Let $1 > \alpha > 0$ and fix some $C > 0$. Consider $\Omega \subset \mathbb{R}^n$ a bounded domain and $Y \subset \Omega$ a discrete (finite) set of points. For $f: Y \to \mathbb{R}$ define $$\|f\|_{C^\alpha(Y)} := \sup_{y \in Y} |f(y)| + \sup_{y\neq z \in Y} \frac{|f(y) - f(z)|}{|y - z|^\alpha}.$$ Is there an extension map $E: C^\alpha(Y) \to H^s(\Omega)$, where $s = n/2 + \alpha$, such that $Ef|_Y = f$, and there is a $C_1 = C_1(C, \Omega) > 0$ (independent of $Y$) $$\|f\|_{C^\alpha(Y)} \leq C \implies \|Ef\|_{H^s(\Omega)} \leq C_1.$$ We require that $C_1$ is independent of $Y$; here $H^s$ denotes the Sobolev space of index $s$. In case of a negative answer, what is the optimal Sobolev exponent $s$ one can get on the right hand side (we always require the extension to be continuous so we may define the restriction)?
Some comments about the problem. Firstly, one can easily check that the extension defined as $$Ef(x) := \min_{y \in Y} \Big(f(y) + |x - y|^\alpha \|f\|_{C^\alpha(Y)}\Big)$$ satisfies the required properties as a map $E: C^\alpha(Y) \to C^\alpha(\Omega) \subset H^{\alpha - \varepsilon}(\Omega)$, for any $\varepsilon > 0$. So the question has positive answer for $s = \alpha - \varepsilon$.
Also, by Sobolev embedding theorems, we know that $H^{n/2 + \alpha} (\Omega) \subset C^\alpha(\Omega)$. So we expect there should be a better Sobolev exponent for some other choice of extension.
Finally, if the answer to my question is negative, it would be interesting to consider the extension map $E$ into spaces $H^s(\Omega)$, where $s \geq 1$ (say for $n \geq 2$). If $s = 1$ one could expect a ``harmonic extension" to be optimal; also, in this case if one takes a piecewise-linear extension on the triangulation determined by points in $Y$, I believe the answer is positive if $\alpha \geq \frac{1}{2}$. However this is not satisfactory, as in the application I have in mind $\alpha$ is very small.
References to literature are very welcome!
