# Special Case of Lipschitz Concave Function

Suppose $$f: [0,1] \rightarrow [0, \infty)$$ is concave and $$f(x) \leq \min(x, 1-x)$$. I read in a paper that this implies that $$f$$ is Lipschitz continuous on $$[0,1]$$ and cannot prove it offhand. I'm sorry to ask such a dumb question, but why is this true?

I am a moron. The left derivative of $$f$$ at $$1$$ is equal to $$\lim_{\delta \rightarrow 0} \frac{f(1) - f(1-\delta)}{\delta} \ge \limsup_{\delta \rightarrow 0} -\min(1/\delta - 1, 1) = -1$$ and the right derivative of $$f$$ at $$0$$ is equal to $$\lim_{\delta \rightarrow 0} \frac{f(\delta) - f(0)}{\delta} \leq \liminf_{\delta \rightarrow 0} \min(1, 1/\delta -1) = 1$$ Since $$f$$ is concave, its left/right derivatives are decreasing and bounded between $$-1$$ and $$1$$, so we have Lipschitz continuity with a constant of $$1$$.
An alternative proof which does not use derivatives. For $$0 < x < y < 1$$ is $$\frac{f(x)-f(0)}{x-0} \ge \frac{f(y)-f(x)}{y-x} \ge \frac{f(1)-f(y)}{1-y} \, .$$ since the slopes of “adjacent” secants decrease with increasing arguments. The restriction $$0 \le f(x) \le \min(x, 1-x)$$ implies that $$\frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x} \le 1$$ and $$\frac{f(1)-f(y)}{1-y} = -\frac{f(y)}{1-y} \ge -1 \, ,$$ so that $$-1 \le \frac{f(y)-f(x)}{y-x} \le 1$$ and that holds for $$x=0$$ and $$y=1$$ as well.