Restriction of fractional Sobolev "function" of negative order to subset

Solution 1:

For your definition of negative order Sobolev spaces, restriction will not always be possible for $s=-1/2,-3/2,\dots$.

This is remedied by slightly altering the definition: For $s\ge 0$ define $H^s_{\bar \Omega}(\Omega)\subset H_0^s(\Omega)$ as the space of functions for which the zero extension lies in $H^s(\mathbb{R^n})$ (here we have $"=" $ if $s\neq 1/2,3/2,\dots$, as you noted correctly). Then define $$ H^{-s}(\Omega):=H_{\bar \Omega}^s(\Omega)^*,\quad s\ge 0.\tag{1} $$ A restriction operator can be constructed by the method you suggest; in particular, we have $$ H^{s}(\Omega)=\{u\in \mathcal{D}'(\Omega):u=U\vert_{\Omega} \text{ for some } U\in H^s(\mathbb{R}^n)\},\quad s\in \mathbb{R}. \tag{2} $$ One can also view (2) as definition (valid for all $s\in \mathbb{R}$), that does not require any interpolation or duality. This viewpoint is taken in Taylor's PDE book; then (1) becomes a theorem (Exercise 18 in Chapter 4.5).

As an illustration, consider $H^{-1/2}(\Omega)$ (with the definition from above) vs. $H^{-1/2}_*(\Omega)=H^{1/2}_0(\Omega)^*$ (your definition). Note that $H^{-1/2}_*(\Omega)\subset H^{-1/2}(\Omega)$ is a proper subset; so for nested opens $U\subset V\subset \mathbb{R}^n$ we can always restrict $f\in H^{-1/2}_*(V)$ to an element $f\vert_U\in H^{-1/2}(U)$, but it is not guaranteed that $f\vert_U$ will lie in the smaller space $H^{-1/2}_*(U)$. This is indeed reasonable: the stronger condition $f\in H_*^{-1/2}(V)$ means that one can make sense of $\langle f, \varphi\rangle$, even for test functions $\varphi\in H^{1/2}_0(V)=H^{1/2}(V)$; this poses a condition on the behaviour of $f$ at the boundary $\partial V$. Of course, there is no reason to assume that $f\vert_{U}$ has a similar boundary behaviour at $\partial U$; after all we are talking about a different boundary now.