Hilbert–Schmidt integral operator

Remarks

The operator is well defined: for almost every $x$, $K(x, \cdot) \in L^2(X)$, and the Cauchy–Schwarz inequality gives $\norm{T_K f} \le \norm{K}_{L^2(X \times X)} \norm{f}$. Its Hilbert–Schmidt norm $\left(\sum_n \norm{T_K e_n}^2\right)^{1/2}$, for any orthonormal basis $(e_n)$, equals $\norm{K}_{L^2(X \times X)}$.

Example: on $X = [0, 1]$ with Lebesgue measure, $K(x, y) = \min(x, y)$ satisfies $\int_0^1 \int_0^1 \min(x, y)^2 \, dx \, dy = 1/6 < \infty$, so $T_K$ is a Hilbert–Schmidt integral operator.

Non-example: on $[0, 1]$, $K(x, y) = \abs{x - y}^{-1/2}$ fails the condition, since $\int_0^1 \int_0^1 \abs{x - y}^{-1} \, dx \, dy = \infty$. Its operator is still bounded and compact on $L^2[0, 1]$, but it is not a Hilbert–Schmidt operator.

Kernels with the condition above are the square-integrable kernels.