How do I prove $f=0$ almost everywhere?
Solution 1:
From Hardy's Inequality for Integrals conclude that the $L_p$ norm of $f$ is zero. This implies $f$ is zero a.e.
Solution 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...