The Banach fixed point theorem is stated in my book (Applied Asymptotic Analysis by Miller) as
Let $\mathcal B$ be a Banach space with norm $\|\cdot\|$. Let $X$ be a nonempty bounded subset of $\mathcal B$ and suppose that $T \colon X \to X$ is a mapping that satisfies, for some $0 < \rho < 1$, the inequality $$ \|T(f) - T(g)\| \leq \rho \|f-g\| $$ for all $f$ and $g$ in $X$. Then there exists a unique element $f^\infty \in X$ such that (i) the sequence of iterates $\{T^k(f)\}_{k \geq 0}$ converges to $f^\infty$ whenever $f \in X$ and (ii) $f^\infty = T(f^\infty)$.
I'm having some trouble understanding this result.
Suppose $X_R$ is the ball of radius $R$, i.e. $$ X_R = \{f \in \mathcal B \colon \|f\| \leq R\}, $$ and suppose that $T$ is a contraction mapping on $X_R$. The theorem says that $T$ has a unique fixed point $f^\infty \in X_R$.
But isn't $T$ also a contraction mapping in every ball $X_S$ with $0 < S < R$? Does the theorem then imply that $T$ has a unique fixed point in $X_S$? It then seems to me that we must have $\|f^\infty\| = 0$, for otherwise it would be outside of some such ball.
Where am I going wrong?
You must have that $T$ is also a contraction mapping in every ball $X_S$, i.e. that $T(X_S) \subseteq X_S$. This part of the criterion seems very implicit but is in fact very important ; it allows you to iterate $T$. That does not follow from the fact that $T$ is a contraction mapping.
Take the example where $T$ just "zooms in" in a sub-ball of your original ball, but that sub-ball closer to the side of the ball than the center (drawing a decreasing sequence of balls to see it is a good idea). Your map will be a contraction mapping, but the limit point will not be zero.
Hope that helps,