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.

  • $\begingroup$ They aren't even isomorphic to each other. See this. I do not know of a simpler argument showing they aren't isometric. $\endgroup$ – David Mitra Dec 25 '13 at 11:09
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    $\begingroup$ 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. $\endgroup$ – Mizar Dec 25 '13 at 16:41
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    $\begingroup$ I find it interesting as well. I imagine someone wanted you to "show your work". (Where did you see this by the way?) $\endgroup$ – David Mitra Dec 25 '13 at 16:58
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    $\begingroup$ I wonder if you can compute the modulus of convexity for these spaces... en.wikipedia.org/wiki/Modulus_of_convexity $\endgroup$ – GEdgar Dec 27 '13 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.]

  • $\begingroup$ Nice! How do you rule out the case $p=q',q=p'$? $\endgroup$ – Mizar Jan 1 '14 at 2:57

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.

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    $\begingroup$ Most of users were aware of this proof. The problem is to find an elementary proof. $\endgroup$ – Norbert Dec 27 '13 at 12:29
  • $\begingroup$ 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!) :) $\endgroup$ – Federico Dec 30 '13 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 $\lvert (x_1, x_2)\rvert_p=\left( \lvert x_1\rvert^p+\lvert x_2\rvert^p\right)^{\frac{1}{p}}.$

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    $\begingroup$ 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? $\endgroup$ – Mizar Dec 30 '13 at 14:12
  • $\begingroup$ @Mizar good catch! $\endgroup$ – Norbert Dec 30 '13 at 16:36
  • $\begingroup$ @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 $(\mathbb{R}^2, \lvert\cdot\rvert_q)$. $\endgroup$ – Giuseppe Negro Dec 30 '13 at 22:08
  • $\begingroup$ 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. $\endgroup$ – Giuseppe Negro Dec 30 '13 at 22:13
  • $\begingroup$ 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.) $\endgroup$ – David Mitra Dec 31 '13 at 15:00

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