Convergence Implies the Existence of a Fixed Point¶
Lemma¶
Let \((X,d)\) be a metric space and \(f:X\to X\) a continuous function. Furthermore, let \(x_0, x\in X\) and \((x_n)_{n\in\mathbb{N}}\) be a sequence in \(X\) such that for all \(n\in\mathbb{N}_0\) the following is true.
\[ x_{n+1} = f(x_n) \ ,\qquad \lim_{n\to\infty} x_n = x \]
In this case, \(x\) is a fixed point of \(f\) and therefore \(f(x) = x\).
Proof¶
To prove the lemma, we directly apply the continuity of \(f\) and the definition of \((x_n)\).
\[ f(x) = f\left( \lim_{n\to\infty} x_n \right) = \lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} x_{n+1} = x \]
Note
The lemma gives us a way of computing the limit of a converging sequence by solving the equation \(f(x)=x\).
Example¶
\[ f:[0,1]\to[0,1] \ ,\qquad f(x) = \begin{cases} x^2 &: x\in (0,1] \\ \frac 12 &: x=0 \end{cases} \]
In this case, \(f\) is not continuous but provides a fixed point \(f(1)=1\). The lemma cannot be applied. For \(x_0=\frac 12\), the sequence \((x_n)\) will converge to \(0\) but \(f(0)\neq 0\).
Last update: August 11, 2020