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$f:[0,\infty)\rightarrow\mathbb R$ is a function that for all $0\leq a<b\in \mathbb R$, $f_{|[a,b]}:[a,b]\rightarrow \mathbb R$ is integrable. assuming that for all $x\in[0,\infty)$ , $f\left(x\right)=\int\limits _{0}^{x}f\left(t\right)dt$ then $f\equiv0$

my attempt:

First we know that $f\left(x\right)=\int\limits _{0}^{x}f\left(t\right)dt$ so $f(x)$ is an anti derivative of itself and continuous on $[0,\infty)$ $$ f\left(x\right) = \int\limits _{0}^{x}f\left(t\right)dt\Rightarrow f\left(x\right)-f\left(0\right)=f\left(x\right) \Rightarrow f\left(0\right)=0 $$ Now according to Mean value theorem because $f$ is continuous and derivative on $[0,\infty)$ there exists $c\in[0,\infty)$ and $$ f(c)=\frac{f\left(x\right)-f\left(0\right)}{x}\Rightarrow f\left(c\right)=\frac{f\left(x\right)}{x} $$

It could be that this attempt will not lead me to a solution but this is all I got right now. if you have any ideas...

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up vote 2 down vote accepted

I think there is a simpler method for that... As far as I know, there is only function satisfying $F(x)=f(x)$ (where F is an antiderivative of f). This function is $f(x)=ae^x$ by definition of the exponential function. Giving your situation, $F(x) = 0$ for $x=0$ but $\forall x\in\Bbb R, e^x > 0$ so $a=0$. Then the only possible function $f$ satisfying $F(x)=f(x)$ is $f(x)=0$.

EDIT : you can also think using the air representation of the integral. $\int\limits_a^b f(x)dx$ is, when $f(x)$ is continuous on $[a;b]$, the algebric measure of the air bounded by $x=a, x=b, y=0$ and $y=f(x)$. So when you write $\int\limits_0^x f(t)dt=0$, this means that the air is null for all $x \in\Bbb R$. Then the only function having a null air over the reals is $f(x)=0$.

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Actually, $f(x) = Ae^x$ will work for any constant $A$. $f(x) = 0$ is then just the special case $A = 0$. – mike4ty4 May 4 '13 at 11:24
Thanks, corrected the answer with your remark. – moray95 May 4 '13 at 11:29
can we prove somehow that if $F(x)=f(x)$ then it has to be that $F(x)=ae^x$? – bob May 4 '13 at 11:34
@bob If $f'=f$, compute $(f/e^x)'$. You'll see it's zero; so $f=Ce^x$ for some constant $C$. moray: one can define the exponential function as a non-zero function that is its own derivative. Showing that such a function exists and that it is unique does require proof. – David Mitra May 4 '13 at 11:55
@moray95 $\exp$ may be defined differently. For example, as the inverse of $\log$, which is defined via integrals, $\displaystyle\ln(t)=\int_1^t\frac{du}u$. Or via the power series expansion, $\displaystyle \exp(x)=\sum_n\frac{x^n}{n!}$. Or as the unique continuous $f$ with $f(1)=e$, $f(x+y)=f(x)f(y)$. Different authors follow different approaches. In any case, under any of these alternative (but standard) approaches, that $\exp(x)'=\exp(x)$ needs to be proved. Anyway, you need to be careful, as the differential equation does not define $\exp$. You also need the initial condition $\exp(0)=1$. – Andrés E. Caicedo May 6 '13 at 6:18

Let $g(x)=\mathrm e^{-x}f(x)$. Then $$g'(x)=f'(x)\mathrm e^{-x}-f(x)\mathrm e^{-x}=0.$$ As $g(0)=0$, we have $g(x)=0$, for all $x$, and the result follows.

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It is related to a recent similar question – Tom-Tom Jan 23 '14 at 14:40

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