Trace and Extension Theorems for Sobolev-Type Functions in Metric

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Lipschitz boundary in Rn. Later, Besov [2] followed up with investigating how Besov functions in a (half)space can be restricted to a hyperplane. In the Euclidean setting, the trace theorems have been substantially generalized by Jonsson– 1,p n 1−θ /p Wallin [25], see also [26], who showed that T : W (R ) → Bp,p (F) is a bounded linear sur- jection provided that F is a compact set that is Ahlfors codimension-θ regular (cf. (2.2) and (2.3) below) and p > θ. However, they did not address the case p = θ > 1. In the metric setting, [32] proved that the trace of a function of bounded variation on Ω (which arise as relaxation of Newton–Sobolev N 1,1(Ω) functions) lies in L1(∂Ω) provided that Ω is a domain of nite perimeter that admits a 1-Poincaré inequality and its boundary is codimension-1 regular. An extension operator for L1 boundary data was constructed in [46] for Lipschitz domains in Carnot–Carathéodory spaces and in [34] for domains with codimension-1 regular boundary in general metric spaces. Recent paper [41] discusses traces of Besov, Triebel–Lizorkin, and Hajłasz–Sobolev functions to porous Ahlfors regular closed subsets of the ambient metric space in case that the Hajłasz gradient is summable in suciently high power. The method there is based on hyperbolic llings of the metric space, cf. [7, 43]. The paper [41] also shows possible relaxation of Ahlfors regularity by replacing it with the Ahlfors codimension-θ regularity. The goal of the present paper is to study trace and extension theorems for Sobolev-type func- tionsindomainsinametricmeasurespacewithalowerand/orupper codimensionbound (possibly unequal) for the boundary, including the case when the upper codimension bound equals the integrability exponent of the gradient. Unlike the trace theorems, the extension theorems make no use of any kind of Poincaré inequality. This gives new results also for weighted classical Sobolev functions in the Euclidean setting. Let us now state the main theorems of this paper. Theorem 1.1. Let (X, d, µ) be a metric measure space. Let Ω ⊂ X be a John domain whose closure is compact. Assume that µ⌊Ω is doubling and admits a p-Poincaré inequality for some p > 1. (If Ω is uniform, then p = 1 is also allowed.) Let ∂Ω be endowed with Ahlfors codimension-θ regular measure H for some θ < p. Then, there is a bounded linear surjective trace operator T : N 1,p (Ω,dµ) → 1−θ /p Ω Bp,p (∂ ,dH), which satises

= , ∂Ω. (1.2) lim+ |u(x)− Tu(z)| dµ(x) 0 for H-a.e. z ∈ r →0 KB(z,r )∩Ω 1−θ /p Ω 1,p Ω Moreover, there exists a bounded linear extension operator E : Bp,p (∂ ,dH) → N ( ,dµ) that = 1−θ /p Ω acts as a right inverse of T , i.e., T ◦ E Id on Bp,p (∂ ,dH). Existence and boundedness of the trace operator is established by Theorems 5.6 (for John domains) and 5.13 (for uniform domains). The linear extension operator is constructed in Section 6, see Lemmata 6.3 and 6.12, and Proposition 6.8.

Theorem 1.3. Let (X, d, µ) be a metric measure space. Let Ω ⊂ X be a domain such that µ⌊Ω is doubling. Let F ⊂ Ω be a bounded set, endowed with a measure ν that satises codimension-θ upper bound, i.e., ν(B(z,r) ∩ F) . µ(B(z,r) ∩ Ω)/r θ for every z ∈ F, r ≤ 2 diam F, and some θ > 0. Then, 1,p Ω 1−θ /p for every p > θ with p ≥ 1, there is a bounded linear trace operator T : M ( ,dµ) → Bp,∞ (F,dν). However, the trace operator need not be surjective and there can be a function u ∈ M1,p (Ω,dµ) such 1−θ /p α that Tu ∈ Bp,∞ (F,dν)\ α >1−θ /p Bp,∞(F,dν). Ω 1,p Ω, 1−θ /p , If, in addition, supportsÐ a p-Poincaré inequality, then T : N ( dµ) → Bp,∞ (F dν). TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 3

The assertion follows from Theorem 4.4, which is however formulatedfor F = ∂Ω; nevertheless, the proof goes through even in the setting of Theorem 1.3. The fact that T need not be surjective 1−θ /p is shown by Examples 8.1 and 8.2. However, it is unclear whether the target space Bp,∞ (F) is optimal under these mild assumptions. 1,p Ω 1−θ /p Open problem. Under the assumptions of Theorem 1.3, is T : M ( ,dµ) → Bp,p (F,dν) 1−θ /p bounded? Note however that even if T were a bounded mapping into Bp,p (F,dν), then it still could lack surjectivity as is seen in Examples 8.1 and 8.2. Theorem 1.4. Let (X, d, µ) be a metric measure space. Let Ω ⊂ X be a bounded domain that admits a θ-Poincaré inequality for some θ ≥ 1. Assume that µ⌊Ω is doubling and that ∂Ω is endowed with θ +ε an Ahlfors codimension-θ regular measure H. Let wε (x) = log(2 diam Ω/dist(x,∂Ω)) for some 1,θ θ ε > 0. Then, there is a bounded linear trace operator T : N (Ω, wε dµ) → L (∂Ω,dH), which satises (1.2). Moreover, there exists a bounded non-linear extension operator E : Lp (∂Ω,dH) → N 1,p (Ω,dµ) that acts as a right inverse of T , i.e., T ◦ E = Id on Lp (∂Ω,dH). The choice of weight wε with an arbitrary ε > 0 is essentially sharp when θ > 1 since for every 1,p −ε < 0 there exists u ∈ N (Ω, w−ε dµ) such that Tu ≡∞ in ∂Ω, i.e., (1.2) fails. Existence and boundedness of the trace operator from the weighted Newtonian space follows from Theorem 4.17. The non-linear extension operator is constructed in Section 7. Essential opti- malityof theweight isproven inProposition8.4, see also Example8.3. Observe that thebehaviorof the trace operator is substantially dierent when H⌊∂Ω is codimension 1 regular since no weight is needed then and L1(∂Ω,dH) is the trace class of N 1,1(Ω,dµ), see [32, 34]. The trace results stated for N 1,1(Ω,dµ) functions under the assumption of 1-Poincaré inequality can be further generalized to functions of bounded variations. Details of such a generalization lie outside of the scope of the main interest and have therefore been omitted. The paper is divided into 8 sections. We start o by preliminaries for the analysis in metric spaces and Sobolev-type functions considered in this paper. Section 3 is then devoted to establishing elementary properties (continuous and compact embeddings, in particular) of Besov spaces in the metric setting. Trace theorems for general domains, whose boundary has an upper codimension-θ bound, are the subject of Section 4. Sharp trace theorems for John and uniform domains that are compactly embedded in X are proven in Section 5. A linear extension operator, i.e., a right inverse of the trace operator, for Besov data is constructed in Section 6. A non-linear extension operator for Lθ boundary data is then constructed in Section 7. Finally, several examples that show sharpness of the hypothesis of the presented trace theorems are provided in Section 8. 2. Preliminaries & Sobolev-type functions in the metric setting Throughout the paper, (X, d, µ) will be a metric space endowed with a σ-nite Borel regular measure that is non-trivial in the sense that µ(B) ∈ (0, ∞) for every ball B ⊂ X. We do not a priori assume that X is (locally) complete, nor that µ is doubling in X. However, given a domain Ω, we will require that µ⌊Ω is doubling and non-trivial, i.e., there is cdbl ≥ 1 such that

0 < µ(B(x, 2r) ∩ Ω) ≤ cdblµ(B(x,r) ∩ Ω) < ∞ for every x ∈ Ω and every r > 0, where B(x,r) denotes an open ball of radius r centered at x. Then, [3, Lemma 3.3] yields that there are Cs > 0 and s > 0 such that µ(B(x,r) ∩ Ω) r s ≥ C µ(B(y, R) ∩ Ω) s R 4 LUKÁŠ MALÝ for all 0 < r ≤ R, y ∈ Ω, and x ∈ B(y, R) ∩ Ω. In particular, if Ω is bounded, then µ⌊Ω has a lower s mass bound µ(B(x,r) ∩ Ω)≥ csr since µ(B(x,r) ∩ Ω) µ(B(x, 2 diam Ω) ∩ Ω) µ(Ω) (2.1) ≥ C = ≕ c , x ∈ Ω. r s s (2 diam Ω)s (2 diam Ω)s s 0 In the rest of the paper, fE denotes the integral mean of a function f ∈ L (E) over a measurable set E ⊂ X of nite positive measure, dened as 1 f = f dµ = f dµ E K µ(E) E ¹E whenever the integral on the right-hand side exists, not necessarily nite though. Given an open ball B = B(x,r) ⊂ X and λ > 0, the symbol λB denotes the inated ball B(x, λr). Throughout the paper, C represents various constants and its precise value is not of interest. Moreover, its value may dier in each occurrence. Given expressions a and b, we write a . b if there is a constant C > 0 such that a ≤ Cb. We say that a and b are comparable, denoted by a ≈ b, if a . b and b . a at the same time. Given a set F ⊂ Ω endowed with a σ-nite Borel regular measure H, we say that H satises a lower codimension-ϑ bound with some ϑ > 0 if µ(B(x,r) ∩ Ω) (2.2) H(B(x,r) ∩ F)≥ cϑ , for all x ∈ F, 0 < r < 2 diam F . r ϑ Analogously, we say that H satises a upper codimension-θ bound with some θ > 0 if µ(B(x,r) ∩ Ω) (2.3) H(B(x,r) ∩ F)≤ cθ , for all x ∈ F,r > 0. r θ Note thatthe trace theorems establishedin Sections 4 and 5 make use only of the upper bound (2.3) with 0 < θ < s, whereas the extension theorems in Sections 6 and 7 assume both (2.2) and (2.3) with 0 < ϑ ≤ θ < s. In the rest of the paper, the conditions (2.2) and (2.3) will be referred to with F = ∂Ω, while the notation H is suggestively used to promote the measure’s relation to the codimension-θ (and codimension-ϑ) Hausdor measure, see (2.5) and Lemma 2.6 below. If both (2.2) and (2.3) hold true with 0 < ϑ = θ < s, then F (as well as H) will be called Ahlfors codimension-θ regular. Observe that (2.2) with (2.1) imply that H has a lower mass bound H(B(x,r) ∩ F) & r s−ϑ . On the other hand, (2.3) guarantees that H⌊F is doubling provided that µ⌊Ω is doubling. If Ω (and α hence F) is bounded, then H⌊F has a lower mass bound H(B(x,r) ∩ F) & r with some α ≥ s − θ by (2.3). However, the lower bound s − θ need not be optimal (as the optimal α might be smaller). The following lemma shows how (2.2) and (2.3) relate H(F) to the measure of an exterior shell of F in Ω.

