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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


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    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
    |F(x) - F_n(x)| = |F(x) - S + S - S_n + S_n - F_n(x)|
    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


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    Scratch that: instead consider that, for all x,
    |F(x) - F_n(x)| = \sum_{k=n+1}^{\infty} f_k(x) \leq \sum_{k=n+1}^{\infty} s_k
    = |S - S_n|
    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.
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