If $f$ is continuous on an interval $I$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\le M$ at every interior point, then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ 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 $\epsilon$-$\delta$ 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 $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ 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

$\mathrm{\mid }{f}^{\mathrm{\prime }}(x)\mathrm{\mid }\le M\text{at every interior point }x\text{ of }I\mathrm{.}$

Then

$\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }\text{for all }x,y\in I,$

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 $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, 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 $\mathrm{\mid }{f}^{\mathrm{\prime }}(x)\mathrm{\mid }\le M$ 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 $f^{\mathrm{\prime }}$ cannot be dropped. A function may be continuous on an interval and differentiable at every interior point with no bound on $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$, 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 $f^{\mathrm{\prime }}$ times the increment of the argument, so whatever bounds $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ 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 $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, 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 $f^{\mathrm{\prime }}$ 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 $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ 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.