Embeddings of Weighted Banach Spaces

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 $p< p '$ ?

• Are you looking for a direct proof, or asking if this follows easily from some more general principle? – Yemon Choi Feb 16 '10 at 0:56
• Hi Yemon, Well, in fact I would like to see a proof of this fact. I tried to write one for myself but I had no success, so I went here to ask for a help or hint, reference or even the proof. – Leandro Feb 16 '10 at 1:36

2 Answers

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".

• Yemon tahnk you very much for the answer, I read Bill's answer also but with my background in functional analysis I took some time to figure out what is happening, but after read your argument everything it was clear for me. Thanks again !! – Leandro Feb 16 '10 at 5:04
• I wasn't very clear in my last paragraph: I'll edit it now. Bill is talking about transforming your original question so that instead of looking at a simple operator (inclusion) between two complicated-seeming spaces, one looks at a slightly (but not very) complicated operator on a simple space, namely l^R. – Yemon Choi Feb 17 '10 at 8:37
• I get stuck about the last remark of your answer. Suppose that $\Omega_p$ now, is given as above but the weights are replaced by a sequence $p(i)\in C_0(\mathbb{Z}^d)$ and for $\Omega_{p'}$, the weights are replaced by $p'(i)\in C_0(\mathbb{Z}^d)$. If $p'(i)\leq p(i)$ then the embedding $\Omega_p\hookrightarrow \Omega_{p'}$ is well defined. Let's suppose also the the sequence decay monotonically, however if $$0<\liminf_{i\in\mathbb{Z}^d} \frac{|p(i)|}{|p'(i)|}$$ Your norm estimates does not works anymore. even in this case the Bill's argument is in hold ? Thanks again – Leandro Feb 17 '10 at 8:46
• Does my updated version answer your question? In the case you refer to, the ratio of the weights does not lie in $c_0$, and so you are right that the estimates don't work: indeed, this case, the embedding is NOT compact. – Yemon Choi Feb 17 '10 at 9:03
• As regards editing the original question: I think you should be able to edit it yourself. Personally I would prefer that you delete as little as possible, and instead add a new part; that way one can see what the original answers of Bill Johnson and myself are referring to. – Yemon Choi Feb 17 '10 at 9:07

If I understand the question, there is almost nothing to prove. It looks to me that you are looking at a diagonal operator from $\ell_R$ to itself where the diagonal entries go to zero. (Each space $\Omega_p$ is isometrically isomorphic to $\ell_R$ via the obvious diagonal map.)

• I agree that this is why the result works, but I've taken the liberty of writing out a more laborious version of this approach, which might be easier to follow for people with (presumably) less func-an background. Hope you don't mind (I started typing it up before I saw your answer). – Yemon Choi Feb 16 '10 at 2:05
• Of course I don't mind. I just assumed that a math physicist, seeing the proper formulation of his question, would understand immediately. – Bill Johnson Feb 16 '10 at 2:20
• You might well be right. In any case: between my verbose version and your economical explanation, the original poster will hopefully extract what he's after. – Yemon Choi Feb 16 '10 at 2:22