real analysis - 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 $k>d/2$?

Answer

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.