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My question is motivated by this one: $\ell_p$ is Hilbert space if and only if $p=2$

Maybe it is a simple thing or im just confused but, suppose we are given any norm in $\ell_{p}$ for $p\neq 2$. How to show that this norm does not come from an inner product?

Thanks

Sorry if I do not post the problem with clarity.

Edit: $\ell_{p}=\{(x_{1},x_{2},...\}:(\sum_{i=1}^{\infty}|x_{i}|^{p})^{\frac{1}{p}}<\infty\}$

So that's my space and it is a vector space. Suppose I define on this space a norm (any norm). How can I show that this norm does not come from a inner product if $p\neq 2$?

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    Perhaps you should consider to rephrase the post what you would like to ask is something like: "Given a norm on $\ell^p$, $p\ne2$, can we prove that the resulting space is not a Hilbert space?".2012-10-19

2 Answers 2

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I think you can turn any separable Banach space $(X,\Vert\cdot\Vert)$ into a Hilbert space. It is known$^1$ that every separable Banach space has a linear basis of cardinality $\mathfrak{c}$. Hence there exists a bijective linear operator $T:X \to\ell_2$. Given this operator, we define a new norm on $X$ by equality $ \Vert x\Vert_\bullet=\Vert T(x)\Vert_{\ell_2} $ It is an easy exercise to check that $(X,\Vert\cdot\Vert_\bullet)$ is a Hilbert space.


$^1$Lacey, H. (1973). The Hamel dimension of any infinite-dimensional separable Banach space is c, Amer. Math. Montly, 80, 298

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    @Tomás I havn't read this article I just believe in mathematical competence of Elton Lacey.2012-10-20
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I think there is a smaller question in the original post. That is supposed that we have an $\ell_p(\mathbb{N})$ space with the norm already defined as $\Vert \{x_k\} \Vert_p = (\sum_{k=1}^{\infty}|x_k|^p)^{1/p} $ where $p \neq 2$. How do we prove that this norm does not come from an inner product?

We can check to see if this norm satisfy the parallelogram law: $ \Vert v+w\Vert^2 + \Vert v-w\Vert^2 = 2(\Vert v \Vert^2 + \Vert w \Vert^2) $

If it doesn't, then according to theorem 4.1.4 in Functions, Spaces, and Expansions - Christensen, Ole, this norm could not come from an inner product. If it does satisfy the parallelogram law then we can also retrieve this hidden inner product by the polarization identity there in the same theorem. The answer however is no, so the $\Vert . \Vert_p$ norm mentioned does not come from an inner product.

And the test also apply for any norms other than the one mentioned.