Lemma 2.4. Let Ω ⊂ X be a domain with µ⌊Ω doubling. Let F ⊂ Ω be endowed with a measure H and set ΩR = {x ∈ Ω : dist(x, F) < R} for R > 0. ϑ (i) If F has a lower codimension-ϑ bound, then µ(ΩR ) . H(F)R . θ (ii) If F has an upper codimension-θ bound, then µ(ΩR ) & H(F)R . It will become apparent from the proof that it would suce to assume the relation between µ −k and H to hold only for all z ∈ E, where E is a dense subset of F, and all rk = 2 diam(Ω), k ∈ N0. Ω = Ω Proof. Since R z ∈F (B(z, R) ∩ ) and µ⌊Ω is doubling, we can apply the simple 5-covering Ω Ω lemma to obtain a collection {zk }k ⊂ F such that R ⊂ k (B(zk , 5R) ∩ ) while the balls Ð Ð TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 5

{B(zk , R)}k are pairwise disjoint. In case of (i), we have that

ϑ ϑ µ(ΩR )≤ µ(B(zk , 5R) ∩ Ω) . µ(B(zk , R) ∩ Ω) . R H(B(zk , R) ∩ F)≤ R H(F). Õk Õk Õk On the other hand, in case of (ii), we obtain that H⌊F is doubling and hence

θ µ(ΩR ) ≥ µ(B(zk , R) ∩ Ω) & R H(B(zk , R) ∩ F) Õk Õk θ θ θ & R H(B(zk , 5R) ∩ F)≥ R H B(zk , 5R) ∩ F = R H(F). Õk Øk Given E ⊂ Ω and θ > 0, we dene its co-dimension θ Hausdor measure H co-θ (E) by µ(B ∩ Ω) (2.5) H co-θ (E) = lim inf i : B balls in Ω, rad(B ) < δ, E ⊂ B . + θ i i i δ →0 rad(Bi ) Õi Øi More generally, it is possible (and natural) to dene H co-θ (E) for a set E ⊂ X by considering the θ Ω limit of inma over i µ(Bi )/rad(Bi ) instead. If µ is globally doubling and satises the measure density condition, i.e., µ(B ∩ Ω) & µ(B) for every ball B centered in Ω, then such a measure would be comparable withÍ (2.5) whenever E ⊂ Ω.

Lemma 2.6. Let Ω ⊂ X be a domain with µ⌊Ω doubling. Let F ⊂ Ω be endowed with a measure H. (i) If F has a lower codimension-ϑ bound, then H co-ϑ (E) . H(E) for every Borel set E ⊂ F. (ii) If F has an upper codimension-θ bound, then H co-θ (E) & H(E) for every Borel set E ⊂ F.

Proof. (i) Let {Bi }i be a cover of E by balls of radius at most δ. As µ⌊Ω is doubling, we can use the simple Vitali 5-covering lemma to extract a collection of pairwise disjoint balls that will cover E after being inated by the factor of 5. Then,

µ(B ∩ Ω) µ(5Bi ∩ Ω) µ(Bi ∩ Ω) (2.7) i ≈ j ≈ j . H(B ∩ F) = H B ∩ F . ϑ ϑ ϑ ij ij rad(Bi ) rad(Bij ) rad(Bij ) Õi Õj Õj Õj Øj In particular,

co-ϑ H (E) ≤ lim inf H Bi ∩ F : Bi balls in Ω, rad(Bi ) < δ, E ⊂ Bi = H(E) δ →0+ Øi Øi since H is outer regular by [35, Theorem 1.10]. (ii) Note that H is doubling. Analogously as in (2.7), we have

µ(Bi ∩ Ω) & H Bi ∩ F ≈ H 5Bi ∩ F ≥H 5Bi ∩ F ≥ H(E) rad(B )θ j j j i i j j j Õ Õ Õ Ø co-θ whenever E ⊂ i Bi . Thus, H (E) & H(E).

2.8. NewtonianÐ functions. In the metric setting, there need not be any natural distributional derivative structure that would allow us to dene Sobolev functions similarly as in the Euclidean setting. Shanmugalingam [42] pioneered an approach to Sobolev-type functions in metric spaces via (weak) upper gradients. The interested reader can be referred to [3, 21]. 6 LUKÁŠ MALÝ

The notion of (weak) upper gradients (rst dened in [20], named “very weak gradients”), corresponds to |∇u|. A Borel function g : X → [0, ∞] is an upper gradient of u : X → R ∪ {±∞} if (2.9) u(γ (b)) − u(γ (a)) ≤ g ds γ ¹ holds for all rectiable curves γ : [a,b] → X whenever u(γ (a)) and u(γ (b)) are both nite and = ∞ γ g ds otherwise. R ∫ If a function u : X → is locally Lipschitz, then |f (y)− f (x)| Lip f (x) = lim sup , x ∈ X, y→x d(y, x) is an upper gradient of u, see for example [18]. 1,p = p For p ≥ 1, one denes the Newtonian space N (X) {u ∈ L (X) : kukN 1,p (X ) < ∞}, where k kp = k kp + k kp . u N 1,p (X ) u Lp (X ) inf g Lp (X ) : g is an upper gradient of u For every u ∈ N 1,p (X), the inmum in the denition of the N 1,p (X) norm is attained by a unique minimal p-weak upper gradient (which is minimal both pointwise and normwise). The phrase “p-weak” refers to the fact that (2.9) may fail for a certain negligible number of rectiable curves. Given a domain Ω ⊂ X, the class N 1,p (Ω) is dened as above with Ω being the underlying metric space (whose metric measure structure is inherited from X). 2.10. Hajłasz functions. Another approach to Sobolev-type functions was introduced in [13]. Given u : X → R ∪ {±∞}, we say that g : X → [0, ∞] is a Hajłasz (α-fractional) gradient if there is E ⊂ X with µ(E) = 0 such that |u(x)− u(y)| ≤ d(x,y)α (g(x) + g(y)) for every x,y ∈ X \ E, where α ∈ (0, 1]. The phrase “α-fractional” is typically dropped in case α = 1. Compared to the n classical Sobolev spaces in R , Hajłasz gradients for α = 1 correspond to cnM|∇u|, where M is the Hardy–Littlewood maximal operator. α,p p Forp > 0 and α ∈ (0, 1], onedenesthe Hajłasz space M (X) = {u ∈ L (X) : kukM α,p (X ) < ∞}, where k kp = k kp + k kp . u M α,p (X ) u Lp (X ) inf g Lp (X ) : g is an α-fractional Hajłasz gradient of u The inmum is attainedfor p > 1, but the minimal Hajłasz gradient is not pointwise minimal. Fora domain Ω ⊂ X, the space Mα,p (Ω) is dened as above with Ω being the underlying metric measure space. Hajłasz gradients are p-weak upper gradients, cf. [24], whence M1,p (Ω) ֒→ N 1,p (Ω) for all p ≥ 1. The inclusion may be proper unless Ω admits a p-Poincaré inequality and p > 1. More information about Hajłasz functions and the motivating ideas can be found in [14]. 2.11. Functions with a Poincaré inequality. Let p ∈ [1, ∞) and q ∈ (0,p]. Then, the space 1,p p Pq (X) consists of those functions u ∈ L (X) for which there is a constant λ ≥ 1 and a non- negative g ∈ Lp (X) such that 1/q (2.12) |u − u | dµ ≤ rad(B) gq dµ K B K B λB holds true for all balls B ⊂ X. 0 1 A non-negative function g ∈ L (X) will be called a q-PI gradient of u ∈ Lloc(X), if the couple (u,g) satises (2.12) for some xed λ ≥ 1. TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 7

1,p 1,p It was observed already in [13] that M (X) ⊂ P1 (X) and there is an absolute constant c such that cg is a 1-PI gradient of u ∈ M1,p (X) with λ = 1, whenever g is a Hajłasz gradient of u. In fact, it follows from [14, Theorems 8.7 and 9.3] that if µ is doubling, then there is q ∈ (0, 1) such that 1,p 1,p M (X) = Pq (X) with a universal dilation factor λ ≥ 1 for every p ≥ 1. Corollary 2.18 below shows that Hajłasz functions can be also characterized by an inmal Poincaré inequality (2.16). The interested reader can refer to [14, 15], where functions with a Poincaré inequality have been thoroughly studied. 1,p Ω Ω Remark 2.13. If u ∈ Pq ( ) for some domain ⊂ X such that µ⌊Ω is doubling, then (2.12) holds true not only for balls with center in Ω, but also for balls whose center lies at ∂Ω possibly at a cost of multiplying the q-PI gradient g by a constant factor. This can be shown by the dominated convergence theorem as follows. If B = B(z,r) with z ∈ ∂Ω, then there is a sequence of points ∞ Ω , < −n Ω = , −n {xn }n=1 ⊂ such that d(xn z) 2 r. Then, χBn → χB a.e. in , where Bn B(xn (1 − 2 )r). The doubling condition ensures that µ(Bn ∩ Ω)≈ µ(B ∩ Ω) with constants independent of n. Thus,

|u − uB∩Ω | dµ ≈ |u(x)− u(y)| dµ(x)dµ(y) KB∩Ω KB∩Ω KB∩Ω µ(B ∩ Ω)2 ≈ lim n |u(x)− u(y)| dµ(x)dµ(y) n→∞ Ω 2 K K µ(B ∩ ) Bn ∩Ω Bn ∩Ω 1/q 1/q . lim (1 − 2−n )r gq dµ ≈ r gq dµ . n→∞ K Ω K Ω λBn ∩ λB∩ Denition 2.14. We say that a metric measure space (X, d, µ) admits a q-Poincaré inequality if there is a constant CPI > 0 and a universal dilation factor λ ≥ 1 such that (2.12) holds true for the , 0 1 couple of functions (u CPIg) whenever g ∈ L (X) is an upper gradient of u ∈ Lloc(X). 1,p 1,p In particular, if X admits a p-Poincaré inequality, then N (X) ⊂ Pp (X). It was shown in Keith–Zhong [27] that spaces that admit p-Poincaré inequality, where p > 1, undergo a self- improvement so that they, in fact, admit a (p −ε)-Poincaré inequality for some ε > 0 provided that 1,p 1,p X is complete and µ doubling. Moreover, one still obtains that N (X) ⊂ Pp−ε (X) also in case X is 1,p Ω 1,p Ω merely locally complete. In view of Corollary 2.17 below, we therefore see that N ( ) ⊂ P1 ( ) for every p ≥ 1 whenever Ω ⊂ X is a domain with a p-Poincaré inequality, µ⌊Ω is doubling, and Ω as a metric space is complete. 1,p Lemma 2.15. Assume that µ doubling. Let 0 < q < p < ∞. Then, u ∈ Pq (X) if and only if there is a non-negative function h ∈ Lp (X) such that

(2.16) |u − uB | dµ ≤ rad(B) ess inf h(x) KB x ∈B for every ball B ⊂ X.

