Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be a function, with the three regularity conditions as in Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction. Then:

1. If $f$ is a contraction, it is Lipschitz.
2. If $f$ is Lipschitz, it is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
3. If $f$ is uniformly continuous, it is continuous (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form).
4. If $f$ is $\alpha$-Hölder for some rational $\alpha$ with $0 < \alpha \le 1$, it is uniformly continuous.
5. Suppose $X$ is nonempty and bounded, and put $R := \operatorname{diam}(X) + 1$, a real with $R > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). If $f$ is Lipschitz with constant $L$, then for every rational $\alpha$ with $0 < \alpha \le 1$ the map $f$ is $\alpha$-Hölder with constant $L\,R^{\,1-\alpha}$.

Claim 5 carries its boundedness hypothesis for a reason, and no unconditional "Lipschitz implies Hölder" is asserted anywhere here. On an unbounded space the implication is false; the witness and its verification are in the first remark below.

Strictness is not claimed. The five implications are asserted and nothing more; that none of them reverses is witnessed elsewhere, and those witnesses are not prerequisites of this theorem. See the second remark.

• Boundedness in claim 5 cannot be dropped, and here is the witness. Take $X = Y = \mathbb{R}$ with the usual metric and $f(x) = x$, which is Lipschitz with constant $1$. Suppose $f$ were $\alpha$-Hölder with constant $C \ge 0$ for some rational $\alpha$ with $0 < \alpha < 1$, so that $|x - y| \le C\,|x-y|^{\alpha}$ for all reals $x,y$. Taking $y = 0$ and $x = t$ with $t > 0$, and writing $t = t^{\alpha} t^{\,1-\alpha}$ (Laws of rational exponents), division by $t^{\alpha} > 0$ gives $t^{\,1-\alpha} \le C$ for every real $t > 0$. At $t = 1$ this reads $1 \le C$, so $C > 0$; and then choosing a natural $n$ with $n > C^{1/(1-\alpha)}$, which exists by the Archimedean property (Every complete ordered field is Archimedean), and raising to the positive rational power $1-\alpha$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$, Laws of rational exponents) gives $n^{\,1-\alpha} > \big(C^{1/(1-\alpha)}\big)^{1-\alpha} = C$, contradicting $t^{\,1-\alpha} \le C$ at $t = n$. So the identity of $\mathbb{R}$ is Lipschitz and $\alpha$-Hölder for no exponent $\alpha < 1$. A chain reading "Lipschitz implies Hölder implies uniformly continuous" is therefore false as stated, which is why claims 4 and 5 are separated here and why claim 5 carries a hypothesis.
• No implication reverses, and two of the witnesses are on the companion page. The square root on $[0,\infty)$ is $1/2$-Hölder and not Lipschitz ($\sqrt{\cdot}$ on $[0,\infty)$ is uniformly continuous and exactly $1/2$-Hölder, and is not Lipschitz ↗), and $x \mapsto 1/x$ on $(0,1)$ is continuous and not uniformly continuous ($x \mapsto 1/x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1/(k+2))_{k \ge 0}$ to an unbounded one ↗). Both are read here as orientation only: this theorem does not depend on them and claims nothing about strictness. That a Lipschitz map need not be a contraction is visible already in the identity map, whose only Lipschitz constants are the reals $\ge 1$.
• The constant $L+1$ and the constant $C+1$ are there to avoid a case split. Dividing by $L$ or by $C$ would require them to be nonzero, and a Lipschitz constant or a Hölder constant is allowed to be $0$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction). Enlarging the constant by $1$ is harmless, since the defining inequalities are preserved by enlarging the constant.
• What claim 5 costs. The constant it produces, $L\,R^{1-\alpha}$, grows with the diameter of $X$, and that dependence is exactly what the unbounded counterexample above exploits: as $R \to \infty$ no single Hölder constant survives.