# Does continous and convex on a closed interval imply Lipschitz?

Let $$f:[a,b] \to \mathbb{R}$$ be a continous, convex function. (By convex, I mean $$f(\lambda x+(1- \lambda)y) \leq \lambda f(x)+(1- \lambda)f(y)$$ for any choice of $$x,y \in [a,b]$$ and $$\lambda \in [0,1]).$$

Q: Is $$f$$ Lipschitz? If not, what would be a counterexample?

I know a theorem that says a convex function on $$(a,b)$$ has to be Lipschitz on each $$[c,d]$$. However, with no extra assumptions, the Lipschitz constant might change.

Will continuity ensure that there's one Lipschitz constant that works for everything?

• square root of $x$ on $[0,1]$ is not a counter-example? Feb 26, 2020 at 19:35
• $\sqrt x$ is concave, but you might be on to something :) Feb 26, 2020 at 19:36
• $x \mapsto -\sqrt{x}$ is convex on $[0,1]$. Feb 26, 2020 at 19:47
• $x^x$ is also interesting.
– zwim
Feb 26, 2020 at 20:12

As a convex function, $$f$$ has a left and right derivative in every point of $$(a, b)$$, and these are monotonically increasing. But these (one-sided) derivatives can approach $$-\infty$$ for $$x \to a$$ or $$+\infty$$ for $$x \to b$$, and then $$f$$ is not Lipschitz continuous on $$[a, b]$$.
An example is $$f(x) = 1 - \sqrt{x}$$ on $$[0, 1]$$. It is convex, but the derivative approaches $$-\infty$$ for $$x \to 0$$, so that it is not Lipschitz continuous.
On the unit circle, look at the arc from $$(0,-1)$$ to $$(1,0).$$ That is the graph of a convex function that has derivative $$+\infty$$ at $$(1,0).$$ Right there you have a counterexample.