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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.

math maniac.
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    Hint: one of them is a Hilbert space, one of them is not. In order to show that the first of them is not a Hilbert space it's possible to use the parallelogram law. – Botnakov N. Dec 17 '20 at 14:33
  • @Botnakov N. another way to think about it is I think isometry takes lines to lines. If my assumption is true then these two spaces are certainly not isometric since if these two spaces are isometric then the open ball in $(\Bbb R^2, |\cdot|_2)$ is a square which is not true. – math maniac. Dec 17 '20 at 14:48

2 Answers2

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

Sam
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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 $p\neq2$.

Nicolas
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