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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...

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

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