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 Post subject: Banach - Mazur
PostPosted: Thu Jul 14, 2016 1:27 pm 
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Suppose that \(\displaystyle{\left(X,||\cdot||\right)\,\,,\left(Y,||\cdot||\right)}\) are two isomorphic normed \(\displaystyle{\mathbb{R}}\) - spaces.

We define \(\displaystyle{d\,(X,Y)=\inf\,\left\{||T||\cdot ||T^{-1}||: T\in GL\,(X,Y)\right\}}\), where :

\(\displaystyle{GL\,(X,Y)}\) is the set of all linear and bounded operators \(\displaystyle{T:X\longrightarrow Y}\) which are

bijections and also \(\displaystyle{T^{-1}\in\mathbb{B}\,(Y,X)}\) .

Prove that \(\displaystyle{d\,(X,Y)\geq 1}\) .


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 Post subject: Re: Banach - Mazur
PostPosted: Thu Jul 14, 2016 1:28 pm 
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Posts: 426
Introduction

If \(\displaystyle{\left(S_{i},||\cdot||_{i}\right)\,,1\leq i\leq 3}\) are normed \(\displaystyle{\mathbb{R}}\) - spaces and

\(\displaystyle{f:S_{1}\longrightarrow S_{2}\,,g:S_{2}\longrightarrow S_{3}}\) are bounded linear operators, then

the function \(\displaystyle{g\circ f: S_{1}\longrightarrow S_{3}}\) is linear (easy) and bounded cause:

\(\displaystyle{\forall\,s_1\in S_{1}: ||(g\circ f)(s_1)||_{3}=g(f(s_1))\leq ||g||\cdot ||f(s_1)||_{2}\leq ||g||\,||f||\cdot ||s_1||_{1}}\) .

Since \(\displaystyle{||g\circ f||=\sup\,\left\{||(g\circ f)(s_1)||_{3}: s_{1}\in S_{1}\,,||s_{1}||_{1}\leq 1\right\}}\), we get :

\(\displaystyle{||g\circ f||\leq ||g||\cdot ||f||}\) .

Solution

Since the normed \(\displaystyle{\mathbb{R}}\) -spaces \(\displaystyle{\left(X,||\cdot||\right)\,,\left(Y,||\cdot||\right)}\) are isomorphic,

there exists \(\displaystyle{f\in GL\,(X,Y)}\), which means that \(\displaystyle{f:X\longrightarrow Y}\) is a linear and bounded operator

and \(\displaystyle{1-1}\) and onto \(\displaystyle{Y}\) with \(\displaystyle{f^{-1}\in \mathbb{B}\,(Y,X)}\) .

Therefore, the set \(\displaystyle{\left\{||T||\cdot ||T^{-1}||: T\in GL\,(X,Y)\right\}}\) is non-empty subset of \(\displaystyle{\mathbb{R}}\)

which is lower bounded by zero because \(\displaystyle{||T||\cdot ||T^{-1}||\geq 0\,,\forall\,T\in GL\,(X,Y)}\) and thus :

\(\displaystyle{d\,(X,Y)\in\mathbb{R}\cap\left[0,+\infty\right)}\) . Let now \(\displaystyle{T\in GL\,(X,Y)}\). Then,

\(\displaystyle{T\circ T^{-1}=Id_{Y}: Y\longrightarrow Y}\) and according to the introduction :

\(\displaystyle{||T||\cdot ||T^{-1}||\geq ||T\circ T^{-1}||=||Id_{Y}||=\sup\,\left\{||y||_{Y}: y\in Y\,,||y||_{Y}\leq 1\right\}=1}\), which means

that the real number \(\displaystyle{1}\) is a lower bound of the set

\(\displaystyle{\left\{||T||\cdot ||T^{-1}||: T\in GL\,(X,Y)\right\}}\), so:

\(\displaystyle{d\,(X,Y)=\inf\,\left\{||T||\cdot ||T^{-1}||: T\in GL\,(X,Y)\right\}\geq 1}\) .

Note

If \(\displaystyle{\left(X,||\cdot||\right)=\left(Y,||\cdot||\right)}\), then \(\displaystyle{d\,(X,Y)=d\,(X,X)=1}\), because the set

\(\displaystyle{GL\,(X,X)}\) contains the identity function of \(\displaystyle{X\,,Id_{X}: X\longrightarrow X}\) and also :

\(\displaystyle{||Id_{X}||\cdot ||Id_{X}^{-1}||=||Id_{X}||\cdot ||Id_{X}||=1\cdot 1=1}\), that is :

\(\displaystyle{d\,(X,X)=\min\,\left\{||T||\cdot ||T^{-1}|| : T\in GL\,(X,X)\right\}=1}\) .


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