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 $$\int_1^{+\infty} A(x^{-1/2}) x^{-5/2}f(x)dx g(x) x^{-3/2} dx < +\infty\ ,$$ and by the same change of variable as before, the latter integral is twice $$2\int_0^1A(x) \int_0^1A(x) g(1/x^2) dx\ .$$
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 $$\int_1^{+\infty} A(x^{-1/2}) x^{-5/2}f(x)dx < +\infty\ ,$$ and by the same change of variable as before, the latter integral is $$2\int_0^1A(x) g(1/x^2) dx\ .$$