# Thread: counting measure space inclusion

1. ## 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.

2. 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.

3. 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 $\displaystyle n \leq N$ ?

4. 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.

5. 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 $\displaystyle 0$.