Is $f$ identically zero on $[0, 1]$?

Let $f$ be a continuous real-valued function on $[0,1]$ such that there is $K>0$ for which $|f(x)|\le K \int_0^x|f(t)|dt$ for all $x\in [0,1]$. Does it follow that $f=0$ on $[0,1]$?

What I know is just $f(0)=0.$

• @MathematicsStudent1122 $K$ is a fixed positive constant here. – bellcircle Jun 16 '17 at 13:20
• Just something that comes to mind: since $f(0) = 0$ you can write: $|f(x)| \leq K \int |f(t) - f(0)| dt \leq K \epsilon x$ for all $x$, where the $\epsilon$ comes from continuity. – Piotr Benedysiuk Jun 16 '17 at 13:36
• But then i think we require some $\delta$ too!.like $|t|<\delta$ ?@PiotrBenedysiuk – BAYMAX Jun 16 '17 at 13:38
• math.stackexchange.com/questions/1168141/… – md2perpe Jun 16 '17 at 14:13
• @md2perpe Both those proof only work if $K \leq 1$. – Sahiba Arora Jun 16 '17 at 14:29

Set $$F(x)=\int_0^x |\,f(t)|\,dt,\quad x\in[0,1],$$ then $F$ is continuously differentiable and satisfies $$F'(x)\le Kf(x), \quad F(x)=0.$$ Hence $$\big(\mathrm{e}^{-Kx}F(x)\big)'=\mathrm{e}^{-Kx}\big(F'(x)- Kf(x)\big)\le 0,$$ and thus $$\mathrm{e}^{-Kx}F(x)\le \mathrm{e}^{-K\cdot 0}F(0)=0,$$ for all $x\in[0,1]$, which implies that $$\int_0^x|\,f(t)|\,dt=F(x)=0, \quad\text{for all x\in[0,1]},$$ and consequently, $\,f\equiv 0$.