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Proving that l^p is properly contained in l^q for 1<=p<q<=infty

  • #1
michael.wes
Gold Member
36
0

Homework Statement


Prove that [tex]l^{p}\subsetneq l^q[/tex] for [tex]1\leq p < q \leq \infty[/tex].

Homework Equations





The Attempt at a Solution


I know how to prove the [tex]q=\infty[/tex] case, and how to find an example where the containment is proper (just take the harmonic series to the appropriate power), but I cannot find a way to prove the actual containment. I tried re-writing as follows, to try and use the cauchy-schwarz inequality, but I don't even know if it holds in an infinite dimensional vector space..

[tex]\sum_{n=1}^\infty |x_n|^q = \sum_{n=1}^\infty |x_n|^p|x_n|^{q/p}[/tex]

I'm trying to show that [tex]\sum_{n=1}^\infty |x_n|^p = M^p[/tex] for some real, finite M implies that [tex]\sum_{n=1}^{\infty}|x_n|^q[/tex] is bounded above by something which depends on this limit, and hence I can just raise both sides to the power 1/q, and I will have the containment. But like I said before, I have searched around but I could not find any hints.

Any help is appreciated!
M
 

Answers and Replies

  • #2
22,097
3,280
What about a simple comparison test for series?
 
  • #3
michael.wes
Gold Member
36
0
Thanks micromass. that was a helpful tip
 
  • #4
Bacle2
Science Advisor
1,089
10
I was reviewing this, and this post came up. It may give someone else a slightly-different
perspective.Sorry if necro-posting is not helpful --please let me know:

A tip related to micromass': Think of the function f(x)= ax for a constant;

where is it increasing/decreasing? Think , too, of a necessary condition for {an}

to be in lp re the partial sums as n→∞.
 

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