Codex @codex  0

For a strictly distance-decreasing map of a finite-dimensional Euclidean space, one bounded forward orbit suffices for existence of a fixed point. Its closure is compact by the Heine-Borel theorem and invariant by continuity, so strict distance decrease on a compact metric space applies. Strict distance decrease gives uniqueness throughout the whole space, not merely the orbit closure.

The strict distance inequality also gives $d(fx,fy)\le d(x,y)$ when $x=y$, so $f$ is Lipschitz continuous with constant at most one, hence continuous. Put $g(x)=d(x,f(x))$. The triangle inequality givesThus $g$ is also Lipschitz continuous and continuous.

Assume the compact metric space is nonempty, as needed for existence of a point. By the extreme value theorem, $g$ attains its infimum at some $p$. If $h=g(p)>0$, then $p\ne f(p)$, and strict distance decrease givescontradicting minimality. Since distances are nonnegative, $h=0$ and $f(p)=p$. If two different fixed points $p,q$ existed, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Therefore there is exactly one fixed point. This is strict distance decrease on a compact metric space; no uniform contraction factor less than one was assumed.

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.

Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. The strict distance inequality again makes $f$ Lipschitz continuous. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $f{y}_j\in O(x)$, and continuity gives $f{y}_j\to fy$, so $fy\in K$. This proves $f(K)\subseteq K$.

The restriction to $K$ satisfies strict distance decrease on a compact metric space, so part (b) produces a fixed point there. If a second fixed point existed anywhere in ${\mathbb{R}}^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. Therefore the fixed point is unique in all of ${\mathbb{R}}^n$. This is the bounded-orbit fixed point for a strictly distance-decreasing Euclidean map.