I'm considering the space $L^2[0,2\pi]$ of square integrable $2\pi$-periodic functions equipped with the inner product $$\langle{f},g\rangle=\int_0^{2\pi}f(x)\overline{g(x)}\,dx$$ and the shift operator $$T_s{f}(x)=f(s+x)$$ for a real number $s$. Specifically, I want to know if $T_\pi$ is compact.
I'm obtaining first that $T_s$ is not a Hilbert-Schmidt operator, so compactness doesn't follow automatically; I'm not sure about this, I'm letting $\{f_n\}_{n\geq1}$ be an orthonormal basis and then I get $$\|T_s\|_{HS}\equiv\sum_{n\geq1}\|T_s{f}_n\|^2=\sum_{n\geq1}\int_0^{2\pi}|f_n(s+x)|^2dx=\sum_{n\geq1}\|f_n\|^2\to\infty$$ using $\int_s^{s+2\pi}|f(y)|^2dy=\int_0^{2\pi}|f(y)|^2dy$. But then I have no idea how to get if $T_\pi$ is compact or not; I have shown that $T_\pi^*=T_\pi$, so that it is self-adjoint and hence it seems that I should invoke the spectral theorem but I can't seem to get anywhere with that.