I have the following problem, which is a variation of problem 2.1.3 in Conway's "A Course in Functional Analysis":
Let $(X,\mathcal{M},\mu)$ be a $\sigma$-finite measure space. Let $k_1,k_2$ be two kernels (i.e., measurable functions on $X\times X$ s.t.: $${ess\sup}_{y\in X} \int_X|k_i(x,y)|d\mu(x),\ {ess\sup}_{x\in X}\int_X|k_i(x,y)|d\mu(y) < \infty$$ We know that the operator: $$ (K_if)(x) = \int_Xk_i(x,y)f(y)d\mu(y)$$ is a well-defined bounded linear operator on $L^p(X,\mathcal{M},\mu)$ for $i = 1,2$ and $ 1 \leq p \leq \infty$. Show that $$k(x,y) = \int_Xk_1(x,z)k_2(z,y)d\mu(z)$$ is a well-defined measurable function on $X\times X$, satisfying: $${ess\sup}_{y\in X} \int_X|k(x,y)|d\mu(x),\ {ess\sup}_{x\in X}\int_X|k(x,y)|d\mu(y) < \infty$$ And the operator $K = K_1K_2$ is given by: $$(Kf)(x)=\int_X k(x,y)f(y)d\mu(y)$$
I'm struggling with understanding measure theory / functional analysis in general. I did notice that if $y$ is a constant, then we get $k(x,c) = (K_1k_2)(x)$, and if $x$ is a constant, then $k(c,y) = (K_2k_1)(y)$. And therefore I could write, for example: $$\int_X|k(x,y)|d\mu(x) = \int_X|(K_1k_2)(x)|d\mu(x)$$ And then maybe I can bound this integral, but I'm not really sure if this is the right approach. Also, I'm not sure how to show $k(x,y)$ is a well-defined or measurable function. I don't know if Fubini's theorem can help me here. I could really use a hint or something to understand how to approach this problem. Thank you all in advance!
You have all the right ideas, you just need to unpack the definitions.
$\bullet$ To show that $k$ is well defined on $X\times X$ : Note that for all $(x,y)\in X\times X$, $$k(x,y):=\int_Xk_1(x,z)k_2(z,y)d\mu(z) = [K_1 k_2(\cdot,y)](x).$$ So for all $(x,y)$, $|k(x,y)| \le \|K_1\|_{L^1\to L^1} \|k_2(\cdot,y)\|_{L^1(d\mu)}$.
Moreover, by our assumption on $k_2$, $k_2(\cdot,y)\in L^1(d\mu)$ for $\mu$-almost all $y\in X$, hence by our assumption on $K_1$, we have that $K_1 k_2(\cdot,y)\in L^1(d\mu)$ for $\mu$-almost all $y$ in $X$ as well. This implies that $k:(x,y)\mapsto [K_1k_2(\cdot,y)](x)$ is almost everywhere equal to an element of $L^1$, hence it is well-defined and an element of $L^1$ itself (and the previous inequality shows it also is an element of $L^\infty$).
$\bullet$ To show that $\int_X|k(x,y)|d\mu(x)$ is essentially bounded over $y\in X$, write $$\begin{align}\int_X|k(x,y)|d\mu(x) &:= \int_X\left|\int_Xk_1(x,z)k_2(z,y)d\mu(z)\right|d\mu(x)\\ &\le\int_X\int_X|k_1(x,z)|\cdot|k_2(z,y)|d\mu(z)d\mu(x)\\ &= \int_X\left[\int_X|k_1(x,z)|d\mu(x)\right]\cdot|k_2(z,y)|d\mu(z)\\ &\le A_1\int_X|k_2(z,y)|d\mu(z)\\ &\le A_1\cdot A_2<\infty, \tag1\end{align} $$ where we have used triangle inequality, Tonelli's theorem and the assumptions on $k_i$ to write for $i=1,2$ $A_i := {\text{ess}\sup}_{y\in X}\int_X|k_i(x,y)|d\mu(y) < \infty$. Since the upper bound $(1)$ does not depend on $y$, the conclusion follows by taking the supremum. Clearly, the exact same argument works for $x$.
I'm guessing you also wanted to show that $K=K_1K_2$ is a bounded linear operator mapping $L^p$ to itself. Can you take inspiration from the above to prove it ?