 Solution

Weak convergence $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^{\ast }$. Weak-star convergence $T_j\stackrel{\ast }{\rightharpoonup }T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The Sequential Banach-Alaoglu theorem for a separable predual says that every norm-bounded sequence in $X^{\ast }$ has a weak-star convergent subsequence. Separability is what turns the bounded weak-star compactness statement into this sequential form.

For the weak-star limit of periodic rescalings, apply it with $X=L^1(\mathbb{R})$, which is separable, and $X^{\ast }=L^{\infty }(\mathbb{R})$. The rescaled functions have uniform essential-supremum bound $\Vert \theta {\Vert }_{\infty }$. Let $q=\theta -\bar{\theta }$ and $F(t)={\int }_0^{t}q(s)ds$. Since $q$ is one-periodic with zero mean, $F$ is a bounded one-periodic absolutely continuous primitive. Thus for every finite interval,Every weak-star subsequential limit $h$ therefore satisfies ${\int }_a^b(h-\bar{\theta })=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and set $v=(h-\bar{\theta })1_{[-R,R]}$. This is in $L^1(\mathbb{R})$, and its integral over any interval is zero, since intersecting with $[-R,R]$ again gives an interval. Hence $v=0$ almost everywhere. Taking integer $R\to \infty$ yields $h=\bar{\theta }$ almost everywhere on the whole line.

Finally, if the full sequence did not converge weak-star to this constant, some $g\in L^1$ and some subsequence would have its pairing with $\theta (jx)-\bar{\theta }$ bounded away from zero. Sequential compactness gives a further weak-star convergent subsequence, whose limit must be the constant just identified, a contradiction. ThereforeThis is averaging against integrable test functions, not a claim of pointwise or strong norm convergence.