The operator is $Tf= g(t)+ \int_{a}^{t}(K(t,s)f(s)ds.$ The proof assumes that $g(t) \in L^2[a,b]$ so I believe I need to only prove that $\int_{a}^{t}(K(t,s)f(s)ds \in L^2[a,b]$.
I started off this way:
$$\|Tf\|^2=\int_{a}^{b}|\int_{a}^{t}(K(t,s)f(s)ds|^2dt$$ However, I'm confused at this point. Can I say by Schwarz inequality, $$\|Tf\|^2 \leq \int_{a}^{b}\bigg(\int_{a}^{t}|(K(t,s)|^2ds \cdot \int_{a}^{t}|f(s)ds|^2ds\bigg)dt$$
How do I proceed?
Assuuming that $K \in L^{2}([a,b]\times [a,b])$ we have
$$\int_{a}^{b}\bigg(\int_{a}^{t}|(K(t,s)|^2ds \cdot \int_{a}^{t}|f(s)ds|^2ds\bigg)dt$$ $$\leq \int_{a}^{b}\bigg(\int_{a}^{b}|(K(t,s)|^2ds \cdot \int_{a}^{b}|f(s)|^2ds\bigg)dt$$
Now just pull out the constant $$\int_{a}^{b}|f(s)|^2ds.$$