix} \) is bounded away from 0 . That is, \( \\begin{Vmatrix}{x}_{n}\\end{Vmatrix} \\in \\left\\l... | Yes |
Lemma 17.6.2 Suppose \( C \in \mathcal{L}\left( {X, X}\right) \) is compact. Then \( \left( {I - C}\right) \left( X\right) \) is closed. | Proof: Let \( \left( {I - C}\right) {x}_{n} \rightarrow y \) . Let \( {z}_{n} \in \ker \left( {I - C}\right) \) such that\n\n\[ \operatorname{dist}\left( {{x}_{n},\ker \left( {I - C}\right) }\right) \leq \begin{Vmatrix}{{x}_{n} - {z}_{n}}\end{Vmatrix} \]\n\n\[ \leq \left( {1 + \frac{1}{n}}\right) \operatorname{dist}\le... | Yes |
Lemma 17.6.3 Suppose \( W \) and \( V \) are closed subspaces of a Banach space \( X \) and \( V \subsetneqq W \) ( \( V \) is a proper subset of \( W \) .) while \( \left( {{\lambda I} - L}\right) \left( W\right) \subseteq V,\lambda \neq 0 \) . Then there exists \( w \in W \smallsetminus V \) such that \( \parallel w\... | Proof: Let \( {w}_{0} \in W \smallsetminus V \) . Then let \( v \in V \) be such that \( \begin{Vmatrix}{\lambda {w}_{0} - v}\end{Vmatrix} \leq 2\operatorname{dist}\left( {\lambda {w}_{0}, V}\right) \) .\n\nThen let\n\[ w = \frac{\lambda {w}_{0} - v}{\begin{Vmatrix}\lambda {w}_{0} - v\end{Vmatrix}} \]\n\nIt follows tha... | Yes |
Lemma 17.6.4 Let \( Y \) be an infinite dimensional Banach space. Then there exists a sequence \( \left\{ {x}_{n}\right\} \) in the unit sphere \( S,\begin{Vmatrix}{x}_{n}\end{Vmatrix} = 1 \), such that \( \begin{Vmatrix}{{x}_{n} - {x}_{m}}\end{Vmatrix} \geq \frac{1}{2} \) whenever \( n \neq m \) . | Proof: Pick \( {x}_{1} \in S \) . Now the span of \( {x}_{1} \) is not everything and so there exists \( {u}_{2} \notin \) \( \operatorname{span}\left( {x}_{1}\right) \) . Let \( {w}_{2} \) be a point of \( \operatorname{span}\left( {x}_{1}\right) \) such that \( \begin{Vmatrix}{{u}_{2} - {w}_{2}}\end{Vmatrix} \leq 2\o... | Yes |
Lemma 17.6.5 Let \( L \) be a compact linear map. Then the eigenspace of \( L \) is finite dimensional for each eigenvalue \( \lambda \neq 0 \) . | Proof: Consider \( {\left( L - \lambda I\right) }^{-1}\left( 0\right) \cap S \) where \( S \) is the unit sphere. The eigenspace is just \( {\left( L - \lambda I\right) }^{-1}\left( 0\right) \) . Let \( Y \) be this inverse image. If \( Y \) is infinite dimensional, then the above Lemma 17.6.4 applies. There exists \( ... | Yes |
Theorem 17.6.6 Let \( L \in \mathcal{L}\left( {X, X}\right) \) with \( L \) compact. Let \( \Lambda \) be the eigenvalues of L. That is \( \lambda \in \Lambda \) means there exists \( x \neq 0 \) such that \( {Lx} = {\lambda x} \) . It is assumed the field of scalars is \( \mathbb{R} \) or \( \mathbb{C} \) . Let \( {R}... | Proof: Consider \( \lambda \neq 0 \) . The \( N\left( {R}_{\lambda }^{k}\right) \) are increasing in \( k \) and \( {R}_{\lambda }\left( {N\left( {R}_{\lambda }^{k + 1}\right) }\right) \subseteq \) \( N\left( {R}_{\lambda }^{k}\right) \) . This follows from the definition. (It isn’t necessary to assume in most of this ... | Yes |
