# $\ell^p$ is not isometric to $\ell^q$

The problem is this: if $1\le p<q<\infty$ then $\ell^p$ and $\ell^q$ are not isometric (as Banach spaces).
This is an exercise but I'd like to see an elegant proof.

• They aren't even isomorphic to each other. See this. I do not know of a simpler argument showing they aren't isometric. Commented Dec 25, 2013 at 11:09
• I'd like to know the reason for the downvote. I know this fact is well-known and that Bertrand's postulate immediately follows from the Prime Number Theorem, but this does not imply that this question isn't interesting on its own. Commented Dec 25, 2013 at 16:41
• I find it interesting as well. I imagine someone wanted you to "show your work". (Where did you see this by the way?) Commented Dec 25, 2013 at 16:58
• I wonder if you can compute the modulus of convexity for these spaces... en.wikipedia.org/wiki/Modulus_of_convexity Commented Dec 27, 2013 at 14:37

Remark As is well-known, the Banach spaces $c_0$ and $\ell^p$ for $1\leq p<\infty$ are mutually non-isomorphic, as a consequence of Pitt's theorem. See Corollary 2.1.6 in Albiac-Kalton's book. It seems a bit easier to prove the weaker result saying that the $\ell^p$ spaces are mutually non-isometric.

Note that $\ell^1$ is not strictly convex, while Clarkson's inequalities show that $\ell^p$ is strictly convex for every $1<p<\infty$. So $\ell^1$ is not isometric to $\ell^p$ for any $1<p<\infty$. We could obtain the same conclusion using reflexivity or uniform convexity instead of strict convexity.

Now assume there exist $1<p,q<\infty$ such that $\ell^p$ and $\ell^q$ are isometric via some isometry $T$: $\|Tx\|_p=\|x\|_q$ for every $x\in \ell^q$. We need to show that $p=q$.

Sketch An elementary manipulation of Clarkson's inequalities (see here for a short proof by R.P. Boas) yields Claim 1 below. Claims 2,3,4 follow immediately. This allows us to rule out every case but $q=p$ and $q=p'$. So it only remains to check that $\ell^2$ is the only $\ell^p$ space which is isometric to its dual (i.e. Claim 5, essentially).

We will denote $p'$ the conjugate exponent associated with $p$, i.e. $\frac{1}{p}+\frac{1}{p'}=1 \iff p'=\frac{p}{p-1}$. Recall that the continuous dual $(\ell^p)'$ is isometric to $\ell^{p'}$ for every $1<p<\infty$.

Claim 1 If $p\geq 2$, then $p'\leq q\leq p$.

Proof For every $x,y \in \ell^q$, we have $$\big\| \frac{x+y}{2} \big\|_q^p+\big\| \frac{x-y}{2} \big\|_q^p=\big\| \frac{Tx+Ty}{2} \big\|_p^p+\big\| \frac{Tx-Ty}{2} \big\|_p^p$$ $$\leq \frac{\|Tx\|_p^p+\|Ty\|_p^p}{2}=\frac{\|x\|_q^p+\|y\|_q^p}{2}$$ where we used Clarkson's inequality for $p\geq 2$. In particular, for $x=(1,1,0,\ldots)$ and $y=(1,-1,0,\ldots)$, this yields $$2=1+1\leq \frac{2^\frac{p}{q}+2^\frac{p}{q}}{2}=2^\frac{p}{q}\quad\Rightarrow\quad q\leq p$$ While for $x=(2,0,\ldots)$ and $y=(0,2,0,\ldots)$, we get $$2\cdot 2^\frac{p}{q}=2^\frac{p}{q}+2^\frac{p}{q} \leq \frac{2^p+2^p}{2}=2^p \quad\Rightarrow\quad p'=\frac{p}{p-1}\leq q$$ Therefore we have $p'\leq q\leq p$. $\Box$

Claim 2 If $p\leq 2$, then $p\leq q\leq p'$.

Proof Taking the dual, $\ell^p\simeq \ell^q$ yields an isometric isomorphism $\ell^{p'}\simeq \ell^{q'}$. Since $p'\geq 2$, Claim 1 applied to $p'$ implies $(p')'\leq q'\leq p'\iff p\leq q\leq p'$. $\Box$

Claim 3 If $p$ and $q$ are both $\geq 2$ or both $\leq 2$, then $p=q$.

