# Functional inequality with integral function [closed]

Given a function $$f:[0;1]\to[0;1]$$ such that $$f(x)\leq2\int_0^x f(t)dt$$, prove that $$f(x)=0$$ $$\forall x\in [0;1]$$.

I've observed that the function has to be concave down in his domain and that $$f(0)=0$$, but nothing more.

• Because $0 \leq f(x) \leq 1,$ can we conclude that $f(x) \leq 2x$? From that, can we conclude $f(x) \leq 2x^2$? What happens when we iterate this logic? Jun 11, 2022 at 13:39
• How do we know that $0 \leq f(x) \leq 1$, and how does that imply that $f(x) \leq 2x$? Jun 11, 2022 at 13:52
• @cpiegore see my answer. Jun 11, 2022 at 13:53
• $0 \leq f(x) \leq 1$ by definition: look at the range Jun 11, 2022 at 13:53
• Does this answer your question? If $f:[0,1]\to[0,1]$ is such that $f(x) \leq 2 \int_0^x f(t) dt$, show that $f(x)=0$ on its domain – found with Approach0 Jun 14, 2022 at 7:09

The other answer by Meowdog is much nicer for continuous $$f,$$ but this is how I would approach the problem for a general $$f.$$

Let's begin by proving that $$f(x) \leq \frac{2^{n}x^n}{n!}$$ for all $$x \in [0,1]$$ and all whole numbers $$n.$$

By the definition of $$f,$$ we have that $$f(x) \leq 1$$ for all $$x,$$ so the base case for $$n = 0$$ is true. Now, given that $$f(x) \leq \frac{2^{k}x^k}{k!},$$ we have that $$f(x) \leq 2\int_0^x f(t) dt \leq 2 \int_0^x \frac{2^{k}t^k}{k!} dt = \frac{2^{k+1}x^{k+1}}{(k+1)!}$$

so by induction our hypothesis is proven.

So, because $$0$$ is the infimum of the sequence $$f_n(x) = \frac{2^{n}x^n}{n!}$$ for all $$x \in [0,1],$$ we must have $$f(x) \leq 0,$$ so we have $$0 \leq f(x) \leq 0 \Rightarrow f(x) = 0.$$

• I think that yours is nicer, though. Mine needs continuity. Jun 11, 2022 at 13:58
• @Meowdog I missed at a first glance that yours required continuity (for FTC) so on that realization I do prefer mine for practical reasons, but I still certainly appreciate your answer, it works out quite beautifully Jun 11, 2022 at 14:01
• Thank you. So does yours. Jun 11, 2022 at 14:04

We must assume that $$f$$ is continuous of course (SO CAUTION).

Set $$g(x) := \exp\left(-2 x \right)\int^x_0 2f(t)~\mathrm{d}t.$$ It follows that $$g'(x) = 2\exp(-2x)\left( \underbrace{-\int^x_0 2f(t)~\mathrm{d}t +f(x)}_{\leq 0} \right) \leq 0 .$$ So $$g$$ is decreasing. Since $$g(0) = 0$$, we have $$\exp(-2x)\int^x_0 2f(t)~\mathrm{d}t~\leq 0$$ for all $$x \in [0, 1]$$ and therefore $$\int^x_0 2f(t)~\mathrm{d}t~\leq 0$$ for all $$x \in [0, 1]$$. In total: $$0 \leq f(x) \leq \int^x_0 2f(t)~\mathrm{d}t \leq 0$$ on $$[0, 1]$$.

This yields $$f(x) = 0$$ as desired. The keyword "Gronwall lemma" is also worth researching.

• Do we actually have to assume $f$ is continuous? I don't see where that appears in my approach Jun 11, 2022 at 13:58
• Exactly. I need it, you don't. I wanted to show this, because it's closely related to Gronwall's Lemma, which I think might be important for @Federico A to learn about. Jun 11, 2022 at 13:59
• Unless I am mistaken, it suffices that $f$ is (Lebesgue) integrable, see my comment here: math.stackexchange.com/questions/3019119/… Jun 14, 2022 at 7:22
• Ah okay, I see... . So should work, yes. Have not thought about it that way, but thank you! Jun 14, 2022 at 7:36

Using @Stephen Donovan 's idea we have that since the function is continuous, for all $$x$$ we have that $$\int_0^xf(t)\,dt=x\cdot f(c)\qquad \text{for some }c\in[0,x]$$ But since the range of $$f$$ is $$[0,1]$$, we have that $$0\leq f(c)\leq 1$$ and hence $$f(x)\leq 2x\cdot f(c)\leq 2x.$$ From here we do a bootstrap-type argument, we insert this inequality back in the original one obtaining $$f(x)\leq 2\int_0^xf(t)\,dt\leq 2 \int_0^x2t\,dt=2x^2$$ and iterating this procedure we obtain that $$f(x)\leq \frac{2^{n}}{n!}\cdot x^n\quad \forall n\geq 1\qquad \overset{n\to\infty}{\longrightarrow}\qquad f(x)\equiv 0\qquad \forall x\in[0,1].$$