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Suppose that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and that $$f(x)=\int_{0}^{x}f(t)dt\qquad\text{for all $x$}$$ Prove that $f(x)=0$ for all $x$.

Attempt Since $f(x)=\int_{0}^{x}f(t)dt$, by the First Fundamental Theorem of Calculus, this gives $f(x)=F(x)-F(0)$. So, $$F(x)=\int_{0}^{x}f(t)dt+F(0)\Longrightarrow F'(x)=f(x)=F(x)-F(0)=\int_{0}^{x}f(t)dt$$

At this step, I don't know how to keep going to work on it. I think my proof doesn't right. Can someone give me a hint or suggestion to the right direction? Thanks

egreg
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Simple
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1 Answers1

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By the fundamental theorem you have $$ f'(x)=f(x) $$ Since $f$ is defined on $\mathbb{R}$, this implies that $f(x)=ke^x$, for some $k\in\mathbb{R}$. But, since $f(0)=0$, we have $k=0$.

Hint for proving $f(x)=ke^x$: consider $g(x)=e^{-x}f(x)$ and compute $g'(x)$.

egreg
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  • Why can we consider $g (x)=e^{-x}f(x) $ ? – Simple Jan 25 '16 at 23:30
  • @Simple it's the standard trick for proving that fact. The idea is that $f(x)$ should be quite similar to $e^x$, so we try seeing whether it really is. If a function has zero derivative over an interval (in particular over $\mathbb{R}$), it is constant. – egreg Jan 25 '16 at 23:32