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?
