Prove that $(\Bbb{R}^n,\| \cdot \|_2)$ and $(\Bbb{R}^n,\| \cdot \|_\infty)$ are not isometrically isomorphic, if $n\geq 2$

Question

Prove that $(\Bbb{R}^n,\| \cdot \|_2)$ and $(\Bbb{R}^n,\| \cdot \|_\infty)$ are not isometrically isomorphic, if $n\geq 2$.

Where, of course, these norms are given by $$\begin{array}{cccc}\| \cdot \|_2:&\Bbb{R}^n&\to&\Bbb{R}\\ &(x_1,\dots,x_n)&\mapsto&\left(\sum\limits_{i=1}^\infty|x_i|^2\right)^{1/2}\end{array}$$ and $$\begin{array}{cccc}\| \cdot \|_\infty:&\Bbb{R}^n&\to&\Bbb{R}\\ &(x_1,\dots,x_n)&\mapsto&\max_{i=1,\dots,n}\{|x_i|\}\end{array}$$

Just in order to simplify notation, suppose $n=2$.

What I have tried: Suppose that exists a isometric isomorphism $T:(\Bbb{R}^2,\| \cdot \|_2)\to(\Bbb{R}^2,\| \cdot \|_\infty)$, say $$T(x,y)=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=(ax+by,cx+dy),$$ such that $\|x\|_2=\|T(x)\|_\infty,\forall x\in \Bbb{R}^2$, i.e. $$\sqrt{x^2+y^2}=\max\{|ax+by|,|cx+dy|\}$$

Then, choosing $x=(1,0),(0,1),(1,1)$, we can see that $$\begin{array}{rcl}1&=&\max\{|a|,|c|\}\\ 1&=&\max\{|b|,|d|\}\\ \sqrt{2}&=&\max\{|a+b|,|c+d|\}\end{array}$$

But I can't see how to reach a contradiction for form there neither keep choosing such $x$'s randomly...

After this, I didn't have any other ideia... Need some hint!

Answer 1

Hint: We can find two linearly independent vectors $x,y$ such that $\|x\|_\infty = \|y\|_\infty = 1$, and $\|\frac 12 (x+y)\|_\infty = 1$.

However, this would mean that $\|T^{-1}x\|_2 = \|T^{-1}y\|_2 = \|\frac 12(T^{-1}x+T^{-1}y)\|_2 = 1$, which is to say that we have three (distinct) colinear points on the (Euclidean) unit ball.

Answer 2

Isometric isomorphisms, being linear, send segments to segments.

A bijective isometry $T:(\Bbb R^n,\lVert\bullet\rVert_\infty)\to (\Bbb R^n,\lVert\bullet\rVert_2)$ must send bijectively $S^n_\infty=\{x\in\Bbb R^n\,:\,\lVert x\rVert_\infty=1\}$ to $S^n=\{x\in\Bbb R^n\,:\,\lVert x\rVert_\infty=1\}$.

But $S^n$ does not contain segments, while $S^n_\infty$, for $n\ge1$, does. So $S^n$ cannot be the image of $S^n_\infty$ under a linear map.