How do I prove $f=0$ almost everywhere?

Question

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.

Answer 1

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

Answer 2

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...

Answer 3

$f(x)=\frac{1}{x}\int^{x}_{0}f(t)dt$, 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.