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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 $f(0)=0.$

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  • $\begingroup$ @MathematicsStudent1122 $K$ is a fixed positive constant here. $\endgroup$ – bellcircle Jun 16 '17 at 13:20
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    $\begingroup$ 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. $\endgroup$ – Piotr Benedysiuk Jun 16 '17 at 13:36
  • $\begingroup$ But then i think we require some $\delta$ too!.like $|t|<\delta$ ?@PiotrBenedysiuk $\endgroup$ – BAYMAX Jun 16 '17 at 13:38
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    $\begingroup$ math.stackexchange.com/questions/1168141/… $\endgroup$ – md2perpe Jun 16 '17 at 14:13
  • $\begingroup$ @md2perpe Both those proof only work if $K \leq 1$. $\endgroup$ – Sahiba Arora Jun 16 '17 at 14:29
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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, $\,f\equiv 0$.

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