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Proposition 4.6. Let \( X \) be a connected \( {C}^{1} \) manifold. If \( X \) is not orientable, then there exists a covering \( {X}^{\prime } \rightarrow X \) of degree 2 such that \( {X}^{\prime } is orientable.
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Sketch of Proof. Suppose first that \( X \) is simply connected. Let \( x \in X \) . Fix a chart \( \left( {{U}_{0},{\varphi }_{0}}\right) \) at \( x \) such that the image of the chart is an open ball in euclidean space. Let \( y \) be any point of \( X \), and let \( \alpha : \left\lbrack {a, b}\right\rbrack \rightarrow X \) be a piecewise \( {C}^{1} \) path from \( x \) to \( y \) . We select a sufficiently fine partition\n\n\[ \left\lbrack {a, b}\right\rbrack = \left\lbrack {{t}_{0},{t}_{1},\ldots ,{t}_{n}}\right\rbrack \]\n\nand open sets \( {U}_{i} \) containing \( \alpha \left( \left\lbrack {{t}_{i},{t}_{i + 1}}\right\rbrack \right) \), such that \( {U}_{i} \) has an isomorphism \( {\varphi }_{i} \) onto an open ball in euclidean space, and such that the charts \( {\varphi }_{i} \) and \( {\varphi }_{i + 1} \) have the same orientation. It is easy to verify that if two paths are homotopic, then the charts which we obtain at \( y \) by \
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No
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Theorem 4.8. Let \( \pi : X \rightarrow Z \) be a submersion. Let \( \Omega \) be a volume form on \( X \) and \( \omega \) a volume form on \( Z \) . Let \( \Omega = \eta \otimes \omega \) . Let \( \widetilde{\eta } \) be a form on \( X \) , of the same degree as \( \eta \), restricting to \( \eta \) on the fibers. Then for all \( f \in {C}_{c}\left( X\right) \), we have\n\n\[ \n{\int }_{X}{f\Omega } = {\int }_{Z}\left( {{\int }_{{Y}_{z}}f\left( y\right) {\widetilde{\eta }}_{z}\left( y\right) }\right) \omega \left( z\right)\n\]
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Proof. The proposition is local, since by using a partition of unity, we are reduced to the case when the support of \( f \) is in a given neighborhood of a point. Then the submersion is represented in a chart as a projection \( U \times W \rightarrow W \), where \( U, W \) are open in \( {\mathbf{R}}^{p} \) and \( {\mathbf{R}}^{q} \) respectively, \( U \) being a chart on \( X \) and \( W \) a chart on \( Z \) . On \( U \times W \) we have the coordinate representation\n\n\[ \n{\left( f\Omega \right) }_{U \times W}\left( {y, z}\right) = f\left( {y, z}\right) \varphi \left( {y, z}\right) \;d{y}_{1} \land \cdots \land d{y}_{p} \land \rho \left( z\right) \;d{z}_{1} \land \cdots \land d{z}_{q},\n\]\n\nwhere \( {y}_{1},\ldots ,{y}_{p} \) are the coordinate functions on \( U,{z}_{1},\ldots ,{z}_{q} \) are the coordinate functions on \( W \) ,\n\n\[ \n{\widetilde{\eta }}_{z}\left( y\right) = \varphi \left( {y, z}\right) d{y}_{1} \land \cdots \land d{y}_{p}\;\text{ and }\;\omega \left( z\right) = \rho \left( z\right) d{z}_{1} \land \cdots \land d{z}_{q}.\n\]\n\nThen the proposition merely amounts to Fubini's theorem, which concludes the proof.
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Yes
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Lemma 5.2. Let \( f \in {C}_{c}\left( G\right) \) . If \( {f}^{H} = 0 \), that is\n\n\[ \n{\int }_{H}f\left( {xh}\right) {dh} = 0 \n\]\n\nfor all \( x \in G \), then\n\n\[ \n{\int }_{G}f\left( x\right) {dx} = 0 \n\]
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Proof. For all \( \varphi \in {C}_{c}\left( G\right) \), we have:\n\n\[ \n{\int }_{G}\varphi \left( x\right) \left( {{\int }_{H}f\left( {xh}\right) {dh}}\right) {dx} = {\int }_{G}\left( {{\int }_{H}\varphi \left( x\right) f\left( {xh}\right) {dh}}\right) {dx} \n\]\n\n\[ \n= {\int }_{H}\left( {{\int }_{G}\varphi \left( x\right) f\left( {xh}\right) {dx}}\right) {dh} \n\]\n\n\[ \n= {\int }_{H}\left( {{\int }_{G}{\Delta }_{G}\left( h\right) \varphi \left( {x{h}^{-1}}\right) f\left( x\right) {dx}}\right) {dh} \n\]\n\n\[ \n= {\int }_{G}\left( {{\int }_{H}{\Delta }_{H}\left( h\right) \varphi \left( {x{h}^{-1}}\right) {dh}}\right) f\left( x\right) {dx} \n\]\n\n\[ \n= {\int }_{G}\left( {{\int }_{H}\varphi \left( {xh}\right) {dh}}\right) f\left( x\right) {dx}. \n\]\n\nBy the surjectivity \( {C}_{c}\left( G\right) \rightarrow {C}_{c}\left( {G/H}\right) \) we can find \( \varphi \) such that \( {\varphi }^{H} = 1 \) on the support of \( f \) . Since by assumption the left side of the equation is 0, this concludes the proof of the lemma.
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Yes
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Corollary 2.3. Let \( X \) be a \( {C}^{2} \) oriented manifold, of dimension \( n \), and let \( \omega \) be an \( \left( {n - 1}\right) \) -form on \( X \), of class \( {C}^{1} \) . Assume that \( \omega \) has almost compact support, and that the measures associated with \( \left| {d\omega }\right| \) on \( X \) and \( \left| \omega \right| \) on \( \partial X \) are finite. Then\n\n\[ \n{\int }_{X}{d\omega } = {\int }_{\partial X}\omega \n\]
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Proof. By our standard form of Stokes' theorem we have\n\n\[ \n{\int }_{\partial X}{g}_{k}\omega = {\int }_{X}d\left( {{g}_{k}\omega }\right) = {\int }_{X}d{g}_{k} \land \omega + {\int }_{X}{g}_{k}{d\omega }. \n\]\n\nWe estimate the left-hand side by\n\n\[ \n\left| {{\int }_{\partial X}\omega - {\int }_{\partial X}{g}_{k}\omega }\right| = \left| {{\int }_{\partial X}\left( {1 - {g}_{k}}\right) \omega }\right| \leqq {\mu }_{\left| \omega \right| }\left( {{U}_{k} \cap \partial X}\right) . \n\]\n\nSince the intersection of the sets \( {U}_{k} \) is empty, it follows for a purely measure-theoretic reason that\n\n\[ \n\mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{\partial X}{g}_{k}\omega = {\int }_{\partial X}\omega \n\]\n\nSimilarly,\n\n\[ \n\mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{X}{g}_{k}{d\omega } = {\int }_{X}{d\omega } \n\]\n\nThe integral of \( d{g}_{k} \land \omega \) over \( X \) approaches 0 as \( k \rightarrow \infty \) by assumption, and the fact that \( d{g}_{k} \land \omega \) is equal to 0 on the complement of \( {\bar{U}}_{k} \) since \( {g}_{k} \) is constant on this complement. This proves our corollary.
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Yes
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Theorem 3.1 (Stokes’ Theorem with Singularities). Let \( X \) be an oriented, \( {C}^{3} \) submanifold without boundary of \( {\mathbf{R}}^{N} \). Let \( \dim X = n \). Let \( \omega \) be an \( \left( {n - 1}\right) \) -form of class \( {C}^{1} \) on an open neighborhood of \( \bar{X} \) in \( {\mathbf{R}}^{N} \), and with compact support. Assume that:\n\n(i) If \( S \) is the set of singular points in the frontier of \( X \), then \( S \cap \operatorname{supp}\omega \) is negligible for \( X \).\n\n(ii) The measures associated with \( \left| {d\omega }\right| \) on \( X \), and \( \left| \omega \right| \) on \( \partial X \), are finite.\n\nThen\n\n\[ \n{\int }_{X}{d\omega } = {\int }_{\partial X}\omega \n\]
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Proof. Let \( U,\left\{ {U}_{k}\right\} \), and \( \left\{ {g}_{k}\right\} \) satisfy conditions NEG 1 and NEG 2. Then \( {g}_{k}\omega \) is 0 on an open neighborhood of \( S \), and since \( \omega \) is assumed to have compact support, one verifies immediately that\n\n\[ \n\left( {\operatorname{supp}{g}_{k}\omega }\right) \cap \left( {X \cup \partial X}\right) \n\]\n\nis compact. Thus Theorem 2.1 is applicable, and we get\n\n\[ \n{\int }_{\partial X}{g}_{k}\omega = {\int }_{X}d\left( {{g}_{k}\omega }\right) = {\int }_{X}d{g}_{k} \land \omega + {\int }_{X}{g}_{k}{d\omega }. \n\]\n\nWe have\n\n\[ \n\left| {{\int }_{\partial X}\omega - {\int }_{\partial X}{g}_{k}\omega }\right| \leqq \left| {{\int }_{\partial X}\left( {1 - {g}_{k}}\right) \omega }\right| \n\]\n\n\[ \n\leqq {\int }_{{U}_{k} \cap \partial X}{1d}{\mu }_{\left| \omega \right| } = {\mu }_{\left| \omega \right| }\left( {{U}_{k} \cap \partial X}\right) .\n\]\n\nSince the intersection of all sets \( {U}_{k} \cap \partial X \) is empty, it follows from purely\n\nmeasure-theoretic reasons that the limit of the right-hand side is 0 as \( k \rightarrow \infty \). Thus\n\n\[ \n\mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{\partial X}{g}_{k}\omega = {\int }_{\partial X}\omega \n\]\n\nFor similar reasons, we have\n\n\[ \n\mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{X}{g}_{k}{d\omega } = {\int }_{X}{d\omega } \n\]\n\nOur second assumption NEG 2 guarantees that the integral of \( d{g}_{k} \land \omega \) over \( X \) approaches 0 . This proves our theorem.
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Yes
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Lemma 3.2. Let \( S \) be a compact subset of \( {\mathbf{R}}^{n} \). Let \( {U}_{k} \) be the open set of points \( x \) such that \( d\left( {x, S}\right) < 2/k \). There exists a \( {C}^{\infty } \) function \( {g}_{k} \) on \( {\mathbf{R}}^{N} \) which is equal to 0 in some open neighborhood of \( S \), equal to 1 outside \( {U}_{k},0 \leqq {g}_{k} \leqq 1 \), and such that all partial derivatives of \( {g}_{k} \) are bounded by \( {C}_{1}k \), where \( {C}_{1} \) is a constant depending only on \( n \).
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Proof. Let \( \varphi \) be a \( {C}^{\infty } \) function such that \( 0 \leqq \varphi \leqq 1 \), and\n\n\[ \varphi \left( x\right) = 0\;\text{ if }\;0 \leqq \parallel x\parallel \leqq \frac{1}{2}, \]\n\n\[ \varphi \left( x\right) = 1\;\text{ if }\;1 \leqq \parallel x\parallel . \]\n\nWe use \( \parallel \parallel \) for the sup norm in \( {\mathbf{R}}^{n} \). The graph of \( \varphi \) looks like this:\n\n\n\nFor each positive integer \( k \), let \( {\varphi }_{k}\left( x\right) = \varphi \left( {kx}\right) \). Then each partial derivative \( {D}_{i}{\varphi }_{k} \) satisfies the bound\n\n\[ \begin{Vmatrix}{{D}_{i}{\varphi }_{k}}\end{Vmatrix} \leqq k\begin{Vmatrix}{{D}_{i}\varphi }\end{Vmatrix} \]\n\nwhich is thus bounded by a constant times \( k \). Let \( L \) denote the lattice of integral points in \( {\mathbf{R}}^{n} \). For each \( l \in L \), we consider the function\n\n\[ x \mapsto {\varphi }_{k}\left( {x - \frac{l}{2k}}\right) \]\n\nThis function has the same shape as \( {\varphi }_{k} \) but is translated to the point \( l/{2k} \). Consider the product\n\n\[ {g}_{k}\left( x\right) = \prod {\varphi }_{k}\left( {x - \frac{l}{2k}}\right) \]\n\ntaken over all \( l \in L \) such that \( d\left( {l/{2k}, S}\right) \leqq 1/k \). If \( x \) is a point of \( {\mathbf{R}}^{n} \) such that \( d\left( {x, S}\right) < 1/{4k} \), then we pick an \( l \) such that\n\n\[ d\left( {x, l/{2k}}\right) \leqq 1/{2k} \]\n\nFor this \( l \) we have \( d\left( {l/2, S}\right) < 1/k \), so that this \( l \) occurs in the product, and\n\n\[ {\varphi }_{k}\left( {x - l/{2k}}\right) = 0. \]\n\nTherefore \( {g}_{k} \) is equal to 0 in an open neighborhood of \( S \). If, on the other hand, we have \( d\left( {x, S}\right) > 2/k \) and if \( l \) occurs in the product, that is\n\n\[ d\left( {l/{2k}, S}\right) \leqq 1/k \]\n\nthen\n\n\[ d\left( {x, l/{2k}}\right) > 1/k \]\n\nand hence \( {g}_{k}\left( x\right) = 1 \). The partial derivatives of \( {g}_{k} \) the bounded in the desired manner. This is easily seen, for if \( {x}_{0} \) is a point where \( {g}_{k} \) is not identically 1 in a neighborhood of \( {x}_{0} \), then \( \begin{Vmatrix}{{x}_{0} - {l}_{0}/{2k}}\end{Vmatrix} \leqq 1/k \) for some \( {l}_{0} \). All other factors \( {\varphi }_{k}\left( {x - 1/{2k}}\right) \) will be identically 1 near \( {x}_{0} \) unless \( \begin{Vmatrix}{{x}_{0} - l/{2k}}\end{Vmatrix} \leqq 1/k \). But then \( \begin{Vmatrix}{l - {l}_{0}}\end{Vmatrix} \leqq 4 \) whence the number of such \( l \) is bounded as a function of \( n \) (in fact by \( {9}^{n} \)). Thus when we take the derivative, we get a sum of a most \( {9}^{n} \) terms, each one having a derivative bounded by \( {C}_{1}k \) for some constant \( {C}_{1} \). This proves our lemma.
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Yes
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Theorem 3.3. Let \( X \) be an open subset of \( {\mathbf{R}}^{n} \). Let \( S \) be the set of singular points in the closure of \( X \), and assume that \( S \) is the finite union of \( {C}^{1} \) images of m-rectangles with \( m \leqq n - 2 \). Let \( \omega \) be an \( \left( {n - 1}\right) \)-form defined on an open neighborhood of \( \bar{X} \). Assume that \( \omega \) has compact support, and that the measure associated with \( \left| \omega \right| \) on \( \partial X \) and with \( \left| {d\omega }\right| \) on \( X \) are finite. Then\n\n\[{\int }_{X}{d\omega } = {\int }_{\partial X}\omega\]
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Proof. Immediate from our two criteria and Theorem 3.2.
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No
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Theorem 1.2. Let \( X \) be a manifold without boundary, of dimension \( n \). Suppose that \( X \) is orientable and connected. Then the map \[ \omega \mapsto {\int }_{X}\omega \] induces an isomorphism of \( {H}_{c}^{n}\left( X\right) \) with \( \mathbf{R} \) itself.
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Proof. By Stokes' theorem (Chapter XVII, Corollary 2.2) the integral vanishes on exact forms (with compact support), and hence induces an R-linear map of \( {H}_{c}^{n}\left( X\right) \) into \( \mathbf{R} \). The theorem amounts to proving the converse statement: if \[ {\int }_{X}\omega = 0 \] then there exists some \( \eta \in {\mathcal{A}}_{c}^{n - 1}\left( X\right) \) such that \( \omega = {d\eta } \). For this, we first have to prove the result locally in \( {\mathbf{R}}^{n} \), which we now do.
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Yes
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Lemma 1.4. Let \( \omega \) be an \( \left( {n - 1}\right) \) -form on \( {I}^{n - 1} \) whose coefficient is a function of \( n \) variables \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) so\n\n\[ \omega \left( x\right) = f\left( {{x}_{1},\ldots ,{x}_{n}}\right) d{x}_{1} \land \cdots \land d{x}_{n - 1}. \]\n\n(Of course, all functions, like forms, are assumed \( {C}^{\infty } \) .) Suppose that \( \omega \) has compact support in \( {I}^{n - 1} \) . Assume that\n\n\[ {\int }_{{I}^{n - 1}}\omega = 0 \]\n\nThen there exists an \( \left( {n - 1}\right) \) -form \( \eta \), whose coefficients are \( {C}^{\infty } \) functions of \( {x}_{1},\ldots ,{x}_{n} \) with compact support such that\n\n\[ \omega \left( {{x}_{1},\ldots ,{x}_{n - 1};{x}_{n}}\right) = {d}_{n - 1}\eta \left( {{x}_{1},\ldots ,{x}_{n - 1};{x}_{n}}\right) . \]\n\nThe symbol \( {d}_{n - 1} \) here means the usual exterior derivative taken with respect to the first \( n - 1 \) variables.
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Proof. By induction. We first prove the theorem when \( n - 1 = 1 \) . First we carry out the proof leaving out the extra variable, just to see what's going on. So let\n\n\[ \omega \left( x\right) = f\left( x\right) {dx} \]\n\nwhere \( f \) has compact support in the open interval \( \left( {0,1}\right) \) . This means there exists \( \epsilon > 0 \) such that \( f\left( x\right) = 0 \) if \( 0 < x \leqq \epsilon \) and if \( 1 - \epsilon \leqq x \leqq 1 \) . We assume\n\n\[ {\int }_{0}^{1}f\left( x\right) {dx} = 0 \]\n\nLet\n\n\[ g\left( x\right) = {\int }_{0}^{x}f\left( t\right) {dt} \]\n\nThen \( g\left( x\right) = 0 \) if \( 0 < x \leqq \epsilon \), and also if \( 1 - \epsilon \leqq x \leqq 1 \), because for instance if \( 1 - \epsilon \leqq x \leqq 1 \), then\n\n\[ g\left( x\right) = {\int }_{0}^{1}f\left( t\right) {dt} = 0. \]\n\nThen \( f\left( x\right) {dx} = {dg}\left( x\right) \), and the lemma is proved in this case. Note that we could have carried out the proof with the extra variable \( {x}_{2} \), starting from\n\n\[ \omega \left( x\right) = f\left( {{x}_{1},{x}_{2}}\right) d{x}_{1} \]\n\nso that\n\n\[ g\left( {{x}_{1},{x}_{2}}\right) = {\int }_{0}^{1}f\left( {t,{x}_{2}}\right) {dt} \]\n\nWe can differentiate under the integral sign to verify that \( g \) is \( {C}^{\infty } \) in the pair of variables \( \left( {{x}_{1},{x}_{2}}\right) \) .
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Yes
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Lemma 1.5. Let \( U \) be an open subset of \( X \), isomorphic to \( {I}^{n} \) . Let \( \psi \in {\mathcal{A}}_{c}^{n}\left( U\right) \) be such that\n\n\[{\int }_{U}\psi \neq 0\]\n\nLet \( \omega \in {\mathcal{A}}_{c}^{n}\left( U\right) \) . Then there exists \( c \in \mathbf{R} \) and \( \eta \in {\mathcal{A}}_{c}^{n - 1}\left( U\right) \) such that\n\n\[ \omega - {c\psi } = {d\eta }\]
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Proof. We take \( c = {\int }_{U}\omega /{\int }_{U}\psi \) and apply Lemma 1.3 to \( \omega - {c\psi } \) .
