Results 1 to 5 of 5

Math Help - counting measure space inclusion

  1. #1
    Junior Member
    Joined
    Mar 2010
    From
    Melbourne
    Posts
    30
    Thanks
    1

    counting measure space inclusion

    How would you prove that for 1\leq p<s<\infty

    L^{p}(\mathbb{N},\cal{S},\mu) \subseteq  L^{s}(\mathbb{N},\cal{S},\mu)

    where \cal{S} is a sigma algebra of \mathbb{N} and \mu is a counting measure.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by willy0625 View Post
    How would you prove that for 1\leq p<s<\infty

    L^{p}(\mathbb{N},\cal{S},\mu) \subseteq  L^{s}(\mathbb{N},\cal{S},\mu)

    where \cal{S} is a sigma algebra of \mathbb{N} and \mu is a counting measure.
    Take elements and notation as in your post, then if (x_n) \in L ^p then, eventually, |x_n|<1, and note that the function f:[1,\infty ) \rightarrow \mathbb{R} given by f(t)=a^t is decreasing iff a\in (0,1) and so, eventually, for n big enough you get |x_n|^s \leq |x_n|^p.

    Another interesting question would be to see if the inclusion mapping is bounded.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    One question:

    So you have |x_n|^s \leq |x_n|^p for n\geq N .

    What if some of the first N-1 terms of x are greater than 1, so that |x_n|^s\not \leq |x_n|^p for some n \leq N ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by mabruka View Post
    One question:

    So you have |x_n|^s \leq |x_n|^p for n\geq N .

    What if some of the first N-1 terms of x are greater than 1, so that |x_n|^s\not \leq |x_n|^p for some n \leq N ?
    That's exactly why I can't conclude the boundedness of the inclusion mapping, but as far as set inclusion goes it suffices because only finitely many terms aren't dominated.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by Jose27 View Post
    Another interesting question would be to see if the inclusion mapping is bounded.
    I think I have an answer so if anyone's interested here it goes:

    Since the inclusion is clearly linear we only have to prove that it's continous at 0. Take (x_n)_m \subset L^p such that (x_n)_m \rightarrow 0 as m\rightarrow \infty. There exists an M such that if m>M then \| (x_n)_m \|_p ^p <1 so we must have |x_{n,m} | <1 for all n\in \mathbb{N} and m>M, but by the argument used to prove the inclusion we then have 0\leq \| (x_n)_m\| _s^s\leq \| x\| _p^p \rightarrow 0 so that it is continous at 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counting using inclusion/exclusion principle.
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 12th 2011, 09:26 AM
  2. [SOLVED] Counting using inclusion/exclusion principle.
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 10th 2011, 08:50 AM
  3. Limit of this counting measure
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 25th 2011, 07:21 PM
  4. Counting, Inclusion-Exclusion Rule
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 11th 2011, 08:17 PM
  5. measure space, sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 6th 2009, 03:06 PM

Search Tags


/mathhelpforum @mathhelpforum