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I'm wondering: What is the tensor product of $L^p({\bf R})$ with $L^q({\bf R})$?

(For p=q=2, the answer clearly should be $L^2({\bf R}^2)$; for other values of $p$ and $q$, it is not at all obvious to me what $L^p({\bf R})\otimes L^q({\bf R})$ should be.)

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    $\begingroup$ there are several common norms on the tensor product of normed algebras. please specify. $\endgroup$
    – Martin Brandenburg
    Commented Feb 6, 2010 at 0:25
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    $\begingroup$ For a hint at just how non-obvious any of the answers are (even for p=q=2$) I recommend a quick look at (the size of) Defant and Floret's bok "Tensor norms and operator ideals" $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2010 at 0:32
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    $\begingroup$ If the tensorproduct of $L^2(\mathbb{R})$ with itself is $L^2(\mathbb{R}^2)$, then perhaps the tensorproduct you're looking for is something like "All f with $(\int (\|f(x,\cdot)\|_p)^q\,dx)^{1/q}=(\int (\|f(\cdot,y)\|_q)^p\,dy)^{1/p}<\infty$" $\endgroup$
    – Johannes Hahn
    Commented Feb 6, 2010 at 0:40
  • $\begingroup$ @Martin: Although I'm obviously not an expert on these thinks, I'm well aware of the variety of possible norms on the tensor product. Part of my question is: Which one is the good definition in this setting? There should be a good notion in this case. In the $L^2$-case, we certainly want the Hilbert space notion of the tensor product, since this one gives a nice answer, and I expect that there is some setting which handles the $L^p \otimes L^q$-case quite well. @Yemon: Thanks. I'll take a look at this book. $\endgroup$
    – Anonymous
    Commented Feb 6, 2010 at 0:41
  • $\begingroup$ @Anonymous: It was your use of the definite article and the word "should" which I was quibbling with - depending on the problem at hand, one may have to use different norms, and so as it stands your question is hard to answer because of the lack of context. FWIW, it sounds like you want something lke one of the Chevet-Saphar tensor norms. R. Ryan has a more accessible though les far-reaching treatment of tensor norms, which might also be in your library or similar. $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2010 at 0:48

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As pointed out in the comments, there are many Banach tensor products, but there is indeed at least one which works nicely for $L^p\otimes L^p$.

In general, the algebraic tensor product $X\otimes Y^*$ can be identified with finite rank operators from $Y$ to $X$. When $X=Y=L^2(\mathbb{R})$, taking the completion in the Hilbert-Schmidt norm gives you the space of Hilbert-Schmidt operators on $L^2(\mathbb{R})$, which can be identified with $L^2(\mathbb{R}^2)$.

Similarly, the space of $q$-summing operators from $L^p(\mathbb{R})$ to $L^q(\mathbb{R})$, when $p^{-1} + q^{-1} = 1$, can be identified with $L^p(\mathbb{R}^2)$. (I don't have the reference for this on hand, and don't recall how much it generalizes; I'll check and update later.)

Added later: I don't know if the anonymous poster is still around, but here is the reference.

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  • $\begingroup$ I think this is one of the Chevet-Saphar norms (cf. comment above). It still isn;t clear to me what the original poster thinks a "sensible answer" should be; the question seems to assume that there is one which is clearly most natural. (From the category-theoretic point of view, the Grothendieck projective tensor product is the only "natural" one; but all the others can be useful and relevant to different problems at hand.) $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2010 at 21:44
  • $\begingroup$ @Yemon: Yes, I think that's true. This is also a special case of what Johannes suggested. As far as I can tell the poster wants something that turns out to be simple and specializes to what was described for p=2; my answer is just meant to suggest one such possibility which is also not obviously ad hoc. $\endgroup$
    – Mark Meckes
    Commented Feb 7, 2010 at 0:26

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