# 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.

One way to see that your intuition is correct (without using any theorems) is to estimate the $W^{2,k}$-norm of the extension radially. Because the extension is a Sobolev function it is absolutely continuous on almost every ray. Let $f$ be a constant function $1$. Now, taking into consideration only the value of the extension and its derivative we already get the desired estimate. Take a ray on which the extension is absolutely continuous and consider the largest radius (here I write only the radial coordinate) $0 \le r < r_0$ for which $Ef(r) = \frac12$ (if it exists). Then by Hölder's inequality $$\int_r^{r_0} (|D(Ef)|^2 + |Ef|^2) \ge \int_r^{r_0} \left(\left|\frac{dEf}{dr}\right|^2 + |Ef|^2\right) \ge \frac{1}{4}\left(|r_0-r|+\frac1{|r_0-r|}\right) \ge \frac14.$$ If there is no such $r$ then $$\int_0^{r_0} (|D(Ef)|^2 + |Ef|^2) \ge \frac{r_0}4.$$ Integrating this over all the directions gives you a constant lower-bound depending only on $r_0$ (or a smaller constant that depends only on $R_0$).
As you noted $$\| 1 \|_{W^{2,k}(\Omega(r_0,r_1))} \to 0, \qquad \text{ as } |r_1 - r_0| \to 0$$ giving you the dependence of the norm of the Sobolev extension operator on the radii $r_0$ and $r_1$.
Using some inequalities you can skip the radial estimates, but I think they show you nicely what is going on. I find it useful to first check radial estimates on any statement involving radial $\Omega$.
The annulus is an $(\epsilon, \delta)$-domain, but the parameters $\epsilon$ and $\delta$ depend on the radii $r_0$ and $r_1$.
• The reasoning simplify somewhat if to take an extention (for constant function 1) which depends on $r$ only? Such an extention can be obtained from an arbitrary one by averaging on shperes. Put $r_0=1$, $r_1=1+\varepsilon$. By stretching arguments (to annulus $1<r<2$) there seems to exisit an extention $f$ with estimates $\|\partial ^l f\|_{L_p(\mathbb R^n)}\le C \varepsilon^{(n-l)/p}$, $|l|\le k$. So norms of some of the derivatives yet tends to zero then $\varepsilon\to+0$. Jun 21, 2011 at 6:56