Solution.

At most one

Suppose \(f(x)=x\) and \(f(y)=y\) with \(x\neq y\). Then \[d(x,y)=d\big (f(x),f(y)\big )<d(x,y),\] which is impossible. So there cannot be two distinct fixed points.

Possibly none

Take \(f(x)=x+\frac 1x\) on \([1,\infty )\), a complete metric space. For \(x\neq y\), the mean value theorem gives \[|f(x)-f(y)|=|f'(c)|\,|x-y|=\left |1-\frac {1}{c^2}\right ||x-y|<|x-y|\] for some \(c>1\), since \(0<1-\frac {1}{c^2}<1\). So the strict contraction condition holds.

But \(f(x)=x\) would require \(\frac 1x=0\), which has no solution. So \(f\) has no fixed point at all.

Why the contraction mapping theorem does not apply

It requires a single constant \(k<1\) with \(d(f(x),f(y))\leq k\,d(x,y)\) for all \(x,y\). Here \(\sup _{c>1}\left |1-\frac {1}{c^2}\right |=1\), approached as \(c\rightarrow \infty \), so no such \(k\) exists. The distinction between ”shrinks every pair” and ”shrinks every pair by a fixed factor” is exactly what the theorem turns on.

Solution.

It maps \([1,\infty )\) into itself

For \(x\geq 1\), \(x+\frac 1x\geq 2\), so \(f(x)\geq 2k\geq 1\) because \(k\geq \frac 12\). This is where the lower bound on \(k\) is used.

It is a contraction

\(f'(x)=k\left (1-\frac {1}{x^2}\right )\), and for \(x\geq 1\) this lies in \([0,k)\). So \(|f'|\leq k<1\) throughout, and by the mean value theorem \[|f(x)-f(y)|\leq k|x-y| \qquad \text {for all }x,y\geq 1 .\]

The fixed point

Solving \(k\left (x+\frac 1x\right )=x\) gives \(kx^2+k=x^2\), so \[x^2=\frac {k}{1-k},\qquad x=\sqrt {\frac {k}{1-k}} .\] Since \(k\geq \frac 12\) we have \(\frac {k}{1-k}\geq 1\), so \(x\geq 1\) and the fixed point does lie in the space. At \(k=\frac 12\) it is exactly \(x=1\).

By the contraction mapping theorem this fixed point is unique and is reached by iterating \(f\) from any starting point in \([1,\infty )\).

Solution.

Contraction

For \(f,g\in C\left [0,\frac 12\right ]\) and \(x\in \left [0,\frac 12\right ]\), \[\big |(Tf)(x)-(Tg)(x)\big |=\left |\int _0^x\big (f(y)-g(y)\big )dy\right | \leq \int _0^x\big |f(y)-g(y)\big |dy\leq x\,||f-g||_\infty .\] Since \(x\leq \frac 12\), taking the supremum over \(x\) gives \[||Tf-Tg||_\infty \leq \tfrac 12\,||f-g||_\infty ,\] so \(T\) is a contraction with constant \(\frac 12\). The space \(C\left [0,\frac 12\right ]\) with the supremum metric is complete, so the contraction mapping theorem applies.

The fixed point

\(Tf=f\) means \(f(x)=1+\int _0^x f(y)\,dy\). Differentiating, \(f'=f\), and setting \(x=0\) in the integral equation gives \(f(0)=1\). Hence \[f(x)=e^{x}.\]

This is Picard’s theorem in miniature: the integral equation is the initial value problem \(y'=y\), \(y(0)=1\) recast so that solving it becomes finding a fixed point. The interval had to be short enough — here \(\frac 12\) — to make the constant less than \(1\).

Solution.

The centres form a Cauchy sequence

For \(m\geq n\) the ball \(B[x_m,r_m]\) is contained in \(B[x_n,r_n]\), and \(x_m\) lies in its own ball, so \(d(x_n,x_m)\leq r_n\). Since \(r_n\rightarrow 0\), given \(\varepsilon >0\) we can take \(N\) with \(r_N<\varepsilon \), and then \(d(x_n,x_m)\leq r_N<\varepsilon \) for all \(m,n\geq N\). So \((x_n)\) is Cauchy, and by completeness it converges to some \(x\).

\(x\) lies in every ball

Fix \(n\). For \(m\geq n\) we have \(x_m\in B[x_n,r_n]\), and a closed ball is a closed set, so the limit \(x\) of these points also lies in \(B[x_n,r_n]\). As \(n\) was arbitrary, \(x\in \bigcap _n B[x_n,r_n]\).

Nothing else does

If \(y\) is also in every ball then \(d(x,y)\leq d(x,x_n)+d(x_n,y)\leq d(x,x_n)+r_n\rightarrow 0\), so \(y=x\).

Both hypotheses are needed. Without completeness the intersection can be empty, and without \(r_n\rightarrow 0\) it can contain more than one point.

Solution.

A counterexample

Take \(f(x)=\frac 1x\) on \(X=(0,1)\) with the usual metric, which is continuous there. The sequence \(x_n=\frac 1n\) is Cauchy in \(X\), as computed earlier. But \[f(x_n)=n ,\] and \(|f(x_n)-f(x_m)|=|n-m|\) does not tend to \(0\), so the image sequence is not Cauchy.

What fixes it

Uniform continuity does. Suppose \(f\) is uniformly continuous and \((x_n)\) is Cauchy. Given \(\varepsilon >0\) choose \(\delta \) from uniform continuity, then \(N\) with \(d(x_n,x_m)<\delta \) for \(n,m\geq N\); then \(\rho \big (f(x_n),f(x_m)\big )<\varepsilon \) for those \(n,m\).

The counterexample is consistent with this: \(\frac 1x\) is continuous on \((0,1)\) but not uniformly so, precisely because the domain is not compact. This is also why uniform continuity, not mere continuity, is what allows a map to be extended to the completion of a space.

Solution.

\(\mathbb {R}\) is complete

Let \((x_n)\) be Cauchy. It is bounded, by the earlier argument, so by the Bolzano–Weierstrass theorem it has a convergent subsequence \(x_{n_k}\rightarrow l\). A Cauchy sequence with a convergent subsequence converges to the same limit: given \(\varepsilon >0\), choose \(N\) with \(|x_n-x_m|<\frac {\varepsilon }{2}\) for \(n,m\geq N\), and \(k\) with \(n_k\geq N\) and \(|x_{n_k}-l|<\frac {\varepsilon }{2}\); then for \(n\geq N\) \[|x_n-l|\leq |x_n-x_{n_k}|+|x_{n_k}-l|<\varepsilon .\]

Complete \(\iff \) closed

Let \(A\subseteq \mathbb {R}\). If \(A\) is closed and \((x_n)\subseteq A\) is Cauchy, then it converges in \(\mathbb {R}\) by the above, and its limit lies in \(A\) because \(A\) is closed; so \(A\) is complete.

Conversely if \(A\) is complete, let \(x\) be a limit point of \(A\) and take \(x_n\in A\) with \(x_n\rightarrow x\). Convergent sequences are Cauchy, so by completeness \((x_n)\) converges in \(A\), and limits are unique, so \(x\in A\). Hence \(A\) contains its limit points and is closed.

The equivalence uses completeness of the ambient space, and fails without it: \((0,1)\) is closed in itself but not complete as a subspace of \(\mathbb {R}\).