Differentiable Bounded Convex Real Function is Constant

Theorem
Let $f$ be a real function which is
 * 1) Differentiable on $\R$;
 * 2) Bounded on $\R$;
 * 3) Either convex or concave on $\R$.

Then $f$ is constant.

Proof
Let $f$ be differentiable and bounded on $\R$.


 * Suppose $f$ is convex on $\R$.

Let $\xi \in \R$.

Suppose $f^{\prime} \left({\xi}\right) > 0$.

Then by Mean Value of Convex and Concave Functions it follows that $f \left({x}\right) \ge f \left({\xi}\right) + f^{\prime} \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to +\infty$, and therefore is not bounded.

Similarly, suppose $f^{\prime} \left({\xi}\right) < 0$.

Then by Mean Value of Convex and Concave Functions it follows that $f \left({x}\right) \ge f \left({\xi}\right) + f^{\prime} \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to -\infty$, and therefore is likewise not bounded.

Hence $f^{\prime} \left({\xi}\right) = 0$ and so, from Zero Derivative means Constant Function, $f$ is constant.


 * Suppose $f$ is concave on $\R$.

By a similar argument, the only way for $f$ to be bounded on $\R$ is for $f$ to be constant.