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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 Discontinuous linear map ： ウィキペディア英語版 Discontinuous linear map In mathematics, linear maps form an important class of "simple" functions which preserve the algebraic structure of linear spaces and are often used as approximations to more general functions (see linear approximation). If the spaces involved are also topological spaces (that is, topological vector spaces), then it makes sense to ask whether all linear maps are continuous. It turns out that for maps defined on infinite-dimensional topological vector spaces (e.g., infinite-dimensional normed spaces), the answer is generally no: there exist discontinuous linear maps. If the domain of definition is complete, it is trickier; such maps can be proven to exist, but the proof relies on the axiom of choice and does not provide an explicit example. == A linear map from a finite-dimensional space is always continuous == Let ''X'' and ''Y'' be two normed spaces and ''f'' a linear map from ''X'' to ''Y''. If ''X'' is finite-dimensional, choose a basis (''e''1, ''e''2, …, ''e''''n'') in ''X'' which may be taken to be unit vectors. Then, :$f(x)=\backslash sum^n\_x\_if(e\_i),$ and so by the triangle inequality, :$\backslash |f(x)\backslash |=\; \backslash left\backslash |\backslash sum^n\_x\_if(e\_i)\backslash right\backslash |\; \backslash le\; \backslash sum^n\_\; |x\_i|\backslash |f(e\_i)\backslash |.$ Letting :$M=\backslash sup\_i\; \backslash ,$ and using the fact that :$\backslash sum^n\_|x\_i|\backslash le\; C\; \backslash |x\backslash |$ for some ''C''>0 which follows from the fact that any two norms on a finite-dimensional space are equivalent, one finds :$\backslash |f(x)\backslash |\backslash le\; \backslash left(\backslash sum^n\_|x\_i|\backslash right)M\backslash le\; CM\backslash |x\backslash |.$ Thus, ''f'' is a bounded linear operator and so is continuous. If ''X'' is infinite-dimensional, this proof will fail as there is no guarantee that the supremum ''M'' exists. If ''Y'' is the zero space , the only map between ''X'' and ''Y'' is the zero map which is trivially continuous. In all other cases, when ''X'' is infinite-dimensional and ''Y'' is not the zero space, one can find a discontinuous map from ''X'' to ''Y''. 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「Discontinuous linear map」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.