= q 1/q ≤ q 1/q Proof. Obviously, (2.16) implies (2.12) since ess inf B h ess inf B h B h dν . ∈ 1,p ( ) ∈ p ( ) ≥ ⨏ ⊂ Let now u Pq X with a q-PI gradient g L X for some λ 1. For every ball B X and ∈ ( ) ≔ ( q )1/q( ) ≥ ( q )1/q every point x B, we have h x Mg x λB g dµ , where M is the non-centered Hardy–Littlewood maximal operator. Thus, ⨏

|u − uB | dµ ≤ rad(B) inf h(x). KB x ∈B 8 LUKÁŠ MALÝ

As µ is doubling, M : L1(X) → weak-L1(X) is bounded due to Coiman–Weiss [9, Theorem III.2.1]. Obviously, M : L∞(X) → L∞(X) is bounded. The Marcinkiewicz interpolation theorem then yields p/q p/q q 1/q that M : L (X) → L (X). Therefore, khkLp (X ) = k(Mg ) kLp (X ) ≤ cp/q kgkLp (X ).

1,p Ω p Ω One can deduce from the proof above that for every u ∈ Pp ( ), there is h ∈ weak-L ( ) such that the couple (u,h) satises the inmal Poincaré inequality (2.16). 1,p = 1,p 1,p Corollary 2.17. Assume that µ is doubling. Let 0 < r ≤ q < p. Then, Pr (X) Pq (X) ⊂ Pp (X). 1,p p = Moreover, if u ∈ Pq (X), then there is h ∈ L (X) such that the couple (u,h) satises (2.12) with λ 1. 1,p 1,p Proof. The inclusion Pq (X) ⊂ Pp (X) follows by the Hölder inequality applied to the right- 1,p 1,p p hand side of (2.12). Equality of Pq (X) and Pr (X) and existence of h ∈ L (X) as desired are an immediate consequence of Lemma 2.15.

1,p 1,p Corollary 2.18. Assume that µ is doubling. Then, M (X) = Pq (X) for every 0 < q < p, where 1 ≤ p < ∞. Moreover, each u ∈ M1,p (X) has a Hajłasz gradient h ∈ Lp (X) such that (u,h) satises the inmal Poincaré inequality (2.16). Proof. Follows from [14, Theorem 9.3], Lemma 2.15, and Corollary 2.17.

3. Besov spaces in the metric setting and their embeddings In this section, we willturn our attentionto spaces of fractional smoothness, in particular, Besov spaces in the metric setting. Besov spaces have been thoroughly studied in the Euclidean setting (see [2, 26, 44, 45] and references therein). They made their rst appearance in the metric setting in [8] and were explored further in [11, 12, 16, 17, 31, 37, 43]. The main goal of this section is to investigate continuous and compact embeddings of Besov spaces with norm based on generalization of modulus of continuity as in [11] under the very mild assumption that the underlying Borel regular measure is non-trivial and doubling. The results obtainedin this section have been proven in some of the papers listed above under more restrictive hypothesis that the measure is Ahlfors regular. An extremely technical approach via frames was employed in [16, 37] to show the embeddings if the measure satises a reverse doubling condition. The great advantage of using the Besov-type norm introduced in [11], see (3.5) below, is that it allows for very elementary proofs, where the main tools are the ordinary Hölder inequality and Sobolev-type embeddings for Hajłasz functions. For our arguments, we will utilize also the zero-smoothness Besov spaces that have been studied only scarcely so far, cf. [34]. Eventually, we will make use of Besov spaces over ∂Ω for a given domain Ω ⊂ X. In order to avoid confusion with the ambient metric space X that is used in the rest of this paper, we will state the denitions and results in this section for a general metric space Z = (Z, d,ν), where ν is a doubling measure. By [3, Lemma 3.3], there are Q > 0 and cQ > 0 such that ν(B(w,r)) r Q (3.1) ≥ c ν(B(z, R)) Q R for all 0 < r ≤ R < ∞, z ∈ Z, and w ∈ B(z, R). In particular, if Z is bounded, then ν has a lower mass bound, i.e., there is c˜Q > 0 such that Q (3.2) ν(B(z,r)) ≥ c˜Qr Q for every z ∈ Z and 0 < r < 2 diam Z. In fact, one can choose c˜Q = cQν(Z)/(diam Z) . TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 9

If Z is connected, then it also satises the reverse doubling condition by [3, Corollary 3.8], i.e., there are σ > 0 and cσ > 0 such that ν(B(y,r)) r σ (3.3) ≤ c ν(B(z, R)) σ R for all 0 < r ≤ R ≤ 2 diamZ, z ∈ Z, and y ∈ B(z, R). Unless explicitly stated otherwise, we will assume neither (3.2), nor (3.3) in this section. Denition 3.4. Fix R > 0 and let α ∈ [0, 1], p ∈ [1, ∞), and q ∈ (0, ∞]. Then, the Besov space α p Bp,q(Z) of smoothness α consists of L -functions of nite Besov (quasi)norm that is given by 1/q R |u(y)− u(z)|p q/p dt (3.5) kuk α = kuk p + dν(z)dν(y) , Bp,q (Z ) L (Z ) K tαp t ¹0 ¹Z B(y,t) ! with a standard modication in case q = ∞. α The function class Bp,q (Z) is in fact independent of the exact value of R ∈ (0, ∞). Moreover, if = α α > 0, then we may also choose R ∞ without changing the function class Bp,q(Z). However, 0 = Bp,q(Z) consists only of (equivalence classes of) constant functions in case R ∞, q < ∞ and Z is 0 = p bounded. Note also that Bp,∞(Z) L (Z) regardless of the value of R ∈ (0, ∞]. 1 Considering the other extreme smoothness value, one can see that Bp,q(Z) consists of constant 1,p 1 1,p 1,p functions if q < ∞. On the other hand, M (Z) ⊂ Bp,∞(Z) ⊂ KS (Z), where M (Z) is the Hajłasz space (see Paragraph 2.10 above) and KS1,p (Z) is the Korevaar–Schoen space introduced in [29]. See Lemma 3.20 below for the former embedding, while the latter follows directly from the denition of KS1,p , cf. [11]. If Z supports a p-Poincaré inequality (see Denition 2.14 above) for some p > 1, then M1,p (Z) = 1 = 1,p = 1,p 1,p Bp,∞(Z) KS (Z) N (Z) by [27, 30], where N (Z) is the Newtonian space (see Para- graph 2.8 above). For more information on the Hajłasz, Korevaar–Schoen, and Newtonian spaces, see [3, 11, 14, 21]. Ûα The corresponding homogeneous seminorm will be denoted by Bp,q(Z). The value of q can be understood as a ne-tuning parameter of the smoothness for Besov functions of equal value of α ∈ [0, 1). By the Fubini theorem, the Besov norm dened by (3.5) with p = q and R > 0 is equivalent to the Besov-type norm considered by Bourdon–Pajot [8], given by |u(w)− u(z)|p 1/p (3.6) kukBα (Z ) = kukLp (Z ) + dν(z)dν(w) p w d(w, z)αpν(B(w, d(w, z))) ¹Z ¹B( ,R) whenever α ∈ [0, 1) and p ∈ [1, ∞), cf. [11, Theorem 5.2]. The setting of [11] has several additional standing assumptions, which however turn out to be superuous for the particular result of Û α equivalence of norms. The corresponding homogeneous seminorm will be denoted by Bp (Z). The Bourdon–Pajot form of the Besov norm (3.6) corresponds very naturally to the Sobolev– Slobodeckij norm used in the classical trace theorems of Gagliardo [10] in the Euclidean setting. In fact, it allows for a cleaner exposition of proofs of trace theorems in John domains and hence will be used in Section 5. Otherwise, we will be using the Gogatishvili–Koskela–Shanmugalingam form of the Besov norm (3.5). For the sake of brevity, we dene

(3.7) E (u, t) = |u(y)− u(z)|p dν(z)dν(y) 1/p, u ∈ Lp (Z), t ∈ (0, ∞). p K ¹Z B(y,t) First, let us establish continuous embedding when varying the second exponent of the Besov norm. 10 LUKÁŠ MALÝ

Lemma 3.8. Let α ∈ [0, 1), p ∈ [1, ∞), and q ∈ (0, ∞) be arbitrary. Then, kuk α . kuk α Bp,q˜ (Z ) Bp,q (Z ) for every q˜ ∈ [q, ∞]. Proof. Let us rst prove the inequality for q˜ = ∞. Then, q t t q Ep (u, t) ds Ep (u, 2s) k kq = . ( , )q . u Û α sup αq sup αq+1 Ep u t sup αq+1 ds Bp,∞(Z ) t s s 0

A careful inspection of the proof above immediately yields that kuk Û α ( ) ≤ Ckuk Û α ( ) with Bp,q˜ Z Bp,q Z C > 0 independent of q˜ ∈ [q, ∞] holds true whenever R > diamZ or R = diamZ = ∞. Next, we will look into simple interpolating properties of the Besov spaces.

Lemma 3.9. Let αj ∈ [0, 1], pj ∈ [1, ∞), and qj ∈ (0, ∞], j = 0, 1. Let λ ∈ [0, 1] and dene α, p, and q by the convex combinations 1 1 − λ λ 1 1 − λ λ α = (1 − λ)α0 + λα1, = + , and = + , p p0 p1 q q0 q1 = = 1−λ λ with standard modication if q0 ∞ or q1 ∞. Then, kuk Û α ≤ kuk α · kuk α . Bp,q (Z ) BÛ 0 (Z ) BÛ 1 (Z ) p0,q0 p1,q1 Proof. The conclusion holds trivially true when λ = 0 or λ = 1. Let us now assume that λ ∈ (0, 1). Suppose also that both q0 and q1 are nite. Then, the desired estimate follows from the Hölder inequality applied twice. R q/p dt k kq = | ( )− ( )|(1−λ)p+λp ( ) ( ) u Û α u y u z dν z dν y αq+1 Bp,q (Z ) K t ¹0 ¹Z B(y,t) R (1−λ)q/p0 ≤ |u(y)− u(z)|p0 dν(z)dν(y) · K ¹0 ¹Z B(y,t) λq/p1 dt · |u(y)− u(z)|p1 dν(z)dν(y) K tαq+1 ¹Z B(y,t) R (1−λ)q λq Ep (u, t) Ep (u, t) dt (3.10) = I ≔ 0 1 tα0 tα1 t ¹0 R q0 (1−λ)q/q0 R q1 (1−λ)q/q1 Ep (u, t) dt Ep (u, t) dt ≤ 0 1 tα0 t tα1 t ¹0 ¹0 = kuk(1−λ)q · kukλq . Û α0 ( ) BÛ α1 (Z ) Bp0,q0 Z p1,q1 TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 11

