# Proving the fact that a continuous function is identically 0 while we are given an inequality. [duplicate]

$$f \in C[a,b]$$. 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.

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 $$f(x)=0$$.