# Trace-class properties of integral operator

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?

$$k$$ 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 $$\sum_{n,m} |\hat k(n,m)|$$.
• 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 $\mathbb{T}^{2}$. – Raphael Feb 11 '19 at 10:31
• 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. – Mikael de la Salle Feb 11 '19 at 10:36