Let $f : [0, \infty) \longrightarrow [0, \infty)$ be concave with $f (0) = 0$ and $f (x) > 0$ for $x > 0$. Show that $f$ is non-decreasing.
It is clear that a concave function $f$ can be decreasing. Intuitively, $f$ must then become negative beyond some point, though. Once it has turned "right" (i.e., downwards) concavity prevents it from avoiding the abscissa.
I'm looking for a reasonably elegant proof. I've tried to formalize my intuition as follows, but I'm afraid my argument isn't very rigorous (if it is right in the first place). Maybe there are also different and better approaches than mine.
Here's my attempt: Concacity implies that $f (\lambda x + (1 – \lambda) z) \ge \lambda f (x) + f ((1 – \lambda) z)$ for $0 \le x < z$ and $\lambda \in (0, 1)$ or, alternatively,
$$\frac{f (z) – f (y)}{z – y} \le \frac{f (z) – f (x)}{z – x} \le \frac{f (y) – f (x)}{y – x}$$
for $0 \le x < y < z$. (Let $1 – \lambda := \frac{y – x}{z – x}$, so that $\lambda x + (1 – \lambda) z = y$ to restate concavity without $\lambda$.) It follows that
$$f (z) \le \frac{f (y) – f (x)}{y – x} (z – y) + f (y).$$
Graphically speaking, $f (z)$ must be on or below the secant through $f (x)$ and $f (y)$ for all $z > y$. Suppose that $f (y) < f (x)$ for some $x < y$. The secant is then strictly decreasing—and so is $f$ for $z > y$.
As a result, $f$ must turn negative where the secant intersects the abscissa or before. This contradicts the assumption. Hence, $f (y) \ge f (x)$ for all $0 \le x < y$.
Note: I feel that the argument requires $f$ to be continuous, although I don't see how to apply the intermediate value theorem rigorously. Continuity follows from concavity, though.
Best Answer
The solution suggested above seems to be valid. The conclusion can be stated in a more compelling way, though.
For any $z > y > x > 0$, concavity implies that $$f (z) \le \frac{f (y) - f (x)}{y - x} (z - y) + f (y).$$ Assuming that $f (y) < f (x)$ for some $0 < x < y$, $\frac{f (y) - f (x)}{y - x}$ is negative. As a result, $$\limsup_{z \rightarrow \infty} f (z) \le \limsup_{z \rightarrow \infty} \frac{f (y) - f (x)}{y - x} (z - y) + f (y) = -\infty.$$ This contradicts the assumption that $f (y) < f (x)$, and it follows that $f$ is non-decreasing.