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Yes
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Lemma 1.6. Assume that \( X \) is connected and oriented. Let \( U,\psi \) be as in Lemma 1.5. Let \( V \) be the set of points \( x \in X \) having the following property. There exists a neighborhood \( U\left( x\right) \) of \( x \) isomorphic to \( {I}^{n} \) such that for every \( \omega \in {\mathcal{A}}_{c}^{n}\left( {U\left( x\right) }\right) \) there exist \( c \in \mathbf{R} \) and \( \eta \in {\mathcal{A}}_{c}^{n - 1}\left( X\right) \) such that
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Proof. Lemma 1.5 asserts that \( V \supset U \) . Since \( X \) is connected, it suffices to prove that \( V \) is both open and closed. It is immediate from the definition of \( V \) that \( V \) is open, so there remains to prove its closure. Let \( z \) be in the closure of \( V \) . Let \( W \) be a neighborhood of \( z \) isomorphic to \( {I}^{n} \) . There exists a point \( x \in V \cap W \) . There exists a neighborhood \( U\left( x\right) \) as in the definition of \( V \) such that \( U\left( x\right) \subset W \) . For instance, we may take \[ U\left( x\right) \approx \left( {{a}_{1},{b}_{1}}\right) \times \cdots \times \left( {{a}_{n},{b}_{n}}\right) \approx {I}^{n} \] with \( {a}_{i} \) sufficiently close to 0 and \( {b}_{i} \) sufficiently close to 1, and of course \( 0 < {a}_{i} < {b}_{i} \) for \( i = 1,\ldots, n \) . Let \( {\psi }_{1} \in {\mathcal{A}}_{c}^{n}\left( {U\left( x\right) }\right) \) be such that \[ {\int }_{U\left( x\right) }{\psi }_{1} = 1 \] Let \( \omega \in {\mathcal{A}}_{c}^{n}\left( W\right) \) . By the definition of \( V \), there exist \( {c}_{1} \in \mathbf{R} \) and \( {\eta }_{1} \in {\mathcal{A}}_{c}^{n}\left( X\right) \) such that \[ {\psi }_{1} - {c}_{1}\psi = d{\eta }_{1} \] By Lemma 1.5, there exists \( {c}_{2} \in \mathbf{R} \) and \( {\eta }_{2} \in {\mathcal{A}}_{c}^{n}\left( X\right) \) such that \[ \omega - {c}_{2}{\psi }_{1} = d{\eta }_{2} \] Then \[ \omega - {c}_{2}{c}_{1}\psi = d\left( {{\eta }_{2} + {c}_{2}{\eta }_{1}}\right) \] thus concluding the proof of Lemma 1.6.
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Yes
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Theorem 2.2. Let \( X \) be a compact, connected oriented manifold of dimension \( n \) . Let \( \omega ,\psi \in {\mathcal{A}}^{n}\left( X\right) \left( { = {\mathcal{A}}_{c}^{n}\left( X\right) }\right) \) be volume forms such that\n\n\[ \n{\int }_{X}\omega = {\int }_{X}\psi \n\]\n\nThen there exists an automorphism \( f : X \rightarrow X \) of \( X \) such that \( \omega = {f}^{ * }\psi \) .
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Proof. Let\n\n\[ \n{\omega }_{s} = \left( {1 - s}\right) \omega + {s\psi }\;\text{ for }\;0 \leqq s \leqq 1. \n\]\n\nThen \( {\omega }_{s} \) is a volume form for each \( s \), and in particular is non-singular. By Theorem 1.1, there exists \( \eta \in {\mathcal{A}}^{n - 1}\left( X\right) \) such that \( \psi - \omega = {d\eta } \) . Note also that \( \psi - \omega = d{\omega }_{s}/{ds} \) . Since \( {\omega }_{s} \) is non-singular, there exists a unique vector field \( {\xi }_{s} \) such that\n\n\[ \n{\omega }_{s} \circ {\xi }_{s} = - \eta \n\]\n\n\nLet \( {\alpha }_{s} \) be the flow of \( {\xi }_{s} \) . Then \( {\alpha }_{s} \) is defined on \( \mathbf{R} \times X \) by Corollary 2.4 of Chapter IV. Then we get:\n\n\[ \n\frac{d}{ds}\left( {{\alpha }_{s}^{ * }{\omega }_{s}}\right) = {\left. \frac{d}{du}\left( {\alpha }_{u}^{ * }{\omega }_{s}\right) \right| }_{u = s} + {\alpha }_{s}^{ * }\left( \frac{d{\omega }_{s}}{ds}\right) \n\]\n\n\[ \n= {\alpha }_{s}^{ * }d\left( {{\omega }_{s} \circ {\xi }_{s}}\right) + {\alpha }_{s}^{ * }\left( {\psi - \omega }\right) \;\text{by Proposition 5.2 of Chapter V} \n\]\n\n\[ \n= - {\alpha }_{s}^{ * }{d\eta } + {\alpha }_{s}^{ * }{d\eta } \n\]\n\n\[ \n= 0\text{.} \n\]\n\nTherefore \( {\alpha }_{s}^{ * }{\omega }_{s} \) is constant as a function of \( s \), so we find\n\n\[ \n\omega = {\alpha }_{0}^{ * }{\omega }_{0} = {\alpha }_{1}^{ * }{\omega }_{1} = {f}^{ * }\psi ,\;\text{ with }f = {\alpha }_{1}, \n\]\n\nthereby proving the theorem.
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Yes
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Theorem 3.1 (Divergence Theorem).\n\n\[ \n{\int }_{X}{\mathcal{L}}_{\xi }\Omega = {\int }_{\partial X}\Omega \circ \xi \n\]
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Suppose that \( \left( {X, g}\right) \) is a Riemannian manifold, assumed oriented for simplicity. We let \( \Omega \) or \( {\operatorname{vol}}_{g} \) be the volume form defined in Chapter XV,\n\n§1. Let \( \omega \) be the canonical Riemannian volume form on \( \partial X \) for the metric induced by \( g \) on the boundary. Let \( {\mathbf{n}}_{x} \) be a unit vector in the tangent space \( {T}_{x}\left( X\right) \) such that \( u \) is perpendicular to \( {T}_{x}\left( {\partial X}\right) \) . Such a unit vector is determined up to sign. Denote by \( {\mathbf{n}}_{x}^{ \vee } \) its dual functional, i.e. the component on the projection along \( {\mathbf{n}}_{x} \) . We select \( {\mathbf{n}}_{x} \) with the sign such that\n\n\[ \n{\mathbf{n}}_{x}^{ \vee } \land \omega \left( x\right) = \Omega \left( x\right) \n\]\n\nWe then shall call \( {\mathbf{n}}_{x} \) the unit outward normal vector to the boundary at \( x \) . In an oriented chart, it looks like this.\n\n\n\nThen by formula CON 3 of Chapter V, §5 we find\n\n\[ \n\Omega \circ \xi = \langle \mathbf{n},\;\xi \rangle \omega - {\mathbf{n}}^{ \vee }\; \land \;\left( {\omega \circ \xi }\right) ,\n\]\n\nand the restriction of this form to \( \partial X \) is simply \( \langle \mathbf{n},\xi \rangle \omega \) . Thus we get:\n\nTheorem 3.2 (Gauss Theorem). Let \( X \) be a Riemannian manifold. Let \( \omega \) be the canonical Riemannian volume form on \( \partial X \) and let \( \Omega \) be the canonical Riemannian volume form on \( X \) itself. Let \( \mathbf{n} \) be the unit outward normal vector field to the boundary, and let \( \xi \) be a \( {C}^{1} \) vector field on \( X \) , with compact support. Then\n\n\[ \n{\int }_{X}\left( {{\operatorname{div}}_{\Omega }\xi }\right) \Omega = {\int }_{\partial X}\langle \mathbf{n},\xi \rangle \omega \n\]
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Yes
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Theorem 3.4 (Green’s Formula). Let \( \left( {X, g}\right) \) be an oriented Riemannian manifold possibly with boundary, and let \( \varphi ,\psi \) be functions on \( X \) with compact support. Let \( \omega \) be the canonical volume form associated with the induced metric on the boundary. Then\n\n\[ \n{\int }_{X}\left( {{\varphi \Delta \psi } - {\psi \Delta \varphi }}\right) {\operatorname{vol}}_{g} = - {\int }_{\partial X}\left( {\varphi {\partial }_{\mathbf{n}}\psi - \psi {\partial }_{\mathbf{n}}\varphi }\right) \omega .\n\]
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Proof. From formula (4) we get\n\n\[ \n{d}^{ * }\left( {\varphi d\psi }\right) = {\varphi \Delta \psi } - \langle {d\varphi },{d\psi }{\rangle }_{g},\n\]\n\nwhence\n\n\[ \n{\varphi \Delta \psi } - {\psi \Delta \varphi } = {d}^{ * }\left( {\varphi d\psi }\right) - {d}^{ * }\left( {\psi d\varphi }\right)\n\]\n\n\[ \n= - \operatorname{div}\left( {\varphi d\psi }\right) + \operatorname{div}\left( {\psi d\varphi }\right)\n\]\n\nWe apply Theorem 3.2 to conclude the proof.
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Yes
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Corollary 3.5 (E. Hopf). Let \( X \) be a Riemannian manifold without boundary, and let \( f \) be a \( {C}^{2} \) function on \( X \) with compact support, such that \( {\Delta f} \geqq 0 \) . Then \( f \) is constant. In particular, every harmonic function with compact support is constant.
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Proof. We first give the proof assuming that \( X \) is oriented. By Green’s formula we get\n\n\[ \n{\int }_{X}{\Delta f}{\operatorname{vol}}_{g} = 0 \n\]\n\nSince \( {\Delta f} \geqq 0 \), it follows that in fact \( {\Delta f} = 0 \), so we are reduced to the harmonic case. We now apply Green’s formula to \( {f}^{2} \), and get\n\n\[ \n0 = {\int }_{X}\Delta {f}^{2}{\operatorname{vol}}_{g} = {\int }_{X}{2f\Delta f}{\operatorname{vol}}_{g} - {\int }_{X}2{\left( \operatorname{grad}f\right) }^{2}{\operatorname{vol}}_{g}. \n\]\n\nHence \( {\left( \operatorname{grad}f\right) }^{2} = 0 \) because \( {\Delta f} = 0 \), and finally grad \( f = 0 \), so \( {df} = 0 \) and \( f \) is constant, thus proving the corollary in the oriented case. For the non-oriented case, by Proposition 4.6 of Chapter XVI, there exists a covering of degree 2 of \( X \) which is oriented, and then one can pull back all the objects from \( X \) to this covering to conclude the proof in this case.
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Yes
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Given \( 1 \leqq r \leqq n \), there exists a unique isomorphism\n\n\[ \n* : \mathop{\bigwedge }\limits^{r}V \rightarrow \mathop{\bigwedge }\limits^{{n - r}}V \]\n\n such that for \( \varphi ,\psi \in \mathop{\bigwedge }\limits^{r}V \) we have\n\n\[ \n\langle \varphi ,\psi {\rangle }_{g}{\operatorname{vol}}_{g} = \varphi \land * \psi \]\n
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The proof will give an explicit determination of the isomorphism on the usual basis for \( \land V \) . Let \( I = \left\lbrack {{i}_{1} < {i}_{2} < \cdots < {i}_{r}}\right\rbrack \) be an ordered set of \( r \) indices. We let\n\n\[ \n{e}_{I} = {e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{r}} \]\n\nIf \( {I}^{\prime } \) is another such ordered set with \( n - r \) elements, and \( I \cup {I}^{\prime } = \) \( \{ 1,\ldots, n\} \) then we let \( {\epsilon }_{I} \) be the sign of the permutation \( \left( {I, J}\right) \) of \( \left( {1,\ldots, n}\right) \) . We then define\n\n\[ \n* {e}_{I} = {\epsilon }_{I}{e}_{{I}^{\prime }} \]\n\nand extend this operation by linearity to all of \( \mathop{\bigwedge }\limits^{r}V \) . Then directly from the definition, we see that if \( J \) is an ordered set of \( r \) indices, then\n\n\[ \n{e}_{I} \land * {e}_{J} = \left\langle {{e}_{I},{e}_{J}}\right\rangle {e}_{1} \land \cdots \land {e}_{n} \]\n\n\[ \n= {\delta }_{IJ}{e}_{1} \land \cdots \land {e}_{n}. \]\n\nThus on the standard basis elements of \( \mathop{\bigwedge }\limits^{r}V \) the desired relation of the proposition is satisfied. The same relation is therefore satisfied for all elements of \( \mathop{\bigwedge }\limits^{r}V \), as desired.
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Yes
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Proposition 4.3. The exterior derivative \( d \) has an adjoint \( {d}^{ * } \) with respect to the scalar product \( \langle \rangle ,{\rangle }_{X} \), namely for \( \varphi \in {\mathcal{A}}_{c}^{r - 1}\left( X\right) \) and \( \psi \in {\mathcal{A}}_{c}^{r}\left( X\right) \) we have\n\n\[ \langle {d\varphi },\psi {\rangle }_{X} = {\left\langle \varphi ,{d}^{ * }\psi \right\rangle }_{X} \]\n\nFurthermore, the adjoint is given by the explicit formula\n\n\[ {d}^{ * } = {\left( -1\right) }^{{nr} + n + 1} * d * \;\text{ on }{\mathcal{A}}_{c}^{r}\left( X\right) . \]\n\n\[ = - * d * \;\text{if}n\text{is even.} \]
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Proof. By Stokes' theorem, we have:\n\n\[ {\int }_{X}{d\varphi } \land * \psi = {\int }_{X}d\left( {\varphi \land * \psi }\right) - {\left( -1\right) }^{r - 1}{\int }_{X} \land d * \psi \]\n\n\[ = {\left( -1\right) }^{r}{\int }_{X}\varphi \land d * \psi \]\n\nNow\n\n\[ {\left( -1\right) }^{r}\varphi \land d * \psi = {\left( -1\right) }^{r}\varphi \land * * \mathbf{w}d * \psi \]\n\n\[ = {\left( -1\right) }^{r}\varphi \land \mathbf{w} * \left( {*d * }\right) \psi \]\n\n\[ = {\left( -1\right) }^{{nr} + n + 1}\varphi \land * \left( {*d * }\right) \psi \]\n\nwhich proves the proposition.
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Yes
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Lemma 1.1. Let \( \\mathbf{F} \) be a subspace of \( \\mathbf{E} \), let \( x \\in \\mathbf{E} \), and let\n\n\[ a = \\inf \\left| {x - y}\\right| \]\n\nthe inf taken over all \( y \\in \\mathbf{F} \) . Then there exists an element \( {y}_{0} \\in \\mathbf{F} \) such that \( a = \\left| {x - {y}_{0}}\\right| \) .
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Proof. Let \( {y}_{n} \) be a sequence in \( \\mathbf{F} \) such that \( \\left| {{y}_{n} - x}\\right| \) tends to \( a \) . We must show that \( {y}_{n} \) is Cauchy. By the parallelogram law,\n\n\[ {\\left| {y}_{n} - {y}_{m}\\right| }^{2} = 2{\\left| {y}_{n} - x\\right| }^{2} + 2{\\left| {y}_{m} - x\\right| }^{2} - 4{\\left| \\frac{1}{2}\\left( {y}_{n} + {y}_{m}\\right) - x\\right| }^{2} \]\n\n\[ \\leqq 2{\\left| {y}_{n} - x\\right| }^{2} + 2{\\left| {y}_{m} - x\\right| }^{2} - 4{a}^{2} \]\n\nwhich shows that \( {y}_{n} \) is Cauchy, converging to some vector \( {y}_{0} \) . The lemma follows by continuity.
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Yes
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Theorem 1.2. If \( \mathbf{F} \) is a subspace properly contained in \( \mathbf{E} \), then there exists a vector \( z \) in \( \mathbf{E} \) which is perpendicular to \( \mathbf{F}\left( {\text{and} \neq 0}\right) \) .
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Proof. Let \( x \in \mathbf{E} \) and \( x \notin \mathbf{F} \) . Let \( {y}_{0} \) be an element of \( \mathbf{F} \) which is at minimal distance from \( x \) (use Lemma 1.1). Let \( a \) be this distance and let \( z = {y}_{0} - x \) . After a translation, we may assume that \( z = x \), so that \( \left| x\right| = a \) . For any complex number \( \alpha \) and \( y \in \mathbf{F} \) we have \( \left| {x + {\alpha y}}\right| \geqq a \) , whence \[ \langle x + {\alpha y}, x + {\alpha y}\rangle = {\left| x\right| }^{2} + \bar{\alpha }\langle x, y\rangle + \alpha \overline{\langle x, y\rangle } + \alpha \bar{\alpha }{\left| y\right| }^{2} \] \[ \geqq {a}^{2}\text{.} \] Put \( \alpha = t\overline{\langle x, y\rangle } \) . We get a contradiction for small values of \( t \) .
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Yes
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Proposition 2.1. A linear map is bounded if and only if it maps the unit sphere on a bounded subset, if and only if it is continuous.
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Proof. Clear.
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No
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Proposition 2.2. If \( A \) is an operator and \( \langle {Ax}, x\rangle = 0 \) for all \( x \), then \( A = O \) .
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Proof. This follows from the polarization identity,\n\n\[ \langle A\left( {x + y}\right) ,\left( {x + y}\right) \rangle - \langle A\left( {x - y}\right) ,\left( {x - y}\right) \rangle = 2\left\lbrack {\langle {Ax}, y\rangle +\langle {Ay}, x\rangle }\right\rbrack .\n\]\nReplace \( x \) by \( {ix} \) . Then we get\n\n\[ \langle {Ax}, y\rangle + \langle {Ay}, x\rangle = 0, \]\n\n\[ i\langle {Ax}, y\rangle - i\langle {Ay}, x\rangle = 0, \]\n\nfor all \( x, y \) whence \( \langle {Ax}, y\rangle = 0 \) and \( A = O \) .
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Yes
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Lemma 2.3. Let \( A \) be an operator, and \( {ca} \) number such that\n\n\[ \n\\left| {\\langle {Ax}, x\\rangle }\\right| \\leqq c{\\left| x\\right| }^{2}\n\]\n\nfor all \( x \\in \\mathbf{E} \) . Then for all \( x, y \) we have\n\n\[ \n\\left| {\\langle {Ax}, y\\rangle }\\right| + \\left| {\\langle x,{Ay}\\rangle }\\right| \\leqq {2c}\\left| x\\right| \\left| y\\right|\n\]
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Proof. By the polarization identity,\n\n\[ \n2\\left| {\\langle {Ax}, y\\rangle +\\langle {Ay}, x\\rangle }\\right| \\leqq c{\\left| x + y\\right| }^{2} + c{\\left| x - y\\right| }^{2} = {2c}\\left( {{\\left| x\\right| }^{2} + {\\left| y\\right| }^{2}}\\right) .\n\]\n\nHence\n\n\[ \n\\left| {\\langle {Ax}, y\\rangle +\\langle {Ay}, x\\rangle }\\right| \\leqq c\\left( {{\\left| x\\right| }^{2} + {\\left| y\\right| }^{2}}\\right) .\n\]\n\n\n\nWe multiply \( y \) by \( {e}^{i\\theta } \) and thus get on the left-hand side\n\n\[ \n\\left| {{e}^{-{i\\theta }}\\langle {Ax}, y\\rangle + {e}^{i\\theta }\\langle {Ay}, x\\rangle }\\right| .\n\]\n\nThe right-hand side remains unchanged, and for suitable \( \\theta \), the left-hand side becomes\n\n\[ \n\\left| {\\langle {Ax}, y\\rangle }\\right| + \\left| {\\langle {Ay}, x\\rangle }\\right| .\n\]\n\n(In other words, we are lining up two complex numbers by rotating one by \( \\theta \) and the other by \( - \\theta \) .) Next we replace \( x \) by \( {tx} \) and \( y \) by \( y/t \) for \( t \) real and \( t > 0 \) . Then the left-hand side remains unchanged, while the right-hand side becomes\n\n\[ \ng\\left( t\\right) = {t}^{2}{\\left| x\\right| }^{2} + \\frac{1}{{t}^{2}}{\\left| y\\right| }^{2}.\n\]\n\nThe point at which \( {g}^{\\prime }\\left( t\\right) = 0 \) is the unique minimum, and at this point \( {t}_{0} \) we find that\n\n\[ \ng\\left( {t}_{0}\\right) = \\left| x\\right| \\left| y\\right|\n\]\n\nThis proves our lemma.
