The extension problem for fractional Sobolev spaces with a partial vanishing trace condition

We construct whole-space extensions of functions in a fractional Sobolev space of order $s\in (0,1)$ and integrability $p\in (0,\infty)$ on an open set $O$ which vanish in a suitable sense on a portion $D$ of the boundary $\partial O$ of $O$. The set $O$ is supposed to satisfy the so-called interior thickness condition in $\partial O \setminus D$, which is much weaker than the global interior thickness condition. The proof works by means of a reduction to the case $D=\emptyset$ using a geometric construction.


Introduction and main results
Let O ⊆ R d be open. For s ∈ (0, 1) and p ∈ (0, ∞) the fractional Sobolev space W s,p (O) consists of those f ∈ L p (O) for which the seminorm whole-space extensions for W s,p (O) were constructed by Zhou [11]. Though the mapping is in general not linear, extensions depend boundedly on the data. The case p ≥ 1 was already treated earlier by Jonsson and Wallin [8], and their extension operator is moreover linear. In fact, Zhou has shown that the interior thickness condition is equivalent for W s,p (O) to admit whole-space extensions. If we impose a vanishing trace condition on ∂O in a suitable sense, zero extension is possible, so in this case no geometric quality of O is needed. It is now natural to ask what happens if a vanishing trace condition is only imposed on a portion D ⊆ ∂O.
To be more precise, we consider the space W s,p The fractional Hardy term in-there models the vanishing trace condition on D, compare for [3][4][5][6]9]. Spaces of this kind were also recently investigated in [2] and have a history of successful application in the theory of elliptic regularity, see for example [7].
The present paper seeks minimal geometric requirements under which functions in W s,p D (O) can be boundedly extended to whole-space functions. We will see in Lemma 2.2 that in (ITC) we could equivalently consider balls centered in ∂O instead of O. Put N := ∂O \ D. In Definition 2.1 we introduce the interior thickness condition in N , which requires that (ITC) holds for balls centered in N . For D = ∅, this is just the usual interior thickness condition in virtue of the aforementioned Lemma 2.2. It is the main result of this article to show that the interior thickness condition in N is sufficient for the W s,p D (O)-extension problem. A major obstacle is that the interior thickness condition in N does not provide thickness in any neighborhood around N , which makes localization techniques not applicable. An example for this is a self-touching cusp, see Example 2.3. Our construction is as follows.
If p ≥ 1, then E is moreover linear.
We will also comment on the sharpness of our result in Section 4.
Finally, a remark on the case p = ∞ is in order. In this situation, the fractional Sobolev space is substituted by the Hölder space of order s ∈ (0, 1). Then the Whitney extension theorem [10, Thm. 3, p.174] provides a linear extension operator without any geometric requirements. In particular, the fractional Hardy term is not needed, though it is easily seen that f d −s D ∞ can only be finite if f vanishes identically on D, and the same is of course true for the extension. for his support, the "Studienstiftung des deutschen Volkes" for financial and academic support, Joachim Rehberg for suggesting the topic and Juha Lehrbäck for valuable discussions.
(Non-)Standard notation. We write B(x, r) for the open ball around x with radius r. The closure of a set A is denoted by A and the Lebesgue measure of A is denoted by |A|. If we integrate with respect to the Lebesgue measure, we write dx, dy and so on. For diameter and distance induced by the Euclidean metric we write diam(·) and d(·, ·). Also, the shorthand notation d E (x) := d({x}, E) is used. We employ the notation and for estimates up to an implicit constant that does not depend on the quantified objects. If two quantities satisfy both and we write ≈.

Geometry
The following lemma shows the equivalence between the (ITC) condition with balls centered in O and (ITC) with balls centered in ∂O already mentioned in the introduction. Though its proof is simple, we include it for good measure.

and only if E satisfies the interior thickness condition in ∂E.
Proof. Assume (ITC) and let x ∈ ∂E, r ∈ (0, 1]. Then pick some y ∈ B(x, r/2) ∩ E and calculate Conversely, let x ∈ E, r ∈ (0, 1] and assume that E is interior thick in ∂E. If B(x, r/2) ⊆ E then the claim follows immediately. Otherwise, pick again some y ∈ B(x, r/2) ∩ ∂E and argue as above.
The following simple example shows that a set can satisfy the thickness condition in some closed subset of the boundary but fails to have it in any neighborhood of it.
This means that O consists of the right half-plane touched by a cusp from the left. Put D to be the boundary of the cusp and N is the y-axis except the origin. Then the (ITC) estimate holds in N since each ball centered in N hits the half-plane with half of its area, but any proper neighborhood around N would contain a region around the tip of the cusp, in which thickness does not hold (consider a sequence that approximates the tip of the cusp and test with balls that do not reach N ).

