Does continous and convex on a closed interval imply Lipschitz?

Question

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?

Answer 1

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.

Answer 2

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.