Is the right-limit of left-derivatives equal to the right-derivative?

Question

Let's say $f$ is a convex function, so that $f_L'$ and $f_R'$ exist everywhere. I was wondering if the following is true: $$\lim_{t\to a^+}f_L'(t)=f_R'(a).$$ We know that $f'_L(t)$ decreases as $t\to a^+$, so $$\lim_{t\to a^+}f'_L(t)=\inf_{t>a}f_L'(t).$$ On the other hand, convexity implies $$f_R'(a)\leq f_L'(t)\qquad\text{for all}\qquad t>a.$$ As such $$f_R'(a)\leq\lim_{t\to a^+}f_L'(t).$$ Should this also be an equality?

Answer 1

For $a < t$ is $$ f_R'(a)\leq f_L'(t) \le f_R'(t) $$ and $f_R'$ is right continuous, see for example How to prove that the right derivative of a convex function is right continuous?.

So $$ f_R'(a)\leq\lim_{t\to a^+}f_L'(t) \le \lim_{t\to a^+}f_R'(t) = f_R'(a) $$ and therefore $$ \lim_{t\to a^+}f_L'(t) = f_R'(a) \, . $$

Answer 2

Turns out the answer is YES (thanks Martin R for posting the first answer).

Here's another way to see this:

We know that $$f_R'(a)\leq f_L'(t)\quad\text{for all}\quad t>a.$$ To see why it is also the greatest lower bound, take $f_R'(a)+\epsilon$. For sufficiently small $h>0$, we have $$\frac{f(a+h)-f(a)}{h}<f_R'(a)+\epsilon.$$ But then (a variant of) the mean value theorem says that $$\frac{f(a+h)-f(a)}{h}\geq f_L'(t)\quad\text{for some}\quad t\in(a,a+h),$$ which proves the claim.