Convex Real Function is Left-Hand and Right-Hand Differentiable

Theorem
Let $f$ be a real function which is either convex or concave on the open interval $\left({a .. b}\right)$.

Then the limits:
 * $\displaystyle \lim_{h \to 0^-} \frac {f \left({x + h}\right) - f \left({x}\right)} h$

and:
 * $\displaystyle \lim_{h \to 0^+} \frac {f \left({x + h}\right) - f \left({x}\right)} h$

both exist for all $x \in \left({a .. b}\right)$.

Proof

 * Let $f$ be convex on $\left({a .. b}\right)$.

Take this definition of convexity:


 * $\displaystyle \forall x_1, x_2, x_3 \in \left({a .. b}\right): x_1 < x_2 < x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

First we show that there exists a right hand limit.

Let $0 < h_1 < h_2$.

Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then:


 * $\displaystyle \frac {f \left({x + h_1}\right) - f \left({x}\right)} {h_1} \le \frac {f \left({x + h_2}\right) - f \left({x}\right)} {h_2}$

Hence the function $\displaystyle F \left({h}\right) = \frac {f \left({x + h}\right) - f \left({x}\right)} h$ increases in some $\left({0 . . \delta}\right)$.

Thus from Limit of Monotone Function it follows that $\displaystyle \lim_{h \to 0^+} F \left({h}\right)$ exists.

A similar argument shows the existence of the left hand limit.


 * Let $f$ be concave on $\left({a .. b}\right)$.

The result follows from a similar argument.