Differentiable Bounded Convex or Concave Function is Constant

Theorem

Let $f$ be a real function which is:

$(1): \quad$ Differentiable on $\R$;

$(2): \quad$ Bounded on $\R$;

$(3): \quad$ 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.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 12.21 \ (5)$