The Continuous Dependence of Solutions to Volterra Equation Can any tell me why the solutions of the lineal _Volterra integral equation_ of second kind, have continuous dependence? $y(t)=g(t)+\int_{0}^{t}k(t,s)y(s)ds$ _k_ and _g_ continuous....

If I understand your question correctly, you want to know why $y$ is continuous. The answer follows from the following $$\begin{align} &y(t+\Delta t)-y(t)=g(t+\Delta t)-g(t)+\int_0^{t+\Delta t} k(t+\Delta t,s)y(s)\mathrm{d}s-\int_0^t k(t,s)y(s) \mathrm{d}s \\\ &=g(t+\Delta t)-g(t)+\int_0^t \left( k(t+\Delta t,s)-k(t,s) \right) y(s) \mathrm{d} s+\int_t^{t+\Delta t} k(t+\Delta t,s) y(s) \mathrm{d}s. \end{align}$$ As $g$ is continuous, $g(t+\Delta t)-g(t) \to 0$ as $\Delta t \to 0$. The continuity of $k$ should help you show that the other terms also approach zero as $\Delta t$ does.