Differentiable Bounded Convex Real Function is Constant

Theorem
Let $$f$$ be a real function which is
 * Differentiable on $$\mathbb{R}$$;
 * Bounded on $$\mathbb{R}$$;
 * Either convex or concave on $$\mathbb{R}$$.

Then $$f$$ is constant.

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


 * Suppose $$f$$ is convex on $$\mathbb{R}$$.

Let $$\xi \in \mathbb{R}$$.

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 $$\mathbb{R}$$.

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