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Yes
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Theorem 2.4. We have:\n\n\[ \n{\left( A + B\right) }^{ * } = {A}^{ * } + {B}^{ * },\;{A}^{* * } = A, \]\n\n\[ \n{\left( \alpha A\right) }^{ * } = \bar{\alpha }{A}^{ * },\;\left| {A}^{ * }\right| = \left| A\right| , \]\n\n\[ \n{\left( AB\right) }^{ * } = {B}^{ * }{A}^{ * },\;\left| {A{A}^{ * }}\right| = {\left| A\right| }^{2}. \]\n\nand the mapping \( A \mapsto {A}^{ * } \) is continuous.
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Proof. Exercise for the reader.
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No
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Proposition 3.1. \( A \) is hermitian if and only if \( \langle {Ax}, x\rangle \) is real for all \( x \) .
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Proof. Let \( A \) be hermitian. Then \( \overline{\langle {Ax}, x\rangle } = \overline{\langle x,{Ax}\rangle } = \langle {Ax}, x\rangle \) . Conversely, \( \langle {Ax}, x\rangle = \overline{\langle {Ax}, x\rangle } = \langle x,{Ax}\rangle = \left\langle {{A}^{ * }x, x}\right\rangle \) implies that\n\n\[\n\left\langle {\left( {A - {A}^{ * }}\right) x, x}\right\rangle = 0\n\]\n\nwhence \( A = {A}^{ * } \) by polarization.
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Yes
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Proposition 3.2. Let \( A \) be a hermitian operator. Then \( \left| A\right| \) is the greatest lower bound of all values \( c \) such that\n\n\[ \left| {\langle {Ax}, x\rangle }\right| \leqq c{\left| x\right| }^{2} \]\n\nfor all \( x \), or equivalently, the sup of all values \( \left| {\langle {Ax}, x\rangle }\right| \) taken for \( x \) on the unit sphere in \( E \) .
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Proof. When \( A \) is hermitian we obtain\n\n\[ \left| {\langle {Ax}, y\rangle }\right| \leqq c\left| x\right| \left| y\right| \]\n\nfor all \( x, y \in E \), so that we get \( \left| A\right| \leqq c \) in Lemma 2.3. On the other hand, \( c = \left| A\right| \) is certainly a possible value for \( c \) by the Schwartz inequality. This proves our proposition.
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No
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Theorem 3.3. Let \( \alpha ,\beta \) be real and \( {\alpha I} \leqq A \leqq {\beta I} \) . Let \( p \) be a real polynomial, semipositive in the interval \( \alpha \leqq t \leqq \beta \) . Then \( p\left( A\right) \) is a semipositive operator.
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Proof. We shall need the following obvious facts.\n\nIf \( A, B \) are hermitian, \( A \) commutes with \( B \), and \( A \geqq O \), then \( A{B}^{2} \) is semipositive.\n\nIf \( p\left( t\right) \) is quadratic, of type \( p\left( t\right) = {t}^{2} + {at} + b \) and has imaginary roots, then\n\n\[ p\left( t\right) = {\left( t + \frac{a}{2}\right) }^{2} + \left( {b - \frac{{a}^{2}}{4}}\right) \]\n\nis a sum of squares.\n\nA sum of squares times a sum of squares is a sum of squares (if they commute).\n\nIf \( p\left( t\right) \) has a root \( \gamma \) in our interval, then the multiplicity of \( \gamma \) is even.\n\nOur theorem now follows from the following purely algebraic statement.\n\nLet \( \alpha \leqq t \leqq \beta \) be a real interval, and \( p\left( t\right) \) a real polynomial which is semipositive in this interval. Then \( p\left( t\right) \) can be written:\n\n\[ p\left( t\right) = c\left\lbrack {\sum {Q}_{i}^{2}+\sum \left( {t - \alpha }\right) {Q}_{j}^{2}+\sum \left( {\beta - t}\right) {Q}_{k}^{2}}\right\rbrack \]\n\nwhere \( {Q}^{2} \) just denotes the square of some polynomial and \( c \) is a number \( \geqq 0 \) .\n\nIn order to prove this, we split \( p\left( t\right) \) over the real numbers into linear and quadratic factors. If a root \( \gamma \) is \( \leqq \alpha \), then we write\n\n\[ \left( {t - \gamma }\right) = \left( {t - \alpha }\right) + \left( {\alpha - \gamma }\right) \]\n\nand note that \( \left( {\alpha - \gamma }\right) \) is a square. If a root \( \gamma \) is \( \geqq \beta \), then we write\n\n\[ \left( {\gamma - t}\right) = \left( {\gamma - \beta }\right) + \left( {\beta - t}\right) \]\n\nwith \( \left( {\gamma - \beta }\right) \) a square. We can then write, after expanding out the factorization of \( p\left( t\right) \) ,\n\n\[ p\left( t\right) = c\left\lbrack {\sum {Q}_{i}^{2}+\sum \left( {t - \alpha }\right) {Q}_{j}^{2}+\sum \left( {\beta - t}\right) {Q}_{k}^{2}+\sum \left( {t - \alpha }\right) \left( {\beta - t}\right) {Q}_{l}^{2}}\right\rbrack \]\n\nwith some constant \( c \) and \( {Q}^{2} \) standing for the square of some polynomial. Note that \( c \) is \( \geqq 0 \) since \( p\left( t\right) \) is semipositive on the interval. Our last step reduces the bad last term to the preceding ones by means of the identity\n\n\[ \left( {t - \alpha }\right) \left( {\beta - t}\right) = \frac{{\left( t - \alpha \right) }^{2}\left( {\beta - t}\right) + \left( {t - \alpha }\right) {\left( \beta - t\right) }^{2}}{\beta - \alpha }. \]
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Yes
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Corollary 3.5. Let \( {\alpha I} \leqq A \leqq {\beta I} \) . Let \( p\left( t\right) \) be a real polynomial. Then \( \left| {p\left( A\right) }\right| \leqq \parallel p\parallel \)
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Proof. Let \( q\left( t\right) = \parallel p\parallel \pm p\left( t\right) \) . Then \( q\left( t\right) \) is \( \geqq 0 \) on the interval. Hence \( q\left( A\right) \geqq O \) and our assertion follows at once.
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Yes
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Corollary 3.5. Let \( {\alpha I} \leqq A \leqq {\beta I} \) . Let \( p\left( t\right) \) be a real polynomial. Then \( \left| {p\left( A\right) }\right| \leqq \parallel p\parallel \)
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Proof. Let \( q\left( t\right) = \parallel p\parallel \pm p\left( t\right) \) . Then \( q\left( t\right) \) is \( \geqq 0 \) on the interval. Hence \( q\left( A\right) \geqq O \) and our assertion follows at once.
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No
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Proposition 3.6. Let \( A \) be a semipositive operator. Then there exists an operator \( B \) in the closure of the algebra generated by \( A \) such that \( {B}^{2} = A \) .
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Proof. The continuous function \( {t}^{1/2} \) maps on \( {A}^{1/2} \) .
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No
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Corollary 3.7. The product of two semipositive, commuting hermitian operators is again semipositive.
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Proof. Let \( A, C \) be hermitian and \( {AC} = {CA} \) . If \( B \) is as in Proposition 3.6 , then\n\n\[ \langle {ACx}, x\rangle = \left\langle {{B}^{2}{Cx}, x}\right\rangle = \langle {BCx},{Bx}\rangle = \langle {CBx},{Bx}\rangle \geqq 0. \]
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Yes
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Lemma 3.8. Let \( X \) be a compact set, \( R \) the ring of continuous functions on \( X \), and \( \mathfrak{a} \) a closed ideal of \( R,\mathfrak{a} \neq R \) . Let \( C \) be the closed set of zeros of \( \mathfrak{a} \) . Then \( C \) is not empty and if a function \( f \in R \) vanishes on \( C \), then \( f \in \mathfrak{a} \) .
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Proof. Given \( \epsilon \), let \( U \) be the open set where \( \left| f\right| < \epsilon \) . Then \( X - U \) is closed. For each point \( t \in X - U \) there exists a function \( g \in \mathfrak{a} \) such that \( g\left( t\right) \neq 0 \) in a neighborhood of \( t \) . These neighborhoods cover \( X - U \), and so does a finite number of them, with functions \( {g}_{1},\ldots ,{g}_{r} \) . Let \( g = \) \( {g}_{1}^{2} + \cdots + {g}_{r}^{2} \) . Then \( g \in \mathfrak{a} \) . Our function \( g \) has a minimum on \( X - U \) and for \( n \) large, the function\n\n\[ f\frac{ng}{1 + {ng}} \]\n\nis close to \( f \) on \( X - U \) and is \( < \epsilon \) on \( U \), which proves what we wanted.
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No
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Theorem 3.9. The map\n\n\\[ \nf\\left( t\\right) \\mapsto f\\left( A\\right) \n\\]\n\ninduces a Banach-isomorphism (i.e. norm-preserving) of the Banach algebra of continuous functions on \\( \\sigma \\left( A\\right) \\) onto the closure of the algebra generated by \\( A \\) .
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Proof. We have already proved that our map is an agebraic isomorphism and that \\( \\left| {f\\left( A\\right) }\\right| \\leqq \\parallel f{\\parallel }_{A} \\) . In order to get the reverse inequality, we shall prove:\n\nIf \\( f\\left( A\\right) \\geqq O \\), then \\( f\\left( t\\right) \\geqq 0 \\) on the spectrum of \\( A \\) . Indeed, if \\( f\\left( c\\right) < 0 \\) for some \\( c \\in \\sigma \\left( A\\right) \\), we let \\( g\\left( t\\right) \\) be a function which is 0 outside a small neighborhood of \\( c \\), is \\( \\geqq 0 \\) everywhere, and is \\( > 0 \\) at \\( c \\) . Then \\( g\\left( A\\right) \\) and \\( g\\left( A\\right) f\\left( A\\right) \\) are both \\( \\geqq 0 \\) by Corollary 3.7. But \\( - g\\left( t\\right) f\\left( t\\right) \\geqq 0 \\) gives \\( - g\\left( A\\right) f\\left( A\\right) \\geqq O \\) whence \\( g\\left( A\\right) f\\left( A\\right) = O \\) . Since \\( g\\left( t\\right) f\\left( t\\right) \\) is not 0 on the spectrum of \\( A \\), we get a contradiction.\n\nLet now \\( s = \\left| {f\\left( A\\right) }\\right| \\) . Then \\( {sI} - f\\left( A\\right) \\geqq O \\) implies that \\( s - f\\left( t\\right) \\geqq 0 \\) , which proves the theorem.
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Yes
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Theorem 3.10. The general spectrum is compact, and in fact, if \( \xi \) is in it, then \( \left| \xi \right| \leqq \left| A\right| \) . If \( A \) is hermitian, then the general spectrum is equal to \( \sigma \left( A\right) \) .
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Proof. The complement of the general spectrum is open, because if \( A - {\xi }_{0} \) is invertible, and \( \xi \) is close to \( {\xi }_{0} \), then \( {\left( A - {\xi }_{0}\right) }^{-1}\left( {A - \xi }\right) \) is close to \( I \), hence invertible, and hence \( A - \xi \) is also invertible. Furthermore, if \( \xi > \left| A\right| \), then \( \left| {A/\xi }\right| < 1 \) and hence \( I - \left( {A/\xi }\right) \) is invertible (by the power series argument). So is \( A - \xi \) and we are done. Finally, suppose that \( \xi \) is in the general spectrum. Then \( \xi \) is real. Otherwise, let\n\n\[ g\left( t\right) = \left( {t - \xi }\right) \left( {t - \bar{\xi }}\right) \]\n\nThen \( g\left( t\right) \neq 0 \) on \( \sigma \left( A\right) \) and \( h\left( t\right) = 1/g\left( t\right) \) is its inverse. From this we see that \( A - \xi \) is invertible.\n\nSuppose \( \xi \) is not in the spectrum. Then \( t - \xi \) is invertible and so is \( A - \xi \) .\n\nSuppose \( \xi \) is in the spectrum. After a translation, we may suppose that 0 is in the spectrum. Consider the function \( g\left( t\right) \) as follows:\n\n\[ g\left( t\right) = \left\{ \begin{array}{ll} 1/\left| t\right| , & \left| t\right| \geqq 1/N \\ N, & \left| t\right| \leqq 1/N \end{array}\right. \]\n\n( \( g \) is positive and has a peak at 0 .) If \( A \) is invertible, \( {BA} = I \), then from \( \left| {\operatorname{tg}\left( t\right) }\right| \leqq 1 \) we get \( \left| {{Ag}\left( A\right) }\right| \leqq 1 \) and hence \( \left| {g\left( A\right) }\right| \leqq \left| B\right| \) . But \( g\left( A\right) \) becomes arbitrarily large as we take \( N \) large. Contradiction.
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Yes
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Theorem 3.11. Let \( S \) be a set of operators of the Hilbert space \( E \), leaving no closed subspace invariant except 0 and \( \mathbf{E} \) itself. Let \( A \) be a Hermitian operator such that \( {AB} = {BA} \) for all \( B \in S \) . Then \( A = {\lambda I} \) for some real number \( \lambda \) .
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Proof. It will suffice to prove that there is only one element in the spectrum of \( A \) . Suppose there are two, \( {\lambda }_{1} \neq {\lambda }_{2} \) . There exist continuous functions \( f, g \) on the spectrum such that neither is 0 on the spectrum, but \( {fg} \) is 0 on the spectrum. For instance, one may take for \( f, g \) the functions whose graph is indicated on the next diagram.\n\n\n\nWe have \( f\left( A\right) B = {Bf}\left( A\right) \) for all \( B \in S \) (because \( B \) commutes with real polynomials in \( A \), hence with their limits). Hence \( f\left( A\right) \mathbf{E} \) is invariant under \( S \) because\n\n\[ \n{Bf}\left( A\right) \mathbf{E} = f\left( A\right) B\mathbf{E} \subset f\left( A\right) \mathbf{E}.\n\]\n\nLet \( \mathbf{F} \) be the closure of \( f\left( A\right) \mathbf{E} \) . Then \( \mathbf{F} \neq 0 \) because \( f\left( A\right) \neq O \) . Furthermore, \( \mathbf{F} \neq \mathbf{E} \) because \( g\left( A\right) f\left( A\right) \mathbf{E} = 0 \) and hence \( g\left( A\right) \mathbf{F} = 0 \) . Since \( \mathbf{F} \) is obviously invariant under \( S \), we have a contradiction.
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Yes
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Corollary 3.12. Let \( S \) be a set of operators of the Hilbert space \( \mathbf{E} \) , leaving no closed subspace invariant except 0 and \( \mathbf{E} \) itself. Let \( A \) be an operator such that \( A{A}^{ * } = {A}^{ * }A,{AB} = {BA} \), and \( {A}^{ * }B = B{A}^{ * } \) for all \( B \in S \) . Then \( A = {\lambda I} \) for some complex number \( \lambda \) .
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Proof. Write \( A = {A}_{1} + i{A}_{2} \) where \( {A}_{1},{A}_{2} \) are hermitian and commute (e.g. \( {A}_{1} = \left( {A + {A}^{ * }}\right) /2 \) ). Apply the theorem to each one of \( {A}_{1} \) and \( {A}_{2} \) to get the result.
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No
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Proposition 1.1 A nonempty set \( X \) is countable if and only if there is a surjection from \( \mathbb{N} \) onto \( X \) .
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Proof. If \( X \) is countably infinite there is a bijection, and thus a surjection, from \( \mathbb{N} \) to \( X \) . If \( X \) is finite with \( n \geq 1 \) elements, there is a bijection \( \varphi : \{ 1,\ldots, n\} \rightarrow X \) . This can be arbitrarily extended to a bijection from \( \mathbb{N} \) to \( X \) .\n\nConversely, suppose there is a surjection \( \varphi : \mathbb{N} \rightarrow X \) and that \( X \) is infinite. Define recursively a sequence \( {\left( {n}_{p}\right) }_{p} \in \mathbb{N} \) by setting \( {n}_{0} = 0 \) and\n\n\[ \n{n}_{p + 1} = \min \left\{ {n : \varphi \left( n\right) \notin \left\{ {\varphi \left( {n}_{0}\right) ,\varphi \left( {n}_{1}\right) ,\ldots ,\varphi \left( {n}_{p}\right) }\right\} }\right\} \;\text{ for }p \in \mathbb{N}.\n\]\n\nThis sequence is well-defined because \( X \) is infinite; by construction, the map \( p \mapsto \varphi \left( {n}_{p}\right) \) is a bijection from \( \mathbb{N} \) to \( X \) .
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Yes
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Corollary 1.2 If \( X \) is countable and there exists a surjection from \( X \) to \( Y \), then \( Y \) is countable.
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Indeed, the composition of two surjections is surjective.
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No
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Corollary 1.4 If \( Y \) is countable and there exists an injection from \( X \) to \( Y \), then \( X \) is countable.
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Proof. An injection \( f : X \rightarrow Y \) defines a bijection from \( X \) to \( f\left( X\right) \). If \( Y \) is countable, so is \( f\left( X\right) \), by the preceding corollary. Therefore \( X \) is countable.
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Yes
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Corollary 1.4 If \( Y \) is countable and there exists an injection from \( X \) to \( Y \), then \( X \) is countable.
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Proof. An injection \( f : X \rightarrow Y \) defines a bijection from \( X \) to \( f\left( X\right) \) . If \( Y \) is countable, so is \( f\left( X\right) \), by the preceding corollary. Therefore \( X \) is countable.
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Yes
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Proposition 1.6 If the sets \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) are countable, the Cartesian product \( X = {X}_{1} \times {X}_{2} \times \cdots \times {X}_{n} \) is countable.
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Proof. It is enough to prove the result for \( n = 2 \) and use induction. Suppose that \( {X}_{1} \) and \( {X}_{2} \) are countable, and let \( {f}_{1},{f}_{2} \) be surjections from \( \mathbb{N} \) to \( {X}_{1},{X}_{2} \) (whose existence is given by Proposition 1.1). The map \( \left( {{n}_{1},{n}_{2}}\right) \mapsto \) \( \left( {{f}_{1}\left( {n}_{1}\right) ,{f}_{2}\left( {n}_{2}\right) }\right) \) is then a surjection from \( {\mathbb{N}}^{2} \) to \( X \) . Since \( {\mathbb{N}}^{2} \) is countable, the proposition follows by Corollary 1.2.
