Show that $\lVert Tf \rVert_{L^2} \leq \sqrt{ab} \lVert f \rVert_{L^2}$ 
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.
 A: 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 $\|Qf\|_{L^2} \leq C\|f\|_{L^2}$
A: 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.
