Dismiss Notice
Join Physics Forums Today!
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

  1. Jun 4, 2016 #1
    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]?

    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?
     
  2. jcsd
  3. Jun 4, 2016 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Does the sequence converge pointswise?
     
  4. Jun 4, 2016 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Examine what happens at [itex]t=\frac{\pi}{4}[/itex]. Value cycles 0,-1,0,1. Sequence can't converge.
     
  5. Jun 4, 2016 #4
    Ah! It is as simple as that. Yes, convergence would have to hold for all [itex] t \in [-1,1] [/itex] so it is enough to find an individual [itex]t[/itex] that won't satisfy it.
     
  6. Jun 5, 2016 #5
    but it is weakly convergent :)
     
  7. Jun 5, 2016 #6

    mathman

    User Avatar
    Science Advisor
    Gold Member

  8. Jun 5, 2016 #7
    No, it does not converge weakly in ##C(-1,1)##. It converge weakly to 0 in ##L^p##, ##1\le p <\infty##, and converges to 0 in weak-* topology in ##L^\infty##, all thanks to Riemann--Lebesgue Lemma.

    But there is no convergence in the weak topology of ##C(-1,1)##: if you consider the functional ##\delta=\delta_0## on ##C(-1,1)##, ##\delta(f)=f(0)##, then ##\delta(f_n)=1## for all ##n##. But for any functional ##\Phi## corresponding to an absolutely continuous measure, $$\Phi(f):= \int_{-1}^1 f(x)\phi(x) dx, \qquad \phi \in L^1(-1,01), $$ we have ##\Phi(f_n)\to 0##, again thanks to Riemann--Lebesgue Lemma.

    PS. And of course, if the sequence diverges in weak topology, it diverges in norm as well.
     
  9. Jun 6, 2016 #8
    O, indeed, even it has already been written:
     
    Last edited: Jun 6, 2016
  10. Jun 6, 2016 #9

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Your statement is confusing. What has the delta function have to do with anything?
     
  11. Jun 6, 2016 #10
    Delta function is a bounded linear functional on ##C(-1,1)##.
     
  12. Jun 7, 2016 #11

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Delta function is not bounded. Linear functionals consist of members of L1(-1,1). [itex]\lim_{n \to \infty}\int_{-1}^{1}g(t)cos(2nt)dt=0[/itex] for any L1 function g(t).
     
  13. Jun 7, 2016 #12
    Delta function is a bounded linear functional on ##C(-1,1)##: $$|\delta (f)| = | f(0)| \le \sup_{x\in(0,1)} |f(x)| =: \|f\|_{C(-1,1)}.$$
     
  14. Jun 8, 2016 #13

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

  15. Jun 8, 2016 #14

    mathman

    User Avatar
    Science Advisor
    Gold Member

    My misunderstanding. There is a concept of weak-* convergence, where L1 is used rather than the measures you described.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convergence of a cosine sequence in Banach space
Loading...