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    证明:显然乘积空间$X imes Y$按范数$left| {left( {x,y} ight)} ight| = sqrt {{{left| x ight|}^2} + {{left| y ight|}^2}} $成为赋范线性空间,且容易证明它是完备的.

    接下来我们定义$T$的图像:[Gleft( T ight) = left{ {left( {x,Tx} ight):x in Dleft( T ight)} ight}]由于$T$为线性算子,则$Gleft( T ight)$为$X imes Y$的线性子空间,由$T$为闭算子知$Gleft( T ight)$为闭集,所以$Gleft( T ight)$本身按范数$left| {left( {x,y} ight)} ight|$成为$f{Banach}$空间.又$Dleft( T ight)$作为$X$的线性子空间也是闭的,即$Dleft( T ight)$本身也是$f{Banach}$空间.

    我们作$Gleft( T ight)$到$Dleft( T ight)$的算子$P$如下:[P:left( {x,Tx} ight) mapsto x,forall x in Dleft( T ight)]这显然是线性算子,且[left| {Pleft( {x,Tx} ight)} ight| = left| x ight| le left| {left( {x,Tx} ight)} ight|]所以$P$是有界的,又易知$P$是$Gleft( T ight)$到$Dleft( T ight)$上的双射,于是由$f逆算子定理$知,${{P^{ - 1}}}$是有界的,即[left| {left( {x,Tx} ight)} ight| = left| {{P^{ - 1}}x} ight| le left| {{P^{ - 1}}} ight|left| x ight|]从而可知[left| {Tx} ight| le left| {left( {x,Tx} ight)} ight| le left| {{P^{ - 1}}} ight|left| x ight|]所以$T$是有界的,即是连续的

    $f注1:$$f(引理)$$T$为闭算子的充要条件是$Gleft( T ight)$为闭集

    方法一

    $f注2:$$f(定理)$定义域为闭集的连续算子是闭算子

    方法一

    $f注3:$$f(引理)$设$X,Y$是两个度量空间,$T$是$Dleft( T ight) subset X$到$Y$中的算子,则$T$为闭算子的充要条件是对任何点列$left{ {{x_n}} ight} subset Dleft( T ight)$,当${x_n} o {x_0},{y_n} = T{x_n} o {y_0}$时,有${x_0} in Dleft( T ight)$,且$T{x_0} = {y_0}$

    方法一

    $f注4:$

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  • 原文地址:https://www.cnblogs.com/ly758241/p/3789897.html
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