The exercise I'm working with is 2.1.6(b) in Conway's A Course in Functional Analysis (2/e):
Let $(X,\,\Omega,\,\mu)$ be a $\sigma$-finite measure space and let $k_1$, $k_2$ be two kernels satisfying \begin{equation} \int_X |k_i(x,\,y)|\,d\mu(y) \le c_1 \quad \text{a.e.}[\mu],\\ \int_X |k_i(x,\,y)|\,d\mu(x) \le c_2 \quad \text{a.e.}[\mu]. \end{equation} Define $$ k:X \times X \to \mathbb F \quad \text{by} \quad k(x,\,y) = \int_X k_1(x,\,z) \, k_2(z,\,y)\,d\mu(z). $$ (a) Show that $k$ also satisfies the same hypothesis as $k_1$ and $k_2$.
(b) If $K$, $K_1$, $K_2$ are the integral operators with kernels $k$, $k_1$, $k_2$, show that $K=K_1K_2$.
The author defines the integral transform by: $$ (Kf)(x) := \int_X k(x,\,y)\,f(y)\,d\mu(y). $$
So easily I solved (a), with $c_i^2$ in the places of $c_i$. I think I also done with (b), but I'm not sure my solution is correct. So please be kind to check whether it is fine or not. A better idea or a more concise solution will always be appreciated.
My solution: We first show that the integral $$ I(x) = \int_X |k_1(x,\,z)| \int_X |k_2(z,\,y)| \cdot |f(y)| \,d\mu(y) \,d\,u(z) $$ is finite for almost all $x$. Use Hölder's inequality for the integrand $\sqrt{|k_2(z,\,y)|} \cdot \sqrt{|k_2(z,\,y)|}|f(y)|$ to get $$ \int_X |k_2(z,\,y)| \cdot \left| f(y) \right| \,d\mu(y) \quad\le\quad \left( \int_X |k_2(z,\,y)| \,d\mu(y) \right)^{1/2} \left( \int_X |k_2(z,\,y)| \cdot |f(y)|^2 \,d\mu(y) \right)^{1/2}. $$ Then we have $$ I(x) \quad\le\quad \sqrt{c_1} \int_X |k_1(x,\,z)| \left( \int_X |k_2(z,\,y)| \cdot |f(y)|^2 \,d\mu(y) \right)^{1/2} \,d\mu(z). $$ The similar Hölder argument for $\sqrt{k_1}$ gives $$ I(x) \quad\le\quad c_1 \left( \int_X |k_1(x,\,z)| \int_X |k_2(z,\,y)| \cdot |f(y)|^2 \,d\mu(y) \,d\mu(z) \right)^{1/2} $$ So we may estimate the $L^2$-norm of $I$ as follows: $$ \|I\|^2 \quad\le\quad {c_1} ^2 \int_X \int_X \int_X |k_1(x,\,z)| \cdot |k_2(z,\,y)| \cdot |f(y)|^2 \,d\mu(y) \,d\mu(z) \,d\mu(x). $$ But Fubini's theorem allows changing the order of the integrals. Hence $$ \text{(RHS)} \quad = \quad {c_1} ^2 \int_X \int_X \int_X |k_1(x,\,z)| \cdot |k_2(z,\,y)| \cdot |f(y)|^2 \,d\mu(x) \,d\mu(z) \,d\mu(y) \quad\le\quad {c_1}^2 {c_2}^2 \|f\|^2 $$ by the assumptions on the kernels.
Since $\|I\| < \infty$, we have $$ \int_X \int_X |k_1(x,\,z)| \cdot |k_2(z,\,y)| \cdot |f(y)| \,d\mu(y) \,d\mu(z) \quad < \quad \infty $$ for almost all $x$. Hence by the Fubini's theorem we justify the change of the order of integrals to conclude that \begin{align} (K_1 K_2 f)(x) =& \int_X \int_X k_1(x,\,z) \, k_2(z,\,y) \, f(y) \,d\mu(y) \,d\mu(z) \\ =& \int_X \int_X k_1(x,\,z) \, k_2(z,\,y) \, f(y) \,d\mu(z) \,d\mu(y) \\ =& (Kf)(x) \end{align} holds for almost all $x$.