On functions satisfying a functional inequality

Question

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.

Answer 1

Let $a \leq C$, recall that $f = \dot{F}$, and consider the following ODE:

$$F - q \dot{F}^2 = a.$$

We can solve this ODE as follows:

$$F -q \dot{F}^2 = a \Rightarrow \dot F^2 = \frac{F - a}{q} \Rightarrow \dot F = \sqrt{\frac{F-a}{q}} \Rightarrow \frac{dF}{\sqrt{F-a}} = \frac{dq}{\sqrt{q}},$$

which as a general solution in the form

$$F(q) = \pm 2b\sqrt{q} +q + b^2 + a,$$

for some $b \in \mathbb{R}$.

Hence:

$$F(x) = \pm 2b \sqrt{x} + x + b^2 + a,$$

and

$$f(x) = \pm \frac{b}{\sqrt{x}} + 1.$$

Since $f$ is monotonically decreasing, then

$$f(x) = \frac{b}{\sqrt{x}} + 1,$$

with $b > 0$.

Answer 2

Let a family of solutions be as follows: $$ f_s(x)=\sum_{n=1}^\infty e^{\frac{ns}{\log x}}=\frac{e^{\frac{s}{\log x}}}{1-e^{\frac{s}{\log x}}}. $$ for $x\in(0,1)$ and parameter $s\in (0,\infty).$

Set $F(1)=C.$

Another family of solutions is:

$$ h(x)=g_k(x)e^{\frac{1}{\log x}} $$

for suitably chosen $g.$ For example: $g_k(x)=(-1)^k(\log x)^k$ for parameter $k\in(1,\infty)$ works.

What I did for the first family of solutions was to start with a function $f:(0,1)\to (0,1)$ and add up many of those functions to get the desired range $(0,\infty).$

Similarly for the second family of solutions I multiplied a function $f:(0,1)\to (0,1)$ by a suitably chosen function to get the desired range $(0,\infty).$