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I'm looking for an example of a concave function $g\colon [0,1]\to\mathbb{R}$, $g(0)=0$ such that: $$\liminf_{x\to 0^+}\frac{g(x)}{-x\ln x}\neq \limsup_{x\to 0^+}\frac{g(x)}{-x\ln x}.$$

Moreover is it possible that the upper limit is infinite while the lower limit is finite?

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1 Answer 1

You can have any pair $-\infty\le a\le b\le+\infty$ as limit inferior and limit superior, choosing a suitable concave function $g$ (and whatever is the nonnegative concave function $h(x)$ in place of $-x\log x$, satisfying $h(0)=0$ and $h'(0)=+\infty$). Idea: construct $g$ piecewise linear. To do so define inductively a strictly decreasing sequence $x_k\to0$, and a decreasing sequence of values $y_k=g(x_k)\to0$, thus defining intervals $J_k:=[x_{k+1},x_k]$ where $g$ is affine. The only constraint to get a concave function is that the slope of $g$ on each interval $J_k$ has to be smaller than $y_k/x_k$, and increasing wrto $k$; this is not an obstruction to approach any limit inferior and limit superior for $g(x)/h(x)$, provided that $x_{k+1} > 0$ is choosen small enough.

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Pietro, thank You very much for suggestion - it has solved many problems –  user27381 Jan 17 '13 at 14:34
    
Another problem which is connected with this limit - is the following equality true for $g$, which fullfills the above assumptions [ \limsup_{x\to 0^+}\frac{g(x)}{-x\ln x}= \sup_{M\in\mathbb{N}}\limsup_{n\to\infty}\frac{g(M^{-n})}{M^{-n}\ln M^{-n}}.] The crucial assuption seems to be concavity of $g$ (of course if the equality is true) –  user27381 Jan 17 '13 at 14:41
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Ok. in this case we are able to show that [\limsup_{z\to 0^+}\frac{g(x)}{-x\ln x}=\limsup_{n\to\infty}\frac{g(M^{-n})}{M^{-n}\ln M^{-n}}] for any $M>1$ and the same is true for the lower limit –  user27381 Mar 19 '13 at 10:47

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