Consider the following second kind Volterra integral equation $$x(t)=\int_0^tk(s,t)x(s)ds+f(t),$$ where $f$ is continuous on $[0,T]$ and $k$ is continuous kernel for all $0\leq s\leq 0 \leq t \leq T$.
Assume that there exists a positive number $m$ such that $$\int_0^t|k(s,t)|ds\leq m<1$$ then there exists a positive number $c$ such that $$sup|x(t)|\leq c|f(t)|.$$
My proof is as the following: If the last estimate is not true, then we have for any $c>0$ $$sup|x(t)|>c|f(t)|.$$ From the integral equation, we have the estimate $$sup|x(t)|\leq sup|x(t)| \int_0^t|k(s,t)|ds+|f(t)|,$$ thus, $$1\leq \int_0^t|k(s,t)|ds+\frac{1}{c}.$$ Now, letting c tends to $\infty$ we get $$1 \leq \int_0^t|k(s,t)|ds,$$ which contradicts the assumption. Is that right?. Thank you in advance.
Your proof is correct. Although, a more fluent way of doing it as follows. \begin{align} \sup |x(t)| &\leq \sup |x(t)|\int_0^t |k(s,t)|\, ds +|f(t)| \\ &\leq \sup |x(t)|m +|f(t)|. \end{align} so \begin{align} (1-m)\sup |x(t)|\leq|f(t)| \\ \implies \sup|x(t)|\leq \frac{|f(t)|}{1-m}. \end{align} This gets out a bound explicitly.