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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.

Idonknow
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  • As you remark they are not isometric. The lengths of the boundary of the unit ball is different for every $p$, but this must be an invariant of the norm. – s.harp Jul 19 '20 at 12:52
  • @s.harp I don't quite follow that. Not knowing exactly how your reasoning goes, I can't tell whether the case $p=1$, $q=\infty$ shows it must be invalid... – David C. Ullrich Jul 19 '20 at 12:55
  • @DavidC.Ullrich if $\gamma$ is a piecewise differentiable curve around the circumference of the unit ball then $\int |\gamma'(t)|_p,dt$ is an invariant. The resulting value is $$4\int_0^1 dt \sqrt[p]{1+(1-t^p)^{1-p}t^{p-p^2}}$$ which is a harder integral than I thought it would end up being, but for example with $p=1$ its 8, whereas for $p=2$ its $2\pi$ and for $p=3$ its something like $6.51..$ – s.harp Jul 19 '20 at 13:42
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    Although numerical testing the integral appears to suggest that it is the same for $\frac1p+\frac1q=1$. – s.harp Jul 19 '20 at 13:52

2 Answers2

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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 $y$.

quid
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David C. Ullrich
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  • But I exclude the possibility that $q=\infty$ as I know that one can be obtained by scaling and rotation from the other. – Idonknow Jul 19 '20 at 14:19
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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.

s.harp
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