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Let $D\subset \mathbb{R}^n$ be a bounded domain. An extension map is $E_D: W^{p,k}(D)\to W^{p,k}(\mathbb{R}^n)$ satisfying:

(1) $E_D(f)(x)=f(x)$ for all $x\in D$,
(2) $\| E_D f\|_{W^{p,k}(\mathbb{R}^n)} \le K(D,p,k) \| f\|_{W^{p,k}(D)}$.

Thus, $K(D,p,k)$ is the norm of $E_D$.

From the answer of Tapio Rajala to this question:

Extension Operators for Sobolev spacesExtension Operators for Sobolev spaces

it would seem that the norm of $E_D$ depends on the domain $D$. I am interested in a family of annuli, $\Omega(r_0,r_1)\subset\mathbb{R}^4$, where the radii satisfy $R_0< r_0< r_1 < R_1$, but the distance $r_1-r_0$ is not bounded away from zero. Can we say that the norm of $E_D$ is independent of $r_0,r_1$ (though not necessarily independent of $R_i$)?

My intuition on this is very weak. For a constant function $c$, $$ \| c \|_{W^{2,k}(\Omega(r_0,r_1))} $$ will go to zero as the volume of $\Omega(r_0,r_1)$ does. I think of $\| E c\|_{W^{2,k}(\mathbb{R}^4)}$ as not going to zero with this volume because the extension needs to go from $c$ to $0$ so the derivative of the extension must be non-zero. However, I realize that the construction of the extension operator is more subtle than just using a cut-off function, so this intuition is probably wrong.

One approach to this question is through the paper "Quasiconformal mappings and extendability of functions in Sobolev spaces" by Peter W. Jones. He introduces the definition of an $(\epsilon,\delta)$ domain and proves, (Thm. 1 in that paper) that the norm of the extension operator for an $(\epsilon,\delta)$ domain depends only on $\epsilon,\delta,p,k,n$. Thus, if one could show that the annulus $\Omega(r_0,r_1)$ were an $(\epsilon,\delta)$ domain with $\epsilon$ and $\delta$ independent of $r_0$ and $r_1$, the independence of the norm of the extension operator would follow. In his paper, "Uniform Domains and the Ubiquitous Quasidisk", F. Gehring states (p.95 in the paragraph after equation 9.2) that a ball is a $(\pi/2,\infty)$ space. He does not state if that is any ball or just the unit ball --if it were any ball, then the discussion above on the constant function could be ignored. I have no intuition about how this $(\epsilon,\delta)$ domain definition works and would be grateful for some guidance on this.

Let $D\subset \mathbb{R}^n$ be a bounded domain. An extension map is $E_D: W^{p,k}(D)\to W^{p,k}(\mathbb{R}^n)$ satisfying:

(1) $E_D(f)(x)=f(x)$ for all $x\in D$,
(2) $\| E_D f\|_{W^{p,k}(\mathbb{R}^n)} \le K(D,p,k) \| f\|_{W^{p,k}(D)}$.

Thus, $K(D,p,k)$ is the norm of $E_D$.

From the answer of Tapio Rajala to this question:

Extension Operators for Sobolev spaces

it would seem that the norm of $E_D$ depends on the domain $D$. I am interested in a family of annuli, $\Omega(r_0,r_1)\subset\mathbb{R}^4$, where the radii satisfy $R_0< r_0< r_1 < R_1$, but the distance $r_1-r_0$ is not bounded away from zero. Can we say that the norm of $E_D$ is independent of $r_0,r_1$ (though not necessarily independent of $R_i$)?

My intuition on this is very weak. For a constant function $c$, $$ \| c \|_{W^{2,k}(\Omega(r_0,r_1))} $$ will go to zero as the volume of $\Omega(r_0,r_1)$ does. I think of $\| E c\|_{W^{2,k}(\mathbb{R}^4)}$ as not going to zero with this volume because the extension needs to go from $c$ to $0$ so the derivative of the extension must be non-zero. However, I realize that the construction of the extension operator is more subtle than just using a cut-off function, so this intuition is probably wrong.

One approach to this question is through the paper "Quasiconformal mappings and extendability of functions in Sobolev spaces" by Peter W. Jones. He introduces the definition of an $(\epsilon,\delta)$ domain and proves, (Thm. 1 in that paper) that the norm of the extension operator for an $(\epsilon,\delta)$ domain depends only on $\epsilon,\delta,p,k,n$. Thus, if one could show that the annulus $\Omega(r_0,r_1)$ were an $(\epsilon,\delta)$ domain with $\epsilon$ and $\delta$ independent of $r_0$ and $r_1$, the independence of the norm of the extension operator would follow. In his paper, "Uniform Domains and the Ubiquitous Quasidisk", F. Gehring states (p.95 in the paragraph after equation 9.2) that a ball is a $(\pi/2,\infty)$ space. He does not state if that is any ball or just the unit ball --if it were any ball, then the discussion above on the constant function could be ignored. I have no intuition about how this $(\epsilon,\delta)$ domain definition works and would be grateful for some guidance on this.