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Yes
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Proposition 1.7 Let \( {\left( {X}_{i}\right) }_{i \in I} \) be a family of countable sets, indexed by a countable set \( I \) . The set \( X = \mathop{\bigcup }\limits_{{i \in I}}{X}_{i} \) is countable.
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Proof. If, for each \( i \in I \), we take a surjection \( {f}_{i} : \mathbb{N} \rightarrow {X}_{i} \), the map \( f : I \times \mathbb{N} \rightarrow X \) defined by \( f\left( {i, n}\right) = {f}_{i}\left( n\right) \) is a surjection. But \( I \times \mathbb{N} \) is countable.
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Yes
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Proposition 2.1 Every compact metric space is separable.
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Proof. If \( n \) is a strictly positive integer, the union of the balls \( B\left( {x,\frac{1}{n}}\right) \) , over \( x \in X \), covers \( X \) . By the Borel-Lebesgue property, \( X \) can be covered by a finite number of such balls: \( X = \mathop{\bigcup }\limits_{{j = 1}}^{{J}_{n}}B\left( {{x}_{j}^{n},\frac{1}{n}}\right) \) . It is then clear that the set\n\n\[ D = \left\{ {{x}_{j}^{n} : n \in {\mathbb{N}}^{ * },1 \leq j \leq {J}_{n}}\right\} \]\n\n is dense in \( X \) .
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Yes
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Proposition 2.2 Every \( \sigma \) -compact metric space is separable.
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This is an immediate consequence of Propositions 2.1 and 1.7.
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No
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Proposition 2.3 If \( X \) is a separable metric space and \( Y \) is a subset of \( X \), then \( Y \) is separable (in the induced metric).
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Proof. Let \( \left( {x}_{n}\right) \) be a dense sequence in \( X \) . Set\n\n\[ \mathcal{U} = \left\{ {\left( {n, p}\right) \in \mathbb{N} \times {\mathbb{N}}^{ * } : B\left( {{x}_{n},1/p}\right) \cap Y \neq \varnothing }\right\} .\n\]\n\nFor each \( \left( {n, p}\right) \in \mathcal{U} \), choose a point \( {x}_{n, p} \) of \( B\left( {{x}_{n},1/p}\right) \cap Y \) . We show that the family \( D = \left\{ {{x}_{n, p},\left( {n, p}\right) \in \mathcal{U}}\right\} \) (which is certainly countable) is dense in \( Y \) . To do this, choose \( x \in Y \) and \( \varepsilon > 0 \) . Let \( p \) be an integer such that \( 1/p < \varepsilon /2 \) ; clearly there exists an integer \( n \in \mathbb{N} \) such that \( d\left( {x,{x}_{n}}\right) < 1/p \) . But then \( x \in B\left( {{x}_{n},1/p}\right) \cap Y \) ; therefore \( \left( {n, p}\right) \in \mathcal{U} \) and \( d\left( {x,{x}_{n, p}}\right) < 2/p < \varepsilon \) .
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Yes
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Proposition 2.5 A normed space is separable if and only if it contains a countable fundamental family of vectors.
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Proof. The condition is certainly necessary, since a dense family of vectors is fundamental. Conversely, let \( D \) be a countable fundamental family of vectors in a normed space \( E \) . Let \( \mathcal{D} \) be the set of linear combinations of elements of \( D \) with coefficients in the field \( Q = \mathbb{Q} \) (if \( \mathbb{K} = \mathbb{R} \) ) or \( \mathbb{Q} + i\mathbb{Q} \) (if \( \mathbb{K} = \mathbb{C} \) ). Then \( \mathcal{D} \) is dense in \( E \), because its closure contains the closure of the vector space generated by \( D \), which is \( E \) . On the other hand, \( \mathcal{D} \) is countable, because it is the image of the countable set \( \mathop{\bigcup }\limits_{{n \in {\mathbb{N}}^{ * }}}\left( {{Q}^{n} \times {D}^{n}}\right) \) under the map \( f \) defined by\n\n\[ f\left( {{\lambda }_{1},\ldots ,{\lambda }_{n},{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{\lambda }_{j}{x}_{j} \]\n\n
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Yes
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Proposition 2.6 A normed space is separable if and only if it has a countable topological basis.
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Proof. The \
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No
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Theorem 3.1 Let \( {\left( {X}_{p},{d}_{p}\right) }_{p \in \mathbb{N}} \) be a sequence of metric spaces, and, for every \( p \in \mathbb{N} \), let \( {\left( {x}_{n, p}\right) }_{n \in \mathbb{N}} \) be a sequence in \( {X}_{p} \) . If, for every \( p \in \mathbb{N} \), the set \( \left\{ {{x}_{n, p} : n \in \mathbb{N}}\right\} \) is relatively compact in \( {X}_{p} \), there exists a strictly increasing function \( \varphi : \mathbb{N} \rightarrow \mathbb{N} \) such that for every \( p \in \mathbb{N} \) the sequence \( {\left( {x}_{\varphi \left( n\right), p}\right) }_{n \in \mathbb{N}} \) converges in \( {X}_{p} \) .
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Proof. Thanks to the assumption of relative compactness, one can inductively construct a decreasing subsequence \( \left( {A}_{n}\right) \) of infinite subsets of \( \mathbb{N} \) such that, for every \( p \in \mathbb{N} \), the sequence \( {\left( {x}_{n, p}\right) }_{n \in {A}_{p}} \) converges in \( {X}_{p} \) . The diagonal procedure consists in defining the map \( \varphi \) by setting\n\n\[ \varphi \left( p\right) = \text{the}\left( {p + 1}\right) \text{-st element of}{A}_{p}\text{.} \]\n\nThus \( \varphi \left( {p + 1}\right) \) is strictly greater than the \( \left( {p + 1}\right) \) -st element of \( {A}_{p + 1} \), which in turn is greater than the \( \left( {p + 1}\right) \) -st element of \( {A}_{p} \), which is \( \varphi \left( p\right) \) . Thus \( \varphi \) is strictly increasing. Moreover, for every \( p \in \mathbb{N} \) the sequence \( {\left( {x}_{\varphi \left( n\right), p}\right) }_{n \geq p} \) is a subsequence of the sequence \( {\left( {x}_{n, p}\right) }_{n \in {A}_{p}} \), because, if \( n \geq p \), we have \( \varphi \left( n\right) \in {A}_{n} \subset {A}_{p} \) . Therefore the sequence \( {\left( {x}_{\varphi \left( n\right), p}\right) }_{n \in \mathbb{N}} \) converges.
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Yes
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Corollary 3.2 (Tychonoff’s Theorem) If \( {\left( {X}_{p}\right) }_{p \in \mathbb{N}} \) is a sequence of compact metric spaces and \( X = \mathop{\prod }\limits_{{p \in \mathbb{N}}}{X}_{p} \) is the product space (with the product distance), \( X \) is compact.
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This follows immediately from the definition of the product metric, from Theorem 3.1, and from the characterization of compact sets by the Bolzano-Weierstrass property.
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No
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Theorem 3.3 Let \( X \) be a metric space. Every relatively compact subset of \( X \) is precompact. The converse is true if \( X \) is complete.
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Proof. The first statement follows directly from the definitions, from the Borel-Lebesgue property of compact sets, and from the fact that \( A \subset X \) implies \( \bar{A} \subset \mathop{\bigcup }\limits_{{x \in X}}B\left( {x,\varepsilon }\right) \) for every \( \varepsilon > 0 \) .\n\nNow suppose that \( X \) is complete and that \( A \subset X \) is precompact. Let \( {\left( {x}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence of points in \( A \) . To prove that it has a convergent subsequence, it is enough to find a Cauchy subsequence. For every \( p \in \mathbb{N} \) , let \( {A}_{1}^{p},\ldots ,{A}_{{N}_{p}}^{p} \) be subsets of \( A \) of diameter at most \( 1/\left( {p + 1}\right) \) and covering \( A \) . We will construct by induction a decreasing sequence \( {\left( {B}_{p}\right) }_{p \in \mathbb{N}} \) of infinite subsets of \( \mathbb{N} \) such that, for every \( p \in \mathbb{N} \), there is an integer \( j \leq {N}_{p} \) for which \( {\left\{ {x}_{p}\right\} }_{p \in {B}_{p}} \subset {A}_{j}^{p}. \n\nConstruction of \( {B}_{0} \) : since all terms of the sequence \( {\left( {x}_{n}\right) }_{n \in \mathbb{N}} \) (of which there are infinitely many) are contained in \( A \), which is the union of the finitely many sets \( {A}_{1}^{0},\ldots ,{A}_{{N}_{0}}^{0} \), there is at least one of these sets, say \( {A}_{{j}_{0}}^{0} \) , containing infinitely many terms \( {x}_{n} \) . (This is the pigeonhole principle.) We then set \( {B}_{0} = \left\{ {n \in \mathbb{N} : {x}_{n} \in {A}_{{j}_{0}}^{0}}\right\} \) .\n\nTo construct \( {B}_{p + 1} \) from \( {B}_{p} \), the idea is the same: the terms of the subsequence \( {\left( {x}_{n}\right) }_{n \in {B}_{p}} \) are all contained in the union of the finitely many sets \( {A}_{1}^{p + 1},\ldots ,{A}_{{N}_{p + 1}}^{p + 1} \) ; therefore at least one of the sets contains infinitely many terms of the subsequence. We define \( {B}_{p + 1} \) as the set of indices of these terms.\n\nHaving constructed the \( {B}_{p} \), we define a strictly increasing function \( \varphi \) : \( \mathbb{N} \rightarrow \mathbb{N} \) by setting\n\n\[ \varphi \left( p\right) = \text{the}\left( {p + 1}\right) \text{-st element of}{B}_{p}\text{.} \]\n\nThen, for every \( p \in \mathbb{N} \) and every integer \( n \geq p \), we have \( \varphi \left( n\right) \in {B}_{p} \) . By the construction of the \( {B}_{p} \), we see that\n\n\[ d\left( {{x}_{\varphi \left( n\right) },{x}_{\varphi \left( {n}^{\prime }\right) }}\right) \leq \frac{1}{p + 1}\;\text{ for all }n,{n}^{\prime } \geq p. \]\n\nThus the sequence \( \left( {x}_{\varphi \left( n\right) }\right) \) is a Cauchy sequence.
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Yes
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Proposition 4.1 Consider a normed space \( E \), a fundamental family \( D \) in \( E \), and a Banach space \( F \) . Consider also a bounded sequence \( {\left( {T}_{n}\right) }_{n \in \mathbb{N}} \) of elements of \( L\left( {E, F}\right) \) . If, for every \( x \in D \), the sequence \( {\left( {T}_{n}x\right) }_{n \in \mathbb{N}} \) converges in \( F \), there exists an operator \( T \in L\left( {E, F}\right) \) such that
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Proof. Let \( M > 0 \) be such that \( \begin{Vmatrix}{T}_{n}\end{Vmatrix} \leq M \) for all \( n \in \mathbb{N} \) . It is clear that the sequence \( \left( {{T}_{n}x}\right) \) converges for any element \( x \) of the vector space \( \left\lbrack D\right\rbrack \) generated by \( D \) . Now take \( x \in E \) and \( \varepsilon > 0 \) . Since \( D \) is a fundamental family, there exists \( y \in \left\lbrack D\right\rbrack \) such that \( \parallel x - y\parallel \leq \varepsilon /\left( {3M}\right) \) . The sequence \( \left( {{T}_{n}y}\right) \) converges; therefore there is a positive integer \( N \) such that \( \begin{Vmatrix}{{T}_{n}y - {T}_{p}y}\end{Vmatrix} \leq \) \( \varepsilon /3 \) for all \( n, p \geq N \) . By the triangle inequality we deduce that, for any \( n, p \geq N \) ,\n\n\[ \begin{Vmatrix}{{T}_{n}x - {T}_{p}x}\end{Vmatrix} \leq \begin{Vmatrix}{{T}_{n}x - {T}_{n}y}\end{Vmatrix} + \begin{Vmatrix}{{T}_{n}y - {T}_{p}y}\end{Vmatrix} + \begin{Vmatrix}{{T}_{p}y - {T}_{p}x}\end{Vmatrix} \leq \varepsilon .\n\]\n\nThus \( \left( {{T}_{n}x}\right) \) is a Cauchy sequence in \( F \), and therefore convergent. For every \( x \in E \) we then set \( {Tx} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{T}_{n}x \) . The map \( T \) thus defined is certainly linear, and, since \( \parallel {Tx}\parallel \leq M\parallel x\parallel \) for all \( x \in E \), it is also continuous.
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Yes
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Proposition 4.3 Consider normed spaces \( E \) and \( F \), a fundamental set \( D \) in \( E \), a bounded sequence \( \left( {T}_{n}\right) \) in \( L\left( {E, F}\right) \) and a map \( T \in L\left( {E, F}\right) \) . If the sequence \( \left( {{T}_{n}x}\right) \) converges toward \( {Tx} \) for every point \( x \in D \), it does also for every \( x \in E \) .
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Proof. By taking differences we can suppose that \( T = 0 \) . Set\n\n\[ M = \mathop{\sup }\limits_{{n \in \mathbb{N}}}\begin{Vmatrix}{T}_{n}\end{Vmatrix} \]\n\nand take \( x \in E \) . For every \( y \in \left\lbrack D\right\rbrack \), we have\n\n\[ \begin{Vmatrix}{{T}_{n}x}\end{Vmatrix} \leq M\parallel x - y\parallel + \begin{Vmatrix}{{T}_{n}y}\end{Vmatrix}. \]\n\nSince \( {T}_{n}y \rightarrow 0 \), we get \( \lim \mathop{\sup }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{{T}_{n}x}\end{Vmatrix} \leq M\parallel x - y\parallel \) . This holds for every \( y \in \left\lbrack D\right\rbrack \), and \( \left\lbrack D\right\rbrack \) is dense in \( E \) ; therefore\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{{T}_{n}x}\end{Vmatrix} = 0 \]
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Yes
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Proposition 1.1 \( C\left( X\right) \) is a separable Banach space.
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Proof. The reader can check that \( C\left( X\right) \) is a Banach space. We show separability. Since \( X \) is precompact, for every \( n \in {\mathbb{N}}^{ * } \) there exist finitely many points \( {x}_{1}^{n},\ldots ,{x}_{{N}_{n}}^{n} \) of \( X \) such that \( X = \mathop{\bigcup }\limits_{{j = 1}}^{{N}_{n}}B\left( {{x}_{j}^{n},1/n}\right) \) . We therefore set, for \( j \leq {N}_{n}, \)\n\n\[ \n{\varphi }_{n, j}\left( x\right) = \frac{{\left( 1/n - d\left( x,{x}_{j}^{n}\right) \right) }^{ + }}{\mathop{\sum }\limits_{{k = 1}}^{{N}_{n}}{\left( 1/n - d\left( x,{x}_{k}^{n}\right) \right) }^{ + }}. \n\]\n\nFrom the choice of the points \( {x}_{j}^{n} \), we see that the denominator does not vanish for any \( x \in X \) . Therefore, \( {\varphi }_{n, j} \in {C}^{ + }\left( X\right) \), \n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}{\varphi }_{n, j} = 1,\;\text{ and }\;{\varphi }_{n, j}\left( x\right) = 0\;\text{ if }d\left( {x,{x}_{j}^{n}}\right) \geq 1/n. \n\]\n\nThe set \( \left\{ {{\varphi }_{n, j} : n \in {\mathbb{N}}^{ * }}\right. \) and \( \left. {1 \leq j \leq {N}_{n}}\right\} \) is certainly countable. We will show that it is a fundamental family in \( C\left( X\right) \) ; this suffices by Proposition 2.5 on page 9.\n\nTake \( f \in C\left( X\right) \) and \( \varepsilon > 0 \) . Since \( X \) is compact, the function \( f \) is uniformly continuous on \( X \) . Take \( \eta > 0 \) such that, for all \( x, y \in X \) with \( d\left( {x, y}\right) < \eta \), we have \( \left| {f\left( x\right) - f\left( y\right) }\right| < \varepsilon \) . Let \( n \in \mathbb{N} \) be such that \( 1/n < \eta \) . For every \( x \in X \), \n\n\[ \n\left| {f\left( x\right) - \mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}f\left( {x}_{j}^{n}\right) {\varphi }_{n, j}\left( x\right) }\right| = \left| {\mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}\left( {f\left( x\right) - f\left( {x}_{j}^{n}\right) }\right) {\varphi }_{n, j}\left( x\right) }\right| \n\]\n\n\[ \n\leq \mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}\left| {f\left( x\right) - f\left( {x}_{j}^{n}\right) }\right| {\varphi }_{n, j}\left( x\right) \n\]\n\nSince \( {\varphi }_{n, j} \) vanishes outside the ball \( B\left( {{x}_{j}^{n},1/n}\right) \), and so outside \( B\left( {{x}_{j}^{n},\eta }\right) \), we see that, for every \( x \in X \), \n\n\[ \n\left| {f\left( x\right) - f\left( {x}_{j}^{n}\right) }\right| {\varphi }_{n, j}\left( x\right) \leq \varepsilon {\varphi }_{n, j}\left( x\right) \n\]\n\nThus, for every \( x \in X \), \n\n\[ \n\left| {f\left( x\right) - \mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}f\left( {x}_{j}^{n}\right) {\varphi }_{n, j}\left( x\right) }\right| \leq \varepsilon \mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}{\varphi }_{n, j}\left( x\right) = \varepsilon . \n\]\n\nIt follows that \n\n\[ \n\begin{Vmatrix}{f - \mathop{\sum }\limits_{{j = 1}}^{{N}_{n}}f\left( {x}_{j}^{n}\right) {\varphi }_{n, j}}\end{Vmatrix} \leq \varepsilon \n\]\n\nwhich concludes the proof.
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Yes
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Proposition 1.2 (Dini’s Lemma) \( \operatorname{Let}{\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) be an increasing sequence in \( {C}^{\mathbb{R}}\left( X\right) \) (this means that \( {f}_{n} \leq {f}_{n + 1} \) for all \( n \) ). If the sequence \( \left( {f}_{n}\right) \) converges pointwise to a function \( f \in C\left( X\right) \), it also converges uniformly to \( f \) .
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Proof. Take \( \varepsilon > 0 \) . For every \( n \in \mathbb{N} \) we set \( {\Omega }_{n} = \left\{ {x \in X : {f}_{n}\left( x\right) > }\right. \) \( f\left( x\right) - \varepsilon \} \) . Clearly, \( \left( {\Omega }_{n}\right) \) is an increasing sequence of open subsets in \( X \) whose union is \( X \) . By the Borel-Lebesgue property, there is an integer \( N \) such that \( {\Omega }_{N} = X \), so that \( {f}_{N}\left( x\right) > f\left( x\right) - \varepsilon \) for all \( x \in X \) . Thus, for every integer \( n \geq N \), we have \( f\left( x\right) - \varepsilon < {f}_{n}\left( x\right) \leq f\left( x\right) \) for all \( x \in X \) . This proves that \( \begin{Vmatrix}{f - {f}_{n}}\end{Vmatrix} \leq \varepsilon \) .