Proof Assume $p,q\geq 2$. Applying Claim 1 to $p$ yields $q\leq p$, while applying it to $q$ gives $p\leq q$. Hence $p=q$. The case $p,q \leq 2$ follows from Claim 2 in a similar manner. Or you can just take the duals and conclude that $p'=q'$ since they are both $\geq 2$. $\Box$

Claim 4 If $p\geq 2$ and $q\leq 2$, then $q=p'$.

Proof By Claim 1 for $p$, we have $p'\leq q$. By Claim 2 for $q$, we get $p\leq q' \iff q\leq p'$. Hence $q=p'$. $\Box$

Claim 5 If $\ell^p$ is isometric to $\ell^{p'}$, then $p=p'=2$.

Proof Still pondering what the easiest argument could be...

[Edit by the OP: see my answer below, which should prove Claim 5 and finish the problem off.]

• Nice! How do you rule out the case $p=q',q=p'$? Commented Jan 1, 2014 at 2:57
• Just curious, is $\ell_{p}$ isomorphic to $\mathbb{R}^n$? Commented Nov 9, 2020 at 20:53

Let me complete the accepted answer by showing that $\ell^p$ is not isometric to $\ell^q$ if $1<p<2$ and $2<q<\infty$. Fix $1<r<\infty$. I claim that

Claim 5' The second derivative $\frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}$ exists for every $x\in\ell^r\setminus\{0\}$ and every $y\in\ell^r$ if and only if $r\ge 2$.

Clearly, the claim implies the first assertion and thus finishes the problem.

Proof If $r<2$ then we take $x:=e_1$, $y:=e_2$ and we see that the second derivative $\frac{d^2}{dt^2}(1+|t|^r)^{1/r}\Big|_{t=0}$ does not exist. On the other hand, let $r\ge 2$: we claim that \begin{aligned}\frac{d}{dt}\|x+ty\|_{\ell^r}\Big|_{t=0}&=\sum_{n=1}^\infty x'_ny_n=:A(x,y), \\ \frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}&=(r-1)\sum_{n=1}^\infty x''_ny_n^2-\frac{(r-1)}{\|x\|}\Big(\sum_{n=1}^\infty x'_ny_n\Big)^2=:B(x,y),\end{aligned} where $x'\in\ell^{r/(r-1)}$ and $x''\in\ell^{r/(r-2)}$ are defined by $$x'_n:=\frac{|x_n|^{r-2}x_n}{\|x\|_{\ell^r}^{r-1}},\quad x''_n:=\frac{|x_n|^{r-2}}{\|x\|_{\ell^r}^{r-1}}.$$ Notice that $x'$ and $x''$ depend continuously on $x\neq 0$ (which is easily seen by dominated convergence) and $\|x'\|_{\ell^{r/(r-1)}}=1$, $\|x''\|_{\ell^{r/(r-2)}}=\|x\|_{\ell^r}^{-1}$. So, by the generalized Holder's inequality, $A(x,y)$ and $B(x,y)$ are continuous and satisfy $|A(x,y)|\le\|y\|_{\ell^r}$, $|B(x,y)|\le 2(r-1)\|x\|_{\ell^r}^{-1}\|y\|_{\ell^r}^2$.

These formulas are a straightforward computation if $x,y\in c_c$ ($c_c$ being the space of sequences where only finitely many terms are nonzero). In order to check that they are true in general, we observe that for $x,y\in c_c$ $$\|x+ty\|_{\ell^r}=\|x\|_{\ell^r}+tA(x,y)+\int_0^t(t-s)B(x+sy,y)\,ds$$ (Taylor's expansion with remainder in integral form) and, by a density argument and dominated convergence, we see that this equation holds for all $x\in\ell^r\setminus\{0\}$ and all $y\in\ell^r$. Hence, \begin{aligned}\frac{d}{dt}\|x+ty\|_{\ell^r}&=A(x,y)+\int_0^t B(x+sy,y)\,ds, \\ \frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}&=B(x,y).\end{aligned} This finishes the proof. $\square$

I think it should work.

Suppose for definiteness $p < q$. Then $\ell^p \subset \ell^q$. Assume that an isometry exists $J: \ell^p \to \ell^q$. Then $J(\ell^p) \subset \ell^q$ and in particular $J(B_{\ell^p}) \subset B_{\ell^q}$, where $B_X$ is the unit ball of $X$. Let us choice $r >0$ s.t. $r B_{\ell^q} \subset J(B_{\ell^p})$ (such an $r$ must exist, since $J(B_{\ell^p})$ is nonempty). Being an isometry, $J$ is an isomorphism from $\ell^p$ onto $J(\ell^p)$, i.e. into $\ell_q$, so by Pitt's theorem it must be compact. But $J(B_{\ell^p})$ can't be compact, since $rB_{\ell^q}$ is not. Hence $J$ can't be an isometry.

