What are the conditions on the kernel $k(x,s)$ that guarantees that the integral operator $T : L_{1}(a,b) \longrightarrow L_{1}(a,b)$ where $Tf(x) = \int_{a}^{b} k(x,s) f(s) ds$ be Lipschitz continuous?
My attempt $k(x,s) : [a,b] \times [a,b] \longrightarrow \mathbb{R}$ should be continuous on $a ≤ x,s ≤ b$, and $M = Sup_{x,s} |k(x,s)|$
\begin{align} \|Tf-Tg\|_1&\le\int_a^b\int_a^b|k(x,s)|\,|f(s)-g(s)|\,ds\,dx\\ &=\int_a^b|f(s)-g(s)|\Bigl(\int_a^b|k(x,s)|\,dx\Bigr)\,ds\\ &\le\Bigl(\sup_{a\le s\le b}\int_a^b|k(x,s)|\,dx\Bigr)\,\|f-g\|_1. \end{align} A suficient condition is $$ \sup_{a\le s\le b}\int_a^b|k(x,s)|\,dx<\infty. $$