= Solution
On $[-a,a]$ let $F=\|f\|_\infty$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)=\int_0^xK(x,s)h(s)\,ds$. The iteration gives $y_n=\sum_{j=0}^nT^jf$. Induction, integrating absolute values also when $x<0$, gives
$$
|T^jf(x)|\leq F\frac{M^j|x|^j}{j!}.
$$
The exponential series is summable uniformly for $|x|\leq a$. The <Weierstrass M-test> therefore proves <uniform convergence> to a <continuous function> $y$. Also $\|T(y_n-y)\|_\infty\leq Ma\|y_n-y\|_\infty\to0$, so passing to the limit yields $y=f+Ty$.
For the indicated kernel, twice differentiating its integral gives
$$
y'(x)=\int_0^x sy(s)\,ds,\qquad y''(x)=xy(x),\qquad y(0)=1,\quad y'(0)=0.
$$
Conversely 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 $\|d\|_\infty\leq\|d\|_\infty(Ma)^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to \b[one solution defined on all of $\mathbb R$].
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