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Yes
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Lemma 2.1 Suppose \( X \) has at least two elements. Let \( H \) be a subset of \( {C}^{\mathbb{R}}\left( X\right) \) satisfying these two conditions:\n\na. For all \( u, v \in H \), the functions \( \sup \left( {u, v}\right) \) and \( \inf \left( {u, v}\right) \) also lie in \( H \) .\n\nb. If \( {x}_{1},{x}_{2} \) are distinct points in \( X \) and \( {\alpha }_{1},{\alpha }_{2} \) are real numbers, there exists \( u \in H \) such that \( u\left( {x}_{1}\right) = {\alpha }_{1} \) and \( u\left( {x}_{2}\right) = {\alpha }_{2} \) .\n\nThen \( H \) is dense in \( {C}^{\mathbb{R}}\left( X\right) \) .
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Proof. Take \( f \in {C}^{\mathbb{R}}\left( X\right) \) and \( \varepsilon > 0 \) . We want to find an element of \( H \) that is \( \varepsilon \) -close to \( f \) . First fix \( x \in X \) . By assumption \( \mathrm{b} \), for every \( y \neq x \) there exists \( {u}_{y} \in H \) such that \( {u}_{y}\left( x\right) = f\left( x\right) \) and \( {u}_{y}\left( y\right) = f\left( y\right) \) .\n\nFor \( y \neq x \), set \( {O}_{y} = \left\{ {{x}^{\prime } \in X : {u}_{y}\left( {x}^{\prime }\right) > f\left( {x}^{\prime }\right) - \varepsilon }\right\} \) . This is an open set that contains \( y \) and \( x \) ; therefore \( X = \mathop{\bigcup }\limits_{{y \neq x}}{O}_{y} \) . By the Borel-Lebesgue property, \( X \) can be covered by finitely many sets \( {O}_{y} : X = \mathop{\bigcup }\limits_{{j = 1}}^{r}{O}_{{y}_{j}} \), with \( {y}_{j} \neq x \) for all \( j \) . Now set \( {v}_{x} = \sup \left( {{u}_{{y}_{1}},\ldots ,{u}_{{y}_{r}}}\right) \) . A simple inductive argument, using assumption a, shows that \( {v}_{x} \in H \) . On the other hand,\n\n\[ \n{v}_{x}\left( x\right) = f\left( x\right) \;\text{ and }\;{v}_{x}\left( {x}^{\prime }\right) > f\left( {x}^{\prime }\right) - \varepsilon \text{ for all }{x}^{\prime } \in X.\n\]\n\nNow make \( x \) vary and set, for each \( x \in X \) ,\n\n\[ \n{\Omega }_{x} = \left\{ {{x}^{\prime } \in X : {v}_{x}\left( {x}^{\prime }\right) < f\left( {x}^{\prime }\right) + \varepsilon }\right\} .\n\]\n\nThus \( {\Omega }_{x} \) is an open subset of \( X \) containing \( x \) ; a new application of the Borel-Lebesgue property allows us to choose finitely many points \( {x}_{1},\ldots ,{x}_{p} \) of \( X \) such that \( {\Omega }_{{x}_{1}},\ldots ,{\Omega }_{{x}_{p}} \) cover \( X \) . Finally, set \( v = \inf \left( {{v}_{{x}_{1}},\ldots ,{v}_{{x}_{p}}}\right) \) . Then \( v \in H \) and \( f - \varepsilon < v < f + \varepsilon \) ; that is, \( \parallel f - v\parallel \leq \varepsilon \) .
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Yes
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Theorem 2.2 If \( H \) is a separating vector subspace of \( {C}^{\mathbb{R}}\left( X\right) \) that is a lattice and contains the constants, then \( H \) is dense in \( {C}^{\mathbb{R}}\left( X\right) \) .
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Proof. If \( X \) has a single element, the result is clear. Suppose \( X \) has at least two elements; we just need to check assumption b of the lemma. Let \( {x}_{1} \) and \( {x}_{2} \) be distinct elements of \( X \) . Since \( H \) is separating, there exists \( h \in H \) such that \( h\left( {x}_{1}\right) \neq h\left( {x}_{2}\right) \) . If \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are real numbers, the system of equations\n\n\[ \left\{ \begin{array}{l} {\lambda h}\left( {x}_{1}\right) + \mu = {\alpha }_{1} \\ {\lambda h}\left( {x}_{2}\right) + \mu = {\alpha }_{2} \end{array}\right. \]\n\nclearly has a unique solution \( \left( {\lambda ,\mu }\right) \in {\mathbb{R}}^{2} \) . For such \( \left( {\lambda ,\mu }\right) \), we see that \( \left( {{\lambda h} + \mu }\right) \left( {x}_{1}\right) = {\alpha }_{1} \) and \( \left( {{\lambda h} + \mu }\right) \left( {x}_{2}\right) = {\alpha }_{2} \) ; moreover, \( {\lambda h} + \mu \in H \), since \( H \) is a vector space containing constants.
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Yes
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Theorem 2.3 (Stone-Weierstrass Theorem, real case) Every separating subalgebra of \( {C}^{\mathbb{R}}\left( X\right) \) containing the constant functions is dense in \( {C}^{\mathbb{R}}\left( X\right) \) .
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Proof. If \( H \) is a separating subalgebra of \( {C}^{\mathbb{R}}\left( X\right) \) containing the constants, so is its closure \( \bar{H} \) . Therefore it suffices to show that \( \bar{H} \) is a lattice and to apply Theorem 2.2. Thus, let \( f \) be a nonzero element of \( \bar{H} \) . We saw in the example on page 29 that there exists a sequence \( \left( {P}_{n}\right) \) of polynomials over \( \mathbb{R} \) that converges uniformly on \( \left\lbrack {-1,1}\right\rbrack \) to the function \( x \mapsto \left| x\right| \) . But then the sequence of functions \( \left( {{P}_{n}\left( {f/\parallel f\parallel }\right) }\right) \) converges uniformly to \( \left| f\right| /\parallel f\parallel \), so \( \left| f\right| \) is the uniform limit of the sequence \( \left( {\parallel f\parallel {P}_{n}\left( {f/\parallel f\parallel }\right) }\right) \) . Since \( \bar{H} \) is a subalgebra of \( {C}^{\mathbb{R}}\left( X\right) \), all terms in this sequence are in \( \bar{H} \) ; therefore so is their uniform limit \( \left| f\right| \) . This shows that \( \bar{H} \) is a lattice.
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Yes
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Theorem 2.4 (Stone–Weierstrass Theorem, complex case) Every separating subalgebra \( H \) of \( {C}^{\mathbb{C}}\left( X\right) \) that is self-conjugate and contains the constant functions is dense in \( {C}^{\mathbb{C}}\left( X\right) \) .
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Proof. Set \( {H}_{\mathbb{R}} = \{ h \in H : h\left( x\right) \in \mathbb{R} \) for all \( x \in X\} \) . Clearly, \( {H}_{\mathbb{R}} \) is a subalgebra of \( {C}^{\mathbb{R}}\left( X\right) \) containing the constants. Now, if \( f \in H \), the real and imaginary parts of \( f \) lie in \( {H}_{\mathbb{R}} \), since \( H \) is self-conjugate and \( \operatorname{Re}f = \) \( \left( {f + \bar{f}}\right) /2,\operatorname{Im}f = \left( {f - \bar{f}}\right) /\left( {2i}\right) \) . If \( {x}_{1} \) and \( {x}_{2} \) are distinct points in \( X \), there exists by assumption \( h \in H \) such that \( h\left( {x}_{1}\right) \neq h\left( {x}_{2}\right) \) . Therefore there exists \( g \in {H}_{\mathbb{R}} \) such that \( g\left( {x}_{1}\right) \neq g\left( {x}_{2}\right) \) : just take \( g = \operatorname{Re}h \) or \( g = \operatorname{Im}h \) as needed. It follows that \( {H}_{\mathbb{R}} \) is separating, hence dense in \( {C}^{\mathbb{R}}\left( X\right) \), by Theorem 2.3. Since \( {C}^{\mathbb{C}}\left( X\right) = {C}^{\mathbb{R}}\left( X\right) + i{C}^{\mathbb{R}}\left( X\right) \) and \( H \) contains \( {H}_{\mathbb{R}} + i{H}_{\mathbb{R}} \), the proof is complete.
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Yes
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Lemma 2.5 The map from \( {C}^{\mathbb{C}}\left( \mathbb{U}\right) \) to \( {C}_{2\pi }^{\mathbb{C}} \) that associates to \( \varphi \in \) \( {C}^{\mathbb{C}}\left( \mathbb{U}\right) \) the function \( f \) given by \( f\left( \theta \right) = \varphi \left( {e}^{i\theta }\right) \) for every real \( \theta \) is a surjective isometry.
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Proof. Only the surjectivity requires proof. For \( z \in \mathbb{U} \), denote by \( \arg z \) some real number such that \( {e}^{i\arg z} = z \) . We know that \( \arg z \) is defined modulo \( {2\pi } \) and that there exist choices of \( \arg z \) that vary continuously in the neighborhood of a given point (for example, if \( {z}_{0} \in \mathbb{U} \) and \( z \in \mathbb{U} \) with \( \left| {z - {z}_{0}}\right| < 1 \), we can take \( \arg z = \arg {z}_{0} + \operatorname{Arccos}\operatorname{Re}\left( {z/{z}_{0}}\right) \) ). Thus, if \( f \in {C}_{2\pi }^{\mathbb{C}} \), the function \( \varphi \) defined by \( \varphi \left( z\right) = f\left( {\arg z}\right) \) is well-defined and continuous in \( \mathbb{U} \), and \( f\left( \theta \right) = \varphi \left( {e}^{i\theta }\right) \) for all \( \theta \in \mathbb{R} \) .
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Yes
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Proposition 3.1 A subset of \( C\left( X\right) \) is equicontinuous if and only if it is uniformly equicontinuous.
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Proof. It is enough to show necessity. Let \( H \) be an equicontinuous subset of \( C\left( X\right) \), and let \( \varepsilon > 0 \) be a real number. By assumption, for every \( x \in X \) there exists \( {\eta }_{x} > 0 \) such that \( \left| {h\left( y\right) - h\left( x\right) }\right| < \varepsilon /2 \) whenever \( h \in H \) and \( d\left( {x, y}\right) < {\eta }_{x} \) . By the Borel-Lebesgue property, we can choose finitely many points \( {x}_{1},\ldots ,{x}_{r} \) such that the balls \( B\left( {{x}_{j},{\eta }_{{x}_{j}}/2}\right) \) cover \( X \) . Now let \( \eta \) be the smallest of the \( {\eta }_{{x}_{j}}/2 \), and let \( x \) and \( y \) be points in \( X \) such that \( d\left( {x, y}\right) < \eta \) . Choosing \( j \) such that \( x \in B\left( {{x}_{j},{\eta }_{{x}_{j}}/2}\right) \), we see that \( x, y \in B\left( {{x}_{j},{\eta }_{{x}_{j}}}\right) \), so\n\n\[ \left| {h\left( y\right) - h\left( x\right) }\right| \leq \left| {h\left( y\right) - h\left( {x}_{j}\right) }\right| + \left| {h\left( x\right) - h\left( {x}_{j}\right) }\right| < \varepsilon \;\text{ for all }h \in H. \]
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Yes
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Proposition 3.2 Let \( \left( {f}_{n}\right) \) be an equicontinuous sequence in \( C\left( X\right) \) and let \( D \) be a dense subset of \( X \) . If, for all \( x \in D \), the sequence of numbers \( \left( {{f}_{n}\left( x\right) }\right) \) converges, the sequence of functions \( \left( {f}_{n}\right) \) converges uniformly to a function \( f \in C\left( X\right) \) .
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Proof. It suffices to show that \( \left( {f}_{n}\right) \) is a Cauchy sequence in \( C\left( X\right) \) . To do this, take \( \varepsilon > 0 \) . By assumption, there exists \( \eta > 0 \) such that, whenever \( d\left( {x, y}\right) < \eta , \)\n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{n}\left( y\right) }\right| < \varepsilon /5\text{ for all }n \in \mathbb{N}. \]\n\nSince \( X \) is precompact, il can be covered by finitely many balls of radius \( \eta \) : \( X = \mathop{\bigcup }\limits_{{j = 0}}^{r}B\left( {{x}_{j},\eta }\right) \) . Since \( D \) is dense, each ball \( B\left( {{x}_{j},\eta }\right) \) contains at least one point \( {y}_{j} \) from \( D \) . Since, by assumption, the sequences \( {\left( {f}_{n}\left( {y}_{j}\right) \right) }_{n \in \mathbb{N}} \) are Cauchy sequences, there exists a positive integer \( N \) such that, for any integer \( j \leq r \), \n\n\[ \left| {{f}_{n}\left( {y}_{j}\right) - {f}_{p}\left( {y}_{j}\right) }\right| < \varepsilon /5\text{ for all }n, p \geq N. \]\n\nNow let \( x \) be a point in \( X \), and let \( j \) be an integer such that \( x \in B\left( {{x}_{j},\eta }\right) \) . Then, for \( n, p \geq N \), \n\n\[ \left| {{f}_{n}\left( x\right) - {f}_{p}\left( x\right) }\right| \leq \left| {{f}_{n}\left( x\right) - {f}_{n}\left( {x}_{j}\right) }\right| + \left| {{f}_{n}\left( {y}_{j}\right) - {f}_{n}\left( {x}_{j}\right) }\right| + \left| {{f}_{n}\left( {y}_{j}\right) - {f}_{p}\left( {y}_{j}\right) }\right| \]\n\n\[ + \left| {{f}_{p}\left( {y}_{j}\right) - {f}_{p}\left( {x}_{j}\right) }\right| + \left| {{f}_{p}\left( x\right) - {f}_{p}\left( {x}_{j}\right) }\right| < \varepsilon . \]\n\nThus, \( \begin{Vmatrix}{{f}_{n} - {f}_{p}}\end{Vmatrix} < \varepsilon \) for all \( n, p \geq N \), and \( \left( {f}_{n}\right) \) is a Cauchy sequence.
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Yes
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Theorem 3.3 (Ascoli) A subset of \( C\left( X\right) \) is relatively compact in \( C\left( X\right) \) if and only if it is bounded and equicontinuous.
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Proof. For the \
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No
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Proposition 3.4 The image under \( T \) of the closed unit ball of \( C\left( Y\right) \) is a relatively compact subset of \( C\left( X\right) \) .
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Proof. It is clear that \( T\left( {\bar{B}\left( {C\left( Y\right) }\right) }\right) \) is bounded by\n\n\[ M = \mu \left( Y\right) \mathop{\max }\limits_{{\left( {x, y}\right) \in X \times Y}}\left| {K\left( {x, y}\right) }\right| .\n\]\n\nOn the other hand, \( K \) is uniformly continuous on \( X \times Y \) ; in particular, for all \( \varepsilon \), there exists \( \eta > 0 \) such that\n\n\[ \left| {K\left( {{x}_{1}, y}\right) - K\left( {{x}_{2}, y}\right) }\right| < \varepsilon \;\text{ for all }y \in Y\text{ and }{x}_{1},{x}_{2} \in X\text{ with }d\left( {{x}_{1},{x}_{2}}\right) < \eta .\n\]\n\nThus, for all \( f \in \bar{B}\left( {C\left( Y\right) }\right) \), we have \( \left| {{Tf}\left( {x}_{1}\right) - {Tf}\left( {x}_{2}\right) }\right| \leq \mu \left( Y\right) \varepsilon \) . Therefore the subset \( T\left( {\bar{B}\left( {C\left( Y\right) }\right) }\right) \) of \( C\left( X\right) \) is equicontinuous, and we can apply Ascoli's Theorem.
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Yes
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Theorem 1.1 (F. Riesz) Let \( X \) be a normed space, with open unit ball \( B \) and closed unit ball \( \bar{B} \). The following properties are equivalent:\n\ni. \( X \) is finite-dimensional.\n\nii. \( X \) is locally compact.\n\niii. \( \bar{B} \) is compact.\n\niv. \( B \) is precompact.
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Proof. Property i implies ii because closed balls in a finite-dimensional normed space are compact. If ii is true, there exists \( r > 0 \) such that \( \bar{B}\left( {0, r}\right) = r\bar{B} \) is compact; this implies iii. That iii implies iv is obvious. Thus the only nontrivial part of the theorem is iv \( \Rightarrow \) i.\n\nSuppose that \( B \) is precompact. Then there is a finite subset \( A \) of \( X \) such that\n\n\[ B \subset \mathop{\bigcup }\limits_{{x \in A}}B\left( {x,\frac{1}{2}}\right) = A + \frac{1}{2}B. \]\n\nLet \( Y \) be the (finite-dimensional) vector space generated by \( A \) ; then \( B \subset \) \( Y + {2}^{-1}B \) . One can easily show by induction that, for any integer \( n \geq 1 \) , we have \( B \subset Y + {2}^{-n}B \), and therefore\n\n\[ B \subset \mathop{\bigcap }\limits_{{n \geq 1}}\left( {Y + {2}^{-n}B}\right) \]\n\nIn particular, if \( x \in B \), there exists for all \( n \geq 1 \) a \( {y}_{n} \in Y \) such that \( \begin{Vmatrix}{x - {y}_{n}}\end{Vmatrix} < {2}^{-n} \) . We deduce that \( B \subset \bar{Y} \) . Since \( Y \) is finite-dimensional, hence complete, hence closed in \( X \), it follows that \( B \subset Y \) and, by homogeneity, \( X = Y \) .
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Yes
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Proposition 1.2 If \( X \) is a locally compact space, there exists for every \( x \in X \) and for every neighborhood \( V \) of \( x \) a real number \( r > 0 \) such that \( \bar{B}\left( {x, r}\right) \) is compact and \( \bar{B}\left( {x, r}\right) \subset V \) .
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Proof. Just choose \( r = \min \left( {{r}^{\prime },{r}^{\prime \prime }}\right) \), where \( {r}^{\prime } \) and \( {r}^{\prime \prime } \) are such that \( \bar{B}\left( {x,{r}^{\prime }}\right) \) is compact and \( \bar{B}\left( {x,{r}^{\prime \prime }}\right) \subset V \) .
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Yes
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Corollary 1.3 Let \( X \) be locally compact. If \( O \) is open in \( X \) and \( F \) is closed in \( X \), the intersection \( Y = O \cap F \) (with the induced metric) is locally compact.
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Proof. Take \( x \in Y \) . By the preceding proposition, there exists \( r > 0 \) such that \( \bar{B}\left( {x, r}\right) \) is compact and contained in \( O \) . Then \( \bar{B}\left( {x, r}\right) \cap Y = \bar{B}\left( {x, r}\right) \cap F \) is compact.
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Yes
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Corollary 1.4 Consider a locally compact space \( X \), a compact subset \( K \) of \( X \), and open subsets \( {O}_{1},\ldots ,{O}_{n} \) of \( X \) covering \( K \) . There exist compact sets \( {K}_{1},\ldots ,{K}_{n} \) with \( {K}_{j} \subset {O}_{j} \) for each \( j \) and such that\n\n\[ K \subset \mathop{\bigcup }\limits_{{j = 1}}^{n}{\overset{ \circ }{K}}_{j} \]
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Proof. By Proposition 1.2, for all points \( x \) of \( K \) there exists \( j \in \{ 1,\ldots, n\} \) and a compact set \( {K}_{x} \) such that \( x \in {\mathring{K}}_{x} \subset {K}_{x} \subset {O}_{j} \) . By the Borel-Lebesgue property, \( K \) can be covered by finitely many of these interiors:\n\n\[ K \subset \mathop{\bigcup }\limits_{{i = 1}}^{p}{\overset{ \circ }{K}}_{{x}_{i}} \]\n\nNow set \( {K}_{j} = \mathop{\bigcup }\limits_{{{K}_{{x}_{i}} \subset {O}_{j}}}{K}_{{x}_{i}} \) for \( 1 \leq j \leq n \) . Then\n\n\[ \mathop{\bigcup }\limits_{{j = 1}}^{n}{\mathring{K}}_{j} \supset \mathop{\bigcup }\limits_{{j = 1}}^{n}\mathop{\bigcup }\limits_{{{K}_{{x}_{i}} \subset {O}_{j}}}{\mathring{K}}_{{x}_{i}} = \mathop{\bigcup }\limits_{{i = 1}}^{p}{\mathring{K}}_{{x}_{i}} \supset K \]\n\nand, sure enough, \( {K}_{j} \subset {O}_{j} \) .
