We are given that $(X,\Omega,\mu)$ is a $\sigma-$finite measure space, with $A:L^2(\mu)\to L^2(\mu)$ given by $$(Af)(x)=\int_X K(x,y)f(y)d\mu(y),$$ where $K\in L^2(\mu\times\mu)$. I am trying to show that $\|A\|\leq\|K\|$. This question is part of an old prelim exam in analysis; I'm studying to take my own prelim.
I have the following: \begin{align} \lVert (Af)(x)\rVert_{L^2(\mu)} &= \left( \int_X\left| \int_X K(x,y)f(y)d\mu(y) \right|^2 d\mu(x)\right)^{1/2}\\ &\leq \left( \int_X \left( \int_X|K(x,y)| \cdot \lVert f\rVert d\mu(y) \right)^2 d\mu(x)\right)^{1/2}\\ &=\lVert f\rVert \cdot\left( \int_X \left( \int_X|K(x,y)| d\mu(y) \right)^2 d\mu(x)\right)^{1/2} \end{align}
Does it look like I'm on the right track? If so, where do I go from here? Does something justify me in just setting my final expression above to be less than or equal to $\lVert f \rVert \cdot \left( \int_X \int_X |K(x,y)|^2d\mu(y)d\mu(x) \right)^{1/2}$?
You are using the infinity norm of $f$, which is irrelevant, and you are not using that $K$ is in $L^2$.
The natural way is, using Cauchy-Schwarz, \begin{align} \|Af\|_2^2&= \int_X\left| \int_X K(x,y)f(y)d\mu(y) \right|^2 d\mu(x)\\ \ \\ &\leq\int_X\left|\left(\int_X|K(x,y)|^2\,d\mu(y)\right)^{1/2}\,\|f\|_2\right|^2\,d\mu(x)\\ \ \\ &=\|f\|_2^2\,\int_X\int_X|K(x,y)|^2\,d\mu(y)\,d\mu(x)\\ \ \\ &=\|f\|_2^2\,\|K\|_2^2. \end{align} So $\|A\|_2\leq\|K\|_2$.