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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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The mean value theorem gives xy1ι(2)xy|\sqrt{x} - \sqrt{y}| \le \tfrac{1}{\iota(2)} |x - y| for x,y1x, y \ge 1, so the square root is Lipschitz with constant 1/21/2 on [1,)[1,\infty)

Example

Write b=b1/2\sqrt{b} = b^{1/2} for the nonnegative square root (Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a, Rational powers ara^r of a positive base) and ι\iota for the canonical natural (The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field).

Claim. Let I:=[1,)I := [1,\infty) and let s:IRs : I \to \mathbb{R}, s(b):=bs(b) := \sqrt{b}. Then

xy    1ι(2)xyfor all x,yI,|\sqrt{x} - \sqrt{y}| \;\le\; \frac{1}{\iota(2)}\,|x - y| \qquad \text{for all } x, y \in I ,

so ss is Lipschitz with constant 1/ι(2)1/\iota(2) on II (Lipschitz map, α\alpha-Hölder map for rational 0<α10 < \alpha \le 1, and contraction, clause 3 of Dictionary: for ARA \subseteq \mathbb{R} with the metric d(x,y)=xyd(x,y) = |x-y|, continuity and uniform continuity of f:ARf : A \to \mathbb{R} agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of R\mathbb{R} is compact in the open-cover sense of R\mathbb{R} exactly when it is a compact metric subspace) and hence uniformly continuous on II (Uniform continuity of f:ARf : A \to \mathbb{R}: one δ\delta serving every pair of points of AA).

The constant is what the derivative bound gives, and the domain is what makes the bound available. On [1,)[1,\infty) the derivative of ss is at most 1/ι(2)1/\iota(2); on (0,1](0,1] it is not bounded at all, and the companion counterexample on this page shows that there the Lipschitz conclusion fails.

Facts & Assumptions

Given: The set I:=[1,)I := [1,\infty), order-convex with at least two elements (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length), and the function s:IRs : I \to \mathbb{R}, s(b):=b1/2s(b) := b^{1/2}.

[L1]

Derivative of the square root (For a natural n1n \ge 1, the derivative of xx1/nx \mapsto x^{1/n} on (0,)(0,\infty) is 1ι(n)x1/n1\frac{1}{\iota(n)}x^{1/n - 1}, obtained from the inverse rule applied to xxnx \mapsto x^{n}; in particular (x)=1/(ι(2)x)(\sqrt{x})' = 1/(\iota(2)\sqrt{x}), at n=2n = 2): the map uu1/2u \mapsto u^{1/2} on (0,)(0,\infty) is differentiable at every b>0b > 0 with derivative 1ι(2)b1/2\frac{1}{\iota(2)}b^{-1/2}.

[L3]

A function differentiable at a point is continuous there (A function differentiable at cc is continuous at cc).

[L4]

Rational powers (Laws of rational exponents, Monotonicity of rarr \mapsto a^{r} and of aara \mapsto a^{r}, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a): ar>0a^{r} > 0 for a>0a > 0; ar=1/ara^{-r} = 1/a^{r}; 11/2=11^{1/2} = 1, since 101 \ge 0 and 12=11^{2} = 1 and the nonnegative square root is unique; and for rational t>0t > 0, a>1a > 1 implies at>1a^{t} > 1 (claim 3 of the monotonicity lemma).

[L5]

Order arithmetic (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication, Canonical naturals are positive and strictly increasing, Multiplying inequalities of positives): ι(2)>0\iota(2) > 0, so 1/ι(2)>01/\iota(2) > 0 and ι(2)0\iota(2) \ne 0; 0<a<b0 < a < b gives 0<1/b<1/a0 < 1/b < 1/a (Inverses of positives are positive, and reciprocation reverses order); a product of two positive reals is positive (Sign rules for products and monotonicity of multiplication); and the NONSTRICT multiplication of inequalities between nonnegatives, 0xy0 \le x \le y and 0uv0 \le u \le v imply xuyvxu \le yv, is Multiplying inequalities of positives and is not stated by Sign rules for products and monotonicity of multiplication, whose multiplicative claims are strict. Also u=u|u| = u for u0u \ge 0 (Basic properties of the absolute value).

[L6]

Interiority (Interior, closure, boundary and exterior of a subset of R\mathbb{R}, The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}): pp is interior to SS exactly when Nε(p)SN_{\varepsilon}(p) \subseteq S for some real ε>0\varepsilon > 0.

Verification

technique · direct
1.1

The interior points of I=[1,)I = [1,\infty) are exactly the reals b>1b > 1. For b>1b > 1 the neighbourhood Nb1(b)N_{b-1}(b) is contained in (1,)I(1,\infty) \subseteq I, so bb is interior; the point 11 is not interior, since 1ε/2Nε(1)1 - \varepsilon/2 \in N_{\varepsilon}(1) and 1ε/2I1 - \varepsilon/2 \notin I for every real ε>0\varepsilon > 0; and every interior point of II lies in II, hence is 1\ge 1.

L6
1.2

For every real b1b \ge 1 one has b1/21b^{-1/2} \le 1. If b=1b = 1 then 11/2=11^{1/2} = 1 by [L4], so 11/2=1/1=11^{-1/2} = 1/1 = 1. If b>1b > 1 then b1/2>1b^{1/2} > 1 by [L4], and b1/2>0b^{1/2} > 0, so b1/2=1/b1/2<1b^{-1/2} = 1/b^{1/2} < 1 by [L4] and [L5].

L4L5
2.1

By [L1] and [L2] the function ss is differentiable at every bIb \in I with s(b)=1ι(2)b1/2s'(b) = \frac{1}{\iota(2)}b^{-1/2}, and by [L3] it is continuous at every point of II, hence continuous on II.

step 1.1L1L2L3
3.1

At every interior point bb of II one has b>1b > 1 by step 1.1, so b1/2>0b^{-1/2} > 0 by [L4] and b1/21b^{-1/2} \le 1 by step 1.2; multiplying the pair 0b1/210 \le b^{-1/2} \le 1 and 01/ι(2)1/ι(2)0 \le 1/\iota(2) \le 1/\iota(2) as in [L5] gives 0<s(b)1/ι(2)0 < s'(b) \le 1/\iota(2) by step 2.1, and therefore s(b)=s(b)1/ι(2)|s'(b)| = s'(b) \le 1/\iota(2) by [L5].

step 1.1step 1.2step 2.1L4L5
4.1

Apply [L7] with J:=IJ := I, h:=sh := s and M:=1/ι(2)M := 1/\iota(2), a real 0\ge 0 by [L5]: the hypotheses hold by step 2.1 for the continuity and differentiability and by step 3.1 for the bound, so s(x)s(y)1ι(2)xy|s(x)-s(y)| \le \frac{1}{\iota(2)}|x-y| for all x,yIx, y \in I, that is xy1ι(2)xy|\sqrt{x} - \sqrt{y}| \le \frac{1}{\iota(2)}|x-y|; ss is Lipschitz with constant 1/ι(2)1/\iota(2) on II; and ss is uniformly continuous on II.

step 2.1step 3.1L5L7

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