Show that $\lVert Tf \rVert_{L^2} \leq \sqrt{ab} \lVert f \rVert_{L^2}$

Question

Suppose $Q(x,y): \Bbb R \times \Bbb R \to [0,\infty)$ is measurable, and $\int_{\Bbb R}Q(x,y) dy \leq a$ for all $x \in \Bbb R$ and $\int_{\Bbb R}Q(x,y) dx \leq b$ for all $y \in \Bbb R$. For $f(x)$ nonnegative and measurable, define $Tf(x) := \int_{\Bbb R}Q(x,y)f(y) dy$. Show that $\lVert Tf \rVert_{L^2} \leq \sqrt{ab} \lVert f \rVert_{L^2}$.

I get that $$\lVert Tf \rVert_{L^2}^2 = \int_{\Bbb R}\left(\int_{\Bbb R}Q(x,y)f(y) dy\right)^2 dx \leq \int_{\Bbb R} \left( \int_{\Bbb R}Q^2(x,y) dy \int_{\Bbb R} f^2(y) dy \right) dx$$ by Hölder's inequality. Not sure what to do next.

Answer 1

Based on the comment by @Mason, this is basically done in Wikipedia. But I'll still write an answer here, just for the clarity of any future user.

We need to change the breaking up of $Q(x,y)f(y)$ from $Q(x,y) \times f(y)$ to $\sqrt{Q(x,y)} \times \sqrt{Q(x,y)} f(y)$ while applying Hölder's inequality (or equivalently, Cauchy-Schwarz inequality) to get $$ \begin{align} \lVert Tf \rVert_{L^2}^2 &= \int_{\Bbb R}\left(\int_{\Bbb R}Q(x,y)f(y) dy\right)^2 dx \\ & \leq \int_{\Bbb R} \left( \int_{\Bbb R}Q(x,y) dy \int_{\Bbb R} Q(x,y) f^2(y) dy \right) dx \\ & \leq a \int_{\Bbb R}\int_{\Bbb R}{Q(x,y)f^2(y) dy} dx \\ & \leq ab \int_{\Bbb R}f^2(y)dy = ab \lVert f \rVert_{L^2}^2 \end{align} $$ This proves our claim.

Answer 2

An alternative way is to get an estimate of the form $$\int\int|Q(x, y)f(y)g(x)|\,dy\,dx \leq C\|f\|_{L^2}\|g\|_{L^2}$$ for $f, g$ simple functions that vanish off a set of finite measure. By "duality", this then implies that $Q$ maps $L^2$ into $L^2$ and $\|Qf\|_{L^2} \leq C\|f\|_{L^2}$