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- Apr 13, 2013

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Hello!!!

I wanted to ask you how I can show that there is exactly one $f$ continuous for $x>0$,so that $f(x)=1+\frac{1}{x}\int_{1}^{x}f(t)dt, \forall x>0$.Is there a theorem that I could use?Or do I have just to suppose that there is a $g\neq f$,which satisfy $g(x)=1+\frac{1}{x}\int_{1}^{x}g(t)dt, \forall x>0$ and then to conclude that it must be $g(x)=f(x)$ ?

I wanted to ask you how I can show that there is exactly one $f$ continuous for $x>0$,so that $f(x)=1+\frac{1}{x}\int_{1}^{x}f(t)dt, \forall x>0$.Is there a theorem that I could use?Or do I have just to suppose that there is a $g\neq f$,which satisfy $g(x)=1+\frac{1}{x}\int_{1}^{x}g(t)dt, \forall x>0$ and then to conclude that it must be $g(x)=f(x)$ ?

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