$\sqrt{\cdot }$ on $[0,\mathrm{\infty })$ is uniformly continuous and exactly $1\mathrm{/}2$-Hölder, and is not Lipschitz

Let $X:=[0,\mathrm{\infty })\subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let $g:X\to X$ be $g(x):=\sqrt{x}=x^{1\mathrm{/}2}$ (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$, Rational powers $a^r$ of a positive base). Then:

1. $\mid \sqrt{x}-\sqrt{y}\mid \le \mathrm{\mid }x-y{\mathrm{\mid }}^{1\mathrm{/}2}$ for all $x,y\ge 0$, so $g$ is $1\mathrm{/}2$-Hölder with constant $1$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction).
2. $g$ is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
3. The constant $1$ cannot be improved: every $1\mathrm{/}2$-Hölder constant $C$ for $g$ satisfies $C\ge 1$.
4. For every rational $\alpha$ with $1\mathrm{/}2<\alpha \le 1$, the map $g$ is not $\alpha$-Hölder. In particular, at $\alpha =1$, $g$ is not Lipschitz.

So $1\mathrm{/}2$ is exactly the Hölder exponent of the square root, and the example separates "Hölder" from "Lipschitz" inside 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.

• Where the failure is located. The obstruction in claims 3 and 4 sits at $0$: the inequality $t^{1\mathrm{/}2}\le C{t}^{\alpha }$ is comfortable for large $t$ and impossible for small $t$ once $\alpha >1\mathrm{/}2$, because $t^{1\mathrm{/}2}$ then dominates $t^{\alpha }$. Away from $0$ the square root is perfectly Lipschitz: on $[a,\mathrm{\infty })$ with $a>0$ one has $\mid \sqrt{x}-\sqrt{y}\mid =\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}(\sqrt{x}+\sqrt{y})\le \mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}(2\sqrt{a})$, since $\sqrt{x}+\sqrt{y}\ge 2\sqrt{a}>0$ and $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y$.
• The exponent is rational, and that is not a restriction here. At this page's position in the reading order, Rational powers $a^r$ of a positive base is the available exponent construction. Both $1\mathrm{/}2$ and every exponent used above are rational, so claim 4 is intentionally local to rational $\alpha$; real powers are introduced later in Real powers for positive bases, with the zero-base positive-exponent convention ↗.
• What this example is for. It is one of the two witnesses named in 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: it shows that "Hölder" is strictly weaker than "Lipschitz", so the implication proved there from Hölder to uniform continuity is not a detour through the Lipschitz condition. The other witness, separating continuity from uniform continuity, is $x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1\mathrm{/}(k+2){)}_{k\ge 0}$ to an unbounded one.
• Claim 1 is sharp in a second sense as well: equality holds whenever one of the two arguments is $0$, since $\mid \sqrt{x}-\sqrt{0}\mid =\sqrt{x}=\mathrm{\mid }x-0{\mathrm{\mid }}^{1\mathrm{/}2}$. So the estimate is attained and not merely approached.