$f : [0,1] \to \mathbb{R}$ is continuous and $f \geq 0$. There is $C>0$ with $|f(x)| < C \int_{0}^{x} |f(t)| dt$ for all $x \in [0,1]$. (so $f(0)=0$)

Is it true that $f = 0$? or is there any counterexamples?


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    $\begingroup$ As @mm-aops pointed out (beneath my deleted answer), this is true and follows from Gronwall's inequality $\endgroup$ – Omnomnomnom Jun 13 '14 at 0:55
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    $\begingroup$ You cannot have strict inequality in $|f(x)|<C\int_0^x\dots$ when $x=0$. $\endgroup$ – user147263 Jun 13 '14 at 4:58

Yes, this is well-known, but I'll write a proof because I don't like the Wikipedia proof in differential form.

Suppose $f$ is nonzero somewhere. Let $a=\inf\{x:f(x)\ne 0\}$ and $b=a+\frac{1}{ 2C}$. Let $m=\max_{[a,b]}|f|$. By the assumption,
$$ m\le C\int_0^{b}|f(t)|\,dt = C\int_a^{b}|f(t)|\,dt \le C m(b-a)=\frac{m}{2} $$ a contradiction.


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