Dini condition and integrability condition - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:40:34Z http://mathoverflow.net/feeds/question/114913 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114913/dini-condition-and-integrability-condition Dini condition and integrability condition djoke 2012-11-29T18:52:49Z 2012-11-30T07:26:49Z <p>Assume that $A$ is an arbitrary positive integrable function on $[0,1]$. Whether exists a convex function $f_A(x)=x g(x)$ of $(0,+\infty)$ into itself (depending on $A$) such that $\lim_{x\to +\infty} g(x)=+\infty$ and $$\int_0^1 A(x) g(1/x^2) dx &lt;+\infty.$$ This question is related to membership of $g$ to some Dini class.</p> http://mathoverflow.net/questions/114913/dini-condition-and-integrability-condition/114920#114920 Answer by Alexandre Eremenko for Dini condition and integrability condition Alexandre Eremenko 2012-11-29T19:55:58Z 2012-11-29T19:55:58Z <p>The answer is yes. First define inductively a sequence $x_k>0$ such that $x_{k+1} &lt; x_k/2$, and $$\int_0^{x_k}A(x)dx&lt;2^{-k}.$$ This is possible because $A$ is integrable. </p> <p>Then define a continuous function $[0,1]$ by $h(x_k)=k$ and $h$ is linear on each interval $[x_{k+1},x_k]$. It is easy to see that this function is convex, decreasing and tends to $+\infty$ as $x\to 0+$. Moreover $$\int_0^1A(x)h(x)dx\leq\sum (k+1)2^{-k}&lt;\infty.$$ Now set $g(x)=h(1/\sqrt{x})$ and it remains to verity that $xg(x)$ is convex. This we do by differentiation: $$(xg(x))^\prime=h(x^{-1/2})-2x^{-1/2}h'(x^{-1/2}).$$ Both summands are increasing, therefore $xg(x)$ is convex.</p> http://mathoverflow.net/questions/114913/dini-condition-and-integrability-condition/114930#114930 Answer by Pietro Majer for Dini condition and integrability condition Pietro Majer 2012-11-29T20:57:18Z 2012-11-29T21:02:23Z <p>Let's change variables: we have: $$\int_0^1A(x)dx=\frac{1}{2}\int_1^{+\infty} A(x^{-1/2}) x^{-3/2}dx &lt; +\infty\ .$$</p> <p>Therefore, by the Dominated Convergence Theorem $$\int_1^{+\infty} A(x^{-1/2}) x^{-5/2}(x-k) _ +dx=o(1),\qquad (\mathrm{as }\ k\to+\infty)\ ,$$ and in particular there is a sequence $k _ n\to +\infty$ such that $$\int_1^{+\infty} A(x^{-1/2}) x^{-5/2}(x-k _ n) _ +dx\le 2^{-n} .$$ The function $f:(0,+\infty)\rightarrow(0,+\infty)$ $$f(x):=x+\sum_{n=1}^\infty(x-k_n) _ +$$ is convex and verifies $g(x):=f(x)/x\to+\infty$; moreover, integrating by series $$\int_1^{+\infty} A(x^{-1/2}) g(x) x^{-3/2} dx &lt; +\infty\ ,$$ and by the same change of variable as before, the latter integral is twice $$\int_0^1A(x) g(1/x^2) dx\ .$$</p>