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

Theorem

Let $f$ be a real function which is convex on the open interval $\openint a b$.

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

Proof

Let $f$ be convex on $\openint a b$.

Then by definition of convexity:

$\forall x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$

Let $0 < h_1 < h_2$.

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

$\dfrac {\map f {x + h_1} - \map f x} {h_1} \le \dfrac {\map f {x + h_2} - \map f x} {h_2}$

Hence the function $\map F h = \dfrac {\map f {x + h} - \map f x} h$ increases in some $\openint 0 \delta$.

Thus from Limit of Increasing Function it follows that both $\ds \lim_{h \mathop \to 0^+} \map F h = \map {f'_+} x$ and $\ds \lim_{h \mathop \to 0^-} \map F h = \map {f'_-} x$ exist.

$\blacksquare$

Also see

• Concave Real Function is Left-Hand and Right-Hand Differentiable

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.14$