Let $p \in [1, \infty]$. Let $f \in L^p_{\text{loc}} (\mathbb R)$ be $T$-periodic, i.e., $f(x+T) = f(x)$ a.e. $x \in \mathbb R$. Let $$ \overline f := \frac{1}{T} \int_0^T f (t) \, dt. $$
We define a sequence $(u_n) \subset L^p([0, 1])$ by $u_n (x) := f(nx)$ for all $x \in [0, 1]$. I'm trying to prove that
$u_n \to \overline f$ in the weak topology $\sigma(L^p, L^{p'})$ where $p'$ is the Hölder conjugate of $p$.
Could you have a check on my below attempt?
Is there another way that avoids such an approximation of $g$?
Let $g \in L^{p'} ([0, 1])$. We need to prove $\int_0^1 g (x) u_n (x) \, dx \to \int_0^1 g (x) \overline f \, dx$. It suffices to assume that $g = 1_B$ where $B$ is a Borel subset of $[0, 1]$ (please see here for more details). On the other hand, we can approximate $1_B$ (in $L^{p'})$ by $1_I$ where $I$ is an open interval of $[0, 1]$ (please see here for more details). So WLOG we assume $g = 1_{(a,b)}$ where $0 \le a < b \le 1$.
Then $$ \int_0^1 g (x) u_n (x) \, dx = \int_a^b f (nx) \, dx = \frac{1}{n} \int_{na}^{nb} f(x) \, dx. $$
Let $\varphi (n) := \lfloor \frac{n(b-a)}{T} \rfloor$. By periodicity of $f$, we have $$ \begin{align} \frac{1}{n} \int_{na}^{nb} f(x) \, dx &= \frac{1}{n} \bigg ( \int_{na}^{na + \varphi (n)T} f(x) \, dx + \int_{na + \varphi (n) T}^{nb} f(x) \, dx \bigg ) \\ &= \frac{1}{n} \bigg ( \varphi(n) \overline f+ \int_{na + \varphi (n)T}^{nb} f(x) \, dx \bigg ). \end{align} $$
Clearly, $\frac{1}{n} \int_{na + \varphi (n)}^{nb} f(x) \, dx \xrightarrow{n \to \infty} 0$ and $\frac{\varphi(n)}{n} \xrightarrow{n \to \infty} \frac{b-a}{T}$. The claim then follows.
