Let $E,F$ be finite dimensional Banach spaces and define $\delta(E,F)=\inf\{\lVert T\rVert\lVert T^{-1}\rVert|\ T:E\to F\text{ is isomorphism}\}$

Let $$E,F$$ be finite dimensional Banach spaces and define $$\delta(E,F):=\inf\{\lVert T\rVert\lVert T^{-1}\rVert|\ T:E\to F\text{ is an isomorphism}\}$$. Prove that, $$\delta(E,F)=1$$ iff $$E$$ and $$F$$ are isometric.

For any isomorphism $$T:E\to F$$, $$1=\lVert id_F\rVert=\lVert T\circ T^{-1}\rVert\le \lVert T\rVert\lVert T^{-1}\rVert$$, this implies $$\delta(E,F)\ge 1$$.

Suppose $$T:E\to F$$ is an isometry then $$\lVert T\rVert=1=\lVert T^{-1}\rVert\implies\delta(E,F)\le1\implies\delta(E,F)=1$$.

I'm stuck with the converse part. But I have observed the following-

If there is an isomorphism $$T:E\to F$$ such that $$\lVert T\rVert\lVert T^{-1}\rVert=1$$, then $$\lVert T^{-1}(Tx)\rVert\le \lVert T\rVert^{-1}\lVert Tx\rVert\implies \lVert T\rVert\lVert x\rVert\le \lVert Tx\rVert\le\lVert T\rVert\lVert x\rVert\implies \lVert Tx\rVert=c\lVert x\rVert\implies \lVert (c^{-1}T)x\rVert=\lVert x\rVert$$ where $$c=\lVert T\rVert>0$$. Hence, $$S:=c^{-1}T$$ is isometry between $$E$$ and $$F$$.

But I don't know whether the set $$\{\lVert T\rVert\lVert T^{-1}\rVert|\ T:E\to F\text{ is isomorphism}\}$$ is closed or not, if yes then $$1=\delta(E,F)$$ belong to the set and the above observation will complete the proof.

Can anyone help me to finish the proof? Thanks for your help in advance.

• Use the (sequential) compactness of the unit ball of the space of linear maps from $E$ to $F$, equipped with the operator norm. Commented Sep 18, 2022 at 21:10
• Got it, Thanks. Commented Sep 18, 2022 at 21:48

Suppose $$\delta(E, F) = 1$$. Then there exists a sequence $$T_n$$ such that $$\delta_n := \|T_n\|\|T_n^{-1}\| - 1 \to 0$$ as $$n \to \infty$$. By replacing $$T_n$$ by $$T_n/\|T_n\|$$ and hence $$T_n^{-1}$$ by $$\|T_n\|T_n^{-1}$$, we can assume without loss of generality that $$\|T_n\| = 1$$ (and so $$\delta_n = \|T_n^{-1}\| - 1$$). Then, $$(\delta_n + 1) \|x\| = \|T_n^{-1}\| \|x\| \ge \|T_n^{-1}x\| \ge \|T_nT_n^{-1}x\| = \|x\|.$$ So, for all $$x$$ and $$n$$, $$0 \le \|T_n^{-1}x\| - \|x\| \le \delta_n \|x\|,$$ or equivalently, $$0 \le \|x\| - \|T_n x\| \le \delta_n \|T_nx\|.$$ As $$\|T_n x\| \le \|x\|$$, $$0 \le \|x\| - \|T_n x\| \le \delta_n \|x\|.$$ This implies that $$\|T_n x\| \to \|x\|$$ as $$n \to \infty$$, for all $$x$$.
Now we just show that $$T_n$$ has a pointwise limit point. It's not difficult to see that the pointwise limit of a sequence of linear functions is linear by definition. The continuity of the norm (on $$F$$) would show that such a limit point would be an isometry.
This is where finite-dimensionality comes in. Consider a basis $$e_1, \ldots, e_m$$ of $$E$$. Since $$\|T_n e_1\| \le \|e_1\|$$, we have a bounded sequence $$(T_n e_1)_n$$. The compactness of the unit ball shows that a subsequence $$(T_{n_k} e_1)_k$$ has a limit.
Then, considering $$(T_{n_k} e_2)_k$$, we can take another subsequence that converges. We do the same for $$e_3$$, etc, up to $$e_m$$. So, without loss of generality (replacing $$T_n$$ by the finest of these subsequences), we can assume $$(T_n e_i)_n$$ converges to some $$f_i \in F$$, for $$i = 1, \ldots, m$$.
Any $$x \in E$$ is a linear combination of $$e_1, \ldots, e_m$$ of the form $$x = \alpha_1 e_1 + \ldots + \alpha_m e_m,$$ so $$T_n x = \alpha_1 T_n e_1 + \ldots + \alpha_m T_n e_m \to \alpha_1 f_1 + \ldots + \alpha_m f_m$$ as $$n \to \infty$$. Thus, we have pointwise convergence of this (sub)sequence, to a map that must be linear, and preserves norms (pointwise), i.e. an isometry.
• Nice answer. As a style comment, the second part of the answer is the proof that closed balls in a finite-dimensional space (in this case, the space of linear maps $E\to F$) is compact. Maybe one can simply mention this fact to make the answer more compact. Commented Sep 19, 2022 at 15:45