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Can you prove or disprove that the sequence $\{\sin (nx)\}$ has a pointwise almost everywhere convergent subsequence with respect to the Lebesgue measure on $\mathbb{R}$ ?

Edit:

I am adding my thoughts here as a motivation, because otherwise this question will hit the bottom. So initially I thought whether $\{\sin{nx}\}$ is convergent in $L_1([0,\pi])$ or on other finite interval in $\mathbb R$. Then I computed the integral $\|\sin{nx}\|_1=\int\limits_{0}^{\pi}{|\sin{nx}|dx}=n\int\limits_{0}^{\pi/n}{\sin{nx}dx}=2$. Therefore $\|\sin{nx}\|_1\rightarrow 2$ but this doesn't tell me to which function converges. Then I thought about convergence in $L_2([0,\pi])$, but obviously it is not convergent there. This is because it is part of the orthonormal basis in $L_2$ (up to a constant), so it is not even Cauchy sequence. But still it is weakly convergent in $L_2$, because from Bessel's inequality it follows that $\langle f,\sin{nx}\rangle\rightarrow 0$ for each $f\in L_2([0,\pi])$. Finally, I decided to check whether $\{\sin{nx}\}$ has a convergent a.e subsequence. If there is no such sequence, then it can not be convergent in $L_1$. And if there is such subsequence, then by Lebesgue DCT it will follow that it converges in $L_1$ (and probably it might be the limit of the whole sequence). I just didn't see that LDCT will work again for $L_2$ : if there is a convergent a.e subsequence, then it should converge in $L_2$ also, but this is impossible since $\{\sin{nx}\}$ is orthonormal (up tp a constant) and no subsequence of $\{\sin{nx}\}$ is Cauchy in $L_2$. This is actually the answer of @Julián Aguirre.

Now I have another Question: Is it possible to prove that there is no convergent pointwise a.e subsequence of $\{\sin{nx}\}$ using only first year calculus and knowing of course what is a set with Lebesgue measure $0$ in $\mathbb R$ ?

What is obvious is that for each $x$ there is a convergent subsequence since $\{\sin{nx}\}$ is bounded. But all $x\in [0,\pi]$ are uncountable set, so we can not apply for example Cantor diagonal argument.

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  • $\begingroup$ How about you show where you've got so far? meta.math.stackexchange.com/questions/1803/… $\endgroup$ – Nate Eldredge Jul 31 '15 at 14:54
  • $\begingroup$ I am in the train and write from my phone, so its very difficult now. $\endgroup$ – Svetoslav Jul 31 '15 at 15:05
  • $\begingroup$ It is not a homework. I just thought about this on my way. $\endgroup$ – Svetoslav Jul 31 '15 at 15:07
  • $\begingroup$ @Svetoslav Your question statement is a bit terse, but also, if you explain what you've tried and where you got stuck, this will help people understand your current math knowledge level and thus give you answers that are appropriate for your level. Also, as a general rule, many of the people who could answer your question won't be enthusiastic to do so unless you show some of your own efforts on how to solve the problem. Or at the very least, efforts on how you arrived at the problem. $\endgroup$ – user2566092 Jul 31 '15 at 15:09
  • $\begingroup$ You are absolutely right@user2566092. I am sorry for this, just I posted , because I was curious. $\endgroup$ – Svetoslav Jul 31 '15 at 15:16
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A different proof. If $\{\sin(n_k\,x)\}$ converges a.e., then $$ (\sin(n_{k+1}\,x)-\sin(n_k\,x))^2 $$ converges to $0$ a.e. By the dominated convergence theorem $$ \lim_{k\to\infty}\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=0, $$ but $$ \int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=\pi. $$

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  • $\begingroup$ Why is the last sentence equal to $1$? $\endgroup$ – kayak Jan 2 '17 at 5:32
  • $\begingroup$ Ups! It should be $\pi$. Thank you for noticing. $\endgroup$ – Julián Aguirre Jan 9 '17 at 12:05
  • $\begingroup$ Is there a way to modify this proof to show that $ \{\sin(nx)\} $ does neither have a pointwise (almost every) convergent subsequence on the open interval $ (0,1) $? $\endgroup$ – Apollo13 May 2 at 10:09
  • $\begingroup$ @Apollo13 It can be adapted if $n_{k+1}-n_k>2$. $\endgroup$ – Julián Aguirre May 2 at 16:05
  • $\begingroup$ @JuliánAguirre Thank you! I would be very curious to see the adaptation of the proof. Would you mind showing it to me? $\endgroup$ – Apollo13 May 9 at 0:12
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In fact $\sin (n_kx)$ diverges a.e. for every subsequence $n_k.$ Proof: Fix a subsequence $n_k$ and let $E= \{x\in \mathbb {R}: \lim_{k\to\infty}\sin (n_kx)\,\,\text {exists}\}.$ Let $f$ be the pointwise limit function on $E.$ Let $a> 0$ and put $E_a=E\cap [-a,a].$ On $E_a$ we have $f\in L^1\cap L^2.$ By the DCT,

$$\int_{E_a}f^2 = \lim_{k\to\infty} \int_{E_a}f\sin(n_kx)\,dx.$$

By the Riemann Lebesgue lemma, this limit is $0.$ But then we get

$$0= \int_{E_a}f^2 = \lim_{k\to\infty} \int_{E_a}\sin^2(n_kx) \,dx = \lim_{k\to\infty} \int_{E_a}\frac{1-\cos (2n_kx)}{2}\, dx = m(E_a)/2.$$

again using DCT and R-L. Therefore $m(E_a) = 0,$ and since $a$ was arbitrary, $m(E)=0.$

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  • $\begingroup$ I don't see where exactly in the second line of equations, starting with $ 0= \int_{E_a}f^2 = ... $, you made use of the DCT. I would be grateful if you could explain it to me! $\endgroup$ – Apollo13 May 2 at 9:56
  • $\begingroup$ The second = in the second line is because of the DCT. $\endgroup$ – zhw. May 2 at 16:26
  • $\begingroup$ Ah yes, of course! Thanks! $\endgroup$ – Apollo13 May 9 at 0:06

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