How to show $\ell^2_p$ not isometric to $\ell^2_{p'}$ unless $p\in\{1,2,\infty\}$?

Question

Let $\ell^n_p$ denote $\mathbb{R}^n$ with the $p$-norm; assume $n>1$. Now it's well-known that for $p\ne q$, $\ell^n_p$ and $\ell^n_q$ are never isometric unless $n=2$ and $\{p,q\}=\{1,\infty\}$, but I realized recently I didn't actually know how to prove this.

I can prove most of it - the only case I still need to rule out is the case where $n=2$ and $q=p'$ (i.e. $\frac{1}{1-1/p}$). Now of course, it is indeed possible that $q=p'$, because $p$ could be $2$ (so $p=q$) or $1$ or $\infty$ (the sole exceptional case). But the problem then is showing that $p$ must actually be one of those three.

Furthermore I can show that if there is such an isometry, then, after applying a translation and some signed permutation matrix, it must be $\lambda R$, where $\lambda=2^{1/p-1/2}=2^{1/2-1/p'}$ and $R$ is rotation by $\pi/4$. So actually all we need to do is, given $p\notin \{1,2,\infty\}$, to find a point $v$ such that $||\lambda R v||_{p'}\ne ||v||_p$.

Now, graphing things, we can see that in fact any point that is not on the axes or the lines $y=\pm x$ will work; but I do not know how to prove it for even a single specified point, even, say, $(1,2)$, because the resulting equation for $p$ (assuming $p\ne 1,\infty$, of course) is just absolutely terrible, and attempting to, say, determine when each side is larger by differentiating seems to only make things worse.

Addendum Nov 21: I should perhaps point out - obviously the point can vary with $p$, but I have no idea how given $p$ one might pick a specific point that makes it easier to prove - especially since it is after all true for nearly any point!

So how is this case handled? And is there by any chance a nicer approach that I'm missing?

Answer 1

You seem to have solved this on your own with "bare hands" but I wanted to point out some connections that this problem has with slightly more general concepts from Banach space theory. It is a little sad that nobody else has pointed these things out yet, but not surprising; Banach space theory does not seem to be very popular on the mathematical Internet. (I suspect if you had posted this problem on Mathoverflow, a few people would vote to close it as "too specialized" out of their own ignorance, and then gone off to write answers to much simpler, much more specific problems in algebraic geometry.) In what follows I write $\ell^p$ for what you have written $\ell^n_p$; suppressing the dependence on $n$ since it can clearly be determined from the dimension.

One way to distinguish these spaces is with the modulus of convexity. For $X$ a Banach space, the modulus of convexity of $X$ is the function $\delta_X: (0,2] \to [0,1]$ given by $$ \delta_X(\epsilon) = \inf \{1 - \frac{1}{2} \|x+y\|: x, y \in X, \|x\|=\|y\|=1, \|x-y\| \geq \epsilon\}. $$ (There is not total agreement on this; you will see the data encoded slightly differently in the literature. For example what I have just written is what some people would call $\delta_X(\epsilon/2)$, and they would define it only for $\epsilon \in (0,1]$. Keep your eyes open.) It is evident from the definition that the modulus of convexity is invariant under isometric isomorphism, that is, if $X$ and $Y$ are Banach spaces and there is an isometric isomorphism $f: X \to Y$, then $\delta_X = \delta_Y$ as functions on $(0,2]$.

To compute the modulus of convexity for $\ell^p$, go and prove Hanner's inequality for $\ell^p$. For $p \in [1,2]$ this is the statement $$ \|x+y\|_p^p + \|x-y\|_p^p \geq (\|x\|_p + \|y\|_p)^p + | \|x\|_p - \|y\|_p|^p, \qquad x, y \in \ell^p, $$ and for $p \in (2, \infty)$ the inequality reverses. There are various nice proofs of Hanner's inequality in the literature--- Hanner's original proof is pretty straightforward, although maybe not the simplest. My advice would be to search Math Reviews for a proof that is to your liking.

A bit of work, that I omit, shows that Hanner's inequality completely determines the modulus of convexity for $\ell^p$, in that you can write down an actual formula for it. And once you have done that, you see that different $p$s give you different moduli of convexity. (For $p \geq 2$ it is something like $1 - (1 - (\epsilon/2)^p)^{1/p}$; for $p \in [1,2)$ the formula is more simply written with $\delta$ as an "implicit" function of $\epsilon$.) The details are in Hanner's original paper, "On the uniform convexity of $L^p$ and $\ell^p$", if you don't want to supply them yourself. (To recover $p$ from the formula in the $[2, \infty)$ case, you can look at the value at $\epsilon = 1$, for example, and use calculus to show that this is strictly decreasing in $p$. Or you could look at the limit of the second $\epsilon$-derivative as $\epsilon$ approaches $0$; I think it's a constant multiple of $(p-1)$. In any case this distinguishes all the $\ell^p$ with $p \in [2, \infty)$ from one another, and doing the calculation for $p \in [1,2)$ you can finish the job in a similar way.)

If you read up on the modulus of convexity you see it has a very geometric interpretation (it quantifies in some sense "how convex" the unit ball of $X$ is) and contains many of the same ideas you were probably getting at with an ad hoc approach.

This problem is also solved in Banach's textbook (Theory of linear operators), I think, in another way. I don't have the book on me at the moment but the idea is that any sequence $(x_k)_{k=1}^{\infty}$ in $\ell^p$ that converges to $0$ has a subsequence $(x_{k_j})_{j=1}^{\infty}$ with the property that the function $n \mapsto \|\sum_{j=1}^n x_{k_j}\|$ is $O(n^{1/p})$, and if you think about this long enough, this can be used to distinguish the $\ell^p$ from one another. There is a more modern or high tech approach that abstracts this idea to an invariant defined more generally in Banach spaces (e.g. you can distinguish the $\ell^p$ using the indices of "Rademacher type" and "Rademacher cotype").

Answer 2

Here's an idea, although I don't have time to try it: draw lines at angles of multiples of $\frac{\pi}{4}$ from the origin to both of the unit circles you're investigating and compute the curvature at the corresponding intersection points. If you can show that the curvatures don't match up, the corresponding unit circles can't be related by a scaling and a rotation.

Answer 3

An elementary approach involving the third mixed partial derivatives shows that such an isometry in the form given in the question is only possible with $p=q=2$.

Assume $1< p,\ q < \infty$ and $1/p + 1/q = 1$. Assume also that there exist an isometry of the given form, i.e., a normalized rotation of $\pi/4$ between $(\mathbb R^2,\|\ \|_p)$ and $(\mathbb R^2,\|\ \|_q)$. Then the corresponding matrix would be $A = \lambda'\left(\matrix{1 & 1\\1 & -1}\right)$ for some scalar $\lambda'$. It follows that $$|x|^p + |y|^p = \lambda ''(|x + y|^q + |x - y|^q)^{p/q}$$ for some constant $\lambda''$.

If we compute¹ the mixed third derivative $\frac {\partial^3}{\partial x \partial y^2}$ of both sides at the point $(1,0)$, after simplifications we obtain

$$0 = (-2+q) (-1+q)+\left(-1+\frac{p}{q}\right) (-1+q) q.$$

This plus the assumptions on $p$ and $q$ gives us that the unique solution is $p=2$ and $q =2$.

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¹ with the help of Mathematica:

D[((x + y)^q + (x - y)^q)^(p/q), x, y, y] /. {x -> 1, y -> 0}