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I am having a problem with a question. Can somebody help me please.

Show that if $f(x)=\int_{0}^x f(t)dt$, then $f=0$

Thank you in advance

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More precise assumptions on the function $f$ are needed here. – Siminore Oct 31 '12 at 15:48
up vote 2 down vote accepted

differentiating both sides gives $f'(x) = f(x)$ which has solutions $f(x) = Ae^x$. But the original statement gives that $f(0) = \int_0^0f(t)dt = 0$ so we have $Ae^0 = 0$ so $A = 0$ and the function is constantly zero.

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And we can "differentiate both sides" because...? – Pedro Tamaroff Oct 31 '12 at 15:10
You are assuming tacitly that $f$ is continuous, so that $x \mapsto \int_0^x f$ is differentiable, and hence $f$ is differentiable. – Siminore Oct 31 '12 at 15:50

If $f$ is continuous, the the argument suggested by Tom Oldfield gives the answer. If $f$ i only locally integrable, you can apply Gronwall's inequality. Choosing $\alpha =0$, you get $f \leq 0$. Applying the same reasoning to $-f$ you get $f \geq 0$. Hence $f=0$ identically.

PS: I think that a more direct proof can be found.

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