Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$ \|K^nx − \bar x\| ≤ \sum_{j=n}^\infty θ_j\cdot \|Kx − x\|,\quad x∈ C. $$

share|cite|improve this question
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 \[ \norm{\bar x - K^n x } \le \sum_{i=n}^\infty \theta_i \cdot \norm{Kx - x}. \]

share|cite|improve this answer
Thank you so much! I'm impressed how quickly you solved it. – Klara Oct 15 '12 at 14:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.