Let $\phi \in C^1_c(\mathbb R)$. Prove that $ \lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0). $
Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. $ \vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}. $ Now, I would use the well-known fact that $ \int_\mathbb R \frac{\sin x}{x} \, dx = \pi. $ On the other hand, by substitution rule, we have also $ \int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi. $ Indeed, I would like to estimate the quantity $ \begin{split} & \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ & = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ & \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert \phi(x)-\phi(0) \right\vert dx \end{split} $ but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.
I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed.