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...