Concave function and little-oh

**batman** · Mar 18th 2011, 12:22 AM

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?

**Opalg** · Mar 18th 2011, 07:36 AM

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 $x\to\infty.$

**batman** · Mar 21st 2011, 02:09 AM

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.

**Opalg** · Mar 21st 2011, 02:48 AM

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.

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