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Yes
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Proposition 1.5 Let \( X \) be a locally compact space. The following properties are equivalent:\n\ni. \( X \) is separable.\n\nii. \( X \) is \( \sigma \) -compact.\n\niii. There exists a sequence \( \left( {K}_{n}\right) \) of compact sets covering \( X \) and such that \( {K}_{n} \subset {\overset{ \circ }{K}}_{n + 1} \) for all \( n \in \mathbb{N} \) .
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Proof. It is clear that iii implies ii. The implication ii \( \Rightarrow \) i is a particular case of Proposition 2.2 on page 8.\n\nNow suppose that \( X \) is separable and let \( \left( {x}_{n}\right) \) be a sequence dense in \( X \) . Set \( A = \left\{ {\left( {n, p}\right) \in \mathbb{N} \times {\mathbb{N}}^{ * } : \bar{B}\left( {{x}_{n},1/p}\right) \text{is compact}}\right\} \) ; we will show that the family \( \mathcal{F} = {\left( \bar{B}\left( {x}_{n},1/p\right) \right) }_{\left( {n, p}\right) \in A} \) covers \( X \) . Take \( x \in X \) and let \( r > 0 \) be such that \( \bar{B}\left( {x, r}\right) \) is compact. Then take \( p \in {\mathbb{N}}^{ * } \) such that \( 1/p < r/2 \) and \( n \in \mathbb{N} \) such that \( d\left( {x,{x}_{n}}\right) < 1/p \) . One sees that \( x \in \bar{B}\left( {{x}_{n},1/p}\right) \subset \) \( B\left( {x,2/p}\right) \subset \bar{B}\left( {x, r}\right) \) . Therefore \( \bar{B}\left( {{x}_{n},1/p}\right) \) is compact and \( x \) belongs to some element of \( \mathcal{F} \) . This shows that i implies ii.\n\nFinally, we show that ii implies iii. Suppose that \( X \) is \( \sigma \) -compact and let \( \left( {L}_{n}\right) \) be a sequence of compact sets that cover \( X \) . We construct the sequence \( \left( {K}_{n}\right) \) by induction, as follows: set \( {K}_{0} = {L}_{0} \) and, for \( n \geq 1 \), choose \( {K}_{n} \) such \( {K}_{n - 1} \cup {L}_{n - 1} \subset {\mathring{K}}_{n} \) (using Corollary 1.4).\n\nA sequence \( \left( {K}_{n}\right) \) of compact sets that covers \( X \) and satisfies \( {K}_{n} \subset {\mathring{K}}_{n + 1} \) for all \( n \) is said to exhaust \( X \) .
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Yes
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Proposition 1.6 Let \( \left( {K}_{n}\right) \) be a sequence of compact sets that exhausts a metric space \( X \) . For every compact \( K \) of \( X \) there exists an integer \( n \) such that \( K \subset {K}_{n} \) .
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Proof. The open sets \( {\mathring{K}}_{n} \) cover \( K \) . By the Borel-Lebesgue property, \( K \) is in fact contained in a finite union of sets \( {\mathring{K}}_{n} \) : but \( \mathop{\bigcup }\limits_{{j \leq n}}{\mathring{K}}_{j} = {\mathring{K}}_{n} \) .
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Yes
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Proposition 1.7 Let \( {\left( {f}_{n}\right) }_{n \in \mathbb{N}} \) be an increasing sequence in \( {C}_{0}^{\mathbb{R}}\left( X\right) \), converging pointwise to a function \( f \in {C}_{0}^{\mathbb{R}}\left( X\right) \) . Then \( \left( {f}_{n}\right) \) converges uniformly to \( f \) .
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Proof. We show that the sequence \( \left( {g}_{n}\right) \) defined by \( {g}_{n} = f - {f}_{n} \) converges uniformly to 0 . Given \( \varepsilon > 0 \), there exists a compact \( K \) such that \( {g}_{0}\left( x\right) \leq \varepsilon \) for all \( x \notin K \) . By Dini’s Lemma, there exists an integer \( n \) such that \( {g}_{n}\left( x\right) \leq \) \( \varepsilon \) for all \( x \in K \) . Since the sequence \( \left( {g}_{n}\right) \) is decreasing, this implies that for all \( p \geq n \) and all \( x \in X \) we have \( 0 \leq {g}_{p}\left( x\right) \leq \varepsilon \) .
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Yes
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Proposition 1.8 (Partitions of unity) Let \( X \) be locally compact. If \( K \) is a compact subset of \( X \) and \( {O}_{1},\ldots ,{O}_{n} \) are open subsets of \( X \) that cover \( K \), there exist functions \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) in \( {C}_{c}^{\mathbb{R}}\left( X\right) \) such that \( 0 \leq {\varphi }_{j} \leq 1 \) and \( \operatorname{Supp}{\varphi }_{j} \subset {O}_{j} \) for each \( j \) and\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{\varphi }_{j}\left( x\right) = 1\;\text{ for all }x \in K \]\n
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Proof. Let \( {K}_{1},\ldots ,{K}_{n} \) be the compact sets whose existence is granted by Corollary 1.4. We just have to set, for \( x \in X \) ,\n\n\[ {\varphi }_{j}\left( x\right) = \frac{d\left( {x, X \smallsetminus {\mathring{K}}_{j}}\right) }{d\left( {x, K}\right) + \mathop{\sum }\limits_{{k = 1}}^{n}d\left( {x, X \smallsetminus {\mathring{K}}_{k}}\right) }.\]\n\nIn particular, \( \operatorname{Supp}{\varphi }_{j} \subset {K}_{j} \subset {O}_{j} \) .
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Yes
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Corollary 1.9 If \( X \) is locally compact, \( {C}_{c}\left( X\right) \) is dense in \( {C}_{0}\left( X\right) \) .
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Proof. Take \( f \in {C}_{0}\left( X\right) \) and \( \varepsilon > 0 \) . Let \( K \) be a compact such that \( \left| {f\left( x\right) }\right| < \varepsilon \) for all \( x \notin K \) . Applying Proposition 1.8 with \( n = 1 \) and \( {O}_{1} = X \), we find a \( \varphi \in {C}_{c}^{\mathbb{R}}\left( X\right) \) such that \( 0 \leq \varphi \leq 1 \) and \( \varphi = 1 \) on \( K \) . Then \( {f\varphi } \in {C}_{c}\left( X\right) \) and \( \parallel f - {f\varphi }\parallel \leq \varepsilon \) .
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Yes
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Corollary 1.10 Let \( X \) be locally compact and separable and let \( O \) be open in \( X \) . There exists an increasing sequence \( \left( {\varphi }_{n}\right) \) of functions in \( {C}_{c}^{ + }\left( X\right) \), each with support contained in \( O \), and such that \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}{\varphi }_{n}\left( x\right) = {1}_{O}\left( x\right) \) for all \( x \in X \) .
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Proof. \( O \) is a locally compact separable space, by Corollary 1.3 above and Proposition 2.3 on page 8. By Proposition 1.5 there exists a sequence of compact sets \( \left( {K}_{n}\right) \) such that \( {K}_{n} \subset {\mathring{K}}_{n + 1} \) for all \( n \) and \( \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{K}_{n} = O \) . By Proposition 1.8 there exists for each \( n \) a map \( {\varphi }_{n} \in {C}_{c}^{\mathbb{R}}\left( X\right) \) such that \( 0 \leq {\varphi }_{n} \leq 1,{\left. {\varphi }_{n}\right| }_{{K}_{n}} = 1 \), and \( \operatorname{Supp}{\varphi }_{n} \subset {\mathring{K}}_{n + 1} \) . The sequence \( \left( {\varphi }_{n}\right) \) clearly satisfies the desired conditions.
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Yes
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Lemma 2.1 \( \mathcal{L} \) is the smallest subset of \( \mathcal{F} \) that contains \( L \) and is closed under pointwise convergence (the latter condition means that the pointwise limit of any sequence in \( \mathcal{L} \) is also in \( \mathcal{L} \) ).
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Proof. It is clear that a minimal set satisfying these conditions exists. Call it \( \mathcal{B} \). - \( \mathcal{B} \) is a vector subspace of \( \mathcal{F} \) and a lattice, and it contains the constants. Proof. If \( \lambda \in \mathbb{R} \), the set \( \{ f \in \mathcal{F} : {\lambda f} \in \mathcal{B}\} \) contains \( L \) and is closed under pointwise convergence, so it contains \( \mathcal{B} \). Therefore \( f \in \mathcal{B} \) and \( \lambda \in \mathbb{R} \) imply \( {\lambda f} \in \mathcal{B}. \) Similarly, for every \( g \in L \), the set \( \{ f \in \mathcal{F} : f + g \in \mathcal{B}\} \) contains \( \mathcal{B} \), so the sum of an element of \( L \) and one of \( \mathcal{B} \) is in \( \mathcal{B} \). Using the same reasoning again we deduce from this that, for every \( f \in \mathcal{B} \), the set \( \{ h \in \mathcal{F} : f + h \in \mathcal{B}\} \) contains \( \mathcal{B} \). Thus the sum of two elements of \( \mathcal{B} \) is in \( \mathcal{B} \), and \( \mathcal{B} \) is a vector space. Since \( L \) is a lattice we see by considering the set \( \{ f \in \mathcal{F} : \left| f\right| \in \mathcal{B}\} \) that \( \mathcal{B} \) is a lattice as well. That \( \mathcal{B} \) contains 1 and therefore all constants follows from condition \( \left( *\right) \). - We now show that \( \mathcal{B} = \mathcal{L} \). Set \( \mathcal{T} = \left\{ {A \subset X : {1}_{A} \in \mathcal{B}}\right\} \). By the preceding paragraphs, \( \mathcal{T} \) is a \( \sigma \) -algebra. If \( f \in L \) and \( a \in \mathbb{R} \), the characteristic function of the set \( \{ f > a\} \) is the pointwise limit of the sequence \( \left( {\inf \left( {n{\left( f - a\right) }^{ + },1}\right) }\right) \), and so belongs to \( \mathcal{B} \), and \( \{ f > a\} \in \mathcal{T} \). Thus the elements of \( L \) are \( \mathcal{T} \) -measurable, which implies that \( \sigma \left( L\right) \subset \mathcal{T} \) ; in other words, \( {1}_{A} \in \mathcal{B} \) for \( A \in \sigma \left( L\right) \). Since every \( \sigma \left( L\right) \) -measurable function is the pointwise limit of \( \sigma \left( L\right) \) -measurable piecewise constant functions, we deduce that \( \mathcal{L} \subset \mathcal{B} \) and, by the minimality of \( \mathcal{B} \), that \( \mathcal{L} = \mathcal{B} \).
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Yes
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Proposition 2.2 If \( X \) is a metric space, the set of Borel functions from \( X \) to \( \mathbb{R} \) is the smallest subset of \( \mathcal{F} \) that contains all continuous functions from \( X \) to \( \mathbb{R} \) and is closed under pointwise convergence.
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Proof. Let \( L \) be the set of continuous functions from \( X \) to \( \mathbb{R} \) . Then \( L \) is a lattice and satisfies \( \left( *\right) \), since \( 1 \in L \) . On the other hand, let \( \mathcal{B} \) be the Borel \( \sigma \) -algebra of \( X \) . Certainly every continuous function on \( X \) is \( \mathcal{B} \) - measurable, so \( \sigma \left( L\right) \subset \mathcal{B} \) . Conversely, every open set \( U \) of \( X \) is contained in \( \sigma \left( L\right) \) : to see this, note, for example, that \( U \) is the inverse image of the open set \( {\mathbb{R}}^{ * } \) under the continuous function \( f \) defined by \( f\left( x\right) = d\left( {x,{U}^{c}}\right) \) . Thus \( \mathcal{B} \subset \sigma \left( L\right) \), which implies \( \mathcal{B} = \sigma \left( L\right) \) . Now apply Lemma 2.1.
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Yes
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Theorem 2.3 (Daniell) Let \( \mu \) be a linear form on \( L \) satisfying these conditions:\n\n1. \( \mu \) is positive, that is, if \( f \in L \) satisfies \( f \geq 0 \) then \( \mu \left( f\right) \geq 0 \) .\n\n2. If a sequence \( \left( {f}_{n}\right) \) in \( L \) satisfies \( {f}_{n} \searrow 0 \), then \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}\mu \left( {f}_{n}\right) = 0 \) .\n\nThen there exists a unique measure \( m \) on the \( \sigma \) -algebra \( \sigma \left( L\right) \) such that\n\n\[ L \subset {\mathcal{L}}^{1}\left( m\right) \;\text{ and }\;\mu \left( f\right) = \int {fdm}\;\text{ for all }f \in L. \]
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Uniqueness of \( m \) . Suppose that two measures \( {m}_{1} \) and \( {m}_{2} \) satisfy the stated properties. Let \( \left( {\varphi }_{n}\right) \) be a sequence satisfying condition \( \left( *\right) \) on page 58 . For every \( n \in \mathbb{N} \) and every real \( \lambda \geq 0 \), the set\n\n\[ \left\{ {f \in \mathcal{L} : \int \inf \left( {{f}^{ + },\lambda {\varphi }_{n}}\right) d{m}_{1} = \int \inf \left( {{f}^{ + },\lambda {\varphi }_{n}}\right) d{m}_{2}}\right\} \]\n\nequals \( \mathcal{L} \), by the minimality of \( \mathcal{L} \) (proved in Lemma 2.1) and the Dominated Convergence Theorem. Making \( n \) go to infinity, then \( \lambda \), we conclude by the Monotone Convergence Theorem that \( \int {f}^{ + }d{m}_{1} = \int {f}^{ + }d{m}_{2} \) for all \( f \in \mathcal{L} \) . Therefore \( {m}_{1} = {m}_{2} \) on \( \sigma \left( L\right) \) .\n\nExistence of \( m \) . The proof of existence is rather long and is carried out in several steps. First of all, let \( \mathcal{U} \) be the set of functions from \( X \) into \( \mathbb{R} \) that are pointwise limits of increasing sequences of elements of \( L \) . The measure \( m \) is constructed by first extending the linear form \( \mu \) to \( \mathcal{U} \) (steps 1-3), then to the space \( {L}^{1} \) defined in step 6 below. Some properties of \( {L}^{1} \) and of \( \mu \) are established in step 7, allowing us to conclude the proof in step 8 .
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Yes
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Proposition 2.4 Under the same assumptions and with the same notation as in Theorem 2.3, the space \( L \) is dense in the Banach space \( {L}^{1}\left( m\right) \) .
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Proof. We maintain the same notation. It suffices to show that if \( A \) is in \( \sigma \left( L\right) \) and \( m\left( A\right) \) is finite then for every \( \varepsilon > 0 \) there exists an element \( \varphi \) of \( L \) such that \( \mu \left( \left| {{1}_{A} - \varphi }\right| \right) < \varepsilon \) . If \( \varepsilon > 0 \), there exists \( \psi \in \mathcal{U} \) such that \( {1}_{A} \leq \psi \) and \( \mu \left( \psi \right) \leq \mu \left( {1}_{A}\right) + \varepsilon /2 \) . Now let \( \varphi \in L \) be such that \( \varphi \leq \psi \) and \( \mu \left( \psi \right) \leq \mu \left( \varphi \right) + \varepsilon /2 \) . Since \( \left| {{1}_{A} - \varphi }\right| \leq \left( {\psi - {1}_{A}}\right) + \left( {\psi - \varphi }\right) \) and \( \psi \in {L}^{1} \) by step 7b, the desired result follows.
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No
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Proposition 3.1 Let \( m \) be a Borel measure on \( X \) . There exists a largest open set \( O \) such that \( m\left( O\right) = 0 \) .
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Proof. Let \( \mathcal{U} \) be the set of all open sets \( \Omega \) of \( X \) such that \( m\left( \Omega \right) = 0 \) . This set is nonempty since it contains \( \varnothing \) . Set \( O = \mathop{\bigcup }\limits_{{\Omega \in \mathcal{U}}}\Omega \) ; this is an open set, which we must prove has \( m \) -measure zero. If \( K \) is compact and contained in \( O \), it can be covered by finitely many elements of \( \mathcal{U} \) . Each of these elements has measure zero, so \( m\left( K\right) = 0 \) . But \( O \) is \( \sigma \) -compact (being locally compact and separable), so it too has measure zero, by the \( \sigma \) -additivity of \( m \) .
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Yes
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Proposition 3.3 Let \( \mu \) be a positive Radon measure on \( X \) . For every compact set \( K \) in \( X \), the restriction of \( \mu \) to \( {C}_{K}^{\mathbb{R}}\left( X\right) \) is continuous; that is, there exists a constant \( {C}_{K} \geq 0 \) such that\n\n\[ \left| {\mu \left( f\right) }\right| \leq {C}_{K}\parallel f\parallel \;\text{ for all }f \in {C}_{K}^{\mathbb{R}}\left( X\right) . \]\n\n(We say that \( \mu \) is continuous on \( {C}_{c}^{\mathbb{R}}\left( X\right) \) .)
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Proof. Let \( K \) be compact in \( X \) . By Proposition 1.8 on page 53, there exists \( {\varphi }_{K} \in {C}_{c}^{ + }\left( X\right) \) such that \( 0 \leq {\varphi }_{K} \leq 1 \) and \( {\varphi }_{K} = 1 \) on \( K \) . Then, for all \( f \in {C}_{K}^{\mathbb{R}}\left( X\right) \), we have \( \left| f\right| \leq \parallel f\parallel {\varphi }_{K} \), and, by Lemma 3.2, \( \left| {\mu \left( f\right) }\right| \leq \mu \left( \left| f\right| \right) \leq \parallel f\parallel \leq \mu \left( {\varphi }_{K}\right) . \)
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Yes
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Theorem 3.4 (Radon-Riesz) For every positive Radon measure \( \mu \) on \( X \) there exists a unique Borel measure \( m \) finite on compact sets and such that \[ \mu \left( f\right) = \int {fdm}\;\text{ for all }f \in {C}_{c}^{\mathbb{R}}\left( X\right) . \] The map \( \mu \mapsto m \) thus defined is a bijection between \( {\mathfrak{M}}^{ + }\left( X\right) \) and the set of Borel measures finite on compact sets, and it commutes with addition and multiplication by nonnegative scalars.