Proposition 17.6.8 Let \( T \in \mathcal{L}\left( {X, Y}\right) \) . Then \( {TX} \) is closed if and only if there exists \( \delta > 0 \) such that\n\n\[ \parallel {Tx}\parallel \geq \delta \operatorname{dist}\left( {x,\ker \left( T\right) }\right) . \]\n | Proof: First suppose \( {TX} \) is closed. Let \( \widehat{T} : X/\ker \left( T\right) \rightarrow Y \) be defined as \( \widehat{T}\left( \left\lbrack x\right\rbrack \right) \equiv {Tx} \) . Then by Theorem 18.7.2, \( \widehat{T} \) is one to one and continuous and \( X/\ker \left( T\right) \) is a Banach space, \( \p... | Yes |
Theorem 17.6.9 If \( T \) is a Fredholm operator, then \( {TX} \) is closed in \( Y \) . | Proof: Recall that \( Y = {TX} \oplus E \) where \( E \) is a closed subspace of \( Y \) . In fact, \( E \) is finite dimensional, but it is only needed that \( E \) is closed. Let \( {T}_{0} \in \) \( \mathcal{L}\left( {X \times E,{TX} \oplus E}\right) \) be given by\n\n\[ {T}_{0}\left( {x, e}\right) \equiv {Tx} + e \... | Yes |
Corollary 17.6.10 If \( {TX} \oplus E \) is closed in \( Y \) and \( E \) is a closed subspace of \( Y \), then \( {TX} \) is closed. Here \( T \in \mathcal{L}\left( {X, Y}\right) \) . | Let \( \mathcal{B} \) be a Hamel basis for \( {TX} \) and consider \( \mathcal{A} \equiv \{ x : {Tx} \in \mathcal{B}\} \) . Then this is a linearly independent set of vectors in \( X \) . Suppose now that \( \ker \left( T\right) = \operatorname{span}\left( {{z}_{1},\cdots ,{z}_{n}}\right) \) where \( \left\{ {{z}_{1},\... | Yes |
Theorem 18.1.2 \( \mathcal{B} \) is the basis for a topology. | Proof: I need to show that if \( {B}_{A}\left( {x,{r}_{1}}\right) \) and \( {B}_{B}\left( {y,{r}_{2}}\right) \) are two elements of \( \mathcal{B} \) and if \( z \in {B}_{A}\left( {x,{r}_{1}}\right) \cap {B}_{B}\left( {y,{r}_{2}}\right) \), then there exists \( U \in \mathcal{B} \) such that\n\n\[ z \in U \subseteq {B}... | Yes |
Theorem 18.1.3 The vector space operations of addition and scalar multiplication are continuous. More precisely,\n\n\[ +\ : X \times X \rightarrow X, \cdot : \mathbb{F} \times X \rightarrow X \]\n\nare continuous. | Proof: It suffices to show \( { + }^{-1}\left( B\right) \) is open in \( X \times X \) and \( { \cdot }^{-1}\left( B\right) \) is open in \( \mathbb{F} \times X \) if \( B \) is of the form\n\n\[ B = \{ y \in X : \rho \left( {y - x}\right) < r\} \]\n\nbecause finite intersections of such sets form the basis \( \mathcal... | Yes |
Theorem 18.1.4 Let \( x \) be given and let \( {f}_{x}\left( y\right) = x + y \) . Then \( {f}_{x} \) is \( 1 - 1 \), onto, and continuous. If \( \alpha \neq 0 \) and \( {g}_{\alpha }\left( x\right) = {\alpha x} \), then \( {g}_{\alpha } \) is also \( 1 - 1 \) onto and continuous. | Proof: The assertions about \( 1 - 1 \) and onto are obvious. It remains to show \( {f}_{x} \) and \( {g}_{\alpha } \) are continuous. Let \( B = {B}_{\rho }\left( {z, r}\right) \) and consider \( {f}_{x}^{-1}\left( B\right) \) . Then it is easy to see that\n\n\[ \n{f}_{x}^{-1}\left( B\right) = {B}_{\rho }\left( {z - x... | Yes |
