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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 Hyperplane separation theorem ： ウィキペディア英語版 Hyperplane separation theorem In geometry, the hyperplane separation theorem is either of two theorems about disjoint convex sets in ''n''-dimensional Euclidean space. In the first version of the theorem, if both these sets are closed and at least one of them is compact, then there is a hyperplane in between them and even two parallel hyperplanes in between them separated by a gap. In the second version, if both disjoint convex sets are open, then there is a hyperplane in between them, but not necessarily any gap. An axis which is orthogonal to a separating hyperplane is a separating axis, because the orthogonal projections of the convex bodies onto the axis are disjoint. The hyperplane separation theorem is due to Hermann Minkowski. The Hahn–Banach separation theorem generalizes the result to topological vector spaces. A related result is the supporting hyperplane theorem. In geometry, a maximum-margin hyperplane is a hyperplane which separates two 'clouds' of points and is at equal distance from the two. The margin between the hyperplane and the clouds is maximal. See the article on Support Vector Machines for more details. == Statements and proof == The proof is based on the following lemma: Proof of lemma: Let $\backslash delta\; =\; \backslash inf\; \backslash .$ Let $x\_j$ be a sequence in ''K'' such that $|x\_j|\; \backslash to\; \backslash delta$. Note that $(x\_i\; +\; x\_j)/2$ is in ''K'' since ''K'' is convex and so $|x\_i\; +\; x\_j|^2\; \backslash ge\; 4\; \backslash delta^2$. Since :$|x\_i\; -\; x\_j|^2\; =\; 2|x\_i|^2\; +\; 2|x\_j|^2\; -\; |x\_i\; +\; x\_j|^2\; \backslash le\; 2|x\_i|^2\; +\; 2|x\_j|^2\; -\; 4\backslash delta^2\; \backslash to\; 0$ as $i,\; j\; \backslash to\; \backslash infty$, $x\_i$ is a Cauchy sequence and so has limit ''x'' in ''K''. It is unique since if ''y'' is in ''K'' and has norm δ, then$|x\; -\; y|^2\; \backslash le\; 2|x|^2\; +\; 2|y|^2\; -\; 4\backslash delta^2\; =\; 0$and ''x'' = ''y''. $\backslash square$ Proof of theorem: Given disjoint nonempty convex sets ''A'', ''B'', let :$K\; =\; A\; +\; (-B)\; =\; \backslash .$ Since $-B$ is convex and the sum of convex sets is convex, ''K'' is convex. By the lemma, the closure $\backslash overline$ of ''K'', which is convex, contains a vector ''v'' of minimum norm. Since $\backslash overline$ is convex, for any ''x'' in ''K'', the line segment :$v\; +\; t(x\; -\; v),\; \backslash ,\; 0\; \backslash le\; t\; \backslash le\; 1$ lies in $\backslash overline$ and so :$|v|^2\; \backslash le\; |v\; +\; t(x\; -\; v)|^2\; =\; |v|^2\; +\; 2\; t\; \backslash langle\; v,\; x\; -\; v\; \backslash rangle\; +\; t^2|x-v|^2$. For , we thus have: :$0\; \backslash le\; 2\; \backslash langle\; v,\; x\; \backslash rangle\; -\; 2\; |v|^2\; +\; t|x-v|^2$ and letting $t\; \backslash to\; 0$ gives: $\backslash langle\; x,\; v\; \backslash rangle\; \backslash ge\; |v|^2$. Hence, for any ''x'' in ''A'' and ''y'' in ''B'', we have: $\backslash langle\; x\; -\; y,\; v\; \backslash rangle\; \backslash ge\; |v|^2$. Thus, if ''v'' is nonzero, the proof is complete since :$\backslash inf\_\; \backslash langle\; x,\; v\; \backslash rangle\; \backslash ge\; |v|^2\; +\; \backslash sup\_\; \backslash langle\; y,\; v\; \backslash rangle.$ In general, if the interior of ''K'' is nonempty, then it can be exhausted by compact convex subsets $K\_n$. Since 0 is not in ''K'', each $K\_n$ contains a nonzero vector $v\_n$ of minimum length and by the argument in the early part, we have: $\backslash langle\; x,\; v\_n\; \backslash rangle\; \backslash ge\; 0$ for any $x\; \backslash in\; K\_n$. We also normalize them to have length one; then the sequence $v\_n$ contains a convergent subsequence with limit ''v'', which has length one and is nonzero. We have $\backslash langle\; x,\; v\; \backslash rangle\; \backslash ge\; 0$ for any ''x'' in the interior of ''K'' and by continuity the same holds for all ''x'' in ''K''. We now finish the proof as before. Finally, if ''K'' has empty interior, the affine set that it spans has dimension less than that of the whole space. Consequently ''K'' is contained in some hyperplane $\backslash langle\; \backslash cdot,\; v\; \backslash rangle\; =\; c$; thus, $\backslash langle\; x,\; v\; \backslash rangle\; \backslash ge\; c$ for all ''x'' in ''K'' and we finish the proof as before. $\backslash square$ The above proof also proves the first version of the theorem mentioned in the lede (to see it, note that ''K'' is closed under the hypothesis below.) Here, the compactness in the hypothesis cannot be relaxed; see an example in the next section. Also, This follows from the standard version since the separating hyperplane cannot intersect the interiors of the convex sets. 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「Hyperplane separation 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.