If $f$ is continuous on an interval $I$ and $|f'| \le M$ at every interior point, then $|f(x) - f(y)| \le M|x-y|$ for all $x,y \in I$, so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$

Let $I \subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), let $f : I \to \mathbb{R}$ be continuous on $I$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) and differentiable at every point of $I$ interior to $I$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$, The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$, and differentiability on a set), and let $M \in \mathbb{R}$ with $M \ge 0$ satisfy

$|f'(x)| \;\le\; M \qquad \text{at every interior point } x \text{ of } I .$

Then

$|f(x) - f(y)| \;\le\; M\,|x - y| \qquad \text{for all } x, y \in I ,$

which is exactly the statement that $f$ is Lipschitz with constant $M$ on $I$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction, clause 3 of Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace). Consequently $f$ is uniformly continuous on $I$ (Uniform continuity of $f : A \to \mathbb{R}$: one $\delta$ serving every pair of points of $A$).

$M \ge 0$ is a hypothesis, not a deduction. It follows from $|f'(x)| \le M$ at any single interior point, absolute values being nonnegative, but $I$ need have no interior point at all, and then the sign condition has to be asked for. With $M \ge 0$ assumed the conclusion is a genuine statement in every case, and at $x = y$ it reads $0 \le 0$.

Boundedness of $f'$ cannot be dropped. A function may be continuous on an interval and differentiable at every interior point with no bound on $|f'|$, and then it need not be Lipschitz there; the companion page's square root on $(0,1]$ is such a function.

• The constant is the bound on the derivative, unchanged. No factor is lost and none is gained: the mean value theorem turns the increment into a single value of $f'$ times the increment of the argument, so whatever bounds $|f'|$ bounds the Lipschitz ratio. That is why this corollary is so much sharper than the mere uniform continuity that follows from it.

• Uniform continuity is obtained through the metric dictionary, not reproved. Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace says that the Lipschitz condition for a real function is the metric one instantiated, and 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 says that a Lipschitz map is uniformly continuous. Neither statement is restated here in an $\mathbb{R}$-native form, because both already exist in the library and duplicating them would create exactly the seam those items were written to close.

• What the converse would say, and why it is not asserted. A Lipschitz function need not be differentiable at any particular point, so no statement about $f'$ follows from a Lipschitz bound alone, and this corollary asserts none.

• The boundedness hypothesis is exactly what is needed. $x \mapsto \sqrt{x}$ on $(0,1]$ is differentiable with unbounded derivative and is not Lipschitz there, so the boundedness hypothesis in the Lipschitz corollary cannot be dropped ↗ on the companion page exhibits a function continuous on $(0,1]$ and differentiable at every interior point, with $|f'|$ bounded by no real at all, which is not Lipschitz there. So the hypothesis cannot be replaced by mere differentiability, and it cannot be replaced by a bound holding only near one end of the interval.