If q1 < q0 = ∞, then λq = q1 and we would proceed from (3.10) as follows:

(1−λ)q R q1 Ep (u, t) Ep (u, t) dt ≤ 0 1 = k k(1−λ)q · k kλq . I sup α α u Û α0 u Û α1 t 0 t 1 t Bp ,q (Z ) Bp ,q (Z ) 0 0. If xz = 0, then the elementary inequality is trivial. Lemma 3.12. Assume that α ∈ [0, 1], p ∈ [1, ∞), and q ∈ (0, ∞]. Let λ ∈ (0, 1). Then,

1−λ λ kuk Û λα ( ) ≤ 2kuk ∞ · kuk Û α . Bp/λ,q/λ Z L (Z ) Bp,q (Z )

p(1/λ−1) Proof. Factoring out ess supy,z |u(y)− u(z)| from the innermost integral yields that

R q/p dt k kq/λ = | ( )− ( )|p/λ ( ) ( ) u Û λα u y u z dν z dν y αq+1 B , (Z ) K t p/λ q/λ ¹0 ¹Z B(y,t) R q/p dt ≤ 2q/λ kuk(1−λ)q/λ |u(y)− u(z)|p dν(z)dν(y) L∞(Z ) K tαq+1 ¹0 ¹Z B(y,t) = q/λ (1−λ)q/λ q 2 kuk ∞ kuk Û α . L (Z ) Bp,q (Z )

1−λ λ Note that similarly as in Corollary 3.11, one obtains kuk λα ( ) ≤ 2kuk ∞ · kuk α . Bp/λ,q/λ Z L (Z ) Bp,q (Z ) Lemma 3.13. Assume that R < ∞. Let 0 ≤ β < α ≤ 1 and suppose that p ∈ [1, ∞), and q ∈ (0, ∞]. Then, kuk Û β . kukBÛ α (Z ). Bp,q (Z ) p,∞ 12 LUKÁŠ MALÝ

Proof. If q < ∞, then R R q (α −β)q dt Ep (u, t) dt R kukq = E (u, t)q = ≤ kukq . Û β p βq+1 α 1−(α −β)q BÛ α (Z ) Bp,q (Z ) t t t p,∞ (α − β)q ¹0 ¹0 β α −β α In case q = ∞, it suces to notice that Ep (u, t)/t ≤ R Ep (u, t)/t for every 0 < t ≤ R.

Remark 3.14. If R = ∞, then one can further use the estimate Ep (u, t) ≤ CkukLp (Z ), where C does > . α < not depend on t 0, to obtain the inequality kuk β kukB , (Z ) for every 0 ≤ β α ≤ 1, Bp,q (Z ) p ∞ p ∈ [1, ∞), and q ∈ (0, ∞]. r Ûα Û0 Lemma 3.15. Suppose that u ∈ L (Z) ∩ Bs,∞(Z) for some 1 ≤ s < r and α > 0. Then, u ∈ Bp,q (Z) for every p ∈ [s,r) and q ∈ (0, ∞) provided that R < ∞ in (3.5). Proof. Case p = s follows immediately from Lemma 3.13. , , = s(r −p) , Fix p ∈ (s r) and q ∈ (0 ∞). Dene η r −s ∈ (0 s). Observe also that ν(B(y, t)) ≤ ν(B(z, t)) ≤ cQ ν(B(y, t)) cQ whenever y, z ∈ Z with d(y, z) < t, where t > 0 is arbitrary. For the sake of brevity, let us dene 2 = 2 Zt {(y, z) ∈ Z : d(y, z) < t}. Then, R q/p q + p−η η dt kuk Û0 ≤ |u(y)| |u(z)| |u(y)− u(z)| dν(z)dν(y) Bp,q (Z ) K t ¹0 ¹Z B(y,t) q/p R dt ≤ 2q(p−η)/p |u(y)|p−η + |u(z)|p−η |u(y)− u(z)|η dν(z)dν(y) K t ¹0 ¹Z B(y,t) q/p R |u(y)|p−η |u(y)− u(z)|η |u(z)|p−η |u(y)− u(z)|η dt ≈ + dν 2(y, z) 2 ν(B(y, t)) ν(B(z, t)) t ¹0 ºZt R |u(y)|p−η |u(y)− u(z)|η q/p dt ≈ dν 2(y, z) ≕ I . 2 ν(B(y, t)) t ¹0 ºZt Next, we apply the Hölder inequality for the inner double integral, which yields R |u(y)|r q(p−η)/pr I ≤ dν 2(y, z) 2 ν(B(y, t)) ¹0 ºZt + |u(y)− u(z)|ηr /(r −p+η) q(r −p η)/pr dt × dν 2(y, z) 2 ν(B(y, t)) t ºZt R q(p−s)/p(r −s) q(r −p)/p(r −s) dt = |u(y)|r dν(y) |u(y)− u(z)|s dν(y)dν(z) K t ¹0 ¹Z ¹Z B(y,t) R dt ≤ kukqr (p−s)/p(r −s) kukqs(r −p)/p(r −s) < ∞ , Lr (Z ) Û α ( ) 1−β B1,∞ Z t ¹0 = r (p−s)/p(r −s) s(r −p)/p(r −s) where β αqs(r − p)/p(r − s) > 0. Hence, kukBÛ 0 (Z ) ≤ Ckuk r kuk Û α . p,q L (Z ) Bs,∞(Z ) The next lemma shows that Besov functions of smoothness α are closely related to α-fractional α α,p α Hajłasz functions. Using also Lemma 3.20 below, we deduce that Bp,p (Z) ⊂ M (Z) ⊂ Bp,∞(Z) and the embeddings are continuous. This fact will be used to prove Sobolev-type embeddings for Besov functions. TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 13

Ûα p Lemma 3.16. Given u ∈ Bp,q (Z) for some α ∈ (0, 1) and 0 < q ≤ p < ∞, there is g ∈ L (Z) and E ⊂ Z with ν(E) = 0 such that

(3.17) |u(y)− u(z)| ≤ d(y, z)α (g(y) + g(z)) for all y, z ∈ Z \ E.

Moreover, kgk p . kuk Û α provided that R > diamZ in (3.5) (or, perhaps, R = diamZ = ∞). L (Z ) Bp,q (Z )

( ) = Q+α −1 | ( )− ( )| −α ( ) Proof. Similarly as in [11, Lemma 6.1], setting g z 2 cQ supr >0 B(z,r ) u z u y r dν y , z ∈ Z, gives the desired function that satises (3.17). Then, [11, Lemma⨏ 6.1] and Lemma 3.8 yield that kgk p . kuk Û α . kuk Û α . L (Z ) Bp,p (Z ) Bp,q (Z )

Corollary 3.18. Let α ∈ (0, 1), p ∈ [1, ∞), and q ∈ (0,p]. Let G ⊂ Z be an arbitrary open ball. Then, there is C > 0 such that: ∗ (i) If αp < Q, then ku − u k p∗ ≤ Ckuk Û α , where p = pQ/(Q − αp). G L (G) Bp,q (Z ) (ii) If αp = Q, then ku − u k ≤ Ckuk Û α . G exp L(G) Bp,q (Z ) κ (iii) If αp > Q, then ku −u k ∞ ≤ Ckuk Û α and |u(y)−u(z)| ≤ Cd(y, z) kuk α for all G L (G) Bp,q (Z ) Bp,q (Z ) y, z ∈ G, where κ = α − Q/p. If (3.2) is satised, then we may choose G = Z in all of the above even if Z is unbounded, in which case uG = 0.

Ûα p Proof. Given u ∈ Bp,q(Z), there is g ∈ L (Z) such that (3.17) is satised. Fix a ball G ⊂ Z. Considering the snow-aked metric d˜(·, ·) = d(·, ·)α , we obtain from (3.1) that , ˜, , Q/α the metric space (G d ν ⌊G ) has lower mass bound with exponent Q/α, i.e., ν(Bd˜(z r)) ≥ cr˜ for , = ˜ , < = Q every z ∈ G and r ≤ diamd˜ G, where Bd˜(z r) {y ∈ G : d(y z) r } and c˜ cQν(G)/radd(G) . Then, all three parts follow from [14, Theorem 8.7], which would give kgk p . kuk Û α on L (Z ) Bp,q (Z ) the right hand side of the inequality in (i) and (ii). As for (iii), we obtain that

1−Q/αp α −Q/p |u(y)− u(z)| ≤ Cd˜(y, z) kgk p ≤ Cd(y, z) kuk Û α for every y, z ∈ G . L (Z ) Bp,q (Z )

By inspecting the proof of [14, Theorem 8.7], one sees that the constant C really depends only on p, Q/α, and c˜ in the lower mass bound condition of G. Hence, if (3.2) holds, then ν(B) ≥ Q/α cQ radd˜(B) for every ball B ⊂ Z and hence we may apply [14, Theorem 8.7] directly to the snow-aked space (Z,d˜,ν).

Remark 3.19. The embedding as stated in Corollary 3.18 (i) is not sharp in case both (3.2) and

k k p∗,q . k k α (3.3) are satised. In fact, one can deduce from [17, Theorem 4.4] that u L (Z ) u Bp,q (Z ) for ∗ all p ∈ [1,Q/α) and q ∈ [1, ∞], where Lp ,q(Z) stands for a Lorentz space (see [1, Section IV.4]). Observe however that [17] assumes that ν is Ahlfors Q-regular, which it fails to be in case σ < Q. Nevertheless, the problem will be circumvented as soon as [17, Lemma 3.4] in the proof of [17, Theorem 4.4] is replaced by [14, Theorem 8.7 (1)].

Lemma 3.20 (cf. [11, Lemma 6.2]). Let p ∈ [1, ∞) and α ∈ (0, 1]. Suppose that the couple of p measurable functions u,g with g ∈ L (Z) satisfy (3.17). Then, kuk Û α . kgk p . Consequently, Bp,∞(Z ) L (Z ) α,p α .(M (Z) ֒→ Bp,∞(Z 14 LUKÁŠ MALÝ

Proof. Plugging in (3.17) into the denition of kuk Û α yields that Bp,∞ |u(y)− u(z)|p k kp = ( ) ( ) u Û α sup αp dν z dν y Bp,∞ K t 0 p will become clear from the proof below). If ′ 1−p′/p ν(Z) < ∞, then the estimate in (ii) is satised also for p < p since Ep′(u, t) ≤ ν(Z) Ep(u,t) by the Hölder inequality.

