# Concave function and little-oh

• Mar 18th 2011, 12:22 AM
batman
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?
• Mar 18th 2011, 07:36 AM
Opalg
Quote:

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 $\displaystyle x=o(f^{-1}(x))$ as $\displaystyle x\to\infty$, or as $\displaystyle x\to0$, or something else?

In any case, I don't think that this result can be true without some additional conditions. For example, the function $\displaystyle f(x) = \sqrt{x^2-1}$ is strictly concave on the interval $\displaystyle [1,\infty)$, but $\displaystyle f^{-1}(x)\approx x$ as $\displaystyle x\to\infty.$
• Mar 21st 2011, 02:09 AM
batman
Sorry, I didn't copy the exercise properly from the blackboard. It should have been:

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

The function you mention is strictly concave, but nog strongly concave. Thank you anyway.
• Mar 21st 2011, 02:48 AM
Opalg
Quote:

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

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

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 $\displaystyle f^{-1}(x)$ will not exist when x is large.