I was working on the following problem,
Prove that $\lim_{\epsilon \downarrow 0} \frac{x - x_0}{(x - x_0)^2 + \epsilon^2} = \mathcal{P}(1/(x - x_0)$, in the weak topology on $\mathcal{S}'(\mathbf{R})$.
Note that $\mathcal{S}'(\mathbf{R})$ is the dual of the rapidly decreasing functions (the Schwartz space) and $\mathcal{P}$ is the principal value integral: $[\mathcal{P}(1/(x-x_0)](f) = \lim_{\epsilon \downarrow 0} \int_{|x| \geq \epsilon} \frac{f(x)}{x - x_0}\, \mathrm{d}x$, for any rapidly decreasing $f \in \mathcal{S}(R)$.
I'm super confused about this. How should I think about the weak topology on $\mathcal{S}'(\mathbf{R})$. I'm quite new to learning about the weak topologies (and I'm not looking for an answer), just some help understanding what this topology is, and possibly a hint about how someone could prove this type of claim.
What you are asked to prove is: $\int f(x) \frac {x-x_0} {(x-x_0)^{2} +\epsilon^{2}} dx \to$ the principal value of f for any f in $\mathcal S(\mathbb R)$ . Consider $\int \{f(x) \frac {x-x_0} {(x-x_0)^{2} +\epsilon^{2}}-\frac {f(x)} {x-x_0}\} dx $ This simplifies to $\int f(x) \frac {\epsilon^{2}} {(x-x_0)^{2} +\epsilon^{2}}dx$. The integrand is domintaed by $|f|$ which is integrable. Since the integrand tends to 0 as $\epsilon \to 0$ we are done. (Assuming that the principal value exists for functions in $\mathcal S(\mathbb R)$).