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 Post subject: Banach - Mazur Posted: Thu Jul 14, 2016 1:27 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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 $\displaystyle{d\,(X,Y)\geq 1}$ .

Top   Post subject: Re: Banach - Mazur Posted: Thu Jul 14, 2016 1:28 pm
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Joined: Mon Nov 09, 2015 1:52 pm
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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