≤ k k p∗ . k k α Proof. Assume that q p. Then, u L (Z ) u Bp,q (Z ) by Corollary 3.18 (i). In particular, kuk 0 ( ) ≈ kuk p∗ ( ) . kukBα (Z ). Hence, Corollary 3.11 yields that Bp∗,∞ Z L Z p,q 1−λ λ kuk λα ≤ kuk 0 kuk α . kuk α ( ) . B (Z ) B (Z ) B , (Z ) Bp,q Z pλ,q/λ p∗,∞ p q Proof of (i) will then be completed by applying Lemma 3.8. Let us now focus on (ii). If p′ = p, then the desired embedding follows immediately from ′ Lemma 3.13. Assume that p ∈ (p,pλ ). Then, there are α˜ ∈ (λα, α) and λ˜ ∈ (λα/α˜, 1) such that 1 1 − λ˜ λ˜ 1 α˜ λ˜ 1 (1 − λ˜)α˜ 1 λ˜α˜ − λα = + ≔ (1 − λ˜) − + = − = + . p′ p˜∗ p p Q p p Q p Q λ α˜ Then, u ∈ B (Z) with kuk α˜ . kuk α by Lemmata 3.8 and 3.13. Next, we obtain from (i) p,p Bp,p (Z ) Bp,q that kuk λ˜ α˜ . kuk α˜ . Finally, Lemma 3.13 concludes the proof of (ii). B (Z ) Bp,p (Z ) p′,∞ Remark 3.22. If αp = Q, then the conclusion of Theorem 3.21(ii) still holds true with p∗ = ∞, i.e., pλ = p/λ and the proof goes through verbatim. If αp > Q, then both parts of Theorem 3.21 ∗ hold true with p = ∞, i.e., pλ = p/λ. One needs to use Corollary 3.18 (iii) and Lemma 3.12 in the proof of (i), and make sure that αp˜ , Q in the proof of (ii). If we assume that ν is not only doubling, but also satises the reverse doubling condition (3.3), ′ then one has a somewhat sharper embedding between Besov spaces in the critical case p = pλ for all q ∈ (0, ∞]. This has been proven by a very dierent (and very technical) approach to Besov spaces via frames. Roughly speaking, (3.5) generalizes the modulus of continuity denition of Besov spaces in Rn, whereas the frames correspond to the Fourier approach to Besov spaces in Rn. TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 15

Theorem 3.23. Assume that both (3.2) and (3.3) are satised. Suppose that αp < Q and dene the ∗ = /( − ) ∈ ( , ) α ( ) ⊂ λα ( ) “Sobolev conjugate” exponent p pQ Q pα . If λ 0 1 , then Bp,q Z Bpλ,q Z , where pλ is ∗ −1 = ∗−1 + −1 given by the convex combination of reciprocals of p and p, viz., pλ (1 − λ)p λp . Proof. The claim follows from [16, Theorem 1.5 (i)], [37, Theorem 4.1], and Lemma 3.8.

Remark 3.24. In the theory of Besov spaces in Rn, it is more common to state the embeddings between spaces of dierent smoothness in terms of their dierential dimension, which is dened Q α > as α − p for Bp,q (Z), where Q 0 is from (3.1), cf. [44, 45] and references therein. Embeddings of the form Bα0 (Z) ֒→ Bα1 (Z), where α − Q = α − Q and α > α are the objective of p0,q0 p1,q1 0 p0 1 p1 0 1 Theorems 3.21 (i) and 3.23, whereas Theorem 3.21 (ii) deals with the case when α − Q > α − Q 0 p0 1 p1 and α0 > α1. Let us now aim our attention to proving compactness of the embeddings. First, we prove a technical lemma that will allow us to run a Rellich–Kondrachev-type argument in the proof of Theorem 3.26 below. ∞ R+ > Lemma 3.25. Let {aj,k }j,k=1 ⊂ . Suppose that there is K 0 such that supk j aj,k ≤ K. Then, for every ε > 0, there are j ∈ N and I ⊂ N such that #I = ∞ and sup a , ≤ ε. 0 k ∈I j0 k Í Proof. For the sake of contradiction, suppose that there is ε > 0 such that Ij ≔ {k ∈ N : aj,k ≤ ε} is = = N nite for every j ∈ N. Let N ⌈K/ε⌉. Then, E j=1(N \ Ij ) , ∅. Hence, if k ∈ E and 1 ≤ j ≤ N , then aj,k > ε. Thus, Ñ ∞ N N K aj,k ≥ aj,k > ε = · ε ≥ K , = = = ε Õj 1 Õj 1 Õj 1 which contradicts the hypothesis that j aj,k ≤ K for every k ∈ N.

Next, will show the compact embeddingÍ of Besov spaces of zero smoothness, which will be later applied to prove compact embeddings for spaces of higher smoothness via continuous embeddings and elementary interpolating properties. , , ∞ Theorem 3.26. Assume that Z is bounded. Let p ∈ (1 ∞) and q ∈ (0 ∞). Let {uk }k=1 be a sequence 0 ∞ of functions with uniformly bounded Bp,q (Z) norm. Then, there is a subsequence {ukj }j=1 and a p p function u ∈ L (Z) such that ukj → u in L (Z). ∞ 0 p > Proof. Let {uk }k=1 be bounded in Bp,q (Z). Then, it is bounded in L (Z), which is reexive as p 1 ∞ p and hence there is a subsequence (still denoted by {uk }k=1) that is weakly convergent in L (Z). We will nd a subsequence that is Cauchy (and hence convergent) with respect to the Lp norm. > 1 Given δ 0, let Aδ be the averaging operator dened for f ∈ Lloc(Z) by

Aδ f (y) = f (z)dν(z), y ∈ Z. KB(y,δ )

For a xed δ > 0, the weak convergence of {uk }k results in the pointwise convergence Aδ uk (y) → Aδ u(y) for every y ∈ Z as k →∞. The lower mass bound (3.2), which is available as Z is bounded −Q/p and ν is doubling, implies that kAδ uk kL∞(Z ) . δ kuk kLp (Z ) ≤ C. The Lebesgue dominated p convergence then yields that Aδ uk → Aδ u in the norm of L (Z). 16 LUKÁŠ MALÝ

1−j ∞ ∞ 2 dt 1 dt E (f , 2−j )q ≈ E (f , 2−j )q . E (f , t)q . k f kq . p p p B0 (Z ) = = 2−j t 0 t p,q Õj 1 Õj 1 ¹ ¹ ≔ −j q q Letting aj,k Ep (uk , 2 ) , we see that supk j aj,k . supk kuk k 0 ≤ C, which allows us Bp,q (Z ) N N to use Lemma 3.25. Thus, for every n ∈ , we canÍ nd jn ≥ 1 and a countably innite set In ⊂ so −jn −n ∞ that Ep (uk , 2 ) < 2 for all k ∈ In. Moreover, it is possible to ensure that the sequence {jn }n=1 is strictly increasing. Using a diagonal argument, we will now construct the desired subsequence ∞ > m {ukm }m=1. Let k1 ∈ I1 be arbitrary. For m 1, we pick km ∈ n=1 In such that km ≥ km−1. By this choice, we have Ñ −jn −n Ep (ukm , 2 ) < 2 whenever m ≥ n ≥ 1.

> N −nε < = −jnε ∞ Let ε 0. Then, there is nε ∈ such that 2 ε. Let δ 2 . Since {Aδ ukm }m=1 is p ′ N p , convergent in L (Z), there is m0 ∈ such that kAδukm − Aδ ukm′ kL (Z ) ≤ ε whenever m m ≥ m0. ′ Thus, for every m,m ≥ max{nε ,m0}, we have

p p + p + p kukm − ukm′ kL (Z ) ≤ kukm − Aδ ukm kL (Z ) kAδ ukm − Aδ ukm′ kL (Z ) kAδukm′ − ukm′ kL (Z ) , + + , < . ≤ Ep (ukm δ) ε Ep (ukm′ δ) 3ε > ∞ p Since ε 0 is arbitrary, we have shown that {ukm }m=1 is a Cauchy sequence in L (Z) and hence p ukm → u in L (Z) as m →∞. , ∞ Corollary 3.27. Assume that Z is bounded. Let p ≥ 1 and α ∈ (0 1]. Let {uk }k=1 be a sequence α ∞ of functions with uniformly bounded Bp,∞(Z) norm. Then, there is a subsequence {ukj }j=1 and a s s ∗ ∗ = function u ∈ L (Z) such that ukj → u in L (Z) for every s ∈ [1,p ), where p pQ/(Q − αp) if αp < Q and p∗ = ∞ otherwise. Moreover, s = ∞ is also possible if αp > Q.

Proof. The claim follows from Lemma 3.13 and Theorems 3.21 and 3.26 (see also Remark 3.22). In view of Corollary 3.18, the assertionin case s = ∞ for αp > Q follows from the compact embedding of Hölder continuous functions C0,κ (Z) ⋐ C(Z) ⊂ L∞(Z). → λα ∈ [ , ) Corollary 3.28. Under the assumptions of Corollary 3.27, ukj u in Bp˜,q for every λ 0 1 , , ˜ , −1 = ∗−1 + −1 ∗ = < q ∈ (0 ∞), and p ∈ [1 pλ ), where pλ (1 − λ)p λp , where p pQ/(Q − αp) if αp Q and p∗ = ∞ otherwise.

Proof. Let λ ∈ [0, 1) and p˜ ∈ [1,pλ ) be given. Without loss of generality, we may assume that p˜ > p since the Hölder inequality will then guarantee norm convergence for smaller values of p˜. Then, we can nd λ˜ ∈ (λ, 1) and s ∈ (p,p∗) such that

1 − λ˜ λ˜ 1 1 1 − λ˜ λ˜ + ≕ < = + . ∗ p p pλ˜ p˜ s p TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 17

∞ s s Let {uk }k=1 be the L (Z)-norm convergent subsequence selected in Corollary 3.27 and u ∈ L (Z) be the limit function. Then, 1−λ˜ λ˜ ku − uk λα . ku − uk λα˜ . ku − uk s ku − uk α k B (Z ) k B (Z ) k L (Z ) k B , (Z ) p˜,q p˜,∞ p ∞ . 1−λ˜ λ˜ kuk − ukLs (Z ) sup kuj kBα (Z ) → 0 as k →∞ j ≥1 p,∞ by Lemma 3.13 (see also Remark 3.14) and Corollary 3.11.

