# Bounded one-sided derivative implies Lipschitz continuity?

Let $$f(x)$$ be a one dimensional increasing and continuous function with support $$x \in [0,1]$$. For any $$x \in (0,1)$$, the left-derivative $$f'(x-)$$ and right derivative $$f'(x+)$$ exists and moreover, they are uniformly bounded, i.e., there exists some constants $$c_1 \geq c_2 > 0$$ such that $$c_2 \leq f'(x-), f'(x+) \leq c_1, \forall x \in (0,1).$$ Can we prove that $$f(x)$$ is Lipschitz continuous on $$[0,1]$$, probably with a parameter $$c_1$$? It is very intuitive to me, but I know some wierd examples can happen in analysis...

• Does "$c_1<X,Y<c_2$" mean $$c_1 < X < c_2 \text{ and }c_1 < Y < c_2$$ or $$c_1< X \\ Y < c_2 ?$$ – Calvin Khor Nov 7 '19 at 4:25
• @CalvinKhor the first one. – Stupid_Guy Nov 7 '19 at 5:39

Yes! Choose $$0\le a < b\le 1$$ and define $$g(x) = f(x)-rx$$, where $$r = \frac{f(b)-f(a)}{b-a}$$. Since $$g(b) = g(a)$$, by the generalized version of Rolle's theorem there exists $$c\in (a,b)$$ such that either ($$g'(c+)\ge 0$$ and $$g'(c-)\le 0$$) or ($$g'(c-)\ge 0$$ and $$g'(c+)\le 0$$). Hence, $$\frac{f(b)-f(a)}{b-a} = r\le f'(c+)\le c_1\qquad\text{or}\qquad \frac{f(b)-f(a)}{b-a} = r\le f'(c-)\le c_1.$$ So, $$c_1$$ is a Lipschitz constant for $$f$$ on $$[0,1]$$.
You don't need $$c_2$$, since $$f'(x\pm)\ge 0$$ for all $$x\in (0,1)$$ anyways.