 Solution

On $[-a,a]$ let $F=\Vert f{\Vert }_{\infty }$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)={\int }_0^{x}K(x,s)h(s)ds$. The iteration gives $y_n={\sum }_{j=0}^n{T}^{j}f$. Induction, integrating absolute values also when $x<0$, givesThe exponential series is summable uniformly for $|x|\le a$. The Weierstrass M-test therefore proves uniform convergence to a continuous function $y$. Also $\Vert T(y_n-y){\Vert }_{\infty }\le Ma\Vert y_n-y{\Vert }_{\infty }\to 0$, so passing to the limit yields $y=f+Ty$.

For the indicated kernel, twice differentiating its integral givesConversely integrating the differential equation twice, with these initial conditions, gives the integral equation. This is valid also for negative $x$, either by oriented integration or by differentiating the double-integral identity.

For every finite $a$ the continuous kernel $(x-s)s$ meets the preceding hypotheses. Solutions on overlapping intervals agree: if $d=Td$ is their difference, iteration gives $\Vert d{\Vert }_{\infty }\le \Vert d{\Vert }_{\infty }(Ma{)}^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to one solution defined on all of $\mathbb{R}$.