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Math Help - Sequence of functions

  1. #1
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    Sequence of functions

    Consider the sequence of functions f_n(t)=\cos^{2n}(n!\pi t), on the space (\mathcal C[0,1],d_\infty). Is f_n(t) convergent?

    Well I think this is a limit problem, but I don't know how to compute it, any ideas?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Consider the sequence of functions f_n(t)=\cos^{2n}(n!\pi t), on the space (\mathcal C[0,1],d_\infty). Is f_n(t) convergent?

    Well I think this is a limit problem, but I don't know how to compute it, any ideas?
    No. Note that if it was convergent then \displaystyle \lim f_n(t) would be continuous...yet I don't think it's hard to show that \lim f_n(t)=\mathbf{1}_{\mathbb{Q}}
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    yet I don't think it's hard to show that \lim f_n(t)=\mathbf{1}_{\mathbb{Q}}
    What do you mean with the right notation?

    How do you solve the problem?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    What do you mean with the right notation?

    How do you solve the problem?
    It is the indicator function--and no I can't show you how to solve the problem. This is your homework, me telling you how to solve it will mean that you might fail your test, drop out of school, etc. I can't handle that kind of guilt.

    Why don't you solve it, and we'll check it.
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    But I'm not actually asking you to solve me the problem completely, a hint could be a help.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    But I'm not actually asking you to solve me the problem completely, a hint could be a help.
    I gave you a big hint. If f_n\to f in C[0,1] then f must be continuous, but a quick check shows that f_n\to\mathbf{1}_{\mathbb{Q}} which is the indicator function on the rationals.
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