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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 Minkowski inequality ： ウィキペディア英語版 Minkowski inequality In mathematical analysis, the Minkowski inequality establishes that the L''p'' spaces are normed vector spaces. Let ''S'' be a measure space, let 1 ≤ ''p'' ≤ ∞ and let ''f'' and ''g'' be elements of L''p''(''S''). Then ''f'' + ''g'' is in L''p''(''S''), and we have the triangle inequality :$\backslash |f+g\backslash |\_p\; \backslash le\; \backslash |f\backslash |\_p\; +\; \backslash |g\backslash |\_p$ with equality for 1 < ''p'' < ∞ if and only if ''f'' and ''g'' are positively linearly dependent, i.e., ''f'' = ''λg'' for some ''λ'' ≥ 0 or ''g'' = 0. Here, the norm is given by: :$\backslash |f\backslash |\_p\; =\; \backslash left(\; \backslash int\; |f|^p\; d\backslash mu\; \backslash right)^\}$ if ''p'' < ∞, or in the case ''p'' = ∞ by the essential supremum :$\backslash |f\backslash |\_\backslash infty\; =\; \backslash operatorname\_|f(x)|.$ The Minkowski inequality is the triangle inequality in L''p''(''S''). In fact, it is a special case of the more general fact :$\backslash |f\backslash |\_p\; =\; \backslash sup\_\; \backslash int\; |fg|\; d\backslash mu,\; \backslash qquad\; \backslash tfrac\; +\; \backslash tfrac\; =\; 1$ where it is easy to see that the right-hand side satisfies the triangular inequality. Like Hölder's inequality, the Minkowski inequality can be specialized to sequences and vectors by using the counting measure: :$\backslash left(\; \backslash sum\_^n\; |x\_k\; +\; y\_k|^p\; \backslash right)^\}\; \backslash le\; \backslash left(\; \backslash sum\_^n\; |x\_k|^p\; \backslash right)^\}\; +\; \backslash left(\; \backslash sum\_^n\; |y\_k|^p\; \backslash right)^\}$ for all real (or complex) numbers ''x''1, ..., ''x''''n'', ''y''1, ..., ''y''''n'' and where n is the cardinality of S (the number of elements in S). ==Proof== First, we prove that ''f''+''g'' has finite ''p''-norm if ''f'' and ''g'' both do, which follows by :$|f\; +\; g|^p\; \backslash le\; 2^(|f|^p\; +\; |g|^p).$ Indeed, here we use the fact that $h(x)=x^p$ is convex over (for ) and so, by the definition of convexity, :$\backslash left|\backslash tfrac\; f\; +\; \backslash tfrac\; g\backslash right|^p\backslash le\backslash left|\backslash tfrac\; |f|\; +\; \backslash tfrac\; |g|\backslash right|^p\; \backslash le\; \backslash tfrac|f|^p\; +\; \backslash tfrac\; |g|^p.$ This means that :$|f+g|^p\; \backslash le\; \backslash tfrac|2f|^p\; +\; \backslash tfrac|2g|^p=2^|f|^p\; +\; 2^|g|^p.$ Now, we can legitimately talk about $(\backslash |f\; +\; g\backslash |\_p)$. If it is zero, then Minkowski's inequality holds. We now assume that $(\backslash |f\; +\; g\backslash |\_p)$ is not zero. Using the triangle inequality and then Hölder's inequality, we find that :$\backslash begin$ \|f + g\|_p^p &= \int |f + g|^p \, \mathrm\mu \\ &= \int |f + g| \cdot |f + g|^ \, \mathrm\mu \\ &\le \int (|f| + |g|)|f + g|^ \, \mathrm\mu \\ &=\int |f||f + g|^ \, \mathrm\mu+\int |g||f + g|^ \, \mathrm\mu \\ &\le \left( \left(\int |f|^p \, \mathrm\mu\right)^} + \left (\int |g|^p \,\mathrm\mu\right)^} \right) \left(\int |f + g|^\right)} \, \mathrm\mu \right)^} && \text \\ &= \left (\|f\|_p + \|g\|_p \right )\frac \end We obtain Minkowski's inequality by multiplying both sides by :$\backslash frac.$ 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「Minkowski inequality」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.