Does the sequence [itex] \{f_n\}=\{\cos{(2nt)}\}[/itex] converge or diverge in Banach space [itex] C(-1,1) [/itex] endowed with the sup-norm [itex] ||f||_{\infty} = \text{sup}_{t\in (-1,1)}|f(t)| [/itex]?(adsbygoogle = window.adsbygoogle || []).push({});

At first glance my intuition is that this sequence should diverge because cosine is a period function. But how to really prove that?

I have tried to approach the problem from the point of view that if the sequence converges it must be a Cauchy sequence, since [itex] C(-1,1) [/itex] is a normed space. It would remain to show that [itex] \{f_n\}[/itex] is not a Cauchy sequence.

For indexes [itex]n[/itex] and [itex]n+1[/itex] it holds:

[itex] ||f_n-f_{n+1}||_{\infty} = ||\cos{(2nt)} - \cos{(2(n+1)t)}||_{\infty} = \text{sup}_{t\in (-1,1)}|\cos{(2nt)} - \cos{(2(n+1)t)}| [/itex].

But how to see that this won't become smaller and smaller for [itex]n [/itex] large enough, exactly? What I tried next was to argue that if [itex]\{\cos{(2nt)}\}[/itex] converges, it must converge towards some constant value that cosine can have (i.e. a number from the interval [itex] [-1,1][/itex]) because there is no limit of the form [itex]\cos{(2kt)}[/itex] for any [itex]k \in N[/itex]. Since [itex] ||\cos{(2nt)}||_{\infty} = 1[/itex] for all indexes, the constant must be [itex] f = 1[/itex] or [itex] f = -1[/itex] because the limit function must have the same norm as all other members of the sequence. But then [itex] ||f_n - f||_{\infty} = ||\cos{(2nt)}\pm 1||_{\infty} = 2[/itex] for all indexes and thus [itex] \{f_n\}[/itex] is not a Cauchy sequence. I'm not sure whether my thinking here is correct.

My other theory was to fix [itex] t \in (-1,1) [/itex] and consider indexes [itex] n [/itex] and [itex] m [/itex], for whom [itex] n< m[/itex], as degrees of freedom. Then it holds that

[itex] ||f_n(t)-f_{m}(t)||_{\infty} = \text{sup}|\cos{(2nt)} - \cos{(2mt)}| [/itex]

and there are arbitrarily many (especially large) [itex] n [/itex] and [itex] m [/itex] for whom [itex]\text{sup}|\cos{(2nt)} - \cos{(2mt)}| = 2 [/itex] for any fixed [itex] t \in (-1,1) [/itex] . Since [itex]n[/itex], the smaller of the indexes, can get larger and larger while [itex] ||f_n(t)-f_{m}(t)||_{\infty} = 2 [/itex] still holds, the sequence [itex] \{f_n\}[/itex] cannot be a Cauchy sequence.

What do you think?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A Convergence of a cosine sequence in Banach space

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**