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

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    **Hint**: Parallelogramm law, $e_1$, $e_2$.2012-10-18
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    Hint: Prove that if $p=2$, the parallelogram law is satisfied. WHen $p\neq 2$ give a counter example to the parallelogram law .2012-10-18
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    @martini what are e1, e2?where to apply?2012-10-18
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    @ 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.2012-10-18
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    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$2012-10-18
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    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 $p=2$.2012-10-18
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    This question gives more details on parallelogram law: [Norms Induced by Inner Products and the Parallelogram Law](http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law)2012-10-18
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    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.2012-10-18
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    Yes @user8268 are right. So what is the norm? hahah2012-10-18
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    related: *[How do you prove that $\ell_p$ is not isomorphic to $\ell_q$?](http://math.stackexchange.com/questions/99086/)* 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](http://math.stackexchange.com/q/97126)*.2012-10-18
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    @cccjay, what norm are you considering in $\ell_{p}$?2012-10-18
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    that's the usual norm Tomas.that is "sigma(modulus Z)^p2012-10-18
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    I see. So it is enough to verify parallelogram law.2012-10-18

1 Answers 1

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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.