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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 Compact operator ： ウィキペディア英語版 Compact operator In functional analysis, a branch of mathematics, a compact operator is a linear operator ''L'' from a Banach space ''X'' to another Banach space ''Y'', such that the image under ''L'' of any bounded subset of ''X'' is a relatively compact subset of ''Y''. Such an operator is necessarily a bounded operator, and so continuous. Any bounded operator ''L'' that has finite rank is a compact operator; indeed, the class of compact operators is a natural generalisation of the class of finite-rank operators in an infinite-dimensional setting. When ''Y'' is a Hilbert space, it is true that any compact operator is a limit of finite-rank operators, so that the class of compact operators can be defined alternatively as the closure in the operator norm of the finite-rank operators. Whether this was true in general for Banach spaces (the approximation property) was an unsolved question for many years; in the end Per Enflo gave a counter-example. The origin of the theory of compact operators is in the theory of integral equations, where integral operators supply concrete examples of such operators. A typical Fredholm integral equation gives rise to a compact operator ''K'' on function spaces; the compactness property is shown by equicontinuity. The method of approximation by finite-rank operators is basic in the numerical solution of such equations. The abstract idea of Fredholm operator is derived from this connection. == Equivalent formulations == A bounded operator ''T : X → Y'' is compact if and only if any of the following is true * Image of the closed unit ball in ''X'' under ''T'' is relatively compact in ''Y''. * Image of any bounded set under ''T'' is relatively compact in ''Y''. * Image of any bounded set under ''T'' is totally bounded in ''Y''. * there exists a neighbourhood of 0, $U\backslash subset\; X$, and compact set $V\backslash subset\; Y$ such that $T(U)\backslash subset\; V$. * For any sequence $(x\_n)\_$ from the unit ball in ''X'', the sequence $(Tx\_n)\_$ contains a Cauchy subsequence. Note that if a linear operator is compact, then it is easy to see that it is bounded, and hence continuous. 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「Compact operator」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.