Is the inverse of a nonnegative convex function being convex/quasi-convex/concave/quasi-concave?

Question

Given a nonnegative and convex function $f: \mathcal{D} \rightarrow \mathbb{R}_{+}$, where the domain $\mathcal{D} \subset \mathbb{R}$. Define the inverse of the function $g(x) = \frac{1}{f(x)}$. Is there a condition that $g(x)$ being either convex, quasi-convex, concave, or quasi-concave?

I tried the following functions:

• $f(x)=1, g(x)=1$ is both convex, quasi-convex, concave, and quasi-concave
• $f(x)=x$ for $x\geq0$, $g(x) = \frac{1}{x}$ is convex, quasi-convex, and quasi-concave
• $f(x)=x^2$, $g(x) = \frac{1}{x^2}$ is quasi-concave for $x\in\mathbb{R}$; $g(x)$ is also convex and quasi-convex for $x>0$.
• $f(x)=x^2+1$, $g(x) = \frac{1}{x^2+1}$ is quasi-concave for $x\in\mathbb{R}$; $g(x)$ is not convex but quasi-convex for $x>0$.
• $f(x)=e^x$, $g(x) = e^{-x}$ is convex, quasi-convex, and quasi-concave.
• $f(x)= x^x$, $g(x) = x^{-x}$ is quasi-concave for $x\in\mathbb{R}$.

I have the following conjectures for the condition of quasi-convexity and quasi-concavity:

1. $g(x)$ is always quasi-concave if $f(x)$ is convex
2. $g(x)$ is quasi-convex if $f(x)$ is monotonically non-decreasing

I wonder if we can prove the above from definition. Also, I could not find a condition so that $g(x)$ being convex or concave.

Answer 1

I think you need to assume that the domain $\cal D$ is convex. It is not very difficult to attempt and show your first conjecture. Please attempt the second one yourself.

When $f(x)$ is convex then $-f(x)$ is concave. Can we show that then $-g(x)=-1/f(x)$ is quasi-convex ?

The answer is yes and this will imply that $g(x)$ is quasi-convave.

Proof.

The sublevel set $S_\alpha=\{x|-g(x)\le \alpha\}$ can be written as $$ S_\alpha=\{x\,|\,g(x)\ge -\alpha\}=\{x\,|\,-g(x)\le \alpha\}=\Big\{x\,\Big|\,\frac{-1}{f(x)}\le \alpha\Big\} $$ For $\alpha\ge 0$ this is obviously the whole (convex) domain $\cal D$. For $\alpha<0$ we get $$ S_\alpha=\Big\{x\,\Big|\,f(x)\le\frac{-1}{\alpha}\Big\} $$ This set is convex because every convex function is quasi-convex. So is $f$. QED

Answer 2

Following Kurt G.'s answer. Here is the proof of the second conjecture.

Conjecture

$g(x)=\frac{1}{f(x)}$ is quasi-convex if $f:\mathcal{D}\rightarrow\mathbb{R}_+$ is monotonically non-decreasing.

Approach

To show $g(x)$ is a quasi-convex function, we show that the set $S_\alpha = \{x | g(x) \leq \alpha \}$ is a convex set for all $\alpha$.

Proof

The set $S_\alpha = \{x | g(x) \leq \alpha \} = \{x | \frac{1}{f(x)} \leq \alpha \}$.

For $\alpha > 0$, the set can be written as $S_\alpha = \{x | f(x) \geq \frac{1}{\alpha} \}$. With $f(x)$ being non-decreasing, $S_\alpha = [x_*, \infty)$ with $f(x_*)=\frac{1}{\alpha}$. Thus $S_\alpha$ is a convex set.

set is a convex set.

As for $\alpha < 0$, the set is written as $S_\alpha = \{x | f(x) \leq \frac{1}{\alpha}\}$. Note that this set is empty as the range of $f(x)$ is nonnegative.

Hence the function $g(x)$ is quasi-convex if $f(x)$ is monotonically non-decreasing.