Existence of an eigenvalue for a nonzero symmetric kernel

Remarks

Since $T_K$ is self-adjoint, $\norm{T_K} = \sup_{\norm{\varphi} = 1} \abs{\ip{T_K \varphi}{\varphi}}$, and $\norm{T_K} > 0$ because a kernel whose operator is $0$ is $0$ almost everywhere. Take unit vectors $\varphi_k$ with $\ip{T_K \varphi_k}{\varphi_k} \to \mu$, where $\mu = \pm \norm{T_K}$. Then $\norm{T_K \varphi_k - \mu \varphi_k}^2 \le 2\mu^2 - 2\mu \ip{T_K \varphi_k}{\varphi_k} \to 0$. Since $T_K$ is compact, a subsequence of $T_K \varphi_k$ converges, so $\varphi_k$ converges to some unit vector $\varphi$, and $T_K \varphi = \mu \varphi$.

Example: for $K(x, y) = \min(x, y)$ on $[0, 1]$, the largest eigenvalue is $4 / \pi^2$, with eigenfunction $\sin(\pi x / 2)$, and $\norm{T_K} = 4 / \pi^2$.

Symmetry is necessary. The Volterra kernel, equal to $1$ for $y \le x$ and $0$ otherwise, is square-integrable and not zero. But its operator $\varphi \mapsto \int_a^x \varphi$ has no eigenfunction at all, so this kernel has no characteristic value. In finite dimension, the statement is the fact that a nonzero Hermitian matrix has an eigenvalue whose absolute value is its operator norm.

Going deeper

For a student

This is the first step of the Hilbert–Schmidt theory. Applying it again on the orthogonal complement of the eigenfunctions already found produces the whole sequence of eigenvalues and eigenfunctions, and leads to the Hilbert–Schmidt expansion theorem. The variational form $\abs{\mu} = \max \abs{\ip{T_K \varphi}{\varphi}}$ is the starting point of the min-max characterization of the further characteristic values.