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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 $$\int_0^1 A(x) g(1/x^2) dx <+\infty.$$ This question is related to membership of $g$ to some Dini class.

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The answer is yes. First define inductively a sequence $x_k>0$ such that $x_{k+1} < x_k/2$, and $$\int_0^{x_k}A(x)dx<2^{-k}.$$ This is possible because $A$ is integrable.

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 $$\int_0^1A(x)h(x)dx\leq\sum (k+1)2^{-k}<\infty.$$ 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.

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Very nice solution. –  djoke Nov 30 '12 at 13:21
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Let's change variables: we have: $$\int_0^1A(x)dx=\frac{1}{2}\int_1^{+\infty} A(x^{-1/2}) x^{-3/2}dx < +\infty\, .$$

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 $$\int_1^{+\infty} A(x^{-1/2}) g(x) x^{-3/2} dx < +\infty\, ,$$ and by the same change of variable as before, the latter integral is twice $$\int_0^1A(x) g(1/x^2) dx\, .$$

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Thanks for the solution. –  djoke Nov 30 '12 at 13:21
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