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.

  • 2
    $\begingroup$ Use the (sequential) compactness of the unit ball of the space of linear maps from $E$ to $F$, equipped with the operator norm. $\endgroup$ Commented Sep 18, 2022 at 21:10
  • $\begingroup$ Got it, Thanks. $\endgroup$
    – MathBS
    Commented Sep 18, 2022 at 21:48

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Sep 19, 2022 at 15:45

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