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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?

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2 Answers 2

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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 $1$.

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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.

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