# counting measure space inclusion

• May 9th 2010, 01:59 PM
willy0625
counting measure space inclusion
How would you prove that for $1\leq p

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

where $\cal{S}$ is a sigma algebra of $\mathbb{N}$ and $\mu$ is a counting measure.
• May 9th 2010, 03:17 PM
Jose27
Quote:

Originally Posted by willy0625
How would you prove that for $1\leq p

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

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

Originally Posted by mabruka
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.
• 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 $0$. 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 $0$.