Let be $d$ a positive integer, $\Omega=\mathbb{R}^{\mathbb{Z}^d}$ and fix $R\geq 2$. We define weighted Banach spaces 


$$ \Omega\_p:=\left\{ x\in \Omega\left| \left[\sum\_{i\in\mathbb{Z}^d}\frac{|x\_i|^R}{(1+|i|)^p}\right]^{\frac{1}{R}} < \infty\right. \right\},\ \ \ \ \ \ \ p>d. $$


My question is why the embeddings 


$\Omega\_p \hookrightarrow \Omega\_{p'}$ are compact whenever $p< p ' $ ?



Embeddings of Weighted Banach Spaces
This is a special case of a much more general phenomenon, so I'm writing an answer which deliberately takes a slightly high-level functional-analytic POV; I think (personally) that this makes it easier to see the wood for the trees, even if it might not be the most direct proof. However, depending on your mathematical background it might not be the most helpful; so apologies in advance.


Anyway, start with a very general observation: let $E$ be a (real or complex) Banach space, and for each $n=1,2,\dots$ let $T\_n:E\to E$ be a bounded linear operator which has finite rank. (In particular, each $T\_n$ is a compact operator.)


Lemma: Suppose that the sequence $T\_n$ converges in the operator norm to some linear operator $T:E\to E$. Then $T$ is compact.


(The proof ought to be given in functional-analytic textbooks, so for sake of space I won't repeat the argument here.)


Now we consider these specific spaces $\Omega\_p$. Let $T:\Omega\_p\to\Omega\_{p'}$ be the embedding that you describe.


For each $n$, define $T\_n: \Omega\_p \to \Omega\_{p'}$ by the following rule:


$$ T\_n(x)\_i = x\_i \ {\rm if } \ |i| \leq n \ {\rm and} \ 0 \ {\rm otherwise} $$ 


Then each $T\_n$ has finite rank (because every vector in the image is supported on the finite set $\{ i \in {\mathbb Z}^d \vert \ \vert i\vert \leq n\}$). I claim that $T\_n$ converges to $T$ in the operator norm, which by our lemma would imply that $T$ is compact, as required.


We can estimate this norm quite easily (and indeed all we need is an upper bound). Let $x\in\Omega\_p$ have norm $\leq 1$; that is,
$$ \sum\_{i\in{\mathbb Z}^d} |x\_i|^R (1+ \vert i\vert)^{-p} \leq 1 $$


Then the norm of $(T-T\_n)(x)$ in $\Omega\_{p'}$ is going to equal $C^{1/R}$, where


$$ \eqalign{
C &:= \sum\_{i\in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p'} \\\\
 & = \sum\_{i \in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert i \vert)^{p-p'} \\\\ 
 & \leq \sum\_{i \in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'} \\\\ 
& \leq \sum\_{i \in {\mathbb Z}^d } |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'} 
 & = (1+\vert n \vert)^{p-p'}
}
$$


This shows that $\Vert T-T\_n\Vert \leq (1+\vert n\vert)^{(p-p')/R}$ and the right hand side can be made arbitrarily small by taking $n$ sufficiently large. That is, $T\_n\to T$ in the operator norm, as claimed, and the argument is complete (provided we take the lemma on trust).


Note that we used very little about the special nature of your weights. Indeed, as Bill Johnson's answer indicates, the only important feature is that in changing weight you in effect multiply your vector $x$ by a "multiplier sequence" which lies in $c\_0({\mathbb Z}^d)$, i.e. the entries "vanish at infinity".


**Edit 17-02-10:** the previous paragraph was perhaps slightly too terse. What I meant was the following: suppose that you have two weights $\omega$ and $\omega'$, such that the ration $\omega/\omega'$ lies in $c\_0({\mathbb Z}^d)$. Then the same argument as above shows that the corresponding embedding will be compact. Really, this is what Bill's answer was driving at: it isn't the weights which are important, it's the fact that the factor involved in *changing* weight is given by something "vanishing at infinity".