The extension operator
In this section we prove Theorem 1.1. First, we construct O and show that it is interior thick. Second, we show that the zero extension to O is bounded using a simple geometric argument. Finally, we patch everything together to conclude. Throughout, O and D are as in Theorem 1.1 and we put N := ∂O \ D for convenience. ∂O ⊆ ∂O and that satisfies (ITC). According to the assumption on N and Lemma 2.2 it suffices to check that O is interior thick in D and the "added" boundary. Of course we could take O as R d \ ∂O in this step but this would make zero extension in Section 3.2 impossible. Therefore, our construction will be in such a way that moreover |x − y| d D (x) whenever x ∈ O and y ∈ O \ O, see Lemma 3.1, which will do the trick in step two.

Embedding into an interior thick set. We construct an open set
Let {Q j } j be a Whitney decomposition for the complement of N , which means that the Q j are disjoint dyadic open cubes such that Using the Whitney decomposition we define Note that for Q ∈ Σ one has Q\D = Q\∂O. Then all claimed properties of O except (ITC) follow immediately by definition. So, let x ∈ ∂O and r ∈ (0, 1]. If x ∈ N then we are done by assumption (keep Lemma 2.2 in mind). Otherwise, either x ∈ D or x ∈ ∂Q for some Q ∈ Σ (to see this, use that the Whitney decomposition is locally finite). But if x ∈ D then x ∈ Q for some Q ∈ Σ by property (i) of the Whitney decomposition and the definition of Σ. Hence, in either case x ∈ Q for some Q ∈ Σ. Now we make a case distinction on the radius size compared to the size of Q. If r ≥ 4 d(Q, N ), pick y ∈ Q and z ∈ N with d(Q, N ) = |y − z|. Then with (ii) we get Proof. We consider y ∈ O \ O and pick some Q ∈ Σ that contains y. We distinguish whether or not x and y are far away from each other in relation to diam(Q).
Case 1 : |x − y| < diam(Q). Fix a point z ∈ ∂O on the line segment connecting x with y. Assume for the sake of contradiction that z ∈ N . Then using (ii) we calculate Case 2 : |x − y| ≥ diam(Q). By definition of Σ and y ∈ O we can pick z ∈ Q ∩ D. Then This enables us to estimate E 0 . Clearly, we only have to estimate the W s,p (O)-seminorm since extension by zero is always isometric on The first term is bounded by f p W s,p (O) , so it only remains to bound the second term. Using Lemma 3.1 and calculating in polar coordinates we find Plugging this back into (1)  The claim for p < 1 then follows already by composition. In the case p ≥ 1, note that E can be constructed to be linear, see also [8].

On the sharpness of our result
In this final section we take a look on how close to a characterization our condition is. We will see in Example 4.1 that the interior thickness condition in N is not necessary for the extension problem, but that our construction might fail without it. Afterwards, we will introduce a degenerate interior thickness condition in N , which is necessary for the extension problem, but is not sufficient for our construction.

Example 4.1.
Consider the upper half-plane in R 2 . A Whitney decomposition can be constructed from layers of dyadic cubes. Let O be a "cusp" that is build from those Whitney cubes which intersect the area below the graph of the exponential function, and let N be its lower boundary given by the real line in R 2 . It is eminent that O is not interior thick in N . Moreover, our construction of O just adds another layer of cubes, so O is of the same geometric quality. Hence, our construction does not work in this situation. But zero extension to the upper half-plane is still possible, so with O chosen as the upper half-plane, we can construct an extension procedure for W s,p D (O) in this configuration. This shows that the interior thickness condition in N is not necessary for W s,p D (O) to admit whole-space extensions, but is "necessary" for our construction to work.
We introduce the aforementioned modified version of the interior thickness condition in N ⊆ ∂O that degenerates near ∂O \ N . In fact, this condition is necessary for the W s,p (O) ∩ L p (O, d −sp D )-extension problem. The technique to show this is due to Zhou [11]. By the restriction in radii, the test functions used in Zhou's proof belong to W s,p D (O), and then his proof applies verbatim, hence we omit the details.  On the other hand, we have seen in that example that in this configuration our construction does not work. Hence, the degenerate interior thickness condition in N is too weak for our proof of Theorem 1.1.