4. Traces of functions with a Poincaré ineality: general domains In this section, we will consider functions that satisfy a Poincaré inequality in a domain Ω ⊂ X, cf. Paragraph 2.11, and we will show that such functions have a measurable trace on ∂Ω in the sense of (1.2) provided that H⌊∂Ω has an upper codimension-θ bound (2.3) and a p-PI gradient of the Sobolev-type function exhibits suciently high summability. Even though the theorems and proofs in this section are stated for boundary traces of functions on a bounded domain Ω, the arguments will also work whenever ∂Ω is replaced by an arbitrary bounded set F ⊂ Ω such that H⌊F has an upper codimension-θ bound. In such a case Ω may very well be unbounded (unlike F). We will apply boundedness of a certain fractional maximal operator. For α ≥ θ, we dene the centered operator 1/p α p (4.1) Mα,p f (z) = sup r |f | dµ , z ∈ ∂Ω, K Ω 0

p p(s−θ ) Lemma 4.2. Let p ∈ [1, ∞). Suppose that α ∈ [θ,s). Then, Mα,p : L (Ω) → weak-L s−α (∂Ω) is ps/α ∞ bounded. Moreover, Mα,p : weak-L (Ω) → L (∂Ω) is bounded. p ∞ If however α ∈ [s, ∞), then Mα,p : L (Ω) → L (∂Ω) is bounded. Proof. Let f ∈ Lp (Ω) and suppose that α ∈ [θ,s). Having xed λ > 0, dene Eλ = {z ∈ ∂Ω : Mα,p f (z) > λ}. For each z ∈ Eλ, there is a = α p p Ω ball Bz B(z,rz ) such that r |f | dµ > λ . Then, Eλ ⊂ ∈ Bz ∩ ∂ . Since µ⌊Ω is z Bz ∩Ω z Eλ doubling and the radii rz are bounded⨏ by 2 diam ∂Ω, we can apply the simple Vitali 5-covering ≔ = Ð lemma to nd pairwise disjoint balls Bk Bzk , k 1, 2,..., for some choice of {zk } ⊂ Eλ such Ω that Eλ ⊂ k 5Bk ∩ ∂ . Then, µ(5B ∩ Ω) µ(B ∩ Ω) H(EÐ)≤H 5B ∩ ∂Ω . H(5B ∩ ∂Ω) . k ≈ k . λ k k θ θ (5rk ) r Øk Õk Õk Õk k By the choice of balls Bz , we have | |p β B f dµ µ(B ∩ Ω) µ(B ∩ Ω) k > k ≥ C k , ∫ λp r α r θ k k for some β ∈ [0, 1] that we will now determine. Recall that the doubling condition of µ⌊Ω gives a lower mass bound, i.e., µ(B(z,r) ∩ Ω) ≥ Cr s whenever r < 2 diam ∂Ω. We strive for µ(B(z,r) ∩ Ω)1−β /r α −θ β ≥ C for every r ∈ (0, 2 diam ∂Ω). This estimate will be satised whenever s(1 − β) − (α − θβ) ≤ 0. Thus, β ≥ (s − α)/(s − θ). 18 LUKÁŠ MALÝ

As α ∈ [θ,s), we have β ≔ (s − α)/(s − θ) ∈ (0, 1]. That gives us that

p (s−θ )/(s−α ) p (s−θ )/(s−α ) p (s−θ )/(s−α ) |f | dµ k |f | dµ Ω |f | dµ H( ) . Bk . Bk ≤ . Eλ p p p ∫ λ Í ∫ λ ∫ λ Õk Therefore, = (s−α )/p(s−θ ) kMα,p f kweak-Lp(s−θ )/(s−α )(∂Ω) sup λH(Eλ) ≤ k f kLp (Ω) . λ>0 The (generalized) Hölder inequality gives us that 1/p p 1/p k |f | χB(z,r ) kweak-Ls/α (Ω) r α |f |p dµ ≤ r α / K , µ(B(z,r) ∩ Ω)α s B(z r ) r α /p = k fχB(z,r ) k ps/α Ω . µ(B(z,r) ∩ Ω)1/s weak-L ( ) Ω 1/s The quantity r/µ(B(z,r) ∩ ) is bounded due to (2.1), and hence Mα,p f (z) . k f kweak-Lps/α (Ω) for every z ∈ ∂Ω. Assume now that α ∈ [s, ∞). Recall that 0 < r < 2 diam ∂Ω. By (2.1), we see that r α r s (diam ∂Ω)s = · r α −s . · r α −s ≤ C. µ(B(z,r) ∩ Ω) µ(B(z,r) ∩ Ω) µ(B(z, 2 diam ∂Ω) ∩ Ω) ∞ This yields that Mα,p f ≤ Ck f kLp (Ω) everywhere on ∂Ω, and hence Mα,p f ∈ L (∂Ω). ∞ q Remark 4.3. If α ∈ [θ,s), we immediately obtain that Mα,p f ∈ L (∂Ω) whenever f ∈ L (Ω) with q ≥ ps/α. On the other hand, if p < q < ps/α, then the Marcinkiewicz interpolation theo- ps q p(s−θ )/( −α ) rem yields that Mα,p : L (Ω) → L q (∂Ω) is bounded. Furthermore, it follows from [1, Theorems IV.4.11 and IV.6.14] that kMθ,p f kLp (∂Ω) . k f log(e + |f |)kLp (Ω). Now, we are ready to prove the rst trace theorem of this section, which shows existence and boundedness of a trace operator for functions with a p-PI gradient in Lp (Ω) with p > θ.

Theorem 4.4. Assume that Ω ⊂ X is a bounded domain such that µ⌊Ω is doubling and H⌊∂Ω has 1,p Ω an upper codimension-θ bound. Let u ∈ Pp ( ) with some p ∈ (θ, ∞). Then, there exists a trace Tu ∈ Lp (∂Ω) that satises (1.2), i.e.,

lim |u(x)− Tu(z)| dµ(x) = 0 for H-a.e. z ∈ ∂Ω. R→0 KB(z,R)∩Ω 1,p Ω 1−θ /p Ω Moreover, T : Pp ( ) → Bp,∞ (∂ ) is a bounded linear operator. In fact, the trace exhibits better integrability than merely Lp (∂Ω) due to the embeddings of Besov spaces and boundedness of the fractional maximal operator, see Corollary 4.14 below. As noted in the introductory paragraph of this section, the proof below can be easily modied 1,p Ω 1−θ /p Ω Ω to obtain that T : Pp ( ) → Bp,∞ (F) is bounded whenever F ⊂ is bounded (while ⊂ X may be unbounded) and H⌊F has an upper codimension-θ bound.

∈ 1,p (Ω) ∈ ∂Ω > ( ) = Proof. Let u Pp be given. For z and r 0, we dene Tr u z B(z,r )∩Ω u dµ. Let us now show thatTru ∈ C(∂Ω) for every r > 0. Forthe sakeofbrevity, let Bz denote⨏ the set B(z,r)∩Ω whenever z ∈ ∂Ω for some xed r > 0. As balls have nite measure, we have µ(Bw △Bz ) → 0 and, TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 19 in particular, µ(Bw ) → µ(Bz ) as w → z, where w, z ∈ ∂Ω. Since we are interested in the behavior of Tr u(w) as w → z, we may also assume that Bw ⊂ 2Bz . Then, we can estimate 1 1 1 1 |Tru(z)− Tru(w)| ≤ u dµ − u dµ + − u dµ µ(B ) µ(Bw ) µ(B ) µ(Bw ) z Bz \Bw Bw \Bz z Bw ∩Bz ¹ ¹ ¹ 1 1 1 1 ≤ + |u| dµ + − |u| dµ, µ(B ) µ(Bw ) µ(B ) µ(Bw ) z Bz △Bw z 2Bz ¹ ¹ w → ∈ 1(Ω) which approaches 0 as z since u L . { }∞ p (∂Ω) → As the next step, we will show that Trk u k=1 is a Cauchy sequence in L whenever rk 0. For z ∈ ∂Ω and ρ > 0, let Bz, ρ = B(z, ρ) ∩ Ω and Ωρ = {x ∈ Ω : dist(x,∂Ω) < ρ}. Suppose that p Ω < R < < g ∈ L ( ) is a p-PI gradient of u with a dilation factor λ ≥ 1, see (2.12). Let 0 2 ≤ r R 2 diam Ω. Then, p p = p . kTRu − Truk p ∂Ω |uBz,R − uBz,r | dH(z) |u(x)− uBz,R | dµ(x) dH(z) L ( ) Ω Ω K ¹∂ ¹∂ Bz,R Rp . Rp g(x)p dµ(x)dH(z)≈ g(x)p dH(z)dµ(x) Ω K Ω Ω µ(B , ) ¹∂ Bz,λR ¹ λR ¹B(x,λR)∩∂ z λR Rp−θ dH(z) . g(x)p dµ(x) . Rp−θ g(x)p dµ(x), Ω Ω H(B(z, λR) ∩ ∂Ω) Ω ¹ λR ¹B(x,λR)∩∂ ¹ λR where in the last inequality we used that H is doubling while B(x, λR) ∩ ∂Ω ⊂ B(z, 2λR) ∩ ∂Ω whenever z ∈ B(x, λR). Let now 0 < r < R ≤ 2 diam Ω. Then, there is N ∈ N such that 2−N R ≤ r < 21−N R. Thus, N + kTRu − Tr ukLp (∂Ω) ≤ kTru − T2−N RukLp (∂Ω) kT21−k R − T2−k R kLp (∂Ω) = Õk 1 N 1−k 1−θ /p 1−θ /p (4.5) . (2 R) kgk p (Ω ) . R kgk p (Ω ). L 21−k λR L λR = Õk 1 { }∞ p (∂Ω) → We have hereby shown that Trk u k=1 is a Cauchy sequence in L whenever rk 0. Thus, p p we can dene the trace of u as the L -limit, i.e., Tu ≔ limr →0 Tru ∈ L (∂Ω). Moreover, it follows Ω from (4.5) that T2−k u(z) → Tu(z) as k →∞ for H-a.e. z ∈ ∂ . Consequently, TRu(z) → Tu(z) as R → 0 for H-a.e. z ∈ ∂Ω since µ⌊Ω is doubling. If R = 2 diam Ω, then TRu ≡ uΩ. Hence, 1−θ /p kTu − uΩ kLp (∂Ω) = lim kTru − TRukLp (∂Ω) . (diam Ω) kgkLp (Ω). r →0

Let E = {z ∈ ∂Ω : Mθ,pg(z) < ∞ and Tru(z) → Tu(z) as r → 0}. It follows from Lemma 4.2 that p p Mθ,p : L (Ω) → weak-L (∂Ω) is bounded, whence H(E) = 0. For every z ∈ ∂Ω \ E and r > 0, we obtain then that

1−θ /p Ω Finally, in order to show that Tu ∈ Bp,∞ (∂ ), we will next nd an estimate for Ep (Tu, R), which was dened in (3.7), for R > 0. If z, w ∈ ∂Ω with d(z, w) ≤ R, then

(4.7) |Tu(z)− Tu(w)| ≤ |Tu(z)− TRu(z)| + |TRu(z)− T2Ru(w)| + |T2Ru(w)− Tu(w)|.