Theorem 18.1.6 The following are equivalent for \( f \), a linear function mapping \( X \) to \( \mathbb{F} \) . \n\n\[ \nf\text{is continuous at 0 .} \n\] \n\n(18.1.1) \n\nFor some \( A \subseteq \Psi, A \) finite, \n\n\[ \n\left| {f\left( x\right) }\right| \leq C{\rho }_{A}\left( x\right) \n\] \n\n(18.1.2) \n\nfor al... | Proof: Clearly 18.1.3 implies 18.1.1. Suppose 18.1.1. Then \n\n\[ \n0 = f\left( 0\right) \in B\left( {0,1}\right) \subseteq \mathbb{F}. \n\] \n\nSince \( f \) is continuous at \( 0,0 \in {f}^{-1}\left( {B\left( {0,1}\right) }\right) \) and there exists an open set \( V \in \tau \) such that \n\n\[ \n0 \in V \subseteq {... | Yes |
Theorem 18.1.7 Let \( X \) be a vector space and let \( Y \) be a vector space of linear functionals defined on \( X \) . For each \( y \in Y \), define\n\n\[ \n{\rho }_{y}\left( x\right) \equiv \left| {y\left( x\right) }\right| .\n\]\n\nThen the collection of seminorms \( {\left\{ {\rho }_{y}\right\} }_{y \in Y} \) de... | Proof: Clearly \( {\left\{ {\rho }_{y}\right\} }_{y \in Y} \) is a collection of seminorms defined on \( X \) ; so, \( X \) supplied with the topology induced by this collection of seminorms is a locally convex topological vector space. Is \( Y = {X}^{\prime } \) ?\n\nLet \( y \in Y \), let \( U \subseteq \mathbb{F} \)... | Yes |
Proposition 18.2.2 Let \( X \) be a locally convex topological vector space. Then \( m \) is defined on \( X \) and satisfies\n\n\[ m\left( {x + y}\right) \leq m\left( x\right) + m\left( y\right) \]\n\n(18.2.4)\n\n\[ m\left( {\lambda x}\right) = {\lambda m}\left( x\right) \;\text{ if }\lambda > 0. \]\n\n\( \left( {18.2... | Proof: Let \( x \in X \) be arbitrary. There exists \( A \subseteq \Psi \) such that\n\n\[ 0 \in {B}_{A}\left( {0, r}\right) \subseteq U \]\n\nThen\n\n\[ \frac{rx}{2{\rho }_{A}\left( x\right) } \in {B}_{A}\left( {0, r}\right) \subseteq U \]\n\nwhich implies\n\n\[ \frac{2{\rho }_{A}\left( x\right) }{r} \geq m\left( x\ri... | Yes |
Lemma 18.2.3 Let \( U \) be an open convex set containing 0 and let \( q \notin U \) . Then there exists \( f \in {X}^{\prime } \) such that\n\n\[ \operatorname{Re}f\left( q\right) > \operatorname{Re}f\left( x\right) \]\n\nfor all \( x \in U \) . | Proof: Let \( m \) be the Minkowski functional just defined and let\n\n\[ F\left( {cq}\right) = {cm}\left( q\right) \]\n\nfor \( c \in \mathbb{R} \) . If \( c > 0 \) then\n\n\[ F\left( {cq}\right) = m\left( {cq}\right) \]\n\nwhile if \( c \leq 0 \) ,\n\n\[ F\left( {cq}\right) = {cm}\left( q\right) \leq 0 \leq m\left( {... | Yes |
Corollary 18.2.4 Let \( U \) be an open nonempty convex set and let \( q \notin U \) . Then there exists \( f \in {X}^{\prime } \) such that\n\n\[ f\left( q\right) > f\left( x\right) \]\n\nfor all \( x \in U \) . | Proof: Let \( {u}_{0} \in U \) and consider \( \widehat{U} \equiv U - {u}_{0} \) . Then \( 0 \in \widehat{U} \) and \( q - {u}_{0} \notin \widehat{U} \) . By separation theorems, Lemma 18.2.3 there exists \( f \in {X}^{\prime } \) such that\n\n\[ f\left( {q - {u}_{0}}\right) > f\left( {x - {u}_{0}}\right) \]\n\nfor all... | Yes |
