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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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    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.2012-10-18
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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.)2012-10-18
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    It is like @LukasGeyer say.2012-10-18
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    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.2012-10-18
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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.2012-10-18
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    An answer to the question seems to be more or less available [via the nLab](http://ncatlab.org/nlab/show/isomorphism+classes+of+Banach+spaces).2012-10-18
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    @HaraldHanche-Olsen, it is any norm. Why i have to add this hypoteshis?2012-10-18
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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!2012-10-18
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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 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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    Is $l_{p}$ complete with this new norm if the operator is unbounded?2012-10-18
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    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?).2012-10-18
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    @MartinArgerami, please wait a little soon I'll improve my answer2012-10-18
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    I can wait :D $ $2012-10-18
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    @Norbert, thanks.2012-10-18
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    is it possible to give an explicit formula for this inner product?2012-10-18
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    I don't think so, because there is no explicit formula for vectors of linear basis of Banach space.2012-10-18
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    Do you really need the assumption that it is Banach? Or even topological? You just need that the dimension is $\frak c$.2012-10-18
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    Also, JSTOR link of the cited paper: http://www.jstor.org/stable/23184582012-10-18
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    @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 article2012-10-18
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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...2012-10-18
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    @AsafKaragila $T$ is a linear isometry by definition of norm $\Vert\cdot\Vert_\bullet$2012-10-18
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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?2012-10-18
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6163/discussion-between-norbert-and-asaf-karagila)2012-10-18
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    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.2012-10-18
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    Although I liked this, it does not answer the OP because $\ell^p$ has another topology.2012-10-18
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    @AD. Why do you think this is not an answer? I gave an example of topology when $\ell_p$ is a Hilbert space.2012-10-18
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    @AsafKaragila then you can ask separate question on MSE.2012-10-18
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    @Norbert: Oh, I don't mind about that. I'm just pointing out that your answer is, well... "cheating".2012-10-18
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    @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?2012-10-18
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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.2012-10-18
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    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.2012-10-19
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    @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?". +12012-10-19
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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.