Example of positive, strictly concave, strictly decreasing function?

Question

I am looking for an example of a positive, strictly concave, strictly decreasing function. The standard ones, exponentials and logs do not seem to work.

Answer 1

The first part of this argument is adapted from https://math.stackexchange.com/a/1959851/213690.

Suppose (i) for some real $a,$ the function $f \colon (a,\infty) \to \mathbb{R}$ is concave. By the definition of concavity, if $a<y<x<z,$ then $$ \frac{f(x)-f(y)}{x-y} \geqslant \frac{f(z)-f(x)}{z-x}, $$ the left hand side of this inequality is a non-increasing function of $y,$ and the right hand side is a non-increasing function of $z.$ Therefore, the left hand derivative $f'_-(x)$ and right hand derivative $f'_+(x)$ exist, and for all $y\in(a,x)$ and $z>x,$ $$ \frac{f(x)-f(y)}{x-y} \geqslant f'_-(x) \geqslant f'_+(x) \geqslant \frac{f(z)-f(x)}{z-x}. $$

Let $x$ be any number ${>}a,$ and let $m_x$ be any number such that $$ f'_-(x) \geqslant m_x \geqslant f'_+(x). $$ Then for all $y\in(a,x)$ and $z>x,$ $$ f(x)-f(y) \geqslant m_x(x-y) \text{ and } f(z)-f(x) \leqslant m_x(z-x). $$ Equivalently, $$ f(y) \leqslant f(x) + m_x(y-x) \text{ for all } y>a. $$

Now suppose also (ii) $f$ is non-increasing on $(a,\infty).$ Then $f(x)-f(y)\leqslant0$ for all $y\in(a,x),$ whence $m_x\leqslant0.$

Finally, suppose also (iii) $f$ is non-negative on $(a,\infty).$ Then for all $z>x,$ $$ 0 \geqslant \frac{f(z)-f(x)}{z-x} \geqslant \frac{-f(x)}{z-x} \to 0 \text{ as } z \to \infty. $$ Therefore, for all $\varepsilon>0,$ $$ 0 \geqslant m_x \geqslant -\varepsilon. $$ Therefore $m_x = 0.$

Therefore $f(y) \leqslant f(x)$ for all $y>a.$ But $x$ was arbitrary. Therefore $f$ is a constant function. Therefore $f$ is not strictly decreasing.

Summarising: a non-negative concave function on $(a,\infty)$ is not strictly decreasing. If it is only hypothesised to be non-increasing, then it is constant.