# $\ell_{p}$ 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
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## 1 Answer

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
–  Norbert 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
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