$\ell_p$ is Hilbert if and only if $p=2$

Question

Can anybody please help me to prove this:

Let $p$ be greater than or equal to $1$.

Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p<\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $<\_\,|\,\_> $ s.t. $||u||^2=<u\,|\,u>$ if and only if $p=2$.

Hint: Prove that if $p=2$, the parallelogram law is satisfied. WHen $p\neq 2$ give a counter example to the parallelogram law . — Tomás, Oct 18 '12 at 13:11
@ Tomas,but this is an if and only if proof...I can prove the case when p=2.that is the sufficient part.but the converse cannot. — ccc, Oct 18 '12 at 13:12
Let $e_1 = (1, 0, \ldots)$, $e_2 = (0, 1, 0, \ldots)$. Now compare $|e_1 + e_2|^2_p + |e_1 - e_2|_p^2$ and $2|e_1|_p^2 + 2|e_2|_p^2$ — martini, Oct 18 '12 at 13:14
Suppose $\ell_p$ is an Hilbert space. So its satisfies the paralelogram law. Write down the formula of the parallelogram law and conclude that the formula is only true for all vector if $p=2$. — Tomás, Oct 18 '12 at 13:15
This question gives more details on parallelogram law: Norms Induced by Inner Products and the Parallelogram Law — Martin Sleziak, Oct 18 '12 at 13:15
I think the question is whether $\ell_p$ admits an inner product making $\ell_p$ to a Hilbert space. The corresponding norm doesn't have to be $\Vert\cdot\Vert_p$ (an equivalent norm would do). None of the comments above seems to answer this question. — user8268, Oct 18 '12 at 13:19
related: How do you prove that $\ell_p$ is not isomorphic to $\ell_q$? and the more general answer in If $1\leq p \lt \infty$ then show that $L_p([0,1])$ and $\ell_q$ are not topologically isomorphic. — , Oct 18 '12 at 14:16
that's the usual norm Tomas.that is "sigma(modulus Z)^p<infinity" the set of all p summable complex sequences. — ccc, Oct 18 '12 at 15:04

score 37 · Accepted Answer · edited May 01 '18 at 00:41

Assuming we are working with the usual norm (as OP said in comments), suppose $\ell_{p}$ is an Hilbert space. So its must satisfy for all $u,v$: $$2\|u\|_{p}^2 + 2\|v\|_{p}^2 = \|u + v\|_{p}^2 + \|u - v\|_{p}^2.$$

As suggested by martini, take $u=e_{1}=(1,0,...,0,...)$ and $v=e_{2}=(0,1,0,...,0,...)$. Hence, by the last equality, we have $$4=2^{\frac{2}{p}}+2^{\frac{2}{p}}$$

Now you can solve the last inequality and verify that $p=2$.

On the other hand, if $p=2$, you can easily check that $\ell_{2}$ is a Hilbert space.

$\ell_p$ is Hilbert if and only if $p=2$

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