Let $D\subset \mathbb{R}^n$ be a bounded domain. An extension map is $E_D: W^{p,k}(D)\to W^{p,k}(\mathbb{R}^n)$ satisfying:

(1) $E_D(f)(x)=f(x)$ for all $x\in D$,
(2) $\| E_D f\|_{W^{p,k}(\mathbb{R}^n)} \le K(D,p,k) \| f\|_{W^{p,k}(D)}$.

Thus, $K(D,p,k)$ is the norm of $E_D$.

From the answer of Tapio Rajala to this question:

Extension Operators for Sobolev spaces

it would seem that the norm of $E_D$ depends on the domain $D$. I am interested in a family of annuli, $\Omega(r_0,r_1)\subset\mathbb{R}^4$, where the radii satisfy $R_0< r_0< r_1 < R_1$, but the distance $r_1-r_0$ is not bounded away from zero. Can we say that the norm of $E_D$ is independent of $r_0,r_1$ (though not necessarily independent of $R_i$)?

My intuition on this is very weak. For a constant function $c$, $$ \| c \|_{W^{2,k}(\Omega(r_0,r_1))} $$ will go to zero as the volume of $\Omega(r_0,r_1)$ does. I think of $\| E c\|_{W^{2,k}(\mathbb{R}^4)}$ as not going to zero with this volume because the extension needs to go from $c$ to $0$ so the derivative of the extension must be non-zero. However, I realize that the construction of the extension operator is more subtle than just using a cut-off function, so this intuition is probably wrong.

One approach to this question is through the paper "Quasiconformal mappings and extendability of functions in Sobolev spaces" by Peter W. Jones. He introduces the definition of an $(\epsilon,\delta)$ domain and proves, (Thm. 1 in that paper) that the norm of the extension operator for an $(\epsilon,\delta)$ domain depends only on $\epsilon,\delta,p,k,n$. Thus, if one could show that the annulus $\Omega(r_0,r_1)$ were an $(\epsilon,\delta)$ domain with $\epsilon$ and $\delta$ independent of $r_0$ and $r_1$, the independence of the norm of the extension operator would follow. In his paper, "Uniform Domains and the Ubiquitous Quasidisk", F. Gehring states (p.95 in the paragraph after equation 9.2) that a ball is a $(\pi/2,\infty)$ space. He does not state if that is any ball or just the unit ball --if it were any ball, then the discussion above on the constant function could be ignored. I have no intuition about how this $(\epsilon,\delta)$ domain definition works and would be grateful for some guidance on this.

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Tom Leness
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Dependence of norm of extension map on Sobolev spaces and $(\epsilon,\delta)$ domains

Let $D\subset \mathbb{R}^n$ be a bounded domain. An extension map is $E_D: W^{p,k}(D)\to W^{p,k}(\mathbb{R}^n)$ satisfying:

(1) $E_D(f)(x)=f(x)$ for all $x\in D$,
(2) $\| E_D f\|_{W^{p,k}(\mathbb{R}^n)} \le K(D,p,k) \| f\|_{W^{p,k}(D)}$.

Thus, $K(D,p,k)$ is the norm of $E_D$.

From the answer of Tapio Rajala to this question:

Extension Operators for Sobolev spaces

it would seem that the norm of $E_D$ depends on the domain $D$. I am interested in a family of annuli, $\Omega(r_0,r_1)\subset\mathbb{R}^4$, where the radii satisfy $R_0< r_0< r_1 < R_1$, but the distance $r_1-r_0$ is not bounded away from zero. Can we say that the norm of $E_D$ is independent of $r_0,r_1$ (though not necessarily independent of $R_i$)?

My intuition on this is very weak. For a constant function $c$, $$ \| c \|_{W^{2,k}(\Omega(r_0,r_1))} $$ will go to zero as the volume of $\Omega(r_0,r_1)$ does. I think of $\| E c\|_{W^{2,k}(\mathbb{R}^4)}$ as not going to zero with this volume because the extension needs to go from $c$ to $0$ so the derivative of the extension must be non-zero. However, I realize that the construction of the extension operator is more subtle than just using a cut-off function, so this intuition is probably wrong.

One approach to this question is through the paper "Quasiconformal mappings and extendability of functions in Sobolev spaces" by Peter W. Jones. He introduces the definition of an $(\epsilon,\delta)$ domain and proves, (Thm. 1 in that paper) that the norm of the extension operator for an $(\epsilon,\delta)$ domain depends only on $\epsilon,\delta,p,k,n$. Thus, if one could show that the annulus $\Omega(r_0,r_1)$ were an $(\epsilon,\delta)$ domain with $\epsilon$ and $\delta$ independent of $r_0$ and $r_1$, the independence of the norm of the extension operator would follow. In his paper, "Uniform Domains and the Ubiquitous Quasidisk", F. Gehring states (p.95 in the paragraph after equation 9.2) that a ball is a $(\pi/2,\infty)$ space. He does not state if that is any ball or just the unit ball --if it were any ball, then the discussion above on the constant function could be ignored. I have no intuition about how this $(\epsilon,\delta)$ domain definition works and would be grateful for some guidance on this.