For a Hilbert space $\mathcal{H} = L^2([0, 1])$ and a linear operator $L: \mathcal{H} \rightarrow \mathcal{H}$, then if $L$ is Hilbert-Schmidt (e.g. $\sum_{i} \lVert L e_i \rVert^2 < \infty$ for any orthonormal basis $\{e_i\}$ for $\mathcal{H}$, where the norm is the one you expect for $\mathcal{H}$) and self-adjoint, then $L$ may be diagonalized and written in the form \begin{equation} L = \sum_i \lambda_i \phi_i \otimes \phi_i^* \tag{1} \end{equation} with respect to eigenvalues $\lambda_i \in \mathbb{R}$ and eigenfunctions $\phi_i \in \mathcal{H}$, and $L$ is trace-class if the sequence of eigenvalues converges. I am used to integral operators of the form $(Lf)(x) = \int dy k(x, y) f(y)$ for positive definite $k$ and $f \in \mathcal{H}$, in which case the above conditions hold for writing this operator in the form of Eq. $(1)$.
I would like to know if it is possible to diagonalize the following operator $L_\delta$ involving the delta function:
$$ (L_\delta f)(x) = \int dy q(x) \delta (y-x) f(y) = f(x)q(x), \tag{2} $$ where $q \geq 0$ and $\lVert q \rVert_1 = 1$ (i.e. $q$ is a probability distribution with respect to its argument.
I think this is not possible because $L_\delta$ doesn't seem to be Hilbert-Schmidt in the first place (but I couldn't prove that explictly).