Note also that the doubling condition for µ⌊Ω leads to the estimate

Comparing the situation with well-known trace theorems for domains in the Euclidean spaces, 1,p Ω 1−θ /p Ω one can expect that the trace class of P∗ ( ) should be the Besov space Bp,p (∂ ) provided that the boundary is suciently regular. In Section 5 below, we will obtain the expected boundedness 1,p Ω 1,p Ω Ω of the trace operator for Pp ( ) and Pq ( ) if is a uniform or a John domain, respectively. 1,p Ω Ω Proposition 4.11. Let u ∈ Pp ( ) for some p ∈ (θ, ∞). Then, there exist a set E ⊂ ∂ with p p H(E) = 0 and a function h ∈ weak-L (∂Ω) with khkweak-Lp (∂Ω) . kgkLp (Ω), where g ∈ L (Ω) is a p-PI gradient of u, such that

|Tu(z)− Tu(w)| . d(z, w)1−θ /p h(z) + h(w) for every z, w ∈ ∂Ω \ E. Proof. Let 0 < R ≤ 2 diam(Ω) and z ∈ ∂Ω. Using the notation established in the proof of Theo- rem 4.4, we have

1/p |T u(z)− T u(z)| . R gp ≤ R1−θ /pM g(z) , R R/2 K θ,p BλR TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 21 where Mθ,p is the fractional maximal operator dened in (4.1). Thus, ∞ ∞ 1−k 1−θ /p 1−θ /p |TRu(z)− Tu(z)| ≤ |T2−k Ru(z)− T21−k Ru(z)| ≤ (2 R) Mθ,pg(z)≈ R Mθ,pg(z). = = Õk 1 Õk 1 Let now z, w ∈ ∂Ω and R = d(z, w). Then,

If we set h = cMθ,pg, then khkweak-Lp (∂Ω) . kgkLp (Ω) by Lemma 4.2 and the proof is complete. If the couple (u,g) satises a Poincaré inequality in the form 1/pz 1/p pz p |u − uB | dµ ≤ CPI rad(B) g dµ K Ω K Ω B∩ λB∩ for some pz > 1, then the trace satises not only (1.2), but also a stronger relation

z (4.13) lim |u(x)− Tu(z)|p dµ(x) = 0 for H-a.e. z ∈ ∂Ω. R→0 KB(z,R)∩Ω Indeed, similarly as in (4.6), one obtains for H-a.e. z ∈ ∂Ω that 1/pz 1/pz z z |u(x)− Tu(z)|p dµ ≤ |u(x)− T u(z)|p dµ(x) + |T u(z)− Tu(z)| K K r r Bz,r Bz,r 1−θ /p . r Mθ,pg(z) + |Tru(z)− Tu(z)| → 0 as r → 0. In view of the Sobolev-type embeddings [15, Theorem 5.1], (4.13) holds for every pz < (s − p)/sp z 1,p z if p < s and for every p < ∞ if p ≥ s. If u ∈ Pq for some q < p < s, then p = (s − p)/sp is also possible by Corollary 2.18 and [13, Theorem 6]. 1,p Ω Corollary 4.14. Suppose that u ∈ Pp ( ) for some p > θ. Then: α ∂Ω < θ < < = p(s−θ ) (i) Tu ∈ Bm,q ( ) for every 0 ≤ α 1 − p , 1 ≤ m pα , and 0 q ≤∞, where pα s−p(1−α ) if p(1 − α) < s and pα = ∞ otherwise. m ∂Ω , ∗ ∗ = p(s−θ ) < ∗ = (ii) Tu ∈ L ( ) for every m ∈ [1 p ), where p s−p if p s, while p ∞ otherwise. > 0,κ ∂Ω = s (iii) If p s, then Tu ∈ C ( ), where κ 1 − p . , θ , , θ < Proof. (i) Given α ∈ [0 1− p ) andm ∈ [1 pα ), we can nd α˜ ∈ (α 1− p ) such thatm pα˜ . A simple modication of the proof of Proposition 4.11 yields an analog of (4.12), viz. |Tu(z) − Tu(w)| . α˜ α˜ d(z, w) (Mp−αp˜ ,pg(z) + Mp−αp˜ ,pg(w)) while |Tu(z)− uΩ | . (diam Ω) Mp−αp˜ ,pg(z). By Lemma 4.2, p m Mp−αp˜ ,pg ∈ weak-L α˜ (∂Ω) ⊂ L (∂Ω). Then, Lemmata 3.20 and 3.13 give the desired result. (ii) The assertion follows immediately from (i) by choosing α = 0. 1−s/p (iii) We use another analog of (4.12), viz. |Tu(z)−Tu(w)| ≤ Cd(z, w) (Ms,pg(z) + Ms,pg(w)). p Ω ∞ ∂Ω s Since Ms,p : L ( ) → L ( ) is bounded by Lemma 4.2, we obtain that Tu has a (1 − p )-Hölder continuous representative. 1,p Ω Corollary 4.15. Assume that H is Ahlfors codimension-θ regular. Suppose that u ∈ Pp ( ) for some p > θ. Then: ∈ α (∂Ω) − s < ≤ − θ = p(s−θ ) (i) Tu Bpα ,∞ for every 1 p α 1 p , where pα s−p(1−α ) . p∗ m ∗ p(s−θ ) ∂Ω ∗ ∂Ω = < (ii) Tu ∈ weak-L ( ) ⊂ m

Proof. Follows from Theorem 4.4, Remark 3.19, and Theorem 3.23.

Proposition 4.16. The trace T : M1,p (Ω) → Lm (∂Ω) is a compact operator for every m < p∗, where ∗ = p(s−θ ) < ∗ = Ω p s−p if p s while p ∞ otherwise. Moreover, if admits a p-Poincaré inequality, then T : N 1,p (Ω) → Lm(∂Ω) is compact for every m < p∗.

Proof. Follows from Corollaries 2.18, 4.14 (i), and 3.27; see also commentary after Denition 2.14.

The proofs of statements about the trace operator above can be easily adapted for functions in α Ω α Ω the Besov class Bp,q( ) provided that αp > θ, q ∈ (0, ∞]. If u ∈ Bp,q( ) for some q ≤ p, then Lemma 3.16 provides us with a function g ∈ Lp (Ω) so that (u,g) satises the fractional Hajłasz α Ω gradient inequality (3.17). By [14, Theorem 9.3], we have u ∈ Pp,q ( ) for some q < p, i.e.,

1/q |u − u | dµ ≤ rad(B)α gq dµ , K B K B λB p Ω α −θ /p Ω where g ∈ L ( ). One obtains that Tu ∈ Bp,∞ (∂ ) by following the steps of the proof of α −θ /p Theorem 4.4. Similarly as in (4.12), we have |Tu(z)−Tu(w)| . d(z, w) (Mθ,pg(z) + Mθ,pg(w)). Open-ended boundedness and compactness of the trace operator can then be obtained similarly as in Corollary 4.14 and Proposition 4.16. α Ω α −ε Ω In case u ∈ Bp,q ( ) with q ∈ (p, ∞], one can rst use Lemma 3.13 to see that u ∈ Bp,p ( ) α −ε−θ /p Ω for every ε > 0 and then proceed as above to show that Tu ∈ Bp,∞ (∂ ) and thus obtain the corresponding open-ended results. Next, we will show a positive result on existence of a trace when a function has a θ-PI gradient in a weighted Lθ (Ω). In Proposition 8.4 below, it is shown that the hypothesis here is essentially sharp for θ = p > 1. On the other hand, if θ = p = 1 and ∂Ω is Ahlfors codimension-1 regular, 1,1 Ω 1 Ω 1 then T : P1 ( ) → L (∂ ) without any extra assumptions on weighted L -integrability of the 1-PI gradient by [32].

Theorem 4.17. Assume that θ = p. Let w : (0, ∞) → [1, ∞) be a decreasing function such that 1 dt < ∞ ∈ 1,p (Ω) ( ) ≔ ( )w( ( ,∂Ω)) ∈ p (Ω) ∈ p (Ω) 0 t w(t) . Suppose that u Pp and g˜ x g x dist x L , where g L is a p-PI gradient of u. Then, Tu satisfying (1.2) exists and kTu − uΩ k p Ω . kg˜k p Ω . ∫ L (∂ ) L ( ) In particular, the proposition can be applied for the weight function

+ + log e 1 ε for 0 < t < 1, 2 diam Ω 1 ε w(t) = t for which g˜(x)≈ g(x) log , x ∈ Ω, 1 for 1 ≤ t < ∞, dist(x,∂Ω) ( where ε > 0 is arbitrary.

Sketch of proof. We can follow the proof of Theorem 4.4 without any modication until (4.5). In what follows, we will be using the notation of the aforementioned proof. < R < < Ω Let 0 2 ≤ r R 2 diam . Then,

1/p k k p Ω p g˜ L ( λR ) kTRu − TrukLp (∂Ω) . g(x) dµ(x) ≤ . Ω w(λR) ¹ λR TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 23

Let now 0 < r < R < 2 diam Ω. Then, there is N ∈ N such that 2−N R ≤ r < 21−N R. Thus,

N + kTRu − Tr ukLp (∂Ω) ≤ kTru − T2−N RukLp (∂Ω) kT21−k R − T2−k R kLp (∂Ω) = Õk 1 N N 21−k λR 2λR kg˜kLp (Ω) kg˜kLp (Ω) dt dt . ≈ ≤ kg˜k p Ω . w 1−k w 1−k L ( ) w = (2 λR) = (2 λR) 2−k λR t 0 t (t) Õk 1 Õk 1 ¹ ¹ ( ) ≔ ρ dt → → { } Since W ρ 0 t w(t) 0 as ρ 0, we have hereby shown that Trk u k is a Cauchy sequence in Lp (∂Ω) whenever r → 0. Thus, the limit function Tu lies in Lp (∂Ω) and T u(z) → Tu(z) as ∫ k R R → 0 for H-a.e. z ∈ ∂Ω. If R = 2 diam Ω, then TRu ≡ uΩ. Hence, kTu − uΩ kLp (∂Ω) . kg˜kLp (Ω)W (2λ diam Ω). Let E = {z ∈ ∂Ω : Mθ,pg˜(z) < ∞ and Tr u(z) → Tu(z) as r → 0}. Then, H(E) = 0 and for every z ∈ ∂Ω \ E, we obtain that

Corollary 4.18. Let p ≥ 1. Assume that both (2.2) and (2.3) are satised for some 0 < ϑ ≤ θ = p. 1,p Ω p Ω + 1+ε p Ω Suppose that u ∈ Pp ( ) has a p-PI gradient g ∈ L ( ) such that g˘ ≔ g log(e g) ∈ L ( ) for some ε > 0. Then, kTu − uΩ kLp (∂Ω) . (1 + kg˘kLp (Ω)).

1+ε Proof. The claim follows from Theorem 4.17 with the weight w(t) = 1 + log+(∆/t) , where ∆ = diam Ω. Let us therefore verify that g˜ ∈ Lp (Ω), where g˜ is as in the theorem above. −k 1−k For the sake of brevity, let δ(x) = dist(x,∂Ω). Dene Ωk = {x ∈ Ω : 2 ∆ < δ(x) ≤ 2 ∆}, −k ϑ k = 1, 2,.... Then, µ(Ωk ) . H(∂Ω)(2 ∆) by Lemma 2.4. Let α ∈ [0, ϑ), β > 0.

