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

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

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:

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

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

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