Does an extension operator in Sobolev spaces commute with derivative operators? Assume that $\Omega\subseteq \mathbb R^d$ is open and has a Lipschitz boundary. Let $\tau\geq0$. Then we know that there exists a linear operator $E:H^\tau(\Omega)\to H^\tau(\mathbb R^d)$ such that for all $u\in H^\tau(\Omega)$ we have 
$Eu = u$, on $\Omega$,
$\|Eu\|_{H^\tau(\mathbb R^d)}\leq C_\tau \|u\|_{H^\tau(\Omega)}$,
where $C_\tau$ is a constant independent of $u$. The same extension $E$
works for all $\tau$. This was proved by E.M. Stein in 1971 for integer $\tau$ and by R.A. DeVore and R.C. Sharpley in 1993 for real $\tau$.
Now my question:
Do $E$ and $D^\alpha$ (weak derivative operator of order $\alpha\in \mathbb N_0^d$ for $|\alpha|\leq \tau$) commute? This means that
$$E(D^\alpha u) = D^\alpha(Eu),~~ for~ all~~ u\in H^{\tau}(\Omega)~~ and~~ |\alpha|\leq\tau.$$
If not, can such alternative extension be proved? 
Thank you for your consideration in advance. 
 A: This can never happen if $\Omega$ is bounded.
As a first remark, the extension operator $E:H^\tau(\Omega)\to H^\tau(\mathbb R^n)$ is not unique.
If $E$ is an extension operator with the desired properties and $\phi$ is a compactly supported smooth function with $\phi\equiv1$ on $\Omega$, then $Fu:=\phi Eu$ also satisfies the desired properties.
It is easy to check that at least $E$ and $F$ cannot both commute with weak partial derivatives.
Therefore the question seems to be whether there is an extension operator commuting with derivatives.
The answer is still no.
As a second remark, there are different extension operators for different orders $\tau$.
Let me restate the question (in the hope that this is what you wanted):

Let $\Omega\subset\mathbb R^n$ be a bounded Lipschitz domain.
  Fix an integer $\tau>0$.
  Does there exist a family of linear extension operators $H^\sigma(\Omega)\to H^\sigma(\mathbb R^n)$, $0\leq\sigma\leq\tau$ so that
  
  
*
  
*$E^\sigma u=u$ on $\Omega$ for all $\sigma\leq\tau$,
  
*$\|E^\sigma u\|_{H^\sigma(\mathbb R^d)}\leq C_\sigma \|u\|_{H^\sigma(\Omega)}$ for all $\sigma\leq\tau$, and
  
*$D^\alpha (E^\sigma u)=E^{\sigma-|\alpha|}(D^\alpha u)$ for all $\sigma\leq\tau$?
  

There is no such family of extension operators, so even the operator that extends at all degrees does not work.
Now let $u(x)=1$ (a constant function).
Since $\Omega$ is bounded, we have $u\in H^\tau(\Omega)$.
Now $D^\alpha u=0$ whenever $|\alpha|>0$.
Thus $D^\alpha(E^{\tau}u)=E^{\tau-1}(D^\alpha u)=0$ whenever $|\alpha|=1$.
This means that the weak gradient of $E^\tau u$ vanishes in $\mathbb R^n$ and thus it is constant.
This constant function must be in $H^\tau(\mathbb R^n)\subset L^2(\mathbb R^n)$, so it must be identically zero.
Since $E^\tau u=0$ in the whole space, we have $1=u=E^\tau u=0$ on $\Omega$.
This is impossible.
If you replace $\mathbb R^n$ with a nice bounded set containing $\Omega$, this argument is no longer valid and the question becomes much harder.
