Exercise 2182Topology › Metric and normed spacesM1Added 3 Nov 2026

Banach fixed point theorem

#Banach#fixed point#contraction

Statement

Let $(X,d)$ be complete and $f:X\to X$ be a contraction ( $d(f(x),f(y))\le kd(x,y)$ with $k<1$ ). Show that $f$ has a unique fixed point and that iterates converge to it.

Hint

Show $(x_n)$ with $x_{n+1}=f(x_n)$ is Cauchy; use $d(x_{n+1},x_n)\le k^n d(x_1,x_0)$ .

Show solutionHide solution

Solution

Cauchy. $d(x_{n+1},x_n)\le k^nd(x_1,x_0)$ ; summing, $d(x_n,x_m)\le\frac{k^n}{1-k}d(x_1,x_0)\to0$ .

Convergence. Complete: $x_n\to x^*$ . Continuity: $f(x^*)=x^*$ .

Uniqueness. If $f(x^*)=x^*$ and $f(y^*)=y^*$ , $d(x^*,y^*)\le kd(x^*,y^*)$ , so $d(x^*,y^*)=0$ .

Conclusion. Banach fixed point theorem holds; iterates converge geometrically.

Banach fixed point theorem

Statement

Solution

More problems in Metric and normed spaces