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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.

Nick Peterson
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Riju
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  • A small warm-up (in preparation for Andre's answer): Do you see that $|f_n|$, $f_n^+$, and $f_n^-$ are bounded, independently of $n$, $p$, $a$, and $b$? – Eric Towers Jul 14 '17 at 23:24
  • Yes $h$ should be bounded – Riju Jul 14 '17 at 23:29
  • OP, I don't have a complete solution yet, but I'd suggest the following for now: we know that $\int_t^{\frac{T}{n} + t} f_n(x)dx = 0$ for every $n$, and we also know that we can approximate any $g \in L^q$ with simple functions $\phi_m$ whose errors $||g -\phi_m||_q$ tend to $0$ as $m \to \infty$. So we can write $\int_a^b f_n\phi_m d\mu + \int_a^b f_n(g - \phi_m)d\mu$, and the latter term we can bound using Hölder's inequality. But the first term we work with explicitly. – Chris Jul 14 '17 at 23:53
  • I'm sure there's something nicer, incidentally, but using $f$'s continuity you can take the maximum value $m$ of $|f|$ and set $M = m(b-a)$; then $(\int_a^b |f_n|^p d\mu)^\frac{1}{p} \le M$ and so we know how small we must make $||g - \phi_m||_q$ explicitly to sufficiently reduce the error term. – Chris Jul 15 '17 at 00:00
  • Ah, okay, I found the missing piece of the puzzle! OP, take a look at this: a proof that you can approximate $g \in L^q$ with step functions. This is what we need to solve your problem: now we can write $g(x) \approx \sum_{k=1}^K c_k \chi_{[a_k, b_k]}$, where we can make the approximation as good as we'd like in terms of the $L^q$ norm, as I mentioned above. Try to solve the rest of this for yourself! – Chris Jul 15 '17 at 00:31

1 Answers1

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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.