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A function $u : U \rightarrow \mathbb R$ is an element of the Hölder space $C^{\alpha}(U)$ if

  • $\sup\limits_{x \in U} |u(x)| < \infty$
  • $\sup\limits_{x,y \in U} \dfrac{|u(x) - u(y)|}{|x-y|^\alpha} < \infty$

A function $u : U \rightarrow \mathbb R$ is an element of the fractional Sobolev space $W^{\alpha,\infty}(U)$ if

  • $\mathrm{esssup}_{x \in U} |u(x)| < \infty$
  • $\mathrm{esssup}_{x,y \in U} \dfrac{|u(x) - u(y)|}{|x-y|^\alpha} < \infty$

Here, $\mathrm{esssup}$ denotes the essentially supremum.

Clearly, $C^\alpha(U)$ is a subspace of $W^{\alpha,\infty}(U)$. Does the converse statement hold, that is, $W^{\alpha,\infty}(U)$ is a subspace of $C^\alpha(U)$?

The claim is true if $\alpha = 1$ when $U$ has sufficient boundary regularity, in which case we just deal with Rademacher's theorem: a function is Lipschitz continuous if and only if it is essentially bounded with essentially bounded derivatives.

The generalization is claimed in the Encyclopedia of Math but no reference is given.

Does there exist a short proof or an explicit reference in the literature (paper, textbook)?

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    $\begingroup$ Technically one has to identify functions that agree a.e. before one gets the identity between $C^\alpha(U)$ and $W^{\alpha,\infty}(U)$ (even in the $\alpha=1$ case), but yes, this statement is true. Given $u \in W^{\alpha,\infty}(U)$, standard mollifiers $u * \varphi_\varepsilon$ will be uniformly bounded in $C^\alpha(U)$ and converge uniformly to a limit $\tilde u \in C^\alpha(U)$ that is equal a.e. to $u$. $\endgroup$
    – Terry Tao
    Commented Aug 23, 2020 at 17:20
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    $\begingroup$ Another way to get this without any boundary regularity is to note that there exists a set of full measure $A$ on which the suprema are finite. The function $u$ is then continuous on $A$. Since $U$ is open, $A$ is dense in $U$ and $u$ has a unique continuous extension to $U$ that satisfies the required properties. $\endgroup$
    – Jean Van Schaftingen
    Commented Aug 24, 2020 at 6:49
  • $\begingroup$ @JeanVanSchaftingen: thanks, but it is not obvious that you get such a set $A$. Let me explain: the function $f(x,y) = |u(x)-u(y)| \cdot |x-y|^{-\alpha}$ is locally integrable over $U \times U$ and hence there exists a dense subset $B \subseteq U \times U$ over which (a representative of) $f$ is bounded. The existence of $A$ is not immediately clear. Is there some handwaving to find it? $\endgroup$
    – shuhalo
    Commented Aug 24, 2020 at 12:08

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