Is tensor product of Sobolev spaces dense? My question is: is $W_2^k(\mathbb{R})\otimes W_2^k(\mathbb{R})$ dense in $W_2^k(\mathbb{R}^2)$, and more generally is this true in $\mathbb{R}^d$?
I found this post:
Tensor products of functions generate dense subspace?
which shows the above type of result for $C_c^\infty$.  So my guess is that the answer should be affirmative, maybe requiring the assumption that $k>d/2$?  
 A: Yes, you can get this result from $C_c^\infty$, because $C^\infty_c(\mathbb R^2)$ is dense in $f\in W^k_2(\mathbb R^2)$ for any $k$. So, any $W^k_2$ function can be approximated by smooth functions with compact support, which in turn are approximated by sums of products of univariate smooth functions (even in the stronger sense, $C^\infty_c$). 
But it may be easier to apply the Fourier transform, which transforms $W^k_2(\mathbb R^2)$  to a weighted $L^2$ space. Since $\widehat{u\otimes v}=\widehat{u}\otimes \widehat{v}$, the question reduces to its analog for Lebesgue spaces. Then we observe that the characteristic functions of cubes have a dense linear span, and they belong to the tensor product. 
A: Is might be of interest that the $d$-fold tensor product of $W^k_2(\mathbb{R})$ is not only dense in $W^k_2(\mathbb{R}^d)$ (for any k and d) but in the space $W^k_{2,\text{mix}}(\mathbb{R}^d):=\{ f : \partial_{\alpha}f\in L^2 \quad \forall  |\alpha|_{\infty}\leq k\}$(equipped with the obvious norm). 
Note that this space has a stronger norm than $W^k_2(\mathbb{R}^d)=\dots \forall |\alpha|_{1}\leq k$, thus this is a stronger result.
To prove this use the method from the accepted answer and note that the product of weight functions, $(1+|x_1|^2)^{k/2}(1+|x_2|^2)^{k/2}$, is a little bit stronger than the $W^k_2$-weight function $(1+|(x_1,x_2)|^2)^{k/2}$.
