1. Limited time only! Sign up for a free 30min personal 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!

Prove that l^1 is complete

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]\ell^1[/tex], the space of all (real) sequences [tex]v = \{v_k\}[/tex] such that [tex]\sum|x_k| < \infty [/tex], is complete.

    2. Relevant equations

    [tex]\ell^1[/tex] is a normed space with the norm [tex]||x|| = \sum |x_k|[/tex]

    3. The attempt at a solution

    Let [tex]v_n[/tex] be a Cauchy sequence of sequences in [tex]\ell^1[/tex]. Then for all [tex]\epsilon > 0[/tex] there exists N > 0 such that for all n,m > N we have [tex]\sum |v_{n,k} - v_{m,k}| < \epsilon[/tex] (here [tex]v_{n,k}[/tex] means the kth term of the nth sequence)

    in particular this means that [tex]|v_{n,k} - v_{m,k}| < \epsilon [/tex] so we can define a sequence [tex]u = \{u_k\}[/tex] as [tex]u_k = \lim v_{n,k}[/tex] as n goes to infinity.

    Now i think the sequence u would be the limit of [tex]v_n[/tex] as n goes to inifnity, but i'm not sure how to prove it. Firstly, I dont know how to prove that u converges absolutely. the problem is by the definition of u, given [tex]\epsilon[/tex] I can find a sequence v_n whose terms are each within [tex]\epsilon[\tex] from the corresponding term in u, but when summing, this is like summin [tex]\epsilon[\tex] infinitly many times.. so how can I do it? any hint would be appreciated :)
    Thanks.
     
  2. jcsd
  3. Jun 22, 2009 #2
    if we do this:
    [tex]\lim \sum |v_{n,k} - v_{m,k}| = \sum \lim |v_{n,k} - v_{m,k}| [/tex] where the limit is taken as m -> infinity, then we have [tex]\sum |v_{n,k} - u_{k}| < \epsilon [/tex] for sufficiently large n. But how can I justify interchanging the sum and the limit?
     
  4. Jun 22, 2009 #3
    Your u is good.

    There is a trick to interchanging the sum and the limit, which roughly speaking is this:

    (1) use epsilon,
    (2) change the bound on the infinite sum to a bound on a finite sum from k=1 to j,
    (3) note that the bound is independent of j,
    (4) take your limit as m approaches infinity (OK to do, since sum is finite),
    (5) bound is still independent of j, so now let j approach infinity.

    Here are the details, applied to the problem of showing u converges absolutely.

    (1) [tex]v_n[/tex] is Cauchy so it's bounded; thus [tex]||v_n||\le M[/tex] (using M instead of epsilon for this example)

    (2) for any j, [tex]\sum_{k=1}^j |v_{n,k}|\le M[/tex]

    (3) previous bound is independent of j

    (4) take limit as n (using n instead of m) approaches infinity, [tex]\sum_{k=1}^j |u_k|\le M[/tex]

    (5) bound is still independent of j, so let j approach infinity, [tex]||u||\le M[/tex].

    Now you can try it for the convergence.
     
  5. Jun 22, 2009 #4
    oh, I see! It's now very clear to me. Thank you so much. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook