Wikipedia:Reference desk/Archives/Mathematics/2012 August 14

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August 14

Convergence Question

Consider the following complex-valued two-way sequences: $S=...,a_{-2},a_{-1},a_{0},a_{1},a_{2},...$ with the property that $\sum _{n\in \mathbb {Z} }|a_{n}|^{2}$ converges. Suppose you have a sequence of such sequences $S_{1},S_{2},S_{3},...$ which and converges to $...,b_{-2},b_{-1},b_{0},b_{1},b_{2}...$ Show that $\sum _{n\in \mathbb {Z} }|b_{n}|^{2}$ converges too. Widener (talk) 12:34, 14 August 2012 (UTC)

How are you defining convergence in the space of sequences? Rckrone (talk) 21:48, 14 August 2012 (UTC)

They're convergent sequences, so presumably you just identify them with their limits. Any other definition would seen strange to me. --Tango (talk) 22:49, 14 August 2012 (UTC)

I think it means pointwise convergence. So if $S_{i}=...,a_{-2}^{i},a_{-1}^{i},a_{0}^{i},a_{1}^{i},a_{2}^{i},...$, then

$b_{n}=\lim _{i}a_{n}^{i}$. Of course, in this interpretation, the stated result is false. Define $a_{n}^{i}=1$ if $|n|<i$ and $=0$ otherwise.--121.73.35.181

Sorry. It does not mean pointwise convergence; it means convergence under the metric $d(S_{i},S_{j})=\sum _{k\in \mathbb {Z} }|a_{k}^{i}-a_{k}^{j}|^{2}$. --Widener (talk) 00:06, 15 August 2012 (UTC)(talk) 23:54, 14 August 2012 (UTC)

You should think of using the triangle inequality with $b_{n}=a_{n}^{i}-(a_{n}^{i}-b_{n})$. Sławomir Biały (talk) 01:40, 15 August 2012 (UTC)

If it helps: a sequence $a_{n}$ such that $\sum |a_{n}|^{2}$ converges must converge to ____? Sławomir Biały (talk) 22:41, 14 August 2012 (UTC)

I misunderstood your question. This observation will not help. Sławomir Biały (talk) 01:37, 15 August 2012 (UTC)

So you are dealing with a convergent sequence $S_{i}$ in the Hilbert space $\ell ^{2}(\mathbb {Z} )$. A convergent sequence is bounded, and the norm is continuous. So $\|S\|_{2}=\lim _{i\to \infty }\|S_{i}\|_{2}<+\infty$ (which you can prove directly by the triangle inequality as suggested above). --pma 09:39, 15 August 2012 (UTC)

Thanks ! 00:28, 16 August 2012 (UTC)Widener (talk)

Example of Limitation of Riemann integration

How do You show That $\sum _{k=2}^{\infty }{\frac {1}{\log(k)}}\sin(kx)$ is Not a Fourier series Of a Riemann integrable Function? Widener (talk) 14:36, 14 August 2012 (UTC)

Why don't you start by trying to explain why

${\frac {\pi ^{2}}{3}}+\sum _{k=1}^{\infty }{\frac {4(-1)^{k}}{k^{2}}}\,\cos(kx)\,$

is a Fourier series of a Riemann integrable function? — Fly by Night (talk) 17:17, 14 August 2012 (UTC)

Is it because the sequence of Fourier coefficients is not in $l_{2}(\mathbb {Z} )$? Widener (talk) 23:06, 14 August 2012 (UTC)

Yes. Sławomir Biały (talk) 00:19, 15 August 2012 (UTC)