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Let $X$ and $Y$ be compact subsets of $\mathbb{R}^d$, let $\mu$ and $\nu$ be finite Borel measures on $X$ and $Y$, resp.

Is it possible to find a dense subset of $L^2(X \times Y)$ consisting of functions of the form $fg$ where $f \in L^2(X)$ and $g \in L^2(Y)$? Can the subset be taken to be countable?

Edit: The answer is yes if for every simple function $h \in L^2(X \times Y)$ and every $\epsilon > 0$, there are simple functions $f \in L^2(X)$ and $g \in L^2(Y)$ such that $\|h-fg\|_2 < \epsilon$.

For the situation, I have in mind $\mu$ is Lebesgue measure.

The following argument gives something close but not quite good enough: If $\{f_n\}$ is an orthonormal basis for $L^2(X)$ and $\{g_m\}$ is an orthonormal basis for $L^2(Y)$, then $\{f_n g_m\}$ is an orthonormal basis for $L^2(X\times Y)$, so that the set of all finite linear combinations $\{f_n g_m\}$ is dense in $L^2(X \times Y)$. See Orthonormal basis for product $L^2$ space

shalop
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RitterSport
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1 Answers1

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By $fg$, I am assuming you mean functions of the form $f(x)g(y)$. In this case the answer is no, such functions are not dense in $L^2(X \times Y)$. I will give an algebraic proof of this fact (though I am sure that analytic proofs exist).

As a Hilbert space, $L^2(X \times Y)$ is isometrically isomorphic to $L^2(X) \otimes L^2(Y)$ and functions of the form $f(x)g(y)$ are precisely the simple tensors $f \otimes g$ in this Hilbert space.

In general, the simple tensors are most certainly not dense in a tensor product of two Hilbert spaces.

Indeed, given two Hilbert spaces $H$ and $K$, we know that $K \simeq K^*$ isometrically, and thus $K \otimes H \simeq K^* \otimes H \simeq \text{HS}(K,H) $ isometrically, where the latter space is the space of Hilbert-schmidt operators from $K$ to $H$ equipped with the Hilbert Schmidt norm $\|A\|_{\text{HS}}^2 :=$ tr$_K(A^*A)=$tr$_H(AA^*)$. One may check that the simple tensors in $\text{HS}(K,H)$ are precisely the bounded linear maps whose image has dimension $1$. In general, a limit of rank-one operators always has rank one or zero (more generally, the rank can only shrink under pointwise convergence of operators; this may be proved by contradiction, first supposing that the limit of some family of operators has rank $k$ but all prelimits have rank $k-1$ or less; then by picking a basis for the image of the limiting operator, one may obtain the contradiction by showing that a certain collection of determinant-zero matrices converges to a full-rank $k\times k$ matrix). Now, as long as $H$ and $K$ both have dimension greater than $1$, there are certainly Hilbert-Schmidt operators from $K\to H$ which have rank greater than $1$, hence these cannot be a limit of simple tensors.

Note: This answer was incorrectly written before 18 May 2019.

shalop
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  • Just to check my understanding, if $H$ and $K$ are infinite dimensional, then there are non-compact operators in $\text{Hom}(K,H)$. But any limit of simple tensors in $K \otimes H \simeq \text{Hom}(K,H)$ must be compact, so you can't approximate the non-compact operators. Right? – RitterSport Aug 17 '17 at 20:42
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    @RitterSport Yes, that is indeed what I'm saying. – shalop Aug 17 '17 at 20:43
  • In terms of an analytic example, I have been thinking about the indicator function of the unit disc. It doesn't seem possible to approximate that by functions f(x)g(y) where $f \in L^2([-1,1])$ and $g \in L^2([-1,1])$. What do you think? – RitterSport Aug 17 '17 at 20:45
  • What happens if $K$ and $H$ are finite dimemsional. Say $K = L^2(G)$ where $G$ is a finite group. – RitterSport Aug 17 '17 at 20:52
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    I also thought about analytic examples like that, but I didn't get anywhere. I guess geometrically product sets are far away from disks, so that should work. In the case that K,H are finite dimensional (say of the same dimension), then any rank-one operator is representable by a square matrix with determinant zero, thus any limit of rank-one matrices must also have determinant zero. So simple tensors are still not dense, even in finite dimensions. – shalop Aug 17 '17 at 21:06
  • I think its a little bit offtopic, what happens if we replace $L^{2}(X \times Y)$ with $C(X\times Y)$ for compact Hausdorff $X,Y$? Does it still cannot be dense? – user124697 Jan 06 '18 at 09:34
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    @user124697 Still not dense in general. Consider finite $X$ and $Y$. Then $L^2$ and $C$ give equivalent topologies, so it's not dense. Even for infinite $X$ and $Y$ I am certain it is false. – shalop Jan 12 '18 at 05:11