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Let $f \colon \mathbb{R}_+ \to \mathbb{R}_+$ be a concave function.

Some academic paper mentioned that "locally Lipschitz continuity of $f$ is always satisfied when $f$ is concave."

I am wondering about if this statement is true.

The reason is that considering a concave function mapping from $\mathbb{R}_+$ to itself which is defined by $f(x) := \sqrt{x}$, this function is not Lipschitz continuous. $f$ becomes infinitely steep as $x$ approaches $0$ since its derivative becomes infinite. It also seems unlikely that $f$ is locally Lipschitz continuous.

Could anyone help to explain that whether the concavity of a function (or an operator) can imply the locally Lipschitz continuity please?

Thank you!

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    $\begingroup$ See here for a proof of the Lipschitz continuity of a convex function on an open interval. Same applies to concave functions (negative of a convex function). Note that a function can be convex on a closed interval but not continuous at an endpoint. $\endgroup$ – RRL May 1 '18 at 4:25
  • $\begingroup$ Thank you so much @RRL. I have seen your proof in that link, and it is a huge help! From the proof, I think the above statement only works when the domain of $f$ is $(0, \infty)$ and not works when the domain is $[0, \infty)$, may I ask that is my understanding correct please? Many thanks again:) $\endgroup$ – Paradiesvogel May 1 '18 at 4:55
  • $\begingroup$ I think you are correct. If convex or concave on an unbounded open interval then it is Lipschitz on any closed and bounded subinterval -- locally in the sense that the Lipschitz constant varies as the interval becomes larger. $\endgroup$ – RRL May 1 '18 at 5:13
  • $\begingroup$ Thank you @RRL . But as we know, the function $f(x) := \sqrt{x}$ defined on $[0,1]$ is not Lipschitz continuous. Is $f$ locally Lipschitz continuous on $[0,1]$? $\endgroup$ – Paradiesvogel May 1 '18 at 5:17
  • $\begingroup$ For any given closed interval, a function could be convex but not Lipschitz or not even continuous (at the endpoint). If we take a convex, continuous and decreasing function on $[0,1]$ and assign a higher value at $0$ it would still be convex on $[0,1]$ but now with a jump discontinuity. No matter how we define the function for $x < 0$ it will not be convex on a bigger interval with $0$ in the interior. But if the closed interval is a subset of an open interval where the function is convex, then we get Lipschitz continuity on that closed interval. $\endgroup$ – RRL May 1 '18 at 5:26
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The domain matters. The word "locally" is normally used when talking about a function defined on some open set. As a function on $(0, \infty)$, $\sqrt{x}$ is indeed locally Lipschitz.

This is true for functions on open subsets of $\mathbb{R}^n$ as well; see Every convex function is locally Lipschitz ($\mathbb{R^n}$)

False in infinite-dimensional spaces. Indeed, every linear functional $f:X\to\mathbb{R}$ is a convex function, but $f$ need not be continuous when $X$ is infinite-dimensional.

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  • $\begingroup$ Well, it's an economics paper; I'm not surprised it fails to handle mathematical edge-cases correctly. $\endgroup$ – user357151 May 2 '18 at 0:07

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