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?

  • $\begingroup$ I don't understand. In the first paragraph, do you mean $(\|\ \|_1, \mathbb R^2)$ is isometric to $(\|\ \|_\infty, \mathbb R^2)$ ? $\endgroup$
    – AgCl
    Nov 20, 2010 at 23:09
  • $\begingroup$ Yes. I wrote it that way because it was shortest way I could think of in context. $\endgroup$ Nov 20, 2010 at 23:15
  • $\begingroup$ I mean I don't see how they are isometric. I.e. $\|(3,4)\|_1 = 7$ but $\|(3,4)\|_\infty = 4$ $\endgroup$
    – AgCl
    Nov 20, 2010 at 23:18
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    $\begingroup$ No, he means that there exists a linear map from $\mathbb{R}^2$ to itself that is an isometry of those two norms. The map doesn't have to be the identity map. If you draw the level-sets of the norm, you see that the level sets of $\ell_1$ are diamonds, and of $\ell_\infty$ are squares, so if you just rotate by 45 degrees and re-scale, you get an isometric map. $\endgroup$ Nov 20, 2010 at 23:48
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    $\begingroup$ This is a great question. $\endgroup$ Nov 21, 2010 at 7:59

3 Answers 3


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").

  • $\begingroup$ So there was a much nicer approach I was missing! Thank you, I will have to check this out! Question: can Hanner's inequality be adapted to $p=\infty$? If not, that leaves that case to be done manually, but I suppose that's no problem since I know how to do so... and of course, once you leave finite dimension, $L^\infty$ becomes very different from the other $L^p$... $\endgroup$ Nov 23, 2010 at 8:00
  • $\begingroup$ I dunno if Hanner has a useful analogue when $p = \infty$, but you can easily compute the modulus of convexity there; it's identically $0$. (When $n=2$ consider $x=(1,1)$ and $y=(1,-1)$ in the infimum, and for larger $n$, just add $0$s.) (Terminology: if $\delta_X > 0$, $X$ is said to be "uniformly convex". $X$ lacks this property when its unit sphere contains a line segment.) As you may have noticed, the above ideas work equally well to distinguish the infinite dimensional $\ell^p$ spaces (or more generally, fixing a nice enough measure space $M$, the spaces $L^p(M)$) from one another. $\endgroup$
    – anon
    Nov 23, 2010 at 8:30
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    $\begingroup$ +1 on this very nice answer. As a side remark: if this were posted on MO, I would've been very vocal against its closure. In fact, if you hadn't provided this wonderful answer I may have asked Harry Altman for permission to repost the question there. $\endgroup$ Nov 23, 2010 at 11:22
  • $\begingroup$ Ah, OK, I didn't go ahead and actually compute that. That does still leave $l^1$ to be manually distinguished from $l^\infty$, of course. $\endgroup$ Nov 23, 2010 at 16:32
  • $\begingroup$ Also: Another nice thing about this is it should work equally well over $\mathbb{C}$. (I was relying on Mazur-Ulam - though I guess if you only count linear isometries in the first place that's not a problem.) $\endgroup$ Nov 24, 2010 at 22:27

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.

  • $\begingroup$ Lines have zero curvature, and one cannot speak of curvature at corners (remember that the unit circles for both $\ell_1$ & $\ell_{\infty}$ are squares). $\endgroup$ Nov 21, 2010 at 23:05
  • $\begingroup$ @J.M.: for the curve |x|^p + |y|^p = 1 where p is not equal to 1, 2, or infinity, the only corners are at the points (\pm 1, 0), (0, \pm 1). This leaves four other points to check the curvature at, which should give the desired result if I'm not horribly mistaken. $\endgroup$ Nov 21, 2010 at 23:09
  • $\begingroup$ Qiaochu, yes, I was noting only that for attempting to match up 1 and $\infty$, you have corners to contend with; otherwise, SFAIK as well that superellipses have no corners at all for $p > 1$ and if $p < 1$ (the case where corners/cusps are guaranteed), you're no longer considering norms anyway. $\endgroup$ Nov 21, 2010 at 23:20
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    $\begingroup$ I'll accept this, since looking at second derivatives ultimately gave me my answer. :) My actual solution: At an angle of $\pi/4$ (for $1<p<\infty$), the curve is infinitely differentiable. However, if 1<p<2, the curve is only once differentiable at the axes. Hence given p, p', assume p<2 WLOG, then the points on the axes for p don't match up with the points on the diagonals for p'. $\endgroup$ Nov 23, 2010 at 4:59

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 $q =2$.

1 with the help of Mathematica:

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


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