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Proof. This will follow as a particular case of Daniell's Theorem. Set \( L = {C}_{c}^{\mathbb{R}}\left( X\right) \) . This space satisfies the assumptions stated on page 58 : in particular, property (*) follows from Corollary 1.10 on page 53. Now take \( \mu \in {\mathfrak{M}}^{ + }\left( X\right) \) ; we will show that assumption 2 of Theorem 2.3 is satisfied. Let \( \left( {f}_{n}\right) \) be a decreasing sequence in \( L \) approaching 0 pointwise. Each \( {f}_{n} \) has support contained in the compact set \( K = \operatorname{Supp}{f}_{0} \) . Thus, by Dini’s Lemma, \( \left( {f}_{n}\right) \) tends to 0 uniformly on \( K \) : in other words, \( {f}_{n} \rightarrow 0 \) in \( {C}_{K}^{\mathbb{R}}\left( X\right) \) . By Proposition 3.3, \( \mu \left( {f}_{n}\right) \rightarrow 0 \) . Next we check that \( \sigma \left( L\right) = \mathcal{B}\left( X\right) \) . Since every continuous function on \( X \) is a Borel function, the smallest \( \sigma \) -algebra that makes all elements of \( L \) measurable is certainly contained in \( \mathcal{B}\left( X\right) \) ; that is, \( \sigma \left( L\right) \subset \mathcal{B}\left( X\right) \) . Conversely, \( \mathcal{B}\left( X\right) \subset \sigma \left( L\right) \) because every open subset \( O \) of \( X \) is \( \sigma \left( L\right) \) -measurable. Indeed, with the notation of Corollary 1.10, an element \( x \in X \) belongs to \( O \) if and only if there exists \( n \in \mathbb{N} \) with \( {\varphi }_{n}\left( x\right) > 0 \) . Thus \( O \) is the (countable) union of the sets \( {\varphi }_{n}^{-1}\left( \left( {0, + \infty }\right) \right) \), which are \( \sigma \left( L\right) \) -measurable since the functions \( {\varphi }_{n} \) are elements of \( L \) . Therefore \( O \) is \( \sigma \left( L\right) \) -measurable and we finally conclude that \( \sigma \left( L\right) = \mathcal{B}\left( X\right) \) . Finally, we see that a Borel measure \( m \) on \( X \) is finite on compact sets if and only if \( L \subset {\mathcal{L}}^{1}\left( m\right) \) . It now suffices to apply Theorem 2.3 to derive the existence and uniqueness of \( m \) . The remaining statements of the theorem are easy to check.
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Yes
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Proposition 3.6 For every positive linear form \( \mathfrak{m} \) on \( {C}_{0}^{\mathbb{R}}\left( X\right) \) there exists a unique positive Radon measure \( \mu \) of finite mass and such that \( \mathfrak{m} = {\mathfrak{m}}_{\mu } \) , or equivalently such that\n\n\[ \mathfrak{m}\left( f\right) = \int {fd\mu }\;\text{ for all }f \in {C}_{0}^{\mathbb{R}}\left( X\right) . \]\n\nThus the map \( \mu \mapsto {\mathfrak{m}}_{\mu } \) is a bijection between \( {\mathfrak{M}}_{f}^{ + }\left( X\right) \) and the set of positive linear forms on \( {C}_{0}^{\mathbb{R}}\left( X\right) \) .
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Proof. The uniqueness of \( \mu \) clearly follows from the inclusion of \( {C}_{c}^{\mathbb{R}}\left( X\right) \) in \( {C}_{0}^{\mathbb{R}}\left( X\right) \) . The important point is existence.\n\nWe first show that \( \mathfrak{m} \) is continuous. If not, there exists a sequence \( \left( {f}_{n}\right) \) in \( {C}_{0}^{\mathbb{R}}\left( X\right) \) such that, for all \( n,\begin{Vmatrix}{f}_{n}\end{Vmatrix} \leq 1 \) and \( \left| {\mathfrak{m}\left( {f}_{n}\right) }\right| \geq n \) . By replacing \( {f}_{n} \) by \( \left| {f}_{n}\right| \), we can assume that \( {f}_{n} \in {C}_{0}^{ + }\left( X\right) \) (note that \( \mathfrak{m}\left( \left| {f}_{n}\right| \right) \geq \left| {\mathfrak{m}\left( {f}_{n}\right) }\right| \geq n \) because \( \mathfrak{m} \) is positive). Now set \( f = \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}{f}_{n}/{n}^{2} \) ; this function is in \( {C}_{0}^{ + }\left( X\right) \) because the series converges absolutely. But, for all integer \( N \geq 1 \) ,\n\n\[ \mathfrak{m}\left( f\right) \geq \mathop{\sum }\limits_{{n = 1}}^{N}\frac{\mathfrak{m}\left( {f}_{n}\right) }{{n}^{2}} \geq \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{n} \]\n\nso \( \mathfrak{m}\left( f\right) = + \infty \), an impossibility. It follows that \( \mathfrak{m} \) is continuous on \( {C}_{0}^{\mathbb{R}}\left( X\right) \) .\n\nIts restriction to \( {C}_{c}^{\mathbb{R}}\left( X\right) \) is a positive Radon measure \( \mu \) . Let \( \left( {\varphi }_{n}\right) \) be an increasing sequence in \( {C}_{c}^{ + }\left( X\right) \) converging pointwise to 1 . By the Monotone Convergence Theorem,\n\n\[ \int {d\mu } = \mathop{\lim }\limits_{{n \rightarrow + \infty }}\int {\varphi }_{n}{d\mu } = \mathop{\lim }\limits_{{n \rightarrow + \infty }}\mathfrak{m}\left( {\varphi }_{n}\right) \leq \parallel \mathfrak{m}\parallel \]\n\nwhere \( \parallel \mathfrak{m}\parallel \) is the norm of \( \mathfrak{m} \) in the topological dual of \( {C}_{0}^{\mathbb{R}}\left( X\right) \) . Thus \( \mu \) has finite mass and \( {\mathfrak{m}}_{\mu }\left( f\right) = \mathfrak{m}\left( f\right) \) for all \( f \in {C}_{c}^{\mathbb{R}}\left( X\right) \) . Since \( {C}_{c}^{\mathbb{R}}\left( X\right) \) is dense in \( {C}_{0}^{\mathbb{R}}\left( X\right) \) and since \( {\mathfrak{m}}_{\mu } \) and \( \mathfrak{m} \) are continuous, we get \( \mathfrak{m} = {\mathfrak{m}}_{\mu } \) .
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Yes
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Lemma 3.7 Let \( \alpha \) be an increasing function from \( \mathbb{R} \) to \( \mathbb{R} \) . If \( a \) and \( b \) are real numbers with \( a < b \), then\n\n\[ \n{d\alpha }(\left( {a, b\rbrack }\right) = \alpha \left( {b}_{ + }\right) - \alpha \left( {a}_{ + }\right)\n\]\n\nwhere \( \alpha \left( {a}_{ + }\right) \) and \( \alpha \left( {b}_{ + }\right) \) denote the right limits of \( \alpha \) at \( a \) and \( b \) .
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Proof. Let \( {\left( {\varphi }_{n}\right) }_{n \geq 1} \) be a sequence in \( {C}_{c}^{\mathbb{R}}\left( X\right) \) such that \( 0 \leq {\varphi }_{n} \leq 1,{\varphi }_{n} = 1 \) on \( \left\lbrack {a + 1/n, b - 1/n}\right\rbrack \), and \( {\varphi }_{n} = 0 \) on \( \mathbb{R} \smallsetminus \left\lbrack {a + 1/\left( {n + 1}\right), b - 1/\left( {n + 1}\right) }\right\rbrack \) . Then\n\n\[ \n\alpha \left( {b - \frac{1}{n}}\right) - \alpha \left( {a + \frac{1}{n}}\right) \leq \int {\varphi }_{n}{d\alpha } \leq \alpha \left( {b - \frac{1}{n + 1}}\right) - \alpha \left( {a + \frac{1}{n + 1}}\right) .\n\]\n\nBy passing to the limit, we get\n\n\[ \n{d\alpha }\left( \left( {a, b}\right) \right) = \alpha \left( {b}_{ - }\right) - \alpha \left( {a}_{ + }\right)\n\]\n\n(*) \n\nwhere \( \alpha \left( {x}_{ - }\right) \) is the left limit of \( \alpha \) at \( x \) . This is true for any \( a \) and \( b \) with \( a < b \) . Applying it to the terms of the sequences \( \left( {a}_{n}\right) ,\left( {b}_{n}\right) \) defined by \( {a}_{n} = b - 1/n,{b}_{n} = b + 1/n \) and taking the limit, we deduce that \( {d\alpha }\left( {\{ b\} }\right) = \) \( \alpha \left( {b}_{ + }\right) - \alpha \left( {b}_{ - }\right) \), which, together with \( \left( *\right) \), yields the desired relation.
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Yes
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Theorem 3.8 Let \( \mu \) be a positive Radon measure on \( \mathbb{R} \) . There exists a unique increasing right-continuous function \( \alpha \) with \( \alpha \left( 0\right) = 0 \) and \( \mu = {d\alpha } .
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Proof. Uniqueness is clear since, by the preceding discussion, if \( \alpha \) is right-continuous and vanishes at 0 , it is determined everywhere:\n\n\[ \alpha \left( x\right) = \left\{ \begin{array}{ll} - \mu \left( {(x,0\rbrack }\right) & \text{ if }x < 0 \\ 0 & \text{ if }x = 0 \\ \mu \left( {(0, x\rbrack }\right) & \text{ if }x > 0 \end{array}\right. \]\n\nConversely, define \( \alpha \) by these relations. Then \( \alpha \) is right-continuous and vanishes at 0 . Also, for \( a < b \) we have \( \alpha \left( b\right) - \alpha \left( a\right) = \mu (\left( {a, b\rbrack }\right) \) (one checks the various possible situations of 0 with respect to \( a \) and \( b \) ).\n\nNow suppose \( f \in {C}_{c}^{\mathbb{R}}\left( \mathbb{R}\right) \) is supported within \( \left\lbrack {a, b}\right\rbrack \), and let \( \Delta = {\left\{ {x}_{j}\right\} }_{0 \leq j \leq n} \) be a subdivision of \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \int {fd\mu } = \mathop{\sum }\limits_{{j = 0}}^{{n - 1}}\int f{1}_{\left( {x}_{j},{x}_{j + 1}\right\rbrack }{d\mu } \]\n\nand so, since \( \mu \left( \left( {{x}_{j},{x}_{j + 1}}\right\rbrack \right) = \alpha \left( {x}_{j + 1}\right) - \alpha \left( {x}_{j}\right) \),\n\n\[ {\mathfrak{S}}_{\Delta }\left( f\right) \leq \int {fd\mu } \leq {S}_{\Delta }\left( f\right) \]\n\nBy taking the limit we deduce that\n\n\[ \int {fd\mu } = {\int }_{a}^{b}{fd\alpha } = \int {fd\alpha } \]\n\nwhich concludes the proof.
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Yes
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Every bounded real Radon measure is the difference of two positive Radon measures of finite mass. More precisely, if \( \mu \in {\mathfrak{M}}_{f}^{\mathbb{R}}\left( X\right) \) , the Radon measures \( {\mu }^{ + } \) and \( {\mu }^{ - } \) defined in Theorem 4.1 have finite mass and\n\n\[ \parallel \mu \parallel = \int d{\mu }^{ + } + \int d{\mu }^{ - } \]\n\nwhere \( \parallel \mu \parallel \) is the norm of \( \mu \) in the dual of \( {C}_{0}^{\mathbb{R}}\left( X\right) \) .
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Proof. We first see that, for any \( f \in {C}_{c}^{ + }\left( X\right) \), \n\n\[ {\mu }^{ + }\left( f\right) + {\mu }^{ - }\left( f\right) = \sup \left\{ {\mu \left( {g - h}\right) : g, h \in {C}_{c}^{ + }\left( X\right) \text{ and }g, h \leq f}\right\} \]\n\n\[ = \sup \left\{ {\mu \left( \varphi \right) : \varphi \in {C}_{c}^{\mathbb{R}}\left( X\right) \text{ and }\left| \varphi \right| \leq f}\right\} . \]\n\nIn particular, \( {\mu }^{ + }\left( f\right) + {\mu }^{ - }\left( f\right) \leq \parallel \mu \parallel \parallel f\parallel \) . Applying this inequality to all terms of an increasing sequence of functions in \( {C}_{c}^{ + }\left( X\right) \) that converges pointwise to 1, we get \( \int d{\mu }^{ + } + \int d{\mu }^{ - } \leq \parallel \mu \parallel \) . Conversely, if \( f \in {C}_{c}^{\mathbb{R}}\left( X\right) \), then\n\n\[ \left| {\mu \left( f\right) }\right| = \left| {{\mu }^{ + }\left( f\right) - {\mu }^{ - }\left( f\right) }\right| \leq {\mu }^{ + }\left( \left| f\right| \right) + {\mu }^{ - }\left( \left| f\right| \right) \leq \left( {\int d{\mu }^{ + }+\int d{\mu }^{ - }}\right) \parallel f\parallel . \]\n\n(Here we used Lemma 3.2.)
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Yes
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Proposition 1.1 (Schwarz inequality) Let \( E \) be a vector space with a scalar semiproduct \( \left( {\cdot \mid \cdot }\right) \) . For every \( x, y \in E \) , \[ {\left| \left( x \mid y\right) \right| }^{2} \leq \left( {x \mid x}\right) \left( {y \mid y}\right) \]
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Proof. One can assume \( \mathbb{K} = \mathbb{C} \) . If \( x, y \in E \) , \( \left( {x + {ty} \mid x + {ty}}\right) = \left( {x \mid x}\right) + {2t}\operatorname{Re}\left( {x \mid y}\right) + {t}^{2}\left( {y \mid y}\right) \geq 0\; \) for all \( t \in \mathbb{R}. \) Consider the expression on the left-hand side of this inequality as a polynomial in \( t \), taking only nonnegative values. If \( \left( {y \mid y}\right) = 0 \), the polynomial is at most of first degree and must be constant, so \( 0 = {\left( \operatorname{Re}\left( x \mid y\right) \right) }^{2} \leq \) \( \left( {x\;|\;x}\right) \left( {y\;|\;y}\right) = 0 \) . If \( \left( {y\;|\;y}\right) \neq 0 \), the polynomial is of second degree and must have negative or zero discriminant; again \( {\left( \operatorname{Re}\left( x \mid y\right) \right) }^{2} \leq \left( {x \mid x}\right) \left( {y \mid y}\right) \) . Now let \( u \) be a complex number of absolute value 1 such that \[ \left| \left( {x \mid y}\right) \right| = u\left( {x \mid y}\right) = \left( {{ux} \mid y}\right) = \operatorname{Re}\left( {{ux} \mid y}\right) . \] We see that \( {\left| \left( x \mid y\right) \right| }^{2} \leq \left( {{ux} \mid {ux}}\right) \left( {y \mid y}\right) = \left( {x \mid x}\right) \left( {y \mid y}\right) \), since \( u\bar{u} = 1 \) .
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Yes
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Corollary 1.2 Let \( E \) be a vector space with a scalar product \( \left( {\cdot \mid \cdot }\right) \) . The expression \( \parallel x\parallel = {\left( x \mid x\right) }^{1/2} \) defines a norm on \( E \) .
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Proof. It is enough to check the triangle inequality. We have\n\n\[ \parallel x + y{\parallel }^{2} = \parallel x{\parallel }^{2} + \parallel y{\parallel }^{2} + 2\operatorname{Re}\left( {x \mid y}\right) \]\n\n\[ \leq \parallel x{\parallel }^{2} + \parallel y{\parallel }^{2} + 2\parallel x\parallel \parallel y\parallel = {\left( \parallel x\parallel + \parallel y\parallel \right) }^{2}. \]
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Yes
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Corollary 1.3 Let \( E \) be a scalar product space. For every \( y \in E \), the linear form \( {\varphi }_{y} = \left( {\cdot \mid y}\right) \) is continuous and its norm in the topological dual \( {E}^{\prime } \) of \( E \) equals \( \parallel y\parallel \) .
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Proof. By the Schwarz inequality, \( \left| {{\varphi }_{y}\left( x\right) }\right| \leq \parallel x\parallel \parallel y\parallel \) for all \( x \in E \), so \( {\varphi }_{y} \in \) \( {E}^{\prime } \) and \( \begin{Vmatrix}{\varphi }_{y}\end{Vmatrix} \leq \parallel y\parallel \) . At the same time, \( {\varphi }_{y}\left( y\right) = \parallel y{\parallel }^{2} \), so \( \begin{Vmatrix}{\varphi }_{y}\end{Vmatrix} = \parallel y\parallel \) .\n\nThus the map \( y \mapsto {\varphi }_{y} \) is an isometry from \( E \) to \( {E}^{\prime } \), linear if \( \mathbb{K} = \mathbb{R} \) and skew-linear if \( \mathbb{K} = \mathbb{C} \) . We will see in Theorem 3.1 below that this isometry is bijective if the space \( E \) is complete.
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Yes
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Proposition 1.4 (Equality in the Schwarz inequality) Two vectors \( x \) and \( y \) in a scalar product space satisfy \( \left| \left( {x \mid y}\right) \right| = \parallel x\parallel \parallel y\parallel \) if and only if they are linearly dependent.
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Proof. The \
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No
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Theorem 2.1 Let \( C \) be a nonempty, closed, convex subset of \( E \). For every point \( x \) of \( E \), there exists a unique point \( y \) of \( C \) such that\n\n\[ \parallel x - y\parallel = d\left( {x, C}\right) . \]\n\nThis point, called the projection of \( x \) onto \( C \) and denoted by \( {P}_{C}\left( x\right) \), is characterized by the following property:\n\n\[ y \in C\;\text{and}\;\operatorname{Re}\left( {x - y \mid z - y}\right) \leq 0\;\text{for all}\;z \in C\text{.} \]\n\n\( \left( *\right) \)
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Proof. Fix \( x \in E \). We first show the existence of the projection of \( x \) onto \( C \). By the definition of \( \delta = d\left( {x, C}\right) \), there exists a sequence \( \left( {y}_{n}\right) \) in \( C \) such that\n\n\[ {\begin{Vmatrix}x - {y}_{n}\end{Vmatrix}}^{2} \leq {\delta }^{2} + \frac{1}{n}\;\text{ for all }n \geq 1. \]\n\nApplying the parallelogram identity to the vectors \( x - {y}_{n} \) and \( x - {y}_{p} \), for \( n, p \geq 1 \), we obtain\n\n\[ {\begin{Vmatrix}x - \frac{{y}_{n} + {y}_{p}}{2}\end{Vmatrix}}^{2} + {\begin{Vmatrix}\frac{{y}_{n} - {y}_{p}}{2}\end{Vmatrix}}^{2} = \frac{1}{2}\left( {{\begin{Vmatrix}x - {y}_{n}\end{Vmatrix}}^{2} + {\begin{Vmatrix}x - {y}_{p}\end{Vmatrix}}^{2}}\right) . \]\n\nSince \( C \) is convex, \( \left( {{y}_{n} + {y}_{p}}\right) /2 \) is in \( C \), so \( \frac{1}{4}{\begin{Vmatrix}{y}_{n} - {y}_{p}\end{Vmatrix}}^{2} \leq \frac{1}{2}\left( {1/n + 1/p}\right) \) , which proves that \( \left( {y}_{n}\right) \) is a Cauchy sequence in \( C \) and so converges to an element \( y \) of \( C \), which must certainly satisfy \( \parallel x - y{\parallel }^{2} = {\delta }^{2} \).\n\nNow let \( {y}_{1} \) and \( {y}_{2} \) be points of \( C \) with \( \begin{Vmatrix}{x - {y}_{1}}\end{Vmatrix} = \begin{Vmatrix}{x - {y}_{2}}\end{Vmatrix} = \delta \). By applying the parallelogram identity as before, we get \( {\begin{Vmatrix}{y}_{1} - {y}_{2}\end{Vmatrix}}^{2} \leq 0 \), which says that \( {y}_{1} = {y}_{2} \). This shows that \( {P}_{C}\left( x\right) \) is unique.\n\nFinally, we check that the point \( y = {P}_{C}\left( x\right) \) satisfies property \( \left( *\right) \). If \( z \in C \) and \( t \in (0,1\rbrack \), the point \( \left( {1 - t}\right) y + {tz} \) belongs to \( C \) (which is convex), so\n\n\[ \parallel x - \left( {1 - t}\right) y - {tz}{\parallel }^{2} \geq \parallel x - y{\parallel }^{2} \]\nor, after expansion,\n\n\[ {t}^{2}\parallel y - z{\parallel }^{2} + {2t}\operatorname{Re}\left( {x - y \mid y - z}\right) \geq 0. \]\n\nDividing by \( t \) and making \( t \) approach 0, we get\n\n\[ \operatorname{Re}\left( {x - y \mid z - y}\right) \leq 0 \]\n\nConversely, suppose a point \( y \) of \( C \) satisfies \( \left( *\right) \). Then, for all \( z \in C \),\n\n\[ \parallel x - z{\parallel }^{2} = \parallel \left( {x - y}\right) + \left( {y - z}\right) {\parallel }^{2} \]\n\n\[ = \parallel x - y{\parallel }^{2} + \parallel y - z{\parallel }^{2} + 2\operatorname{Re}\left( {x - y \mid y - z}\right) \geq \parallel x - y{\parallel }^{2}, \]\n\nso \( y = {P}_{C}\left( x\right) \).
