Weak convergence for $L^2$ on the torus It is well-known that if we consider, for example, $L^2(\mathbb{R})$, we can pick any function $f\in L^2(\mathbb{R})$ with norm $\Vert f\Vert_{L^2}=1$, and then the sequence $$
f_n:=f(x-n)
$$
will weakly converge to zero in $L^2(\mathbb{R})$, even when it is not converging to anything in the strong sense. On the other hand, I was thinking that, for example, if we change the domain to the torus $\mathbb{T}$, this trick doesn't work anymore. I was wondering what kind of "weird behaviors" could we have by considering the weak topology on $L^2(\mathbb{T})$. In other words, what kind of sequence could we have weakly-converging to something, but not strongly (for this particular case). For me it is much more easier to think about examples of this kind for unbounded domains like $\mathbb{R}^n$, but I am struggling a little bit to think about the case of the torus.
Edit: For example I was wondering if there is any analogous trick for constructing a sequence of functions weakly converging to zero, by considering some group (?) acting on a fixed function (like in the example, just translations of a fixed function)?
 A: Typically there are three kinds of behaviors which prevent a weakly convergent sequence $(f_n) \subset L^2$ from converging strongly:


*

*Evanescence, i.e. escape of the mass to infinity. As you point out, this does not occur in the case of the torus $\mathbb{T}$ as it is compact.

*Oscillation. Take for example $f_n(x) = e^{inx}$. Then for every $g \in L^2(\mathbb{T})$,
$$\langle g,f_n\rangle = \frac{1}{2\pi} \int_{-\pi}^\pi g(x)e^{-inx}\, dx = \hat{g}(n) \xrightarrow[n\to\infty]{} 0
$$
by the Riemann-Lebesgue lemma, so that $f_n \to 0$ weakly in $L^2(\mathbb{T})$. However,
$$\lVert f_n \rVert_{L^2}^2 = \frac{1}{2\pi} \int_{-\pi}^\pi |e^{inx}|^2\, dx = 1,
$$
and $f_n$ cannot converge strongly to $0$.

*Concentration. Set $f(x) = \mathbf{1}_{[-\pi,\pi]}$ and take $f_n(x) = \sqrt{n}f(nx)$. Then $\lVert f_n \rVert_{L^2} = 1$ so $f_n$ cannot converge strongly to $0$. For every $g \in C(\mathbb{T})$, we have
$$\langle g,f_n \rangle = \frac{1}{2\pi \sqrt{n}} \int_{-\pi}^{\pi} g\left(\frac{y}{n}\right)\, dy \xrightarrow[n\to\infty]{} 0
$$
by the dominated convergence theorem. I'll leave it to you as an exercise to show that this, together with the density of $C(\mathbb{T})$ in $L^2(\mathbb{T})$ and the boundedness of $f_n$ in $L^2(\mathbb{T})$ implies the weak convergence of $f_n$ to $0$. Bonus: notice that in this case, the sequence of measures $f_n^2(x)\, dx$ converges weakly (in the sense of weak convergence of measures) to $\delta_0$.

