= Solution
Let $B=\{p:\|p-p_0\|\leq r\}$ and write $\varepsilon=\|A(p_0)-p_0\|$. Integrating the <Fréchet derivative> along the line segment between two points in the <convex set> gives
$$
A(p)-A(q)=\int_0^1 DA(q+t(p-q))(p-q)\,dt,
$$
and the <operator norm> bound implies
$$
\|A(p)-A(q)\|\leq\kappa\|p-q\|.
$$
In particular, for $p\in B$,
$$
\|A(p)-p_0\|\leq\|A(p)-A(p_0)\|+\varepsilon
\leq\kappa r+\varepsilon\leq r.
$$
Thus the <a posteriori contraction ball> is mapped into itself. The ball is a closed subset of a <Banach space>, hence is a <complete metric space>. Applying the <contraction mapping theorem> to $A:B\to B$ gives existence and uniqueness:
$$
\boxed{\varepsilon\leq(1-\kappa)r\quad\Longrightarrow\quad
\text{exactly one fixed point }p_*\in B.}
$$
The estimate also gives $\|p_*-p_0\|\leq\varepsilon/(1-\kappa)$, by applying the same triangle inequality to $A(p_*)=p_*$. Equality in the residual bound is allowed; strict <contraction mapping> comes from $\kappa<1$, and a <fixed point> on the closed ball's boundary is included.
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