Mean value theorem for convex functions

Question

Let $f$ be a real function with left and right derivatives $f'_-$ and $f'_+$ on the open interval $(a,b)$, and continuous on $[a,b]$ (e.g., let $f$ be convex on $[a,b]$). Then,

Is there something like the mean value theorem for $f$?

Answer 1

Not quite an answer:

For locally Lipschitz functions $f$ there is a notion of generalised gradient $\partial f$ that you can view as a generalisation of the subdifferential in convex analysis. The (generalised) mean value theorem then states that there is some $ t \in (a,b)$ such that ${f(b)-f(a) \over b-a } \in \partial f (a + t(b-a))$.

This is not as strong as the corresponding differentiable result.

As an example, take $f(x) = |x|$, $a=-1, b=1$, then $\partial f(x) = \begin{cases} \{-1\}, & x <0 \\ [-1,1], & x = 0 \\ \{1\}, & x > 0 \end{cases}$, so we see that ${|1|-|-1| \over 1 - (-1) } = 0 \in \partial f(0) = [-1,1]$.

Answer 2

By the mean value version of Taylor's theorem we have: \begin{align} f(y) &=f(x)+f'(x)(y-x)+\dfrac{1}{2}f''(z)(y-x)^2, \text{for some }z\in[x,y].\\ \end{align}