= Solution
The <Green's function> is nonpositive, so $f\geq0$ gives $y\leq0$. If also $f\leq m$, then
$$
\int_0^x\zeta f(\zeta)d\zeta\leq\frac{mx^2}{2},\qquad\int_x^\infty\frac{f(\zeta)}{\zeta^2}d\zeta\leq\frac mx.
$$
The integral solution therefore gives
$$
\boxed{-\frac m2=-\frac13\left(\frac m2+m\right)\leq y(x)\leq0.}
$$
Both bounds are attained, by the constant forcings $f=m$ and $f=0$, respectively. In particular, boundedness of the solution does not require it to approach zero.
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