Suppose $f \colon (0,1) \to (0,\infty)$ is monotonically decreasing and integrable on (0,1). Denote $F(x) = \int_0^x f(y) dy $. Suppose that there exists a constant $C>0$ such that $$F(q) - qf(q)^2 \le C$$ for all $q \in (0,1)$. Is there anything that can be said about $f$? Is there such an $f$, and if so, what does it look like?
If one makes the ansatz that $f(q) = aq^k$ for some $k \in [-1/2,0)$ and $a>0$, then this inequality holds true by Young's inequality. I wonder if these are already all possible functions. Any comment or insight is welcome.
As $F(0) = 0$ we deduce that the limit $\lim_{q \to 0} qf(q)^2 $ has to exist and needs to be finite.
\infty
, to render $\infty$ $\endgroup$