Completeness in l^\infty Norm

  • Thread starter Euclid
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  • #1
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Show that the space [tex]c_0[/tex] of all sequences of real numbers that converge to
0 is a complete space with the [tex]l^\infty[/tex] norm. First I let [tex] A^j=\{a_k^j\}_{k=1}^\infty[/tex] be a sequence of sequences converging to zero and I assume that it is norm summable:
[tex]\sum \limits_{j=1}^\infty ||A^j||_\infty < \infty[/tex]
I argue that [tex] S= \sum \limits_{j=1}^\infty A^j[/tex] converges componentwise.
Then I want to show that the sequence S converges to 0 at infinity (and hence is in [tex]c_0[/tex]). Fix [tex]\varepsilon >0[/tex]. For each j, choose a [tex]K_j[/tex] such that [tex]k \geq K_j[/tex] implies [tex]|a_k^j| < \frac{\varepsilon}{2^j}[/tex]. Given [tex] N>0[/tex] let [tex]K=\sup\limits_{k\leq N} K_k[/tex]. Then for [tex]k\geq K[/tex] we have
[tex] |S_k|=|\sum \limits_{j=1}^{\infty}a_k^j| \leq \sum \limits_{j=1}^{N}|a_k^j|+\sum \limits_{j=N+1}^{\infty}|a_k^j|\leq \varepsilon + \sum \limits_{j=N+1}^{\infty}|a_k^j|[/tex].
Here is where I am stuck. I know the quantity on the right goes to 0 if I make N large enough, but putting that down rigorously presents a difficulty. Any ideas?
 
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  • #2
CarlB
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Can you choose another integer, say L_j, so that if j is larger than L_j, it makes that annoying summation on the end small enough?

Carl
 
  • #3
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Not really.... The best I can do is this:
[tex] |S_k|\leq 2\varepsilon + \sum \limits_{j=N+1}^{L}|a_k^j|[/tex]
but here L will depend on K, so if I want to make those finite number of terms small I'll have to increase K, and this may force me to change L, which will force me to increase K, etc.

Edit: I think I figured it out. I showed that l^\infty is complete by showing norm summable implies summable. Then I went back to the Cauchy sequence definition. Given a Cauchy sequence of convergent sequences, we know it has a limit in l^\infty. From there it's not too bad to show that the limit is itself convergent.
 
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