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Let $k$ be a smooth, compactly supported function defined on $\mathbb{R}^{2}$ and let $Op(k)$ denote the integral operator on $L^{2}(\mathbb{R})$ defined as $$Op(k)f(\cdot)=\int_{\mathbb{R}}k(y,\cdot)f(y)dy,\quad f\in L^{2}(\mathbb{R}).$$ It is known that such an operator is not only bounded, but also it is Hilbert-Schmidt, and clearly, it is in any Schatten Class $S_{p}$ for any $p\geq 2$.

I am sure that I have also seen somewhere that such an operator is trace-class however I cannot find a reference to this anywhere, nor can I prove it. At some point in my proof, which uses polynomials restrited to the $supp(k)$, I get stuck and cannot go forward. I have also tried adapting the proof in P. Lax's book on Functional analysis, but I cannot make it work.

Can someone point me in the right direction with either a reference or a proof of this fact?

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$k$ being compactly generated, you can as well assume that $k$ is a smooth function defined on $\mathbb{T}^2$ and $Op(k)$ acts on $L^2(\mathbb T)$ (for $\mathbb{T} = \mathbb R/\mathbb Z$ the unit circle). Then $k$ being smooth, its Fourier coefficients are summable. So writing $k(y,x) = \sum_{n,m \in \mathbb Z} \hat k(n,m) e^{2i\pi (nx+my)}$ you see that $Op(k) = \sum_{n,m \in \mathbb Z} \hat k(n,m) Op(e^{2i\pi (nx+my)})$, where $Op(e^{2i\pi (nx+my)})$ has rank $1$ and norm $1$. This implies that $Op(k)$ is trace class with norm bounded by $\sum_{n,m} |\hat k(n,m)|$.

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  • $\begingroup$ Hi and thank you. This really helps! Also, I was wondering if the same would work if onu changes slightly the assumptions to saying that $k$ is smooth on $\mathbb{R^{2}}$ and all its derivatives are continuous on $\widehat{\mathbb{R}}$? Here $\widehat{\mathbb{R}}$ is the real-line compactified by adding the point at infinity. In this case, it seems that you can still do the same, since $k$ can be considere as a smooth function on $\mathbb{T}^{2}$. $\endgroup$ – Raphael Feb 11 at 10:31
  • $\begingroup$ No, this does not work ($Op(k)$ need not even be Hilbert-Schmidt). The problem is that the Lebesgue measures on $\mathbb R$ and $\mathbb T$ are not the same. $\endgroup$ – Mikael de la Salle Feb 11 at 10:36
  • $\begingroup$ Thank you for having clarified it! $\endgroup$ – Raphael Feb 11 at 10:39

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