Dini condition and integrability condition 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.
 A: 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.
A: 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
$$\int_0^1A(x)  g(1/x^2)   dx\, .$$
