Counter-example concerning isometry.

Question

Can $(\Bbb R^2 , \|\cdot\|_1)$ and $(\Bbb R^2, \|\cdot\|_2)$ be isometric?

I think that these two spaces can't be isometric. But I can't able to show that. Would anybody please help me in this regard?

Thanks in advance.

Answer 1

(Apologies, I cannot comment). This question is a duplicate, and already has answers here.

Answer 2

Assume that $U:(\mathbb{R}^d,\|\cdot\|_1)\to(\mathbb{R}^d,\|\cdot\|_2)$ is an isometry. Then \begin{align*} \|Ux+Uy\|_2^2+\|Ux-Uy\|_2^2&=2\|Ux\|_2^2+2\|Uy\|_2^2 \end{align*} by the parallelogram law. On the other hand, as $U$ is an isometry,\begin{align*} \|x+y\|_1^2+\|x-y\|_1^2&=2\|x\|_1^2+2\|y\|_1^2. \end{align*} Thus $\|\cdot\|_1$ also satisfies the parallelogram law, thus comes from a scalar product, a contradiction. The conclusion is the same if we replace $\|\cdot\|_1$ by any $\|\cdot\|_p$ with $p\neq2$.