Schur test

Remarks

Proof idea: write $\abs{K} \abs{f} = \big(\abs{K} q\big)^{1/2} \big(\abs{K} \abs{f}^2 / q\big)^{1/2}$ and apply the Cauchy–Schwarz inequality in $y$. The first hypothesis gives $\abs{T f(x)}^2 \le \alpha \, p(x) \int_Y \abs{K(x, y)} \abs{f(y)}^2 q(y)^{-1} \, d\nu(y)$. Integrating in $x$ and exchanging the integrals by Tonelli's theorem, the second hypothesis gives $\norm{T f}_2^2 \le \alpha \beta \norm{f}_2^2$.

With $p = q = 1$ the test says: if $\int_Y \abs{K(x, y)} \, d\nu(y) \le \alpha$ for almost every $x$ and $\int_X \abs{K(x, y)} \, d\mu(x) \le \beta$ for almost every $y$, then $\norm{T} \le \sqrt{\alpha \beta}$ on $L^2$. For $\alpha = \beta$ this is the case $p = 2$ of Young's inequality for integral operators. The freedom to choose the weights $p$ and $q$ is what makes the test useful for kernels such as $1/(x + y)$, which have no finite row or column integrals.

Example: let $X = Y = (0, \infty)$ with Lebesgue measure and $K(x, y) = 1/(x + y)$. Take $p(x) = x^{-1/2}$ and $q(y) = y^{-1/2}$. The substitution $y = x u$ gives $\int_0^\infty \frac{y^{-1/2}}{x + y} \, dy = x^{-1/2} \int_0^\infty \frac{u^{-1/2}}{1 + u} \, du = \pi x^{-1/2}$, and by symmetry the same holds for the second hypothesis. So $\alpha = \beta = \pi$ and $\norm{T} \le \pi$, which is the continuous form of Hilbert's inequality.

Going deeper

For a student

The Schur test is the main elementary tool to prove that an integral operator is bounded on $L^2$ when its kernel is not square-integrable. Such an operator is not Hilbert–Schmidt. It is used for kernels with a singularity or slow decay, such as homogeneous kernels on $(0, \infty)$ and kernels in the study of Bergman projections. It is the continuous analogue of the bound of the norm of a matrix by the geometric mean of the largest row sum and the largest column sum. For $L^p$ with $p \ne 2$, the same idea with Hölder's inequality in place of Cauchy–Schwarz gives a version with weights.