Theorem 18.2.5 Let \( K \) be closed and convex in a locally convex topological vector space and let \( p \notin K \) . Then there exists a real number \( c \), and \( f \in {X}^{\prime } \) such that\n\n\[ \operatorname{Re}f\left( p\right) > c > \operatorname{Re}f\left( k\right) \]\n\nfor all \( k \in K \) . | Proof: Since \( K \) is closed, and \( p \notin K \), there exists a finite subset of \( \Psi, A \), and a positive \( r > 0 \) such that\n\n\[ K \cap {B}_{A}\left( {p,{2r}}\right) = \varnothing . \]\n\nPick \( {k}_{0} \in K \) and let\n\n\[ U = K + {B}_{A}\left( {0, r}\right) - {k}_{0}, q = p - {k}_{0}. \]\n\nIt follo... | Yes |
Corollary 18.2.6 In the situation of the above theorem, there exist real numbers \( c, d \) such that \( \operatorname{Re}f\left( p\right) > d > c > \operatorname{Re}f\left( k\right) \) for all \( k \in K \) . | Proof: From the theorem, there exists \( \widehat{c} \) such that \( \operatorname{Re}f\left( p\right) > \widehat{c} > \operatorname{Re}f\left( k\right) \) for all \( k \in K \) . Thus \( \operatorname{Re}f\left( p\right) > \widehat{c} \geq \mathop{\sup }\limits_{{k \in K}}\operatorname{Re}f\left( k\right) \) . Now cho... | Yes |
Lemma 18.2.8 Let \( X, Y \) be two Banach spaces. Then letting \[ \parallel \left( {x, y}\right) \parallel \equiv \max \left( {\parallel x{\parallel }_{X},\parallel y{\parallel }_{Y}}\right) , \] it follows \( X \times Y \) is a Banach space and \( \phi \in {\left( X \times Y\right) }^{\prime } \) if and only if there ... | Proof: Most of these conclusions are obvious. In particular it is clear \( X \times Y \) is a Banach space with the given norm. Let \( \phi \in {\left( X \times Y\right) }^{\prime } \) . Also let \( {\pi }_{X}\left( {x, y}\right) \equiv \left( {x,0}\right) \) and \( {\pi }_{Y}\left( {x, y}\right) \equiv \left( {0, y}\r... | Yes |
Lemma 18.2.9 Let \( \phi \) be a functional as described in Definition 18.2.7. Then \( \phi \) is lower semicontinuous if and only if the epigraph of \( \phi \) is closed in \( X \times \mathbb{R} \) with the strong topology. Here the epigraph is defined as \[ \operatorname{epi}\left( \phi \right) \equiv \{ \left( {x, ... | Proof: First suppose \( \operatorname{epi}\left( \phi \right) \) is closed and suppose \( {x}_{n} \rightarrow x \) . Let \( l < \phi \left( x\right) \) . Then \( \left( {x, l}\right) \notin \operatorname{epi}\left( \phi \right) \) and so there exists \( \delta > 0 \) such that if \( \left| {x - y}\right| < \delta \) an... | Yes |
Theorem 18.2.11 Let \( \phi \) be a lower semicontinuous convex functional as described in Definition 18.2.7 and let \( X \) be a real Banach space. Then \( \phi \) is also weakly lower semicontinuous. | Proof: By Lemma 18.2.9 \( \operatorname{epi}\left( \phi \right) \) is closed in \( X \times \mathbb{R} \) with the strong topology as well as being convex. Letting \( \left( {z, l}\right) \notin \operatorname{epi}\left( \phi \right) \), it follows from Theorem 18.2.5 and Lemma 18.2.8 there exists \( \left( {{x}^{ * },\... | Yes |
