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During one of the problems in Rudin I was asked to show $f=0$ a.e. Here $f$ satisfies this condition:

$f(x)=\frac{1}{x}\int^{x}_{0}f(t)dt$ almost everywhere and is in $L^{p}(0,\infty)$. So constant functions would not work. I tried to prove by contradiction, and a few imaginary counter-examples' failure convinced me this is true. But what is a good way of proving this statement? Since we know $f\in L^{p}$ I am thinking about using Holder's inequality, but in our case it is difficult to apply (since the other side is larger ). We can assume $f\in C_{c}(0,\infty)$ since this is dense in $L^{p}$, but I still do not know how to prove this statement.

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    I see. If $f$ is continuous then I can show this. Now we have $D(\log[G(x)])=1/x$, which gives $log[G(x)]=\log[x]$, so $G[x]$ can only be a constant, and hence $f=0$ almost everywhere.2012-12-27

3 Answers 3

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From Hardy's Inequality for Integrals conclude that the $L_p$ norm of $f$ is zero. This implies $f$ is zero a.e.

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    Thanks a lot for the nice proof!2012-12-27
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Let us put

$F'(x)=f(x)\Longrightarrow \int_0^xf(t)dt=F(x)-F(0)\Longrightarrow$

$F'(x)=f(x)=\frac{1}{x}\int_0^xf(t)dt=\frac{F(x)-F(0)}{x}\Longrightarrow$

$\int\frac{dF}{F(x)-F(0)}=\int\frac{dx}{x}\Longrightarrow\log|F(x)-F(0)|=\log|x|+K\Longrightarrow$

$F(x)=C_1x+C_2\ldots.$

But then I get $\,f\,$ is a constant...

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    I see. This could work since we can pass from $L^{k}$ to $C_{c}$, then this holds for $C_{c}$ means $f$ can only be constant almost everywhere, which would then imply the $L^{p}$ case is impossible unless the constant is 0.2012-12-27
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$f(x)=\frac{1}{x}\int^{x}_{0}f(t)dt$, so $xf(x)=\int^{x}_{0}f(t)dt$. Differentiating, $f(x)+x f'(x) = f(x)$, so $x f'(x) = 0$. Therefore $f'(x) = 0$ (except at $0$), so $f(x)$ is a constant. Since the only possible constant is $0$, we are done.

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    Oh, OK! I d$i$dn't think things through before.2012-12-27