Let $f:(0,\infty)\to [0,\infty)$ be a concave function such that $f$ is not indentically zero. It seems to me that the following statements are true:
I - For every $a>0$ fixed, $f_{|[a,\infty)}$ is bounded below by a positive constant depending on $a$,
II - $f$ cannot attain a maximum value in $[a,\infty)$.
Am I seeing things or those statements are true?
Thank you
The first is true, the second false.
Simple counterexample for the second, constants, or $f(x) = \min \lbrace x, 1\rbrace$. If you meant strictly concave, the second is also true, because then $f$ must be strictly monotonically increasing.
The first is true because a concave function that is non-negative must be monotonic(ally nondecreasing), and cannot have a zero larger than $0$ unless it is identically $0$.
A concave function that is not monotonically non-decreasing cannot be bounded below:
Let $x_1 < x_2$ with $f(x_1) > f(x_2)$. Let $h = x_2 - x_1$ and $\delta = f(x_1) - f(x_2)$. Then
$$ f(x_1 + k\cdot h) \leqslant f(x_1) - k \cdot \delta\;, k \geqslant 1.$$
That is because by the definition of concaveness,
$$f(x_1) - \delta = f(x_2) = f\left(\frac{k-1}{k}x_1 + \frac{1}{k}(x_1 + k\cdot h)\right) \geqslant \frac{k-1}{k}f(x_1) + \frac{1}{k}f(x_1 + k\cdot h)$$
Thus (subtracting the first summand of the right hand side)
$$ \frac{1}{k}f(x_1) - \delta \geqslant \frac{1}{k}f(x_1 + k \cdot h) \iff f(x_1) - k\cdot \delta \geqslant f(x_1 + k\cdot h). $$
By assumption $\delta > 0$, so $f$ is not bounded below.
Geometrically, concaveness means the slope of the graph is monotonically non-increasing ($f'' \leqslant 0$, if by $f''$ we understand the distribution derivative, if the classical derivative doesn't exist), thus once the slope is negative, it remains so, and can only become more negative, never less.
