counting measure space inclusion

Mar 2010
32
3
Melbourne
How would you prove that for \(\displaystyle 1\leq p<s<\infty\)

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

where \(\displaystyle \cal{S}\) is a sigma algebra of \(\displaystyle \mathbb{N}\) and \(\displaystyle \mu\) is a counting measure.
 

Jose27

MHF Hall of Honor
Apr 2009
721
310
México
How would you prove that for \(\displaystyle 1\leq p<s<\infty\)

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

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

Another interesting question would be to see if the inclusion mapping is bounded.
 
  • Like
Reactions: willy0625
Jan 2010
150
29
Mexico City
One question:

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

What if some of the first N-1 terms of x are greater than 1, so that \(\displaystyle |x_n|^s\not \leq |x_n|^p\) for some \(\displaystyle n \leq N \) ?
 

Jose27

MHF Hall of Honor
Apr 2009
721
310
México
One question:

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

What if some of the first N-1 terms of x are greater than 1, so that \(\displaystyle |x_n|^s\not \leq |x_n|^p\) for some \(\displaystyle n \leq N \) ?
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.
 

Jose27

MHF Hall of Honor
Apr 2009
721
310
México
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 \(\displaystyle 0\). Take \(\displaystyle (x_n)_m \subset L^p\) such that \(\displaystyle (x_n)_m \rightarrow 0\) as \(\displaystyle m\rightarrow \infty\). There exists an \(\displaystyle M\) such that if \(\displaystyle m>M\) then \(\displaystyle \| (x_n)_m \|_p ^p <1\) so we must have \(\displaystyle |x_{n,m} | <1\) for all \(\displaystyle n\in \mathbb{N}\) and \(\displaystyle m>M\), but by the argument used to prove the inclusion we then have \(\displaystyle 0\leq \| (x_n)_m\| _s^s\leq \| x\| _p^p \rightarrow 0\) so that it is continous at \(\displaystyle 0\).