- #1
Jaggis
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Does the sequence \{f_n\}=\{\cos{(2nt)}\} converge or diverge in Banach space C(-1,1) endowed with the sup-norm ||f||_{\infty} = \text{sup}_{t\in (-1,1)}|f(t)|?
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 C(-1,1) is a normed space. It would remain to show that \{f_n\} is not a Cauchy sequence.
For indexes n and n+1 it holds:
||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)}|.
But how to see that this won't become smaller and smaller for n large enough, exactly? What I tried next was to argue that if \{\cos{(2nt)}\} converges, it must converge towards some constant value that cosine can have (i.e. a number from the interval [-1,1]) because there is no limit of the form \cos{(2kt)} for any k \in N. Since ||\cos{(2nt)}||_{\infty} = 1 for all indexes, the constant must be f = 1 or f = -1 because the limit function must have the same norm as all other members of the sequence. But then ||f_n - f||_{\infty} = ||\cos{(2nt)}\pm 1||_{\infty} = 2 for all indexes and thus \{f_n\} is not a Cauchy sequence. I'm not sure whether my thinking here is correct.
My other theory was to fix t \in (-1,1) and consider indexes n and m, for whom n< m, as degrees of freedom. Then it holds that
||f_n(t)-f_{m}(t)||_{\infty} = \text{sup}|\cos{(2nt)} - \cos{(2mt)}|
and there are arbitrarily many (especially large) n and m for whom \text{sup}|\cos{(2nt)} - \cos{(2mt)}| = 2 for any fixed t \in (-1,1) . Since n, the smaller of the indexes, can get larger and larger while ||f_n(t)-f_{m}(t)||_{\infty} = 2 still holds, the sequence \{f_n\} cannot be a Cauchy sequence.
What do you think?
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 C(-1,1) is a normed space. It would remain to show that \{f_n\} is not a Cauchy sequence.
For indexes n and n+1 it holds:
||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)}|.
But how to see that this won't become smaller and smaller for n large enough, exactly? What I tried next was to argue that if \{\cos{(2nt)}\} converges, it must converge towards some constant value that cosine can have (i.e. a number from the interval [-1,1]) because there is no limit of the form \cos{(2kt)} for any k \in N. Since ||\cos{(2nt)}||_{\infty} = 1 for all indexes, the constant must be f = 1 or f = -1 because the limit function must have the same norm as all other members of the sequence. But then ||f_n - f||_{\infty} = ||\cos{(2nt)}\pm 1||_{\infty} = 2 for all indexes and thus \{f_n\} is not a Cauchy sequence. I'm not sure whether my thinking here is correct.
My other theory was to fix t \in (-1,1) and consider indexes n and m, for whom n< m, as degrees of freedom. Then it holds that
||f_n(t)-f_{m}(t)||_{\infty} = \text{sup}|\cos{(2nt)} - \cos{(2mt)}|
and there are arbitrarily many (especially large) n and m for whom \text{sup}|\cos{(2nt)} - \cos{(2mt)}| = 2 for any fixed t \in (-1,1) . Since n, the smaller of the indexes, can get larger and larger while ||f_n(t)-f_{m}(t)||_{\infty} = 2 still holds, the sequence \{f_n\} cannot be a Cauchy sequence.
What do you think?
