# Extending Hölder functions

I originally asked this question on MathStackExchange some time ago, but it seems that MathOverflow would be more appropriate. Essentially, I would like to find references for extension theorems for (higher order) Hölder spaces that would not come from the extension theorem for Sobolev spaces.

If $$(X, d)$$ is a metric space and $$A \subseteq X$$, McShane's Extension Theorem states that every $$L$$-Lipschitz function $$f \colon A \to \mathbb{R}$$ can be extended to $$L$$-Lipschitz function $$f' \colon X \to \mathbb{R}$$. By changing metric $$d$$ to $$d^{\alpha}$$, where $$\alpha \in (0,1)$$, we can make this statement about $$\alpha$$-Hölder functions. We see that there are no requirements on the regularity of $$A$$ (as in, we do not require anything about the boundary of $$A$$) to perform this extension.

Now, suppose $$X = \mathbb{R}^n$$ and we would like to extend $$f$$ which is of class $$C^{k, \alpha}(A)$$ to an element $$C^{k, \alpha}(X)$$. I'm aware that if $$A$$ is open then we can think of elements of $$C^{k, 1}(A)$$ as elements of the Sobolev space $$W^{k+1, \infty}(A)$$. Hence, we can have an extension for sufficiently regular $$A$$'s. However, I'm more interested in whether this regularity condition can be relaxed much like in the case $$k = 0$$. Is there an extension theorem for Hölder functions with weaker conditions on the regularity of $$A$$ than the ones from the Sobolev extension theorem?

• By changing metric $d$ to $d^{\alpha}$, where $\alpha \in (0,1)$, we can make this statement about $\alpha$-Holder functions. -- Perhaps I'm missing a subtle issue here, but I don't understand the purpose of this observation, since McShane proved his extension theorem for an arbitrary concave (down) modulus of continuity. And the result for Holder functions is not buried unnoticed within this more general result -- he explicitly states the result for Holder functions in Corollary 1 on p. 840. May 22 at 17:07
• I've written it mostly because I've usually seen the name of this theorem invoked in the context of Lipschitz functions (so I wasn't aware of the more general statement about the moduli --- thank you for pointing it out). Because of that, I've usually seen the result for the Holder functions derived as a consequence of the fact that when $d$ is a metric, then so is $d^\alpha$. (Hence the $\alpha$-Holder functions with respect to $d$ can be seen as Lipschitz with respect to $d^\alpha$). May 22 at 17:55
• A big part of the issue is that $C^{k,\alpha}$ functions need not have extensions to the boundary! According to standard definitions, the Heaviside function restricted to the open set $\mathbb{R}\setminus \{0\}$ is in $C^{1,1}$ but not in $C^{0,1}$. (Note a $C^{0,1}(A)$ function in a metric space automatically has an extension to $\bar{A}$ with the same Lipschitz constant.) And once you have extensions to the boundary for $A\subseteq\mathbb{R}^n$, you can use Whitney extension to finish the job. May 22 at 20:45
• Whitney extension theorem for Holder functions can be found in e.g. Stein's Singular Integrals book. // I think if you modify the definition of $C^{k,\alpha}$ to require all lower derivatives to be Lipschitz and the top derivative to be Holder, then there's a good chance that you can get an extension to the boundary with minimal regularity assumptions on $A$. May 22 at 20:52

I think that for the $$C^{k, \alpha}(A)$$ space, where $$A$$ is bounded and the regularity of its boundary is $$C^{k, \alpha}$$ as well, an extension theorem holds: see for example Theorem 4 in Brian Krumme's 2016 Hölder spaces lecture notes.