Functional inequality with integral function 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.
 A: 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.
A: 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 $0 \leq f(x) \leq 0 \Rightarrow f(x) = 0.$
A: 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
$$f(x)\leq \frac{2^{n}}{n!}\cdot x^n\quad \forall n\geq 1\qquad \overset{n\to\infty}{\longrightarrow}\qquad f(x)\equiv 0\qquad \forall x\in[0,1].$$
