If $T:(\mathbb{R}^2,\|\cdot\|_p) \to (\mathbb{R}^2,\|\cdot\|_q)$ is an onto linear isometry, then must it be $p=q$?

Question

Question: Let $p,q\in [1,\infty)$ and suppose that that $T:(\mathbb{R}^2,\|\cdot\|_p) \to (\mathbb{R}^2,\|\cdot\|_q)$ is an onto linear isometry. Must it be $p=q$?

I think it is true as isometry preserves extreme points. However, it would be good if there is an elementary arguments.

Answer 1

At first I thought the answer was clearly yes; now I'm not sure. If we allow $q=\infty$ the answer is no: the unit balls of $L^1$ and $L^\infty$ are both squares, so you can transform one to the other. Sure enough, $T(x,y)=(x+y,x-y)$, $p=1$, $q=\infty$ is a counterexample.

Lemma. If $x,y\in\Bbb R$ then $\max(|x+y|,|x-y|)=|x|+|y|$.

Proof: If you don't see anything more elegant just consider the four cases determined by the sign of $x$ and $y$.

Answer 2

To provide an answer the question by reference: $\ell^p_n$ and $\ell^q_n$ are not isometric for $p\neq q$ and $1≤p<q<\infty$.

Check out this reference, Corollary 3.2 on page 15 states that for any $n$ and $2≤p≤q≤\infty$ or $1≤p≤q≤2$ you have $$d_{BM}(\ell_p^n, \ell_q^n) = n^{1/p-1/q}$$ where $d_{BM}(X,Y)=\inf\{ \|T\|\cdot\|T^{-1}\|\ \mid T:X\to Y\text{ linear isomorphism}\}$ is the Banach-Mazur distance, which is $1$ if two spaces are isometric. In particular if $p$ and $q$ lie on the same side of $2$ their $\ell_p$ spaces cannot be isometric. Additionally if they are on different sides and $\frac1p+\frac1q\neq2$ you get that (wlog $p<2$) $$d_{BM}(\ell^n_p,\ell_2^n)=n^{1/p-1/2}\neq n^{1/2-1/q}=d_{BM}(\ell^n_q,\ell^n_2)$$ which gives every case except for when $p,q$ are dual. For this case we can steal the proof that $\ell_p$ and $\ell_q$ are not isometric when $p$, $q$ are dual (and neither is $\infty$) from yet another source. Namely one can shows that $\frac{d^2}{dt^2}\|x+t\,y\|_p\lvert_{t=0}$ exists for all $x,y$ ($x\neq0$) iff $p≥2$, implying that $\ell_p$ and $\ell_q$ cannot be isometric for $p,q$ dual and not equal to $2$. See here, the computation also works for finite dimensions.

These considerations are a bit like killing an ant with a bomb, so I'm sure there is a simpler way.

In a comment above I mentioned that you could look at the length of the unit ball, which is given by the integral $I(p)=4\int_0^1 \sqrt[p]{1+t^{p-p^2}(1-t^p)^{1-p}}dt$. Numerics suggest that this is strictly decreasing in $p$ when $p≤2$, strictly increasing in $p$ when $p≥2$ and that when $p,q$ are dual you have that $I(p)=I(q)$. These properties (if they were true) would give that $\ell_2^p$ and $\ell_2^q$ cannot be isometric unless $p,q$ are dual, and the case when they are dual can then be covered by the computation with the derivative.