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!
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.