FALSE: $d(fx, fy) < d(x,y)$ for all $x \ne y$ on a complete metric space forces a fixed point

The following statement is FALSE.

Let $(X,d)$ be a nonempty complete metric space (Complete metric space: every Cauchy sequence converges in the space) and let $f : X \to X$ satisfy $d\big(f(x), f(y)\big) < d(x,y) \qquad \text{for all } x, y \in X \text{ with } x \ne y .$ Then $f$ has a fixed point.

The condition displayed above is what many texts call contractive; it is strictly weaker than being a contraction (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction), which demands a single constant $q < 1$ with $d(f(x),f(y)) \le q\,d(x,y)$ for all pairs at once. Banach's theorem (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point) assumes the latter, and the difference between the two hypotheses is exactly what this item is about.

• What goes wrong, quantitatively. The factor by which $f$ shrinks distances is $1 - 1/(xy)$, which is below $1$ at every pair but approaches $1$ as $x$ and $y$ grow. No single $q < 1$ dominates all of them, so $f$ is not a contraction and Banach's theorem does not apply. The failure is therefore not an accident of this example but the exact difference between a pointwise inequality and a uniform one, which is the same difference as between continuity and uniform continuity (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
• Compactness would repair it; boundedness would not. On a compact space the strict condition does force a fixed point, by minimising $x \mapsto d(x,f(x))$; compactness of metric spaces is a later page in this library and nothing of the sort is claimed at this point. Adding boundedness to completeness, by contrast, is not enough, and the witness is small: on $\mathbb{N}$ put $d(m,n) := 1 + 2^{-m-1} + 2^{-n-1}$ for $m \ne n$ and $d(n,n) := 0$. Every nonzero distance lies in $(1, 7/4]$, so the triangle inequality is automatic and $d$ is a bounded metric; the space is complete because distinct points are more than $1$ apart, so a Cauchy sequence is eventually constant; and $n \mapsto n+1$ satisfies $d(m+1,n+1) = 1 + 2^{-m-2} + 2^{-n-2} < d(m,n)$ for $m \ne n$ and has no fixed point. What the present item establishes is only that completeness alone is not enough.
• The unboundedness of $[1,\infty)$ is doing the work, and the map is pushing every point to the right by a shrinking but always positive amount. The worked-out version of this witness, including the verification that no contraction constant exists, is $x \mapsto x + 1/x$ on $[1,\infty)$ strictly decreases every distance and has no fixed point ↗ on the companion page.