How can we show this property of a symmetric integral operator?

Question

Let

• $D\subseteq\mathbb R^d$ be bounded and open;
• $H:=L^2(D)$
• $Q\in\mathfrak L(H)$ be nonnegative and self-adjoint with finite trace and hence $$Qf_n=\lambda_nf_n\tag1$$ for some orthonormal basis $(f_n)_{n\in\mathbb N}$ of $H$ and nonincreasing $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$.

Assume $$(Qf)(x)=\int_Dq(x,y)f(y)\:{\rm d}y\;\;\;\text{for }f\in H$$ for some symmetric Borel measurable $q:D^2\to[0,\infty)$. How can we show that $$\sum_{n\in\mathbb N}\lambda_nf_n(x)f_n(y)=q(x,y)\tag2$$ for all $x,y\in D$?

We can clearly rewrite the left-hand side of $(2)$ as $$\left\langle q(x,\;\cdot\;),\sum_{n\in\mathbb N}\lambda_n\langle q(y,\;\cdot\;),f_n\rangle_Hf_n\right\rangle_H\tag3,$$ but I'm not able to proceed from here ...

The desired claim is implicitly made on p. 450 of An Introduction to Computational Stochastic PDEs:

Answer 1

In this answer I assume that all equalities in the question are meant to be interpreted as equalities as elements of $L^2(D)$ (i.e. holding up to sets of measure $0$). I don't think you should expect to do better than this since $Qf(x)$ is anyway only defined only for almost all $x$, and the assumption that $$Qf(x) = \int q(x,y) f(y) dy$$ for fixed $x$ only says something about $q(x,y)$ for almost everywhere $y$. I will also use the important feature of $q$ appearing in the text that $q \in L^2(D \times D)$.

Note that since $\{f_n: n \in \mathbb{N}\}$ is an orthonormal basis of $L^2(D)$, $\{f_n \otimes f_m: n, m \in \mathbb{N}\}$ is an orthonormal basis of $L^2(D\times D) \simeq L^2(D) \otimes L^2(D)$. As a result, we can decompose $q$ according to this orthonormal basis and get that as an element of $L^2(D \times D)$, $$q(x,y) = \sum_{n,m} \langle q, f_n \otimes f_m \rangle_{L^2(D \times D)} f_n(x) f_m(y).$$ Now \begin{align*}\langle q, f_n \otimes f_m \rangle_{L^2(D \times D)} & = \iint q(x,y) f_n(x) f_m(y) dy dx = \int Qf_m(x) f_n(x) dx \\ &= \lambda_m \int f_m(x) f_n(x) dx = \lambda_n \delta_{n,m}.\end{align*} Therefore for almost all $(x,y) \in D \times D$, $q(x,y) = \sum_n \lambda_n f_n(x) f_n(y)$ as desired.