# counting measure space inclusion

• May 9th 2010, 01:59 PM
willy0625
counting measure space inclusion
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.
• May 9th 2010, 03:17 PM
Jose27
Quote:

Originally Posted by willy0625
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.
• May 9th 2010, 03:53 PM
mabruka
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 $\displaystyle n \leq N$ ?
• May 9th 2010, 04:00 PM
Jose27
Quote:

Originally Posted by mabruka
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.
• May 9th 2010, 05:45 PM
Jose27
Quote:

Originally Posted by Jose27
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 $\displaystyle 0$.