# Prove that l^1 is complete

1. Jun 22, 2009

### winter85

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$\ell^1$$ is a normed space with the norm $$||x|| = \sum |x_k|$$

3. The attempt at a solution

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

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

Now i think the sequence u would be the limit of $$v_n$$ 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 $$\epsilon$$ I can find a sequence v_n whose terms are each within $$\epsilon[\tex] from the corresponding term in u, but when summing, this is like summin [tex]\epsilon[\tex] infinitly many times.. so how can I do it? any hint would be appreciated :) Thanks. 2. Jun 22, 2009 ### winter85 if we do this: [tex]\lim \sum |v_{n,k} - v_{m,k}| = \sum \lim |v_{n,k} - v_{m,k}|$$ where the limit is taken as m -> infinity, then we have $$\sum |v_{n,k} - u_{k}| < \epsilon$$ for sufficiently large n. But how can I justify interchanging the sum and the limit?

3. Jun 22, 2009

### Billy Bob

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) $$v_n$$ is Cauchy so it's bounded; thus $$||v_n||\le M$$ (using M instead of epsilon for this example)

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

(3) previous bound is independent of j

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

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

Now you can try it for the convergence.

4. Jun 22, 2009

### winter85

oh, I see! It's now very clear to me. Thank you so much. :)