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Yes
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Proposition 2.2 Under the assumptions of Theorem 2.1,\n\n\\begin{Vmatrix}{{P}_{C}\\left( {x}_{1}\\right) - {P}_{C}\\left( {x}_{2}\\right) }\\end{Vmatrix} \\leq \\begin{Vmatrix}{{x}_{1} - {x}_{2}}\\end{Vmatrix}\\;\\text{ for all }{x}_{1},{x}_{2} \\in E.
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Proof. Set \\( {y}_{1} = {P}_{C}\\left( {x}_{1}\\right) \\) and \\( {y}_{2} = {P}_{C}\\left( {x}_{2}\\right) \\) . First,\n\n\\begin{align*}\n\\operatorname{Re}\\left( {{x}_{1} - {x}_{2} \\mid {y}_{1} - {y}_{2}}\\right) &= \\operatorname{Re}\\left( {{x}_{1} - {y}_{2} \\mid {y}_{1} - {y}_{2}}\\right) + \\operatorname{Re}\\left( {{y}_{2} - {x}_{2} \\mid {y}_{1} - {y}_{2}}\\right) \\\\\n&= \\operatorname{Re}\\left( {{x}_{1} - {y}_{1} \\mid {y}_{1} - {y}_{2}}\\right) + {\\begin{Vmatrix}{y}_{1} - {y}_{2}\\end{Vmatrix}}^{2} + \\operatorname{Re}\\left( {{y}_{2} - {x}_{2} \\mid {y}_{1} - {y}_{2}}\\right) \\\\\n&\\geq {\\begin{Vmatrix}{y}_{1} - {y}_{2}\\end{Vmatrix}}^{2}\n\\end{align*}\n\nThus, by the Schwarz inequality, \\( {\\begin{Vmatrix}{y}_{1} - {y}_{2}\\end{Vmatrix}}^{2} \\leq \\begin{Vmatrix}{{x}_{1} - {x}_{2}}\\end{Vmatrix}\\begin{Vmatrix}{{y}_{1} - {y}_{2}}\\end{Vmatrix} \\), and finally \\( \\begin{Vmatrix}{{y}_{1} - {y}_{2}}\\end{Vmatrix} \\leq \\begin{Vmatrix}{{x}_{1} - {x}_{2}}\\end{Vmatrix} \\).
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Yes
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Proposition 2.3 Let \( F \) be a closed vector subspace of \( E \) . Then \( {P}_{F} \) is a linear operator from \( E \) onto \( F \) . If \( x \in E \), the image \( {P}_{F}\left( x\right) \) is the unique element \( y \in E \) such that\n\n\[ y \in F\\text{ and }x - y \in {F}^{ \\bot }.\]
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Proof. Condition \( \\left( *\\right) \) of Theorem 2.1 becomes\n\n\[ y \\in F\\;\\text{and}\\;\\operatorname{Re}\\left( {x - y \\mid z - y}\\right) \\leq 0\\;\\text{for all}z \\in F.\\]\n\nNow, if \( y \\in F \) and \( \\lambda \\in {\\mathbb{C}}^{ * } \), the map \( {z}^{\\prime } \\mapsto z = y + \\bar{\\lambda }{z}^{\\prime } \) is a bijection from \( F \) onto \( F \) . Condition \( \\left( *\\right) \) is therefore equivalent to\n\n\[ y \\in F\\;\\text{and}\\;\\operatorname{Re}\\left( {\\lambda \\left( {x - y \\mid {z}^{\\prime }}\\right) }\\right) \\leq 0\\;\\text{for all}{z}^{\\prime } \\in F\\text{and}\\lambda \\in \\mathbb{C},\\]\n\nand this in turn is obviously equivalent to\n\n\[ y \\in F\\;\\text{ and }\\;x - y \\in {F}^{ \\bot }.\]\n\nThat \( {P}_{F} \) is linear follows easily.
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Yes
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Corollary 2.4 For every closed vector subspace \( F \) of \( E \), we have\n\n\[ E = F \oplus {F}^{ \bot } \]\n\nand the projection operator on \( F \) associated with this direct sum is \( {P}_{F} \) .
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Proof. For \( x \in E \), we can write \( x = {P}_{F}\left( x\right) + \left( {x - {P}_{F}\left( x\right) }\right) \) and, by Proposition \( {2.3},{P}_{F}\left( x\right) \in F \) and \( x - {P}_{F}\left( x\right) \in {F}^{ \bot } \) . On the other hand, if \( x \in F \cap {F}^{ \bot } \) , then \( \left( {x \mid x}\right) = 0 \) and so \( x = 0 \) .
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Yes
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Corollary 2.5 For every vector subspace \( F \) of \( E \) , \[ E = \bar{F} \oplus {F}^{ \bot }. \] In particular, \( F \) is dense in \( E \) if and only if \( {F}^{ \bot } = \{ 0\} \) .
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Proof. Just recall that \( {F}^{ \bot } = {\bar{F}}^{ \bot } \).
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No
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Proposition 2.6 Let \( \mu \) be a positive Radon measure on a locally compact, separable metric space \( X \) . Then \( {C}_{c}\left( X\right) \) is dense in \( {L}^{2}\left( \mu \right) \) .
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Proof. We write \( F = {C}_{c}\left( X\right) \) . If \( f \) is an element of \( {F}^{ \bot } \), then \( \int \varphi \bar{f}{d\mu } = 0 \) for all \( \varphi \in {C}_{c}\left( X\right) \) . Thus, for all \( \varphi \in {C}_{c}^{\mathbb{R}}\left( X\right) \) ,\n\n\[ \n\int \varphi {\left( \operatorname{Re}f\right) }^{ + }{d\mu } = \int \varphi {\left( \operatorname{Re}f\right) }^{ - }{d\mu }\n\]\n\n\[ \n\int \varphi {\left( \operatorname{Im}f\right) }^{ + }{d\mu } = \int \varphi {\left( \operatorname{Im}f\right) }^{ - }{d\mu }.\n\]\n\nBy the uniqueness part of the Radon-Riesz Theorem (page 69), these equalities hold for any nonnegative Borel function \( \varphi \) . Applying them to the characteristic functions of the sets \( \{ \operatorname{Re}f > 0\} ,\{ \operatorname{Re}f < 0\} ,\{ \operatorname{Im}f > 0\} \), and \( \{ \operatorname{Im}f < 0\} \), we conclude that \( f = {0\mu } \) -almost everywhere; that is, \( f = 0 \) as an element of \( {L}^{2}\left( \mu \right) \) . We finish by using Corollary 2.5.
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Yes
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Corollary 2.7 If \( E \) is a Hilbert space and \( F \) is a vector subspace of \( E \) , then \( \bar{F} = {F}^{ \bot \bot } \) .
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Proof. Clearly \( F \subset {F}^{ \bot \bot } \) . Therefore, since \( {F}^{ \bot \bot } \) is closed, \( \bar{F} \subset {F}^{ \bot \bot } \) . On the other hand, we have \( E = \bar{F} \oplus {F}^{ \bot } \) and \( E = {F}^{ \bot \bot } \oplus {F}^{ \bot } \) . The result follows immediately.
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Yes
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Theorem 3.1 (Riesz) The map from \( E \) to \( {E}^{\prime } \) defined by \( y \mapsto {\varphi }_{y} = \left( {\cdot \mid y}\right) \) is a surjective isometry. In other words, given any continuous linear form \( \varphi \) on \( E \), there exists a unique \( y \in E \) such that\n\n\[ \varphi \left( x\right) = \left( {x \mid y}\right) \;\text{ for all }x \in E, \]\n\nand, furthermore, \( \parallel \varphi \parallel = \parallel y\parallel \) .
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Proof. That this map is an isometry was seen in Corollary 1.3. We now show it is surjective. Take \( \varphi \in {E}^{\prime } \) such that \( \varphi \neq 0 \) . We know from Corollary 2.4 that \( E = \ker \varphi \oplus {\left( \ker \varphi \right) }^{ \bot } \), since, \( \varphi \) being continuous, \( \ker \varphi \) is closed. Now, \( \varphi \) is a nonzero linear form, so \( \ker \varphi \) has codimension 1 . The space \( {\left( \ker \varphi \right) }^{ \bot } \) therefore has dimension 1; it is generated by a vector \( e \), which we can choose to have norm 1 . Set \( y = \overline{\varphi \left( e\right) }e \) if \( \mathbb{K} = \mathbb{C} \), or \( y = \varphi \left( e\right) e \) if \( \mathbb{K} = \mathbb{R} \) . Then \( {\varphi }_{y}\left( e\right) = \varphi \left( e\right) \) and \( {\varphi }_{y} = 0 \) on \( \ker \varphi \) . It follows that \( {\varphi }_{y} \) and \( \varphi \) coincide on \( {\left( \ker \varphi \right) }^{ \bot } \) and on \( \ker \varphi \), so \( \varphi = {\varphi }_{y} \) .
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Yes
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Proposition 3.2 Given \( T \in L\left( E\right) \), there exists a unique operator \( {T}^{ * } \in \) \( L\left( E\right) \) such that\n\n\[ \left( {{Tx} \mid y}\right) = \left( {x \mid {T}^{ * }y}\right) \;\text{ for all }x, y \in E. \]\n\nMoreover, \( \begin{Vmatrix}{T}^{ * }\end{Vmatrix} = \parallel T\parallel \) .\n\n\( {T}^{ * } \) is called the adjoint of \( T \) .
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Proof. Take \( y \in E \) . The map \( {\varphi }_{y} \circ T : x \mapsto \left( {{Tx} \mid y}\right) \) is an element of \( {E}^{\prime } \), so by Theorem 3.1 there exists a unique element of \( E \), which we denote by \( {T}^{ * }y \), such that\n\n\[ \left( {{Tx} \mid y}\right) = \left( {x \mid {T}^{ * }y}\right) \;\text{ for all }x \in E \]\n\nmoreover \( \begin{Vmatrix}{{T}^{ * }y}\end{Vmatrix} = \begin{Vmatrix}{{\varphi }_{y} \circ T}\end{Vmatrix} \leq \parallel y\parallel \parallel T\parallel \) . The uniqueness of such a \( {T}^{ * }y \) easily shows that \( {T}^{ * } \) is linear; at the same time, by the preceding inequality, \( \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \leq \parallel T\parallel \) . Moreover, if \( x \in E \),\n\n\[ \parallel {Tx}{\parallel }^{2} = \left( {{Tx} \mid {Tx}}\right) = \left( {x \mid {T}^{ * }{Tx}}\right) \leq \parallel x\parallel \begin{Vmatrix}{T}^{ * }\end{Vmatrix}\parallel {Tx}\parallel \]\n\nwhich implies that \( \parallel {Tx}\parallel \leq \parallel x\parallel \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \), and so that \( \parallel T\parallel \leq \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) .
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Yes
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Proposition 3.4 For every \( T \in L\left( E\right) \), we have \( \begin{Vmatrix}{T{T}^{ * }}\end{Vmatrix} = \begin{Vmatrix}{{T}^{ * }T}\end{Vmatrix} = \parallel T{\parallel }^{2} \) .
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Proof. Certainly \( \begin{Vmatrix}{{T}^{ * }T}\end{Vmatrix} \leq \parallel T{\parallel }^{2} \) . On the other hand,\n\n\[ \parallel {Tx}{\parallel }^{2} = \left( {{Tx} \mid {Tx}}\right) = \left( {x \mid {T}^{ * }{Tx}}\right) \leq \parallel x{\parallel }^{2}\begin{Vmatrix}{{T}^{ * }T}\end{Vmatrix}, \]\n\nwhich shows that \( \parallel T{\parallel }^{2} \leq \begin{Vmatrix}{{T}^{ * }T}\end{Vmatrix} \) . Therefore \( \begin{Vmatrix}{{T}^{ * }T}\end{Vmatrix} = \parallel T{\parallel }^{2} \) and, applying this result to \( {T}^{ * } \), we get \( \begin{Vmatrix}{T{T}^{ * }}\end{Vmatrix} = {\begin{Vmatrix}{T}^{ * }\end{Vmatrix}}^{2} = \parallel T{\parallel }^{2} \) .
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Yes
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Proposition 3.5 Assume \( E \neq \{ 0\} \) . For every selfadjoint operator \( T \in \) \( L\left( E\right) \) , \n\n\[ \n\parallel T\parallel = \sup \{ \left| \left( {{Tx} \mid x}\right) \right| : x \in E\text{ and }\parallel x\parallel = 1\} . \n\]
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Proof. Let \( \gamma \) be the right-hand side of the equality. Clearly \( \gamma \leq \parallel T\parallel \) and, for all \( x \in E,\left| \left( {{Tx} \mid x}\right) \right| \leq \gamma \parallel x{\parallel }^{2} \) . Assume for example that \( \mathbb{K} = \mathbb{C} \), and take \( y, z \in E \) and \( \lambda \in \mathbb{R} \) . Then \n\n\[ \n\left| \left( {T\left( {y \pm {\lambda z}}\right) \;|\;y \pm {\lambda z}}\right) \right| = \left| {\left( {{Ty}\;|\;y}\right) \pm {2\lambda }\;\mathrm{{Re}}\left( {{Ty}\;|\;z}\right) + {\lambda }^{2}\left( {{Tz}\;|\;z}\right) }\right| \leq \gamma \;\| y \pm {\lambda z}{\| }^{2}. \n\] \n\nWe deduce, by combining the two inequalities, that \n\n\[ \n4\left| \lambda \right| \left| {\operatorname{Re}\left( {{Ty} \mid z}\right) }\right| \leq \gamma \left( {\parallel y + {\lambda z}{\parallel }^{2} + \parallel y - {\lambda z}{\parallel }^{2}}\right) = {2\gamma }\left( {\parallel y{\parallel }^{2} + {\lambda }^{2}\parallel z{\parallel }^{2}}\right) , \n\] \n\nand this holds for any real \( \lambda \) . We conclude that \( \left| {\operatorname{Re}\left( {{Ty} \mid z}\right) }\right| \leq \gamma \parallel y\parallel \parallel z\parallel \) , from the condition for a polynomial function on \( \mathbb{R} \) of degree at most 2 to be nonnegative-valued. Now it is enough to choose \( z = {Ty} \) to obtain \( \parallel {Ty}\parallel \leq \gamma \parallel y\parallel \) for all \( y \in E \), and hence \( \parallel T\parallel \leq \gamma \) .
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Yes
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Proposition 3.6 Let \( \left( {x}_{n}\right) \) be a sequence in \( E \) that converges weakly to \( x \) . Then\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow + \infty }}\begin{Vmatrix}{x}_{n}\end{Vmatrix} \geq \parallel x\parallel \]\n\nMoreover, the following properties are equivalent:\n\n1. The sequence \( \left( {x}_{n}\right) \) converges (strongly) to \( x \) .\n\n2. \( \lim \mathop{\sup }\limits_{{n \rightarrow + \infty }}\begin{Vmatrix}{x}_{n}\end{Vmatrix} \leq \parallel x\parallel \) .\n\n3. \( \mathop{\lim }\limits_{{n \rightarrow + \infty }}\begin{Vmatrix}{x}_{n}\end{Vmatrix} = \parallel x\parallel \) .
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Proof. First,\n\n\[ \parallel x{\parallel }^{2} = \mathop{\lim }\limits_{{n \rightarrow + \infty }}\left| \left( {x \mid {x}_{n}}\right) \right| \leq \parallel x\parallel \mathop{\liminf }\limits_{{n \rightarrow + \infty }}\begin{Vmatrix}{x}_{n}\end{Vmatrix}, \]\n\nwhich proves the first statement. At the same time, \( {\begin{Vmatrix}x - {x}_{n}\end{Vmatrix}}^{2} = \parallel x{\parallel }^{2} + \) \( {\begin{Vmatrix}{x}_{n}\end{Vmatrix}}^{2} - 2\operatorname{Re}\left( {{x}_{n} \mid x}\right) \), so\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow + \infty }}{\begin{Vmatrix}x - {x}_{n}\end{Vmatrix}}^{2} \leq {\left( \mathop{\limsup }\limits_{{n \rightarrow + \infty }}\begin{Vmatrix}{x}_{n}\end{Vmatrix}\right) }^{2} - \parallel x{\parallel }^{2} \]\n\nwhich yields the equivalence between 1 and 2 . The equivalence between 2 and 3 follows immediately from the first statement.
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Yes
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Theorem 3.7 Any bounded sequence in \( E \) has a weakly convergent subsequence.
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Proof. Suppose first that \( E \) is separable. Let \( \left( {x}_{n}\right) \) be a bounded sequence in \( E \) . In the notation of Theorem 3.1, the Banach–Alaoglu Theorem (page 19) applied to the sequence \( \left( {\varphi }_{{x}_{n}}\right) \) guarantees the existence of a subsequence \( \left( {x}_{{n}_{k}}\right) \) and of a \( \varphi \in {E}^{\prime } \) such that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow + \infty }}{\varphi }_{{n}_{k}}\left( y\right) = \varphi \left( y\right) \;\text{ for all }y \in E. \]\n\nBy Theorem 3.1, there exists an element \( x \in E \) such that \( \varphi = {\varphi }_{x} \), which proves the theorem in the separable case.\n\nWe turn to the general case. Let \( \left( {x}_{n}\right) \) be a bounded sequence in \( E \) and let \( F \) be the closure of the vector subspace of \( E \) spanned by \( {\left\{ {x}_{n}\right\} }_{n \in \mathbb{N}} \) . By construction, this is a separable Hilbert space. The first part of the proof says that there exists a subsequence \( \left( {x}_{{n}_{k}}\right) \) and a point \( x \in F \) such that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow + \infty }}\left( {{x}_{{n}_{k}} \mid y}\right) = \left( {x \mid y}\right) \;\text{ for all }y \in F. \]\n\nSince this equality obviously takes place also if \( y \in {F}^{ \bot } \), it suffices now to apply Corollary 2.4.
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Yes
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