# counting measure space inclusion

#### 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.

#### Jose27

MHF Hall of Honor
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.

willy0625

#### 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 $$\displaystyle n \leq N$$ ?

#### Jose27

MHF Hall of Honor
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
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 $$\displaystyle 0$$.