most recent 30 from http://mathoverflow.net 2013-05-20T19:23:54Z http://mathoverflow.net/feeds/question/110081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110081/concave-function-with-sublinear-growth FF 2012-10-19T09:10:32Z 2012-10-19T13:09:17Z

<p>Does there exist a concave, increasing function $h\colon[0,\infty)\to\mathbb{R}$ such that </p> <ol> <li><p>$\lim_{x\to\infty} h(x)=\infty$</p></li> <li><p>$\lim_{x\to\infty} h(x)/x=0$</p></li> <li><p>There exist sequences of positive numbers $a_n,b_n,c_n,d_n$ which converge to infinity such that:</p></li> </ol> <p>3a. $\infty>\lim_{n\to\infty} a_n/b_n=\lim_{n\to\infty} c_n/d_n>0$ but</p> <p>3b. $\lim_{n\to\infty} h(a_n)/h(b_n) \neq \lim_{n\to\infty} h(c_n)/h(d_n)$</p> <p>Is it possible?</p>

http://mathoverflow.net/questions/110081/concave-function-with-sublinear-growth/110089#110089 pgassiat 2012-10-19T11:31:57Z 2012-10-19T12:23:25Z

<p>The negation of your property 3. is called <em>regular variation</em>. </p> <p>Here is an example of a concave not regularly varying function (taken from <a href="http://users.unicyb.kiev.ua/~iksan/library/ui/upload/articles-16.pdf" rel="nofollow">this paper</a> by Iksanov and Rösler, p.10) :</p> <p>Take $$f(x) = 2^{-k} x + 2^{k+1} - 3, \;\;\;\; x \in [4^k,4^{k+1})$$</p> <p>Then for $x_n=4^n$, $y_n=3\cdot4^n$, $$\lim_n \frac{f(2x_n)}{f(x_n)} = 2 \neq \frac{7}{5} = \lim_n \frac{f(2y_n)}{f(y_n)}.$$</p>