Square-integrable kernel

Remarks

Example: $K(x, s) = \abs{x - s}^{-1/4}$ for $x \ne s$ on $[0, 1]$. Here $\abs{K}^2 = \abs{x - s}^{-1/2}$, and $\int_0^1 \abs{x - s}^{-1/2} \, ds \le 2 \int_0^{1} t^{-1/2} \, dt = 4$ for each $x$, so $\norm{K}_2^2 \le 4$.

Non-example: $K(x, s) = \abs{x - s}^{-1/2}$ on $[0, 1]$. For $0 < x < 1$, $\int_0^1 \abs{x - s}^{-1} \, ds = \infty$, so the double integral of $\abs{K}^2$ is infinite.

In other words, $K$ lies in the Lp space $L^2([a, b] \times [a, b])$. Every continuous kernel on $[a, b]$ is square-integrable. For $\varphi \in L^2(a, b)$, the function $(K \varphi)(x) = \int_a^b K(x, s) \varphi(s) \, ds$ lies in $L^2(a, b)$ and $\norm{K \varphi}_2 \le \norm{K}_2 \norm{\varphi}_2$, by the Cauchy–Schwarz inequality. The same definition is used on a measurable subset of $\R^m$.