# Thread: Concave function and little-oh

1. ## Concave function and little-oh

The problem is:

Suppose f(x) is a strictly concave function. Prove that x=o(f^(-1)(x)).

Do you have any suggestions on how to prove this?

2. Originally Posted by batman
The problem is:

Suppose f(x) is a strictly concave function. Prove that x=o(f^(-1)(x)).

Do you have any suggestions on how to prove this?
Do you mean $x=o(f^{-1}(x))$ as $x\to\infty$, or as $x\to0$, or something else?

In any case, I don't think that this result can be true without some additional conditions. For example, the function $f(x) = \sqrt{x^2-1}$ is strictly concave on the interval $[1,\infty)$, but $f^{-1}(x)\approx x$ as $x\to\infty.$

3. Sorry, I didn't copy the exercise properly from the blackboard. It should have been:

Suppose $f(x)$ is strongly concave. Then, $x=o(f^{-1}(x))$ as $x\rightarrow \infty$.

The function you mention is strictly concave, but nog strongly concave. Thank you anyway.

4. Originally Posted by batman
Sorry, I didn't copy the exercise properly from the blackboard. It should have been:

Suppose $f(x)$ is strongly concave. Then, $x=o(f^{-1}(x))$ as $x\rightarrow \infty$.
That certainly makes a difference, but I still don't understand what's going on here. As far as I can see, a strongly concave function has to be bounded above, so $f^{-1}(x)$ will not exist when x is large.