# proving that continuous function smaller than integral is identically zero

$f : [0,1] \to \mathbb{R}$ 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?

Thanks.

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

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.