The Banach Fixed-Point Theorem

From Repeated Estimates to a Fixed Point

A fixed point of a map \(T:X\to X\) is a point \(p\in X\) such that \(T(p)=p\). Finding such a point directly can be difficult. The Banach fixed-point theorem gives a reliable alternative: start at any point, repeatedly apply \(T\), and use a uniform shrinking condition to show that the resulting sequence converges to the unique fixed point.

Throughout, a metric space \((X,d)\) is given its metric topology. It is complete if every Cauchy sequence in \(X\) converges to a point of \(X\). A sequence \((x_n)\) is Cauchy if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x_m)<\varepsilon\) whenever \(n,m\ge N\). The nonemptiness of \(X\) matters: an empty space has no point to serve as a fixed point.

The condition says more than that distances do not increase: they shrink by a common factor strictly less than one. The map must also take every point of \(X\) back into \(X\), so its iterates are defined.

A Contraction Is Continuous

Proof. Let \(x\in X\), and let \(\varepsilon>0\). If \(d(y,x)<\varepsilon\), then the contraction inequality gives \(d(Ty,Tx)\le qd(y,x)<\varepsilon\), since \(q<1\) (and the conclusion also holds when \(q=0\)). Thus points sufficiently close to \(x\) have images within \(\varepsilon\) of \(Tx\); in fact, \(d(y,x)<\varepsilon\) itself suffices. This is continuity at \(x\). Since \(x\) was arbitrary, \(T\) is continuous. \(\square\)

We will also use the elementary finite geometric-sum identity \(\sum_{k=n}^{m-1}q^k=q^n(1-q^{m-n})/(1-q)\) for \(0\le q<1\) and \(m>n\). Since \(0\le 1-q^{m-n}\le1\), this sum is at most \(q^n/(1-q)\). These estimates include \(q=0\), with \(q^0=1\).

The Banach Fixed-Point Theorem

Proof. Choose any \(x_0\in X\), which is possible because \(X\) is nonempty, and define the iterates \(x_{n+1}=T(x_n)\). Fix a contraction constant \(q\) with \(0\le q<1\), and put \(a=d(x_0,x_1)\). We first claim that \(d(x_n,x_{n+1})\le q^n a\) for every \(n\ge0\). For \(n=0\) this is equality. If it holds for \(n\), then \(d(x_{n+1},x_{n+2})=d(Tx_n,Tx_{n+1})\le qd(x_n,x_{n+1})\le q^{n+1}a\). This proves the claim by induction.

For \(m>n\), repeated use of the triangle inequality now gives \(d(x_n,x_m)\le\sum_{k=n}^{m-1}d(x_k,x_{k+1})\le a\sum_{k=n}^{m-1}q^k\le aq^n/(1-q)\). If \(a=0\), this estimate says \(d(x_n,x_m)=0\) for every \(m>n\), so the sequence is constant and hence Cauchy. If \(a>0\), then \(aq^n/(1-q)\to0\) as \(n\to\infty\); therefore, for every \(\varepsilon>0\), choosing \(N\) with \(aq^N/(1-q)<\varepsilon\) gives \(d(x_n,x_m)<\varepsilon\) for all \(m>n\ge N\). Symmetry of the metric gives the same condition when \(n>m\). Thus \((x_n)\) is Cauchy in either case.

Completeness supplies a point \(p\in X\) with \(x_n\to p\). To show that \(p\) is fixed, use the contraction inequality and the triangle inequality: \(d(Tp,p)\le d(Tp,Tx_n)+d(x_{n+1},p)\le qd(p,x_n)+d(x_{n+1},p)\). Both terms on the right tend to zero, since \(x_n\to p\) and the shifted sequence \(x_{n+1}\) also converges to \(p\). Hence \(d(Tp,p)=0\), so \(Tp=p\).

Finally, if \(p\) and \(r\) are fixed points, then \(d(p,r)=d(Tp,Tr)\le qd(p,r)\). If \(d(p,r)>0\), dividing by this positive number would give \(1\le q\), contrary to \(q<1\). Therefore \(d(p,r)=0\), so \(p=r\). The fixed point is unique. The argument started from an arbitrary \(x_0\), so the iterates from every starting point converge to this same point. \(\square\)

The proof has two separate jobs. The contraction estimate makes the iterates Cauchy; completeness ensures their limit belongs to \(X\). The contraction inequality then makes that limit fixed, and the same strict inequality forces any two fixed points to coincide.

Why Completeness and Strict Shrinking Matter

Common pitfall. Replacing the strict condition \(q<1\) by \(q\le1\) makes the theorem false. On the complete space \(\mathbb{R}\), the identity map \(T(x)=x\) satisfies \(d(Tx,Ty)=d(x,y)\), but its only available Lipschitz factor is \(1\), not a number strictly below \(1\). Every real number is a fixed point, so uniqueness fails. The strict inequality is what makes the uniqueness argument work and the geometric distances shrink.

The theorem is useful whenever a problem can be expressed as finding a fixed point and the map can be shown to send a complete space into itself while uniformly shrinking distances. Its proof is also a model of how completeness converts an estimate on successive approximations into an actual point in the space. Each hypothesis has a distinct role: nonemptiness allows a starting point, the self-map condition allows iteration, strict contraction controls the iterates and uniqueness, and completeness keeps their limit inside the space.