(Reposted from [MSE](https://math.stackexchange.com/questions/4720776/weighted-sobolev-spaces-and-decay) after no responses)
Introduce the following weighted Sobolev space norm on $\mathbb{R}^n$ (common in the study of hyperbolic PDE):
$$
\|u\|\_{H\_{k,\delta}}^2 = \sum\_{0 \leq i \leq k} \int\_{\mathbb{R}^n} \langle x \rangle^{2(\delta + i)}|\nabla^i u|^2 \, dx.
$$
This is found (for instance; I'm sure they're found in other places) in Christodoulou & Klainerman's stability of Minkowski spacetime, and in Choquet-Bruhat's book on general relativity and the Einstein equations. Here, $\langle x \rangle = (1 + |x|^2)^{1/2}$.


They behave nicely and have their own set of embedding theorems (see Choquet-Bruhat, *General Relativity and the Einstein Equations*, Appendix I, Theorem 3.4). The idea of their construction seems to be that $|u|$ has a certain (integrated) decay rate (related to $\delta$), and each derivative of $u$ behaves one power of decay better.


Let us focus on the case $\delta = 0$, so $u \in H\_{k, 0}$. This means, in particular, that $\langle x \rangle^\delta u \in L^2$. (However, this is not the same as the usual Sobolev space $H^k$, since recall each derivative of $u$ gains a power of decay in our case. This does not happen with classical Sobolev spaces.). Now, from learning about power functions, integration, and the $p$-test in calculus class, one might expect this (heuristically) to imply that $|u| \lesssim \langle x \rangle^{-n/2}$, as this is the "borderline" case of integrability. Of course, this is in general false. But the weighted Sobolev embeddings tell us that if $k > n/2$, then in fact
$$
\sup\_{x \in \mathbb{R}^n} \langle x \rangle^{n/2} |u| \lesssim \|{u}\|\_{H\_{k,\delta}}.
$$
So we recover the decay one might heuristically have expected!


**My question:** It seems to me like the derivative estimates (i.e. the fact that $\langle x \rangle^{i}\nabla^i u \in L^2$) did not come into the estimate here: in fact, if $u \in H\_{100, 0}$ versus $u \in H\_{100000,0}$, one cannot improve the decay of $u$ just by having "more derivatives". (For instance let $u(x) = \langle x \rangle^{-q}$.) It seemed like once you had $u \in H\_{k,0}$ with $k > n/2$, that was all you need, and any more derivatives are "useless." Does this seem right?


And on a related note, the decay could be guessed at from the zeroth-order behavior, i.e. just from $u \in L^2$ one expects $|u| \lesssim \langle x \rangle^{-n/2}$. The higher derivatives seemed like they were just there to rule out highly irregular behavior, like narrow/tall spikes that might spoil this. So, my **second question:** is the same embedding true if one just assumes $u \in L^2$ and $u \in H^k$ with $k > n/2$, in the sense of classical Sobolev spaces? That is, can the Sobolev embeddings for classical Sobolev spaces (without weights) be improved to give *decay*?



Weighted Sobolev Spaces and Decay
Question 1 (that higher derivatives are not used) is **yes**.


Question 2 (getting decay without weights) is **no**.


Without weights, let $u$ be a compactly supported smooth function. Let $f\_k(x) = u(x - k v) + u(x + kv)$ where $v$ is a unit vector. The family $f\_k$ is uniformly bounded in any classical $H^s$ space. But the family $f\_k$ is NOT uniformly decaying.




---


To better understand the behavior of these weighted Sobolev spaces, you want to first split it into a compact part ($|x| < 1$) and the remainder. In the compact part, you just use the standard Sobolev theory, since decay does not care about the compact region.


Outside, it is more convenient to think of these spaces as defined with respect to the differential operators $|x|\partial$ instead of with respect to $\partial$. The advantage of $|x|\partial$ is that they are scale invariant: if you consider mappings of $\mathbb{R}^n$ to itself given by $x\mapsto \lambda x$, then the operator $|x|\partial$ pushes forward to itself. This means that the $H\_{s,\delta}$ spaces are *scaling homogeneous*. (For standard Sobolev spaces, the $\mathring{H}^k$ portion has a different scaling from the $L^2$ portion.) It is this scaling homogeneity that enables us to get decay from Sobolev embedding.


(For comparison, the standard $H^s$ spaces are translation invariant, which the $H\_{s,\delta}$ spaces are not. Which you want to leverage depends on which problem you are solving.)


