# $\ell_{p}$ space is not Hilbert for any norm if $p\neq 2$

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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If you want a non-standard norm on $\ell^p$, it's probably best to include the explicit definition of the set $\ell^p$ you want to consider. –  Lord_Farin Oct 18 '12 at 15:31
$\ell^p$ is pretty standard, it is the space of all sequences $(a_k)$ such that $\sum_k |a_k|^p < +\infty$. (Whether real or complex, or whether the index set is all integers or only positive integers won't matter for the answer to this question.) –  Lukas Geyer Oct 18 '12 at 15:50
I think you forgot to mention that the given norm on $\ell_p$ should be equivalent to the usual one. Or put differently, you want to prove that $\ell_p$ is not isomorphic (as a Banach space) to a Hilbert space. –  Harald Hanche-Olsen Oct 18 '12 at 16:13
An answer to the question seems to be more or less available via the nLab. –  Harald Hanche-Olsen Oct 18 '12 at 16:23
Because of @Norbert's answer below. Note that the norm he defines, or rather whose existence he shows, will not be equivalent to the original norm! –  Harald Hanche-Olsen Oct 18 '12 at 19:00

## 2 Answers

I think you can turn any separable Banach space $(X,\Vert\cdot\Vert)$ into Hilbert space. It is known$^1$ that every separable Banach space have linear basis of cardinality $\mathfrak{c}$. Hence there exist bijective linear operator $T:X \to\ell_2$. Given this operator we define new norm on $X$ by equality $$\Vert x\Vert_\bullet=\Vert T(x)\Vert_{\ell_2}$$ It is 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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Do you really need the assumption that it is Banach? Or even topological? You just need that the dimension is $\frak c$. –  Asaf Karagila Oct 18 '12 at 20:14
@Norbert: I'm not arguing that the Hamel basis of $\ell_2$ is of size $\frak c$. I'm just saying that you just need the assumption that $\dim V=\frak c$, not that it is a separable Banach space. Even if your space is not a topological space. The question is what if you require this $T$ to be continuous, or even a homeomorphism, or a linear isometry... –  Asaf Karagila Oct 18 '12 at 20:20
@Norbert: No. $T$ is a transport of structure from $\ell_2$ to your original space. Suppose I give you a topological vector space of dimension $\frak c$, for simplicity sakes assume it is a separable Banach space. Now I require that this $T$ is a linear isometry between the pre-existing norm, and the new one. Can this be done without the space being a Hilbert space to begin with? –  Asaf Karagila Oct 18 '12 at 20:26
–  userNaN Oct 18 '12 at 20:32
Although I liked this, it does not answer the OP because $\ell^p$ has another topology. –  AD. Oct 18 '12 at 20:37

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.

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