# Weissinger's Theorem. How to prove?

Theorem (Weissinger). Let $C$ be a (nonempty) closed subset of a Banach space $X$. Suppose $K : C → C$ satisfies $$\|K^nx − K^ny\| ≤ θ_n\|x − y\|, \quad x,y∈ C$$ with $\sum_n θ_n < ∞$. Then $K$ has a unique fixed point $\bar x$ such that $$\|K^nx − \bar x\| ≤ \sum_{j=n}^\infty θ_j\cdot \|Kx − x\|,\quad x∈ C.$$

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What is $K_n$? Do you know how to use Tex in this site? – Davide Giraudo Oct 15 '12 at 14:02
@DavideGiraudo: Perhaps it is to be read as $K^n$? – martini Oct 15 '12 at 14:04
Davide, No I do not know how to use Tex yet. Kn is K^n the nth iteration of K. – Klara Oct 15 '12 at 14:11
@Klara Please check if I did the TeXification correclty. – martini Oct 15 '12 at 14:13
@ Martini you did great! – Klara Oct 15 '12 at 14:20

$\def\norm#1{\left\|#1\right\|}$Let $x, y \in C$ be fixed points of $K$, we then have, as $\theta_n \to 0$, that $$\norm{x-y} = \norm{K^nx - K^n y} \le \theta_n \|x-y\| \to 0.$$ So $x=y$ and $K$ has at most one fixed point.
To prove existence, let $x \in C$, we have for $n,k \ge 0$ \begin{align*} \norm{K^{n+k}x - K^nx} &\le \norm{K^{n+k}x - K^{n+k-1} x} + \norm{K^{n+k-1}x - K^nx} \\ &\le \theta_{n+k-1} \norm{Kx - x} + \norm{K^{n+k-1}x -x}\\ &\le \cdots\\ &\le \sum_{i=n}^{n+k-1} \theta_{i} \norm{Kx - x}\\ &\le \sum_{i=n}^\infty \theta_i \cdot \norm{Kx - x} \end{align*} As $\sum_i \theta_i < \infty$, we have $\sum_{i=n}^\infty \theta_i \to 0$ for $n\to\infty$. So $\norm{K^{n+k} x - K^nx} \to 0$, $n \to \infty$ uniformly in $k$. So $(K^n x)_n$ is Cauchy, hence convergent (as $C$ is a closed subspace of a complete space, so complete). Let $\bar x := \lim_n K^n x$. Then, as $K$ is continuous $K\bar x = K(\lim_n K^n x) = \lim_n K^{n+1} x = \bar x$ So $\bar x$ is a fixed point, this proves existence. Returning to our above estimate $\norm{K^{n+k} x - K^n x } \le \sum_{i=n}^\infty \theta_i \cdot \norm{Kx - x}$ No for $k \to \infty$, we have $K^{n+k}x \to \bar x$, so $\norm{\bar x - K^n x } \le \sum_{i=n}^\infty \theta_i \cdot \norm{Kx - x}.$