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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Hilbert projection theorem ： ウィキペディア英語版 Hilbert projection theorem In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every point $x$ in a Hilbert space $H$ and every nonempty closed convex $C\; \backslash subset\; H$, there exists a unique point $y\; \backslash in\; C$ for which $\backslash lVert\; x\; -\; y\; \backslash rVert$ is minimized over $C$. This is, in particular, true for any closed subspace $M$ of $C$. In that case, a necessary and sufficient condition for $y$ is that the vector $x-y$ be orthogonal to $M$. ==Proof== : * ''Let us show the existence of ''y'':'' Let δ be the distance between ''x'' and ''C'', (''y''''n'') a sequence in ''C'' such that the distance squared between ''x'' and ''y''''n'' is below or equal to δ2 + 1/''n''. Let ''n'' and ''m'' be two integers, then the following equalities are true: : $\backslash |\; y\_n\; -\; y\_m\; \backslash |^2\; =\; \backslash |y\_n\; -x\backslash |^2\; +\; \backslash |y\_m\; -x\backslash |^2\; -\; 2\; \backslash langle\; y\_n\; -\; x\; \backslash ,\; ,\; \backslash ,\; y\_m\; -\; x\backslash rangle$ and : $4\; \backslash left\backslash |\; \backslash frac2\; -x\; \backslash right\backslash |^2\; =\; \backslash |y\_n\; -x\backslash |^2\; +\; \backslash |y\_m\; -x\backslash |^2\; +\; 2\; \backslash langle\; y\_n\; -\; x\; \backslash ,\; ,\; \backslash ,\; y\_m\; -\; x\backslash rangle$ We have therefore: : $\backslash |\; y\_n\; -\; y\_m\; \backslash |^2\; =\; 2\backslash |y\_n\; -x\backslash |^2\; +\; 2\backslash |y\_m\; -x\backslash |^2\; -\; 4\backslash left\backslash |\; \backslash frac2\; -x\; \backslash right\backslash |^2$ By giving an upper bound to the first two terms of the equality and by noticing that the middle of ''y''''n'' and ''y''''m'' belong to ''C'' and has therefore a distance greater than or equal to ''δ'' from ''x'', one gets : : $\backslash |\; y\_n\; -\; y\_m\; \backslash |^2\; \backslash ;\; \backslash le\; \backslash ;\; 2\backslash left(\backslash delta^2\; +\; \backslash frac\; 1n\backslash right)\; +\; 2\backslash left(\backslash delta^2\; +\; \backslash frac\; 1m\backslash right)\; -\; 4\backslash delta^2=2\backslash left(\; \backslash frac\; 1n\; +\; \backslash frac\; 1m\backslash right)$ The last inequality proves that (''y''''n'') is a Cauchy sequence. Since ''C'' is complete, the sequence is therefore convergent to a point ''y'' in ''C'', whose distance from ''x'' is minimal. : * ''Let us show the uniqueness of ''y'' :'' Let ''y''1 and ''y''2 be two minimizers. Then: : $\backslash |\; y\_2\; -\; y\_1\; \backslash |^2\; =\; 2\backslash |y\_1\; -x\backslash |^2\; +\; 2\backslash |y\_2\; -x\backslash |^2\; -\; 4\backslash left\backslash |\; \backslash frac2\; -x\; \backslash right\backslash |^2$ Since $\backslash frac2$ belongs to ''C'', we have $\backslash left\backslash |\; \backslash frac2\; -x\; \backslash right\backslash |^2\backslash geq\; \backslash delta^2$ and therefore : $\backslash |\; y\_2\; -\; y\_1\; \backslash |^2\; \backslash leq\; 2\backslash delta^2\; +\; 2\backslash delta^2\; -\; 4\backslash delta^2=0\; \backslash ,$ Hence $y\_1=y\_2$, which proves uniqueness. : * ''Let us show the equivalent condition on'' ''y'' ''when'' ''C'' = ''M'' ''is a closed subspace.'' The condition is sufficient: Let $z\backslash in\; M$ such that $\backslash langle\; z-x,\; a\; \backslash rangle=0$ for all $a\backslash in\; M$. $\backslash |x-a\backslash |^2=\backslash |z-x\backslash |^2+\backslash |a-z\backslash |^2+2\backslash langle\; z-x,\; a-z\; \backslash rangle=\backslash |z-x\backslash |^2+\backslash |a-z\backslash |^2$ which proves that $z$ is a minimizer. The condition is necessary: Let $y\backslash in\; M$ be the minimizer. Let $a\backslash in\; M$ and $t\backslash in\backslash mathbb\; R$. : $\backslash |(y+t\; a)-x\backslash |^2-\backslash |y-x\backslash |^2=2t\backslash langle\; y-x,a\backslash rangle+t^2\; \backslash |a\backslash |^2=2t\backslash langle\; y-x,a\backslash rangle+O(t^2)$ is always non-negative. Therefore, $\backslash langle\; y-x,a\backslash rangle=0.$ QED 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Hilbert projection theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.