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

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. $$||u||^2=$$ if and only if $$p=2$$.

• Hint: Parallelogramm law, $e_1$, $e_2$. – martini Oct 18 '12 at 13:10
• 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
• 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

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.