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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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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). --pm a 09:39, 15 August 2012 (UTC)[reply]

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)[reply]