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?

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# A Convergence of a cosine sequence in Banach space

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