Wikipedia:Reference desk/Archives/Mathematics/2012 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 $\backslash sum\_\{n\; \backslash in\; \backslash mathbb\{Z\}\}\; |a\_n|^2$ 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 $\backslash sum\_\{n\; \backslash in\; \backslash mathbb\{Z\}\}\; |b\_n|^2$ 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

$b\_n\; =\; \backslash lim\_i\; a\_n^i$. Of course, in this interpretation, the stated result is false. Define $a\_n^i\; =\; 1$ if and $=0$ otherwise.--121.73.35.181

::::Sorry. It does not mean pointwise convergence; it means convergence under the metric $d(S\_i,S\_j)\; =\; \backslash sum\_\{k\; \backslash in\; \backslash mathbb\{Z\}\}\; |a\_k^i-a\_k^j|^2$. --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 $\backslash sum\; |a\_n|^2$ 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 $\backslash ell^2(\backslash Z)$. A convergent sequence is bounded, and the norm is continuous. So (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)

How do You show That $\backslash sum\_\{k=2\}^\backslash infty\; \backslash frac\{1\}\{\backslash log(k)\}\backslash sin(kx)$ 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

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

: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 $l\_2(\backslash mathbb\{Z\})$? Widener (talk) 23:06, 14 August 2012 (UTC)

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