TITLE: $\ell^p$ is not isometric to $\ell^q$
QUESTION [13 upvotes]: 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.

REPLY [4 votes]: 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. $\square$