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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 Cauchy functional equation ： ウィキペディア英語版 Cauchy's functional equation Cauchy's functional equation is the functional equation :$f(x+y)=f(x)+f(y).\; \backslash$ Solutions to this are called additive functions. Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely for any arbitrary rational number . Over the real numbers, this is still a family of solutions; however there can exist other solutions that are extremely complicated. Further constraints on ''f'' sometimes preclude other solutions, for example: * if is continuous (proven by Cauchy in 1821). This condition was weakened in 1875 by Darboux who showed that it was only necessary for the function to be continuous at one point. * if is monotonic on any interval. * if is bounded on any interval. On the other hand, if no further conditions are imposed on , then (assuming the axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by Georg Hamel using Hamel bases. Such functions are sometimes called ''Hamel functions''.〔Kuczma (2009), p.130〕 The fifth problem on Hilbert's list is a generalisation of this equation. Functions where there exists a real number $c$ such that $f(cx)\; \backslash ne\; cf(x)\; \backslash$ are known as Cauchy-Hamel functions and are used in Dehn-Hadwiger invariants which are used in the extension of Hilbert's third problem from 3-D to higher dimensions.〔V.G. Boltianskii (1978) "Hilbert's third problem", Halsted Press, Washington〕 == Proof of solution over rationals == We wish to prove that any solution $f$ to Cauchy's functional equation, $f(x+y)\; =\; f(x)\; +\; f(y)$, takes the form $f\backslash left(q\backslash right)\; =\; q\; f\backslash left(1\backslash right),\; q\; \backslash in\; \backslash mathbb$. Case 1: q=0 Set $x=1,\; y=0$. :$\backslash Rightarrow\; f(0)\; =\; 0$. Case 2: q>0 By repeated application of Cauchy's equation to $f\backslash left(x\; +\; x\; +\; ...\; +\; x\backslash right)\; =\; f\backslash left(\backslash alpha\; x\backslash right)$: :$\backslash alpha\; f\backslash left(x\backslash right)\; =\; f\backslash left(\backslash alpha\; x\backslash right),\; \backslash quad\; \backslash alpha\; \backslash in\; \backslash mathbb$ Replacing $x$ by $\backslash frac$, and multiplying by $\backslash frac$: :$\backslash beta\; f\backslash left(\backslash frac\backslash right)\; =\; \backslash frac\; f\backslash left(x\backslash right),\; \backslash quad\; \backslash alpha\; \backslash in\; \backslash mathbb$ By the first equation: :$f\backslash left(\backslash fracx\backslash right)\; =\; \backslash frac\; f\backslash left(x\backslash right),\; \backslash quad\; \backslash alpha,\; \backslash beta\; \backslash in\; \backslash mathbb$ :$\backslash Rightarrow\; f\backslash left(qx\backslash right)\; =\; q\; f\backslash left(x\backslash right),\; \backslash quad\; q\; \backslash in\; \backslash mathbb,\; q\; >\; 0$ :$\backslash Rightarrow\; f\backslash left(q\backslash right)\; =\; q\; f\backslash left(1\backslash right),\; \backslash quad\; q\; \backslash in\; \backslash mathbb,\; q\; >\; 0$. Case 3: q<0 Set $y=-x$. :$\backslash Rightarrow\; f(-x)\; =\; -f(x)$. Combining this with the result from case 2: :$-f\backslash left(q\backslash right)\; =\; -q\; f\backslash left(1\backslash right),\; \backslash quad\; q\; \backslash in\; \backslash mathbb,\; q\; >\; 0$ :$\backslash Rightarrow\; f\backslash left(-q\backslash right)\; =\; -q\; f\backslash left(1\backslash right),\; \backslash quad\; q\; \backslash in\; \backslash mathbb,\; q\; >\; 0$ Replacing -q with q: : 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「Cauchy's functional equation」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.