Show that the distribution defined by $f_n(x)=\sin(nx)$ converges to $T_0$

Solution 1:

By definition, if $\varphi$ is a test function, $$\langle T_{f_n}, \varphi\rangle = \int_{\mathbb R} f_n(x)\varphi(x)dx=\int_{\mathbb R} \sin(nx)\varphi(x)dx$$ If you integrate by parts, noting that the test function evaluated at $\pm \infty$ is $0$: $$\langle T_{f_n}, \varphi\rangle = \int_{\mathbb R} \frac{\cos(nx)}n\varphi^\prime(x)dx = \frac 1 n\int_I \cos(nx)\varphi^\prime(x)dx$$ where $I$ is the support of $\varphi$. Finally, you can conclude by noticing that $$\left|\langle T_{f_n}, \varphi\rangle \right|\leq \frac{\|\varphi^\prime\|_\infty |I|}{n}\xrightarrow[n\rightarrow+\infty]{}0$$

As noted in the comments, this is known as the Riemann-Lebesgue lemma.