Timeline for Reference request: sufficiently smooth functions on the plane belong to the projective tensor square of $L^2$ of the line
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 2, 2020 at 8:18 | comment | added | Mikael de la Salle | @DirkWerner : thanks, I also like this argument. | |
| Jun 2, 2020 at 0:29 | comment | added | Yemon Choi | @DirkWerner thanks! I like this argument, since in one version of the intended application, integral operators occur quite naturally, and this factorization trick might be useful for independent reasons | |
| Jun 1, 2020 at 21:20 | comment | added | Dirk Werner | Another way to see this (for $k=1$) is by rephrasing your question in terms of integral operators; what you are asking is whether an integral operator on $L_2$ with a $C^1$-kernel is nuclear. By compactness of the support one can assume that $k\in C^1([0,1]^2)$ vanishes at the boundary; let $T_k$ be the associated integral operator. Let $k_2(s,t) = \frac{\partial}{\partial t} k(s,t)$ and $V(f)(t)=\int_0^t f(s)\,ds$. Then $T_k = T_{k_2}V$ by integration by parts; hence $T_k$ factors as a product of two Hilbert-Schmidt operators and is therefore nuclear. | |
| Jun 1, 2020 at 15:37 | comment | added | Yemon Choi | Hi @MikaeldelaSalle - yes, your first comment was more or less the argument I had in mind, but it is good to get confirmation (I was taking $k\geq2$ just to play safe but as you observe this is unnecessary). I had overlooked your second point that the group structure is unnecessary and that one can localize to $L^2$ of a $d$-cube (hence a $d$-torus) | |
| Jun 1, 2020 at 7:28 | comment | added | Mikael de la Salle | For your second question, the fact that you have a Lie group is not relevant: in any compact manifold (or any compactly supported function on a manifold), working in local coordinates with a partition of unity reduces the problem to $(\mathbf{R}/\mathbf{Z})^d$, in which case the above Fourier transform argument shows that $C^k$ functions work whenever $k>d/2$. | |
| Jun 1, 2020 at 7:21 | comment | added | Mikael de la Salle | I do not know of a good reference, but the Fourier transform argument works very well for every $k>1$: by the compact support assumption, you can as well periodize the problem and work with a $C^k$ function $(\mathbf{R}/{\mathbf{Z}})^2 \to \mathbf{R}$. Then the Fourier expansion gives that $\|f\|_{L^2 \hat \otimes L^2} \leq \sum_{i,j} |\hat f(i,j)| \leq C(\varepsilon) \|f\|_{H^{1+\varepsilon}}$ for every $\varepsilon >0$ (here $H^s$ is the usual Sobolev of functions such that $((1+|i|+|j|)^s\hat f(i,j))_{i,j}$ belongs to $\ell^2$). | |
| May 31, 2020 at 22:12 | history | asked | Yemon Choi | CC BY-SA 4.0 |