What is the tensor product of $L^p(\bf R)$ with $L^q(\bf R)$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:49:50Z http://mathoverflow.net/feeds/question/14329 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14329/what-is-the-tensor-product-of-lp-bf-r-with-lq-bf-r What is the tensor product of $L^p(\bf R)$ with $L^q(\bf R)$? Anonymous 2010-02-06T00:05:27Z 2010-02-19T15:34:59Z <p>I'm wondering: What is the tensor product of $L^p({\bf R})$ with $L^q({\bf R})$?</p> <p>(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.)</p> http://mathoverflow.net/questions/14329/what-is-the-tensor-product-of-lp-bf-r-with-lq-bf-r/14429#14429 Answer by Mark Meckes for What is the tensor product of $L^p(\bf R)$ with $L^q(\bf R)$? Mark Meckes 2010-02-06T21:23:05Z 2010-02-19T15:34:59Z <p>As pointed out in the comments, there are many Banach tensor products, but there is indeed at least one which works nicely for $L^p\otimes L^p$.</p> <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 $L^2(\mathbb{R}^2)$.</p> <p>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.)</p> <p>Added later: I don't know if the anonymous poster is still around, but <a href="http://www.ams.org/mathscinet-getitem?mr=247504" rel="nofollow">here</a> is the reference.</p>