# The eigenvalues of a certain integral operator are square-summable

Let $(X,\Omega,\mu)$ be a measure space, and $k\in L^2(X\times X, \Omega\times \Omega,\mu\times\mu)$. Then it is well-known that

$$(Kf)(x)=\int k(x,y)f(y)\ d\mu(y)$$

is a compact operator with norm at most $\|k\|_2$.

Suppose that $k(x,y)=\overline k(y,x)$. Show that $K$ is self-adjoint and if $\{\mu_n\}$ are the eigenvalues of $K$, each repeated $\dim \ker (K-\mu_n)$ times, then $\sum_1^n |\mu_n|^2 <\infty.$

This is problem II.6.3 on page 49 of Conway's A Course in Functional Analysis, 2nd edition. A proof that the operator is compact is proposition II.4.7 on page 43.

Hint:

Apply the spectral theorem for compact self-adjoint operators. The operator diagonalizes.

Working through the definitions shows that $K$ is self-adjoint.

Using the spectral theorem for compact self-adjoint operators, we diagonalize the operator and write

$$Kv=\sum_1^\infty \mu_k\langle v,e_k \rangle e_k,$$

where the $e_k$ form a basis for $(\ker K)^\perp$.

It is now fairly clear that we have

$$k(x,y)=\sum_1^\infty \mu_k e_k(x)e_k(y).$$

But $k$ was assumed to be in $L^2$, and we have

$$\|k\|_2 = \sum_1^\infty |\mu_k|^2.$$

In the present context, the $k$ has a $L^{2}(X\times X,\mu\times \mu)$-convergent series representation $$k=\sum_{i=1}^{\infty}\mu_{i}e_{i}\otimes\overline{e}_{i},$$ where $e_{i}\otimes\overline{e}_{i}(x,y):=e_{i}(x)\overline{e_{i}(y)}$, $x,y \in X$. Since $\{e_{i}\}$ is an orthonormal set in $L^{2}(X,\mu)$ it is clear that $e_{i}\otimes\overline{e}_{i}$ is an orthonormal set in $L^{2}(X\times X,\mu\times \mu)$. Now Parseval's identity ensures $$||k||^{2}_{2}=\sum_{i}|\mu_{i}|^{2}<\infty.$$