Let $h$ be a continuous function defined on all of $\mathbb{R}$ that is periodic of period $T$ and $\int_{0}^{T} h=0$. Let $[a,b]$ be a closed bounded interval and for each natural number $n$, define the function $f_{n}$ on $[a,b]$ as $f_{n}(x)=h(nx)$. Define $f\equiv 0$ on $[a,b]$ and show that for $1\leq p <\infty$, $f_{n}$ weakly converges to $f$.
I have no idea how to show this. There are a lots of theorem at hand but nothing is helping. I tried using the definition , so if I can prove
$$ \lim_{n\to \infty} \int g(x)h_{n}(x)dx \to 0, \forall g\in L^{q},$$ we are done but I don't get how to use the periodicity to show that.
Thanks in advance for any help.
As pointed out in the comments, $\|h_n\|_p\le M$ ($1\le p\le\infty$), so it suffices to verify the condition for weak convergence, $$ \int_a^b g(x)h(nx)\, dx \to 0 , $$ for continuous $g$. Then, since $g$ is uniformly continuous on $[a,b]$, $$ \int_a^{a+T/n} g(x)h(nx)\, dx = g(a)\int_a^{a+T/n}h(nx)\, dx + o(MT/n) = o(1/n) . $$ Of course, we can do this on every interval $[a+jT/n, a+(j+1)T/n]$, and there are $\simeq n$ of these, so $\int gh_n= o(1)$, as claimed.