I want to proof that $\langle f,\phi \rangle=\int_0^{\infty} \frac{\phi(x)-\phi(-x)}{x} dx$ defines a Distribution.
Because $\phi$ has a compact support one can find a constant C such that $\int_0^{\infty} \frac{\phi(x)-\phi(-x)}{x} dx=\int_0^{C} \frac{\phi(x)-\phi(-x)}{x} dx$
Linearity follows from the properties of the integral. Continuity:
Let be $\phi_n \rightarrow \phi$ and let $S$ be a compact subset such that $supp(\phi_n) \subset S$ and $\phi_n$ converges uniformly to $\phi$ on $S$
Now consider the limit: $lim_{n \rightarrow \infty} \int_{0}^{C} \frac{\phi_n(x)-\phi_n(-x)}{x} dx$
My Question is: Can I use that $\phi_n \rightrightarrows \phi$ and change limit and integration such that $lim_{n \rightarrow \infty} \int_{0}^{C} \frac{\phi_n(x)-\phi_n(-x)}{x} dx= \int_{0}^{C} lim_{n \rightarrow \infty} \frac{\phi_n(x)-\phi_n(-x)}{x} =\int_{0}^{C} \frac{\phi(x)-\phi(-x)}{x}$
Or do I need to show that $|\langle f,\phi_m \rangle - \langle f,\phi_n \rangle|<\epsilon$
It is important to note that $\phi_n \to \phi$ in the space of test functions implies that the supports of $\phi_n$ and $\phi$ are contained in one fixed compact set, hence in one compact interval, say $[0,c]$. Now you can take the limit inside because of uniform convergence.