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.



Dini condition and integrability condition
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.


