# On functions satisfying a functional inequality

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.

• What is the motivation for this question? Is it a set problem, or did it occur in a wider context e.g. a problem you’re working on? Oct 15, 2022 at 12:34
• Oh and by the way, every monotonic function is Riemann integrable Oct 15, 2022 at 12:35
• Apologies, I did not take in the fact that $f$'s domain is $(0,1)$ rather than $[0,1)$. The lack of an $f(0)$ does mean it is possible for the function not to be integrable, and also for my answer to be wrong Oct 16, 2022 at 13:23
• Btw it's \infty, to render $\infty$ Oct 16, 2022 at 13:32

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 $$b > 0$$.

• Great Answer. Thank you. This also shows that if a function $g$ satisfies the inequality it must be dominated by $f$ by a comparison principle. Oct 28, 2022 at 8:32
• @Rooibos Is the condition $F(0) = 0$ mandatory? Nov 2, 2022 at 17:22
• By definition of $F$, yes. Nov 4, 2022 at 12:37
• Well, then $b^2 + a = 0 \Rightarrow b^2 = -a \ldots$ Nov 4, 2022 at 16:09

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 $$(0,\infty).$$

• How does this answer the question? Oct 16, 2022 at 21:07
• What do you mean? I gave two different solutions. They asked if such an $f$ exists and what does it look like Oct 17, 2022 at 12:56
• Can you please provide proof that those functions have the desired property. Oct 25, 2022 at 14:41