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I'm reading something and the consider the following example:

Define: $$\chi(x) = \mathbb{1}_{[0,1/2]}-\mathbb{1}_{[1/2,1]}$$ and extend periodically to $\mathbb{R}$. Let $u_j = \chi(jx)$ on $(0,1)$. Then $u_j \rightharpoonup 0$ in $L^p(0,1)$ but $u_j$ does not converge in norm to $0$ on $L^p(0,1)$.

I'm getting this mixed up though, I am not getting weak convergence. What am I doing wrong in the following calculations?

For weak convergence, let $g \in L^q(0,1)$ consider: $$\int_{[0,1]}u_jg = \int_{\{x|u_j(x) > 0\}}u_jg+\int_{\{x|u_j(x) = 0\}}u_jg+ \int_{\{x|u_j(x) < 0\}}u_jg$$ If I choose $g$ to be the indicator function of all irrationals on $(0,1)$, as $j$ increments, won't the value of this integral keep bouncing around without a limit? (Consider an irrational number like $1/\pi$, every new dyadic approximation will send the number to 1 or -1, this is happening for all irrationals)

yoshi
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    Unless I am missing something, it looks like $u_j(x) = 0$ whenever $x>1/j$, while $|u_j(x)|\leq 1$ if $x \in (0,1/j]$. – Michael May 15 '19 at 16:55
  • In other words, clearly $\lim_{j\rightarrow\infty}\int_0^1 |u_j(x)|^pdx = 0$. Notice that I am taking an absolute value, so it does not matter if $p$ is even or odd. Usually it is assumed that $p$ is a real number (not necessarily an integer) that is greater than or equal to 1. It might help for you to write what you view "converge in norm" to mean and how you see rationals/irratinoals/dyadics bouncing around relates to this problem. – Michael May 15 '19 at 17:01
  • @Michael sorry, I forgot a key detail, "extend periodically to $\mathbb{R}$". This has been added. I am referring to example 2.8 here: https://people.cs.kuleuven.be/~marcus.webb/pdfs/webbyoungmeasures.pdf – yoshi May 15 '19 at 17:55
  • @Michael ah yes, I have forgetten the absolute value sign in the defn of the norm. I have to think about my question some more. – yoshi May 15 '19 at 17:59
  • @Michael, I have edited my question. Indeed, the norms all $u_j$ in any $L^p$ space are $1$. – yoshi May 15 '19 at 18:50

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The integral $\int u_jg$ indeed converges to zero as $j\to\infty$. You are correct that the functions $u_jg$ will fail to converge almost everywhere (even for $g$ identically $1$), but this fact doesn't prevent the integrals from converging. The reason is that the sequence of functions $(u_n)$, while oscillating faster and faster with $n$, consists of rescales of a single integrable "base" function $\chi$. Hint: This should remind you of the Riemann-Lebesgue lemma, specifically in this form.

grand_chat
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  • This was helpful, there's one part of the calculation I didn't understand -- I commented on it in the post there. Could you take a look? (if it hasn't already been answered by the time you see this) – yoshi May 18 '19 at 15:54