Remark 1. Take with care! It seems to me that it works but wait for community peer-review!

Remark 2. It is well known that every Banach separable space is isometrically embedded into a closed subspace of $\ell^\infty$, so we are correctly considering only the case $p,q < \infty$. The proof of this fact relies on the Hahn-Banach theorem; maybe one can adapt it to our case.

• Most of users were aware of this proof. The problem is to find an elementary proof. Commented Dec 27, 2013 at 12:29
• What do we extacly mean by "elementary"? A proof based only upon definitions? In my opinion, a proof is elementary if it involves only concepts usually taught in basic courses. In both senses, however, I agree that the demonstration I wrote is not elementary (but I believed we were looking for an "elegant" proof, that is not the same) and obviously I was expecting that a reasoning was widely known. (is too easy not to be!) :) Commented Dec 30, 2013 at 15:41

WARNING: This answer is based on a (most probably) false assumption and so it is (most probably) wrong. See comments

Here's a partial answer which works when $p$ and $q$ are even integers. This extra assumption is used in the shaded area below.

If a linear isometry of $\ell^p$ onto $\ell^q$ existed, its restriction to the subspace $$\left\{ (x_1, x_2, 0, 0 \ldots)\ :\ x_1, x_2\in \mathbb{R}\right\}\subset \ell^p$$ would yield a linear isometry of $(\mathbb{R}^2, \lvert\cdot\rvert_p)$ onto $(\mathbb{R}^2, \lvert\cdot\rvert_q)^{[1]}$. Therefore we only need to show that the latter cannot exist.

Since $p$ is an even integer, the $\lvert\cdot\rvert_p$-unit circle is an algebraic curve of degree $p$: $$S_p=\left\{ (x_1, x_2)\ :\ x_1^p+x_2^p=1\right\},$$ and any nonsingular linear operation transforms it into an algebraic curve of the same degree. This rules out the existence of a linear isometry of $(\mathbb{R}^2, \lvert\cdot\rvert_p)$ onto $(\mathbb{R}^2, \lvert\cdot\rvert_q)$ because such a mapping would transform $S_p$ into $S_q$ and the latter has degree $q$.

$^{[1]}$ Where $\lvert (x_1, x_2)\rvert_p=\left( \lvert x_1\rvert^p+\lvert x_2\rvert^p\right)^{\frac{1}{p}}.$

• I don't understand what you mean. $S_p$ is carried into a subset of $\sum |x_i|^q=1$: your answer seems to imply that actually its image is of the form $(0,\dots,x_m,0,\dots,x_n,0,\dots)$ (for two distinct $m<n$, which wlog are $1,2$) but why is it so? Commented Dec 30, 2013 at 14:12
• @Mizar good catch! Commented Dec 30, 2013 at 16:36
• @Mizar: You are right that I was being careless: actually I was implicitly using the following fact, which I think is true but I cannot prove. Fact? Every 2-dimensional subspace of $\ell^q$ is isometric to $(\mathbb{R}^2, \lvert\cdot\rvert_q)$. Commented Dec 30, 2013 at 22:08
• If the "Fact?" is true, assuming by contradiction that a linear isometry $J\colon \ell^p\to\ell^q$ exists, and setting $$E_p=\mathrm{span}(e_1, e_2)\subset\ell^p,$$ $$E_q=J(E_p),$$ we have an induced isometry of $E_p$ onto $E_q$. Now $E_p$ is isometric to $(\mathbb{R}^2, \lvert\cdot\rvert_p)$ and, if the "Fact?" is true, $E_q$ is isometric to $(\mathbb{R}^2, \lvert\cdot\rvert_q)$. We can then proceed to prove that an isometry of $(\mathbb{R}^2, \lvert\cdot\rvert_p)$ onto $(\mathbb{R}^2, \lvert\cdot\rvert_q)$ cannot exist. Commented Dec 30, 2013 at 22:13
• I'm not sure the "fact" is true. For instance over the reals, $\ell_2^2$ embeds isometrically in $\ell_4^6$. This is shown in Proposition 18, Chapter 21 of the Handbook of the Geometry of Banach Spaces, vol 1. So, $\ell_4$ contains a two-dimensional subspace isometric to $\ell_2^2$. Of course, $\ell_2^2$ is not isometric to $\ell_4^2$. (Superscripts are dimensions.) Commented Dec 31, 2013 at 15:00