I have asked this [question](https://math.stackexchange.com/questions/1668628/every-self-adjoint-trace-class-operator-on-l2-has-integral-kernel) on MSE but did not receive an answer. I thought I could try it here.


Let $T$ be a self-adjoint trace-class operator on $L^2(\mathbb{R})$. Is is true that it can be represented as an integral operator. 


I thought the kernel would be $$k\_T(x,y) =\sum\_{i=1}^\infty \lambda\_i \phi\_i(x) \bar\phi\_i(y).$$


Here $\{\phi\_i\}$ is an eigenbasis of $T$, i.e. $T=\sum\_i \lambda\_i |\phi\_i\rangle\langle\phi\_i|$. Then, we have $$\int k\_T(\cdot,y) f(y) = \int\sum\_i \lambda\_i \phi\_i(\cdot) \bar\phi\_i(y) f(y) dy = \sum\_i \lambda\_i \phi\_i \langle \phi\_i, f\rangle=\sum\_i \lambda\_i |\phi\_i\rangle\langle\phi\_i|f\rangle = Tf.$$


**Is this correct?**



Every self-adjoint trace class operator on $L^2$ has integral kernel
Yes, this is correct. Actually, "self-adjoint trace-class" is more than you need; any Hilbert-Schmidt operator can be represented as an integral operator. The Hilbert-Schmidt operators from $L^2(X)$ to $L^2(Y)$ are precisely the integral operators with kernel in $L^2(X\times Y)$ (at least for $\sigma$-finite $X$ and $Y$). This should be in a standard reference, probably Dunford-Schwartz, for example.


