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Willie Wong
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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.)

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.

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.)

Source Link
Willie Wong
  • 39.1k
  • 4
  • 94
  • 176

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.