Mean Value of Convex Real Function

Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.

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

Then:

$\forall \xi \in \openint a b: \map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$

Proof

By the Mean Value Theorem:

$\exists \eta \in \openint x \xi: \map {f'} \eta = \dfrac {\map f x - \map f \xi} {x - \xi}$

From Real Function is Convex iff Derivative is Increasing, the derivative of $f$ is increasing.

Thus:

$x > \xi \implies \map {f'} \eta \ge \map {f'} \xi$

$x < \xi \implies \map {f'} \eta \le \map {f'} \xi$

Hence:

$\map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$

$\blacksquare$

Also see

• Mean Value of Concave Real Function

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.21 \ (4)$