Corollary 18.2.12 Let \( \phi \) be a lower semicontinuous convex functional as described in Definition 18.2.7 and let \( X \) be a real Banach space. Then if \( {x}_{n} \) converges weakly to \( x \), it follows that \[ \phi \left( x\right) \leq \lim \mathop{\inf }\limits_{{n \rightarrow \infty }}\phi \left( {x}_{n}\r... | Proof: Let \( l < \phi \left( x\right) \) so that \( \left( {x, l}\right) \notin \operatorname{epi}\left( \phi \right) \) . Then by Theorem 18.2.11 there exists \( B \times \left( {-\infty, l + \delta }\right) \) such that \( B \) is a weakly open set in \( X \) containing \( x \) and \[ B \times \left( {-\infty, l + \... | Yes |
Proposition 18.2.13 Let \( A \) be a linear operator which maps a real normed linear space \( \left( {X,\parallel \cdot {\parallel }_{X}}\right) \) to a real normed linear space \( \left( {Y,\parallel \cdot {\parallel }_{Y}}\right) \) . Then \( {x}_{n} \rightarrow x \) strongly implies \( A{x}_{n} \rightarrow {Ax} \) i... | Proof: \( \Rightarrow \) Define \( \phi \left( x\right) \equiv f\left( {Ax}\right) \) where \( f \in {Y}^{\prime } \) . Then \( \phi \) is convex and continuous. Therefore, if \( {x}_{n} \rightarrow x \) weakly, then\n\n\[ \phi \left( x\right) = f\left( {Ax}\right) \leq \lim \mathop{\inf }\limits_{{n \rightarrow \infty... | Yes |
Theorem 18.2.14 Let \( A \) and \( B \) be disjoint, convex and nonempty sets with \( B \) open. Then there exists \( f \in {X}^{\prime } \) such that\n\n\[ \operatorname{Re}f\left( a\right) < \operatorname{Re}f\left( b\right) \]\n\nfor all \( a \in A \) and \( b \in B \) . | Proof: Let \( {b}_{0} \in B,{a}_{0} \in A \) . Then the set\n\n\[ B - A + {a}_{0} - {b}_{0} \]\n\nis open, convex, contains 0, and does not contain \( {a}_{0} - {b}_{0} \) . By Lemma 18.2.3 there exists \( f \in {X}^{\prime } \) such that\n\n\[ \operatorname{Re}f\left( {{a}_{0} - {b}_{0}}\right) > \operatorname{Re}f\le... | Yes |
Lemma 18.2.15 If \( B \) is convex, then \( \operatorname{int}\left( B\right) \equiv \) union of all open sets contained in \( B \) is convex. Also, if \( \operatorname{int}\left( B\right) \neq \varnothing \), then \( B \subseteq \overline{\operatorname{int}\left( B\right) } \) . | Proof: Suppose \( x, y \in \operatorname{int}\left( B\right) \) . Then there exists \( r > 0 \) and a finite set \( A \subseteq \Psi \) such that\n\n\[ \n{B}_{A}\left( {x, r}\right) ,{B}_{A}\left( {y, r}\right) \subseteq B.\n\]\n\nLet\n\n\[ \nV \equiv { \cup }_{\lambda \in \left\lbrack {0,1}\right\rbrack }\lambda {B}_{... | Yes |
Corollary 18.2.16 Let \( A, B \) be convex, nonempty sets. Suppose \( \operatorname{int}\left( B\right) \neq \varnothing \) and \( A \cap \operatorname{int}\left( B\right) = \varnothing \) . Then there exists \( f \in {X}^{\prime }, f \neq 0 \), such that for all \( a \in A \) and \( b \in B \) , | Proof: By Theorem 18.2.14, there exists \( f \in {X}^{\prime } \) such that for all \( b \in \operatorname{int}\left( B\right) \) , and \( a \in A \) ,\n\n\[ \operatorname{Re}f\left( b\right) > \operatorname{Re}f\left( a\right) . \]\n\nThus, in particular, \( f \neq 0 \) . By Lemma 18.2.15, if \( b \in B \) and \( a \i... | Yes |
