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

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Theorem

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


Then the left-hand derivative $f'_- \left({x}\right)$ and right-hand derivative $f'_+ \left({x}\right)$ both exist for all $x \in \left({a \,.\,.\, b}\right)$.


Proof

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

Then by definition of convexity:

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


Let $0 < h_1 < h_2$.

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

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

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

Thus from Limit of Increasing Function it follows that both $\displaystyle \lim_{h \to 0^+} F \left({h}\right) = f'_+ \left({x}\right)$ and $\displaystyle \lim_{h \to 0^-} F \left({h}\right) = f'_-\left({x}\right)$ exist.

$\blacksquare$


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