For clearly deliver my purpose, I rewrite this question.
Consider first kind Volterra integral equations $$ \int_0^t k(t,s)f(s)ds=g(t) \quad 0\leq t\leq T $$ where $k(t,s)$ is continuous but not differentiable and $g(t)$ is sufficiently smooth.
More specifically, I consider the following Volterra integral equation $$ \int_0^t \sqrt{t-s}f(s)ds = t $$
Assume that there exists a unique continuous solution $f(s)$. (Indeed, by differentiating, the above integral equation is converted to Abel integral equation so that we assure that there exists a unique continuous solution $f(s)$ satisfying the above first kind Volterra integral equation. )
We can simply obtain approximate solutions for the integral equation by discretization. Simplest rectangular approximation leads to $$ h\sum_{i=0}^{n-1} \sqrt{t_n-t_i}F(t_i)=t_n \quad 0\leq n\leq N $$ where $t_i=hi$ and $h=T/N$. approximate solution $F(t_i)s$ can be recursively computed.
My question is whether the following relation holds $$ |F(t_i)-f(t_i)|\rightarrow 0 \quad h\rightarrow0 $$ for all $i$.
For differentiable $k(t,s)$ , it is shown that the above numerical methods are convergent in Linz(1962)Numerical methods for volterra integral equations of the first kind. But our case is not differentiable.
If convergent, how to prove?
If not, What kind of counterexamples are?
Volterra equations are typically convolution function and therefore are suitable for use Laplace transform. In particular, this approach can be applied to the equation $$\int_0^t\sqrt{t-s}\,f(s)\,ds = t,$$ or $$\sqrt t*f(t) = t.$$ Let $$F(p) = \mathcal L(f(t)) = \int_0^\infty f(s)e^{-ps}\,ds.$$ Using data from the table of Laplace transforms, we get: $$\mathcal L(\sqrt t) = \dfrac{\sqrt\pi}{2p^{3/2}},\quad \mathcal L(t) = \dfrac1{p^2},\quad\mathcal L\left(\dfrac1{\sqrt t}\right) = \sqrt{\dfrac\pi {p}},$$ so Laplace transform for the considering equation takes form: $$\dfrac{\sqrt\pi}{2p^{3/2}}F(p) = \dfrac1{p^2},$$ then $$F(p) = \dfrac2{\sqrt{\pi p}},$$ and $$\boxed{f(t) = \dfrac2{\pi\sqrt t}}.$$ Check it: $$\sqrt t*f(t) = \dfrac2\pi\int_0^t\sqrt{\dfrac{t-s}s}\,ds = \genfrac{|}{|}{0pt}{0}{s=t\sin^2u,}{ds = 2t\sin u \cos u} = \dfrac{4t}\pi\int_0^{\pi/2}\cos^2u\,du = t.$$