Lemma 18.2.17 If \( X \) is a topological Hausdorff space then compact implies closed. | Proof: Let \( K \) be compact and suppose \( {K}^{C} \) is not open. Then there exists \( p \in {K}^{C} \) such that\n\n\[{V}_{p} \cap K \neq \varnothing\]\n\nfor all open sets \( {V}_{p} \) containing \( p \) . Let\n\n\[\mathcal{C} = \left\{ {{\left( {\bar{V}}_{p}\right) }^{C} : {V}_{p}\text{ is an open set containing... | Yes |
Lemma 18.2.18 If \( X \) is a locally convex topological vector space, and if every point is a closed set, then the seminorms and \( {X}^{\prime } \) separate the points. This means if \( x \neq y \) , then for some \( \rho \in \Psi \) ,\n\n\[ \rho \left( {x - y}\right) \neq 0 \]\n\nand for some \( f \in {X}^{\prime } ... | Proof: Let \( x \neq y \) . Then by Theorem 18.2.5, there exists \( f \in {X}^{\prime } \) such that \( f\left( x\right) \neq f\left( y\right) \) . Thus \( {X}^{\prime } \) separates the points. Since \( f \in {X}^{\prime } \), Theorem 18.1.6 implies\n\n\[ \left| {f\left( z\right) }\right| \leq C{\rho }_{A}\left( z\rig... | Yes |
Lemma 18.3.1 The functions \( {\delta }_{G} \) for \( G \) a finite subset of \( {X}^{\prime } \) are seminorms and the sets\n\n\[ \n{B}_{G}\left( {x, r}\right) \equiv \left\{ {y \in X : {\delta }_{G}\left( {x - y}\right) < r}\right\} \n\]\n\nform a basis for a topology on \( X \) . Furthermore, \( X \) with this topol... | Proof: It is obvious that the functions \( {\delta }_{G} \) are seminorms and therefore the proof that the sets \( {B}_{G}\left( {x, r}\right) \) form a basis for a topology is the same as in Theorem 18.1.2. To see every point is a closed set in this new topology, assuming this is true for \( X \) with the original top... | Yes |
Lemma 18.3.2 The functions \( {\gamma }_{F} \) for \( F \) a finite subset of \( X \) are seminorms and the sets\n\n\[ \n{B}_{F}\left( {f, r}\right) \equiv \left\{ {g \in {X}^{\prime } : {\gamma }_{F}\left( {f - g}\right) < r}\right\} \n\]\n\nform a basis for a topology on \( {X}^{\prime } \) . Furthermore, \( {X}^{\pr... | Proof: The proof is similar to that of Lemma 18.3.1 but there is a difference in the part where every point is shown to be a closed set. Let \( f \in {X}^{\prime } \) and let \( g \neq f \) . Thus there exists \( x \in X \) such that \( f\left( x\right) \neq g\left( x\right) \) . Let \( F = \{ x\} \) . Then\n\n\[ \n{B}... | Yes |
Theorem 18.3.3 Let \( K \) be closed and convex in a Banach space \( X \) . Then it is also weakly closed. Furthermore, if \( p \notin K \), there exists \( f \in {X}^{\prime } \) such that\n\n\[ \operatorname{Re}f\left( p\right) > c > \operatorname{Re}f\left( k\right) \]\n\nfor all \( k \in K \) . If \( {K}^{ * } \) i... | Proof: By Theorem 18.2.5 there exists \( f \in {X}^{\prime } \) such that 18.3.8 holds. Therefore, letting \( A = \{ f\} \), it follows that for \( r \) small enough, \( {B}_{A}\left( {p, r}\right) \cap K = \varnothing \) . Thus \( K \) is weakly closed. This establishes the first part.\n\nFor the second part, the semi... | Yes |
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