I have $x_n(t)=\sin(nt)$ with $t \in [0,2\pi]$, and I have to prove that there's no $ \sigma:\mathbb N \to \mathbb N$ such that $x_{\sigma(n)}(t)$ converges for every $t$.
I have proved that there's no pointwise convergence, but I'm stuck at uniform convergence.
You can try proof by contradiction: supposing that theres a $\sigma$ you get: $$x_{\sigma(n)}(t)\rightarrow x_0$$ Now you pick a function $f \in C^1$ with compact support . Then, using chain rule you get $$\int\limits_{0}^{2\pi}{\sin(nt)f(t)}dt=\frac{1}{n}\int \cos(nt)f'(t)dt$$ then you have:$$\int\limits_{0}^{2\pi}{\sin(nt)f(t)}dt\rightarrow 0$$so $$x_{\sigma(n)}(t)\rightarrow x_0(t)$$ $$\int\limits_{0}^{2\pi}{x_0(t)f(t)}dt= 0 \Longrightarrow x_0(t)\equiv0 \ a.e.$$ On the other hand you have: $$\int\limits_{0}^{2\pi}{\sin^2(nt)}dt=\pi\Longrightarrow $$ so $$\int\limits_{0}^{2\pi}{x_0^2(t)}dt=\pi $$ and this is impossible because $x_0(t)\equiv0 \ a.e$