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

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?

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 $$\lim_{t\to a^+}f_L'(t) = f_R'(a) \, .$$
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 $$\frac{f(a+h)-f(a)}{h} But then (a variant of) the mean value theorem says that $$\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.