Whitney's extension theorem states that if $D \subset \mathbb{R}^n$ is closed and $f: D \to \mathbb{R}$ is $C^k$ in some sense to be specified below, then $f$ can be extended to $\mathbb{R}^n$ so that it is real analytic on $D^c$. Now let me specify what it means to be $C^k$ in the sense that we want - we will call this Whitney $C^k$.

We say that $f: D \to \mathbb{R}$ is Whitney $C^k$ (on $D$) if there exist functions $f_\alpha: D \to \mathbb{R}$, $|\alpha| \le k$ (note here $\alpha$ is a multindex with $n$ entries) s.t. for all $x,y \in D$ and $\alpha \in \mathbb{Z}^n$, $$f_\alpha (x) = \sum_{|\beta| \le k-|\alpha|} \frac{f_{\alpha + \beta}(y)}{\beta !} (x-y)^\beta +R_\alpha(x,y)$$ with $R_\alpha$ having the following property: Given any point $z \in D$ and any $\epsilon > 0$ there exists a $\delta > 0$ s.t. if $x, y \in A$ and $|z-x|< \delta$ and $|z-y|< \delta$ then $|R_\alpha (x,y)| \le \epsilon |x-y|^{k-|\alpha|}$.

Now for my question. I am wondering for what closed sets can we say that if $f$ is $C^k$ (see the Edit below for definition) then it is Whitney $C^k$. Phrased differently, I am wondering for what closed sets (with $C^k$ functions defined on these sets) can we apply Whitney's extension theorem.

I have been able to show that if $D$ is the closure on an open bounded convex set and $f: D \to \mathbb{R}$ is $C^k$ then $f$ is Whitney $C^k$. To see this one applies the multivariable Taylor's theorem (integral version- see here https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_multivariate_functions). Notice that the proof, and therefore the estimates, rely on integrating along a line segment - this is where convexity is used. I suspect this can be generalized to $D$ being the closure of a connected open set with suitably smooth boundary (i.e. $C^\infty$).

References are also appreciated.

Edit: For completeness, let me add the definition of $C^k$ on the closure of an open set. Let $U$ be an open set, then $f: \overline{U} \to \mathbb{C}$ is $C^k$ if all the partial derivative of $f$ up to order $k$ extend continuously to $\overline{U}$.

  • $\begingroup$ A bit vague the question. Say your set is $[0,1]$. What are the two definitions for a $C^{k}$ functions? $\endgroup$ – Orest Bucicovschi Jun 22 '15 at 20:21
  • $\begingroup$ @orangekid Let U be an open set, then $f: \overline{U} \to \mathbb{C}$ is $C^k$ if all the partial derivative of $f$ up to order $k$ extend continuously to $\overline{U}$. As for the particular example of $U = (0,1)$ we have a open bounded convex set, so if $f$ is $C^k$ on $\overline{U}$ then it is Whitney $C^k$. $\endgroup$ – wellfedgremlin Jun 23 '15 at 1:15
  • $\begingroup$ I see. I would rather consider $\bar U = C$ as a starting (closed) set , postulate that $\overline{Int C} = C$ and define a function to be $C^k$ on $C$ if it is is in the usual sense on $Int C$ and moreover the partials have a continuous extension to $C$ (clearly unique). As for Whitney $C^k$, a function is $C^k$ if it is the restriction of a usual $C^k$ function from a larger open subset. $\endgroup$ – Orest Bucicovschi Jun 23 '15 at 5:37
  • $\begingroup$ @orangeskid I'm really only interested in closed sets that are the closure of a domain. And yes, you're right. If $C$ is closed and $C \subset U$ where $U$ is open and $f$ is $C^k$ on $U$ then $f$ is Whitney $C^k$ on $C$. (I think that's what you meant to say.) But what I'm really looking for is a boundary condition for the closed set (such as $C^\infty$) so that if $f$ is $C^k$ then it is Whitney $C^k$. $\endgroup$ – wellfedgremlin Jun 23 '15 at 6:19

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