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$f \in C[a,b]$. There exists $M \geqslant 0$ such that $|f(x)| \leqslant M \int_a^x|f(t)|dt$ $\forall x \in [a,b]$. It is be proved that $f(x) = 0$ $\forall x \in [a,b]$.

If $M = 0$, nothing to prove. Otherwise, let $K$ be such that $|f(x)| \leqslant K$. We observe that $|f(x)| \leqslant MK(x-a)$. I am not able to go any further.

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You have $f(a)=0$. If $g(x)=\int_a^x|f(t)|\,dt$, the inequality is $$\tag1g'(x)\leq M\,g(x), $$with $g(a)=g'(a)=0$ and $g(x)\geq0$. If $h(x)=e^{Mx}$, then multiplying both sides by $h$ the inequality becomes $$\tag2g'(x)h(x)\leq g(x)h'(x).$$ So with $k(x)=g(x)h(x)$, now $(2)$ can be rewritten as $$\tag3 k'(x)\leq0.$$ Together with $k(x)\geq0$ and $k(a)=0$, this gives $k(x)=0$ on $[a,b]$. Then $g(x)=0$, and so $f(x)=0$.

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