1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bounded Sequence of Functions

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data
    fn is a sequence of functions and sn is a sequence of reals such that 0 ≤ fn(x) ≤ sn for all x.
    I want to show that if [itex]\sum_{k=0}^{n}s_k[/itex] is Cauchy then [itex]\sum_{k=0}^{n}f_k[/itex] is uniformly Cauchy and that if [itex]\sum_{k=0}^{\infty}s_k[/itex] converges then [itex]\sum_{k=0}^{\infty}f_k[/itex] converges uniformly.

    2. Relevant equations



    3. The attempt at a solution
    If [itex]\sum_{k=0}^{n}s_k[/itex] is Cauchy then that means there exists an N such that [itex]\left|\sum_{k=0}^{n}s_k-\sum_{k=0}^{m}s_k\right|<\epsilon[/itex] for all [itex]\epsilon[/itex] where m,n >N.
    Also [itex]\sum_{k=0}^{n-1}f_k\leq \sum_{k=0}^{n}f_k[/itex] for all n because every fn is at least zero and [itex]\sum_{k=0}^{n}f_k\leq \sum_{k=0}^{n}s_k[/itex].
    I guess I'm missing how to put these pieces together.
     
  2. jcsd
  3. Dec 15, 2012 #2

    pasmith

    User Avatar
    Homework Helper

    There's a fact about real Cauchy sequences which you should know: a real sequence converges if and only if it is Cauchy (if you don't know that, try to prove it for yourself).

    If [itex]\sum s_n[/itex] converges, then [itex]\sum f_n(x)[/itex] converges for all x, because for all x every term is positive and less than or equal to the corresponding term of [itex]\sum s_n[/itex].

    Just for convenience I'll define [itex]S = \sum_{k=0}^{\infty} s_k[/itex], [itex]S_n = \sum_{k=0}^n s_k[/itex], [itex]F(x) = \sum_{k=0}^{\infty} f_k(x)[/itex] and [itex]F_n(x) = \sum_{k=0}^n f_k(x)[/itex].

    You want to show that [itex]F_n \to F[/itex] uniformly, the definition of which is that for all [itex]\epsilon > 0[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]x[/itex], if [itex]n \geq N[/itex] then [itex]|F(x) - F_n(x)| < \epsilon[/itex]. So you might like to consider
    [tex]
    |F(x) - F_n(x)| = |F(x) - S + S - S_n + S_n - F_n(x)|
    [/tex]
    and recall the definition of convergence of [itex]S_n \to S[/itex].

    You may then want to satisfy yourself that if [itex]F_n \to F[/itex] uniformly then [itex]F_n[/itex] is uniformly Cauchy (and vice versa).
     
  4. Dec 16, 2012 #3

    pasmith

    User Avatar
    Homework Helper

    Scratch that: instead consider that, for all x,
    [tex]
    |F(x) - F_n(x)| = \sum_{k=n+1}^{\infty} f_k(x) \leq \sum_{k=n+1}^{\infty} s_k
    = |S - S_n|
    [/tex]
    and recall the definition of convergence of [itex]S_n \to S[/itex].
     
  5. Dec 16, 2012 #4
    That was very helpful. Thanks for reminding me about the relationship between Cauchy and convergence of real sequences. Don't know how I overlooked that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Bounded Sequence of Functions
  1. Bounded sequences (Replies: 4)

  2. Bounded sequence (Replies: 1)

  3. Bound Sequences (Replies: 26)

  4. Sequence Bound (Replies: 13)

Loading...