Banach Fixed-Point Theorem

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Let $\struct {M, d}$ be a complete metric space.

Let $f: M \to M$ be a contraction.

That is, there exists $q \in \hointr 0 1$ such that for all $x, y \in M$:

$\map d {\map f x, \map f y} \le q \, \map d {x, y}$

Then there exists a unique fixed point of $f$.



Let $f$ have two fixed points $p_1, p_2 \in M$.

It is to be proved that $p_1 = p_2$.

\(\ds \map d {p_1, p_2}\) \(=\) \(\ds \map d {\map f {p_1}, \map f {p_2} }\) Definition of Fixed Point
\(\ds \) \(\le\) \(\ds q \, \map d {p_1, p_2}\) Definition of Contraction Mapping
\(\ds \leadsto \ \ \) \(\ds \paren {1 - q} \map d {p_1, p_2}\) \(\le\) \(\ds 0\) subtracting $q \, \map d {p_1, p_2}$ from both sides

Then from $\paren {1 - q} > 0$ and $\map d {p_1, p_2} \ge 0$:

$\map d {p_1, p_2} = 0$

From Metric Space Property $(\text M 4)$:

$p_1 = p_2$


The fixed point $p$ will be found from an arbitrary member $a_0$ of $M$ by iteration.

The plan is to obtain $\displaystyle p = \lim_{n \mathop \to \infty} a_n$ with definition $a_{n + 1} = \map f {a_n}$.

The sequence of iterates converges in complete metric space $M$ because it is a Cauchy sequence in $M$, as is proved in the following.

Induction on $n$ applies to obtain the contractive estimate:

$\map d {a_{n + 1}, a_n} \le q^n \, \map d {a_1, a_0}$

Induction details $n = 1$:

\(\ds \map d {a_2, a_1}\) \(=\) \(\ds \map d {\map f {a_1}, \map f {a_0} }\) Definition of $\sequence {a_n}$
\(\ds \) \(\le\) \(\ds q \, \map d {a_1, a_0}\) Definition of Contraction Mapping

Assume the contractive estimate for $n = k$.

Induction details for $n = k + 1$:

\(\ds \map d {a_{k + 2}, a_{k + 1} }\) \(=\) \(\ds \map d {\map f {a_{k + 1} }, \map f {a_k} }\)
\(\ds \) \(\le\) \(\ds q \, \map d {a_{k + 1}, a_k}\) Definition of Contraction Mapping
\(\ds \) \(\le\) \(\ds q \, q^k \, \map d {a_1, a_0}\) Induction hypothesis applied. Induction complete.

Next it is to be proved that $\sequence {a_n}$ is a Cauchy sequence in $M$ by showing that $\displaystyle \lim_{m \mathop \to \infty} \map d {a_{n + m}, a_n} = 0$ for $n$ large.

\(\ds \map d {a_{n + m}, a_n}\) \(\le\) \(\ds \sum_{j \mathop = n}^{n + m - 1} \map d {a_{j + 1}, a_j}\) Metric Space Property $(\text M 2)$: Triangle Inequality and telescoping sum ideas
\(\ds \) \(\le\) \(\ds \sum_{j \mathop = n}^{n + m - 1} q^j \, \map d {a_1, a_0}\) Apply the contractive estimate.
\(\ds \) \(\le\) \(\ds q^n \paren {\dfrac {1 - q^m} {1 - q} } \, \map d {a_1, a_0}\) Sum of Geometric Sequence on right hand side

We have that:

$\displaystyle \lim_{n \mathop \to \infty} q^n = 0$


$\dfrac {1 -q^m} {1 - q} \le \dfrac 1 {1 - q}$

Then $\sequence {\map d {a_{n + m}, a_n} }$ has limit zero at $m = \infty$ for large $n$.

Sequence $\sequence{a_n}$ is a Cauchy sequence convergent to some $p$ in $M$.


\(\ds \map d {\map f p, p}\) \(\le\) \(\ds \map d {\map f p, \map f {a_n} } + \map d {\map f {a_n}, p}\) Metric Space Property $(\text M 2)$: Triangle Inequality
\(\ds \) \(\le\) \(\ds q \, \map d {p, a_n} + \map d {a_{n + 1}, p}\) Definition of Contraction Mapping, and $a_{n + 1} = \map f {a_n}$

The right hand side has limit zero at $n = \infty$.


$\map d {\map f p, p} = 0$

Then by Metric Space Property $(\text M 4)$:

$\map f p = p$.


Also known as

Also known as:

the Contraction Mapping Theorem
the Contraction Theorem
the Banach Contraction Theorem
the Contraction Lemma.

Source of Name

This entry was named for Stefan Banach.