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Let $X,Y$ be isomorphic Banach spaces.

The Banach-Mazur distance:

$d(X,Y)=\inf \{\| T\| \| T^{-1}\| : T:X\rightarrow Y \text{is an isomorphism} \}$

can be rewritten as:

$d(X,Y)=\inf \{\| T^{-1}\| : T:X\rightarrow Y \text{is an isomorphism}, \|T\|=1 \}$.

If $X,Y$ are finite dimensional, show that the infimum is reached.

After thinking a bit, my thoughts are as follows. If $X,Y$ are finite dimensional Banach spaces, then so is $L(X,Y)$. Choose $T_n\in L(X,Y)$ such that $\|T_n^{-1}\|\rightarrow d(X,Y)$ with $\|T_n\|=1$ for all $n$. Then each $T_n$ is in the closed unit ball of $L(X,Y)$, which is compact since it is finite-dimensional. Then there exists a subsequence $T_{n_k}$ that converges in the unit ball in $L(X,Y)$; say $\|T_{n_k}-T\| \rightarrow 0$, where $T$ is in the closed unit ball in $L(X,Y)$.

The last thing to show is that $T$ is invertible, which is likely true by taking $\|T^{-1} \|=\lim_{k\rightarrow \infty}\|T_{n_k}^{-1}\|=d(X,Y)$, but I have not justified this yet.

Any comments would be helpful. Thank you.

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1 Answer 1

The idea is good. I prefer to think in terms of lower bound $m(T)=\inf\{\|Tx\|:\|x\|=1 \}$, which is the same thing as $\|T^{-1}\|^{-1}$ when $T$ is invertible, but does not require us to consider the inverse explicitly.

The definition of lower bound implies that $|m(T)-m(S)|\le \|T-S\|$ for any two operators $T,S$. Therefore, $T_{n_k}\to T$ implies $m(T_{n_k})\to m(T)$. In particular, $m(T)>0$ which makes $T$ invertible. And since $\|T\|/m(T)=d(X,Y)$, you are done.

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