Let \(Tf(x) = \int_{\mathbb{R}^n} K(x,y) f(y) \, dy\), where \(K(x,y)\) is a nonnegative measurable function on \(\mathbb{R}^n \times \mathbb{R}^n\). Suppose that there are measurable functions \(p(x) > 0\) and \(q(x) > 0\) on \(\mathbb{R}^n\) and real numbers \(\alpha, \beta > 0\) for which
\[\int K(x,y) q(y) \, dy \leq \alpha p(x),\]
for almost all \(x\) and
\[\int p(x) K(x,y) \, dx \leq \beta q(y)\]
for almost all \(y\). Show that for \(f \in L^2(\mathbb{R}^n)\) we have
\[\|Tf\|_{L^2} \leq \sqrt{\alpha \beta} \|f\|_{L^2}.\]
(This is called Schur’s test.)
Solution
Let \(f\in L^2(\mathbb R^n)\).
1. The integral defining \(Tf\) converges absolutely almost everywhere.
Set \[ B(x)=\int_{\mathbb R^n}K(x,y)\frac{|f(y)|^2}{q(y)}\,dy. \] This is a nonnegative measurable function, possibly initially infinite. Tonelli and the second kernel inequality give [@Fol13] \[ \begin{aligned} \int_{\mathbb R^n}p(x)B(x)\,dx &=\int_{\mathbb R^n}\frac{|f(y)|^2}{q(y)} \left(\int_{\mathbb R^n}p(x)K(x,y)\,dx\right)dy\\ &\leq\beta\int_{\mathbb R^n}|f(y)|^2\,dy<\infty. \end{aligned} \] Since \(p(x)>0\), this implies \(B(x)<\infty\) almost everywhere. The first kernel inequality also gives \(A(x):=\int K(x,y)q(y)\,dy\leq\alpha p(x)<\infty\) for almost every \(x\). For points where both bounds hold, Cauchy–Schwarz applied to \(K^{1/2}q^{1/2}\) and \(K^{1/2}|f|q^{-1/2}\) yields \[ \int K(x,y)|f(y)|\,dy\leq A(x)^{1/2}B(x)^{1/2}<\infty \] [@Fol13]. Thus \(Tf\) is defined almost everywhere, and the usual parameter-integral measurability follows by Tonelli applied to positive and negative real and imaginary parts. Set \(Tf=0\) on the exceptional null set.
2. The same weighted inequality gives the claimed norm bound.
For almost every \(x\), Cauchy–Schwarz in the \(y\) variable gives \[ \begin{aligned} |Tf(x)|^2 &=\left|\int K(x,y)^{1/2}q(y)^{1/2} \,K(x,y)^{1/2}q(y)^{-1/2}f(y)\,dy\right|^2\\ &\le \left(\int K(x,y)q(y)\,dy\right) \left(\int K(x,y)\frac{|f(y)|^2}{q(y)}\,dy\right)\\ &\le \alpha p(x) \int K(x,y)\frac{|f(y)|^2}{q(y)}\,dy. \end{aligned} \] Integrating in \(x\) and using Tonelli’s theorem gives [@Fol13] \[ \begin{aligned} \|Tf\|_2^2 &\le \alpha \int_{\mathbb R^n} p(x) \int_{\mathbb R^n}K(x,y)\frac{|f(y)|^2}{q(y)}\,dy\,dx\\ &=\alpha \int_{\mathbb R^n}\frac{|f(y)|^2}{q(y)} \left(\int_{\mathbb R^n}p(x)K(x,y)\,dx\right)dy\\ &\le \alpha\beta\int_{\mathbb R^n}|f(y)|^2\,dy =\alpha\beta\|f\|_2^2. \end{aligned} \] Taking square roots yields \[ \|Tf\|_{L^2}\le \sqrt{\alpha\beta}\,\|f\|_{L^2}. \]