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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 $p\neq 2$?

Community
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Tomás
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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
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    $\ell^p$ 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
  • It is like @LukasGeyer say. – Tomás Oct 18 '12 at 15:58
  • Note that the $\ell_2$ norm is also a norm on any $\ell_p$ for $1 \le p < 2$, since $\ell_p \subset \ell_2$ in that case. However, I am pretty sure that these spaces will not be complete with the $\ell_2$ norm. – Lukas Geyer Oct 18 '12 at 16:08
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    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
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    An answer to the question seems to be more or less available via the nLab. – Harald Hanche-Olsen Oct 18 '12 at 16:23
  • @HaraldHanche-Olsen, it is any norm. Why i have to add this hypoteshis? – Tomás Oct 18 '12 at 16:26
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    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
  • 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?". – AD - Stop Putin - Oct 19 '12 at 05:26

2 Answers2

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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

Norbert
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  • Is $l_{p}$ complete with this new norm if the operator is unbounded? – Tomás Oct 18 '12 at 18:31
  • Very nice. Of course, the example itself shows that any separable Banach space admits a norm that makes it a Hilbert space (and nothing remains of the original structure). And you can remove the "separable" condition for each Banach space where you can construct a Hilbert space with the same algebraic dimension (are they all?). – Martin Argerami Oct 18 '12 at 18:35
  • @MartinArgerami, please wait a little soon I'll improve my answer – Norbert Oct 18 '12 at 18:41
  • I can wait :D $ $ – Martin Argerami Oct 18 '12 at 18:43
  • @Norbert, thanks. – Tomás Oct 18 '12 at 18:53
  • is it possible to give an explicit formula for this inner product? – Tomás Oct 18 '12 at 19:00
  • I don't think so, because there is no explicit formula for vectors of linear basis of Banach space. – Norbert Oct 18 '12 at 19:19
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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
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    Also, JSTOR link of the cited paper: http://www.jstor.org/stable/2318458 – Asaf Karagila Oct 18 '12 at 20:18
  • @AsafKaragila Well one can weaken assumption of completeness but, for me it is simpler to find linear dimension of $\ell_p$ via theorem in Lacey's article – Norbert Oct 18 '12 at 20:18
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    @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
  • @AsafKaragila $T$ is a linear isometry by definition of norm $\Vert\cdot\Vert_\bullet$ – Norbert Oct 18 '12 at 20:24
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    @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
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    let us continue this discussion in chat – Norbert Oct 18 '12 at 20:32
  • I have a self-imposed exile from the chat service here. I don't like it, and I have vowed not to return to it. – Asaf Karagila Oct 18 '12 at 20:34
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    Although I liked this, it does not answer the OP because $\ell^p$ has another topology. – AD - Stop Putin - Oct 18 '12 at 20:37
  • @AD. Why do you think this is not an answer? I gave an example of topology when $\ell_p$ is a Hilbert space. – Norbert Oct 18 '12 at 20:40
  • @AsafKaragila then you can ask separate question on MSE. – Norbert Oct 18 '12 at 20:57
  • @Norbert: Oh, I don't mind about that. I'm just pointing out that your answer is, well... "cheating". – Asaf Karagila Oct 18 '12 at 21:01
  • @AsafKaragila, excuse me, due to my poor English I didn't understand your question, and I'm interested what you were asking? Using notation from my answer we are given isometry $T:(X,\Vert\cdot\Vert)\to(X,\Vert\cdot\Vert_\bullet)$. And you are asking whether $(X,\Vert\Vert)$ is a Hilbert space. Right? – Norbert Oct 18 '12 at 21:13
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    @Norbert: It's quite simple. You can strip all the topology from a vector space, and your approach still works. In fact you can even strip any structure and just take a set of the proper size ($\frak c$). The question is how much structure you can preserve while still having this. So we know that you can begin with no structure, or with a vector space structure -- and it works. But can you start with a topological vector space and have $T$ continuous, or even a homeomorphism with $\ell_2$? What you did is to disregard the preexisting structure of $\ell_p$, and I was merely pointing this out. – Asaf Karagila Oct 18 '12 at 21:19
  • I think Norbert's answer addresses exactly what the OP asked. The question was, given a norm in $\ell^p$, who to prove it is not an inner product norm. The answer provides an example that such method cannot exist because there is indeed a norm in $\ell^p$ coming from an inner product. – Martin Argerami Oct 19 '12 at 01:39
  • @Norbert I guess I was too tired when I wrote my answer.. Thanks for correcting me! Also, today I can interpret the OP as "Given a norm on $\ell^p$, $p\ne2$, can we prove that it is not a Hilbert space?". +1 – AD - Stop Putin - Oct 19 '12 at 05:20
  • @Tomás I havn't read this article I just believe in mathematical competence of Elton Lacey. – Norbert Oct 20 '12 at 16:30
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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.

Khoa
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