Global Picard iteration with a decaying kernel ($4|b|<c$)

Consider the Volterra integral equationwhere $h$ is continuous, $|h|\le 1$, $h(0)=1$, and the continuous kernel satisfies $0\le K(t,s)\le e^{-c(t-s)}$ for $0\le s\le t$, with $c>0$. If $0<|b|<c/4$, there is exactly one continuous solution on the whole half-line. It is bounded, and Picard iteration beginning at $y_0=bh$ converges in supremum norm with contraction ratio at most $\gamma =4|b|/c<1$.

For a proof, the integral of the kernel is at most $1/c$. Thus $T$ maps the closed ball of radius $2|b|$ in the Banach space of bounded continuous functions into itself, since $\Vert Ty{\Vert }_{\infty }\le |b|+4b^2/c<2|b|$. Factoring $y^2-v^2$ gives $\Vert Ty-Tv{\Vert }_{\infty }\le \gamma \Vert y-v{\Vert }_{\infty }$ on this ball. The contraction mapping theorem gives its unique fixed point. A first-exit argument shows that every continuous solution, even one not initially assumed bounded, stays strictly inside the same ball: a first time with $|y|=2|b|$ would contradict $|y|\le |b|+4b^2/c$. Reapplying the estimate gives the strict global bound $\Vert y{\Vert }_{\infty }\le |b|(1+\gamma )<2|b|$, so uniqueness is unrestricted within the continuous solution class. If $b=0$, zero is the only solution by the Gronwall inequality on each finite interval.