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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.

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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 $0 \leq f(x) \leq 0 \Rightarrow f(x) = 0.$

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  • $\begingroup$ I think that yours is nicer, though. Mine needs continuity. $\endgroup$ Jun 11, 2022 at 13:58
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    $\begingroup$ @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 $\endgroup$ Jun 11, 2022 at 14:01
  • $\begingroup$ Thank you. So does yours. $\endgroup$ Jun 11, 2022 at 14:04
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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.

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  • $\begingroup$ Do we actually have to assume $f$ is continuous? I don't see where that appears in my approach $\endgroup$ Jun 11, 2022 at 13:58
  • $\begingroup$ 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. $\endgroup$ Jun 11, 2022 at 13:59
  • $\begingroup$ Unless I am mistaken, it suffices that $f$ is (Lebesgue) integrable, see my comment here: math.stackexchange.com/questions/3019119/… $\endgroup$
    – Martin R
    Jun 14, 2022 at 7:22
  • $\begingroup$ Ah okay, I see... . So should work, yes. Have not thought about it that way, but thank you! $\endgroup$ Jun 14, 2022 at 7:36
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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 $$f(x)\leq \frac{2^{n}}{n!}\cdot x^n\quad \forall n\geq 1\qquad \overset{n\to\infty}{\longrightarrow}\qquad f(x)\equiv 0\qquad \forall x\in[0,1].$$

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