Does continous and convex on a closed interval imply Lipschitz?

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.