∞ ∆ α ∆ 1 α 1 logβ dµ ≤ logβ dµ Ω δ(x) δ(x) Ω 2−k 2−k ¹ k=1 ¹ k Õ∞ ∞ kα β β k(α −ϑ ) ≈ 2 k µ(Ωk ) . H(∂Ω) k 2 < ∞. = = Õk 1 Õk 1 We can split Ω into two parts, namely E = x ∈ Ω : g(x) ≤ ∆/δ(x) ϑ /2p and F = Ω \ E. Then,

ϑ /2 + ∆ ∆ + ∆ g(x)p logp(1 ε) dµ ≤ logp(1 ε) dµ Ω δ(x) E δ(x) δ(x) ¹ ¹ + 2p p(1 ε) + g(x)p log g(x) dµ . 1 + kg˘kp . ϑ Lp (Ω) ¹F Ω > ≔ 2 diam( ) 1/p p Ω Proposition 4.19. Assume that θ p. Suppose that g˜(x) g(x) log( dist(x,∂Ω) ) ∈ L ( ), where p Ω 1,p Ω 1−θ /p Ω g ∈ L ( ) is a p-PI gradient of u ∈ Pp ( ). Then, Tu ∈ Bp,p (∂ ). 1−θ /p Ω In particular, in order to obtain Tu ∈ Bp,p (∂ ), it suces to assume that both (2.2) and (2.3) are satised for some 0 < ϑ ≤ θ < p and that g˘ ≔ g log(e + g)1/p ∈ Lp (Ω) . 24 LUKÁŠ MALÝ

Sketch of proof. Let (4.10) from the proof of Theorem 4.4 be the starting point of the current proof. Let K = 2 diam Ω. Then, K p K Ep (Tu, R) dR dR kTukp = . kgkp Û1−θ /p Ω p−θ Lp (Ω ) B , (∂ ) R R 2λR R p p ¹0 ¹0 K = p dR = p 2λK p g(x) dµ(x) g(x) log dµ(x) ≈ kg˜k p Ω . Ω Ω R Ω dist{x,∂Ω} L ( ) ¹ ¹dist{x,∂ }/2λ ¹ If g˘ ∈ Lp (Ω), then g˜ ∈ Lp (Ω), which can be proven similarly as in Corollary 4.18. Remark 4.20. Instead of assuming that g˜ ∈ Lp (Ω) in Theorem 4.17 and 4.19, it is possible to obtain the desired boundedness of the trace operator by logarithmically rening the codimension relation between H⌊∂Ω and µ⌊Ω. Namely, if there is η > 1 such that µ(B(z,r) ∩ Ω) H(B(z,r) ∩ ∂Ω) . , z ∈ ∂Ω, 0 < r < 2 diam Ω, θ + 1 + + 1 η r log e r log e log 1 r 1,p Ω 1−θ /p Ω then T : Pp ( ) → Bp,p (∂ ) is bounded whenever p >θ. Similarly, if there is η > 1 such that µ(B(z,r) ∩ Ω) H(B(z,r) ∩ ∂Ω) . , z ∈ ∂Ω, 0 < r < 2 diam Ω, + 1 + + 1 η θ r log e r log e log 1 r 1,θ Ω θ ∂Ω then T : Pθ ( ) → L ( ) is bounded. Both these statements can be proven as Theorem 4.4, mutatis mutandis. This section will now be concluded by establishing local estimates for the trace operator. Such estimates can be applied when studying regularity of solutions of the Neumann problem for the p-Laplace equation using the de Giorgi method, cf. [33]. If a domain Ω ⊂ X is quasiconvex, then a biLipschitz change of the metric will turn it into a length space. Recall that Ω is called quasiconvex if every two distinct points x,y ∈ Ω can be connected by a curve inside Ω whose length is at most Ld(x,y) for some universal constant L ≥ 1. If L = 1, then we say that Ω a geodesic space. If Ω is quasiconvex for every L > 1, then it is called a length space. A biLipschitz modication of the metric does not change the Hajłasz space M1,p (Ω) and hence 1,p the Pq (Ω) space also remains unchanged. Thus, we do not lose generality by assuming that Ω is a length space (rather than merely quasiconvex) in the proposition below. Proposition 4.21. Assume that a domain Ω ⊂ X is a length space. Suppose that θ < p < s while p ≥ 1. Let p˜ ∈ (p,p∗), where p∗ = p(s − θ)/(s − p). Then, the trace operator T constructed in Theorem 4.4 satises the local estimate 1 1 ( p˜ − p∗ )(s−θ ) kTu − uB∩Ω kLp˜ (B∩∂Ω) ≤ C rad(B) kgkLp (B∩Ω) 1,p for every u ∈ Pq (Ω) with q < p and every ball B centered at an arbitrary boundary point, where g ∈ Lp (Ω) is a q-PI gradient of u. In particular, the conclusion of the proposition holds true for every u ∈ N 1,p (Ω) with g ∈ Lp (Ω) being its p-weak upper gradient provided that Ω admits a p-Poincaré inequality and p > 1. Proof. Fix a couple of Lp functions (u,g) that satisfy (2.12) with q < p and λ = 1. By the Hölder s θ > s−θ inequality, we may assume that q is suciently close to p so that p − q p˜ . Fix also a ball = , ∂Ω = 1 , 1 ,∂Ω Ω B B(z R). For every point x ∈ B ∩ , dene rx 2 (R − d(x z)) ≤ 2 dist({x} \B). Since is TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 25 a length space, we can nd an arc-length parametrized curve γx : [0,lx ] → Ω such that γx (0) = z, γx (lx ) = x, γx (0,lx ) ⊂ Ω, and lx ≤ (1 + δ)d(x, z), where the constant δ = δx ∈ (0, 1) is chosen such that (1 + δ)lx < R. Next, we construct a nite decreasing sequence of balls whose centers lie on γx and all the balls = = contain the point x and are contained in B(z, R). Let N ⌈log2(2R/rx )⌉. For each k 0, 1,..., N , k+1 −k −k let rk = (δ + 2 )lx and xk = γx (1 − 2 )lx . Then, we dene Bk = B(xk ,rk ). It follows from the triangle inequality that Bk+1 ⊂ Bk ⊂ B(z, R), and x ∈ Bk for all k = 0, 1,... N . For the sake of clarity, let us see detailed calculations:

d(y, z) ≤ d(y, xk ) + d(xk , z) k+1 −k −k k+1 < (δ + 2 )lx + (1 − 2 )lx = (1 + δ )lx < R whenever y ∈ Bk ; −(k+1) d(w, xk ) ≤ d(w, xk+1) + d(xk , xk+1) < rk+1 + 2 lx k+2 −(k+1) k+1 −k ≤ (δ + 2 · 2 )lx < (δ + 2 )lx = rk whenever w ∈ Bk+1; −k −k d(x, xk ) ≤ lx − (1 − 2 lx ) = 2 lx < rk .

−k For k > N , we dene Bk = B(x, 2 lx ) ⊂ B(x,rx ). The dierence |Tu(x)− uB∩Ω | can be estimated = using the chain of balls {Bk }k since Tu(x) limk→∞ uBk ∩Ω as follows: ∞ + |uB∩Ω − Tu(x)| ≤ |uB∩Ω − uB0∩Ω | |uBk ∩Ω − uBk−1∩Ω | k=1 Õ ∞ . + |u − uB∩Ω | dµ |u − uBk ∩Ω | dµ K Ω K Ω B∩ = Bk ∩ Õk 0 1/q ∞ 1/q ( 1 − 1 )(s−θ ) q + −k q p˜ p∗ ∗ . R g dµ 2 R g dµ . R Mα,qg(x), K Ω K Ω B∩ = Bk ∩ Õk 0 = 1 1 ∗ where α q − (p˜ − p∗ )(s − θ)q and Mα,q denotes a restricted non-centered fractional maximal operator, dened by

1/q ∗ = α q Ω q Ω Mα,q f (x) sup rad(U ) |f | dµ , x ∈ B ∩ ∂ , f ∈ L (B ∩ ). x ∋U KU ∩Ω ball U ⊂B s θ > s−θ > The condition p − q p˜ establishedat thebeginning of theproof yieldsthat α θ. Boundedness of the restricted non-centered fractional maximal operator can be proven similarly as for the non- p p˜ restricted centered operator in Lemma 4.2. In particular, Mα,q : L (B ∩ Ω) → L (B ∩ ∂Ω) as in Remark 4.3 (note however that the roles of q and p are switched in that remark). Thus,

( 1 − 1 )( − ) ( 1 − 1 )( − ) p˜ p∗ s θ ∗ p˜ p∗ s θ kuB∩Ω − TukLp˜ (B∩∂Ω) . R kMα,qgkLp˜ (B∩∂Ω) . R kgkLp (B∩Ω) . Remark 4.22. Under the assumptions of Proposition 4.21, one can show that

( 1 − 1 −ε)(s−θ ) p˜ p∗ Ω ε kTu − uB∩Ω kLp˜ (B∩∂Ω) ≤ Cε rad(B) H(B ∩ ∂ ) kgkLp (λB∩Ω) 1,p Ω , 1 1 for every u ∈ Pp ( ), ε ∈ 0 p˜ − p∗ , and every ball B centered at an arbitrary boundary point. Indeed, one can follow the steps of the proof above, but pick p˘ ∈ (p˜,p∗) and use the boundedness p Ω p˘ ∂Ω ֒ p˜ ∂Ω ˘ = 1 1 : ∗ of Mα˘,p L (B ∩ ) → weak-L (B ∩ ) → L (B ∩ ), where α p − p(p˜ − p∗ − ε)(s − θ). 26 LUKÁŠ MALÝ

5. Traces of functions with a Poincaré ineality: John and uniform domains In this section, we will focus on showing that it is in fact possible to obtain a closed-ended 1,p Ω 1−θ /p Ω smoothness result for the trace, namely that T : Pq ( ) → Bp,p (∂ ), where q < p and p ≥ 1, under the additional assumption that Ω is a John domain with compact closure. If Ω ⋐ X is a 1,p Ω 1−θ /p Ω uniform domain, then also T : Pp ( ) → Bp,p (∂ ) is bounded. Lemma 5.1. Let z ∈ ∂Ω be xed. Suppose that a ∈ (0, diam(∂Ω)) and α ∈ (0, ∞). Then, dH(w) 1 . Ω d(z, w)α H(B(z, d(z, w)) ∩ ∂Ω) aα ¹∂ \B(z,a) Proof. Let us split the integration domain dyadically with respect to the distance from z. For k+1 k k = 0, 1,..., let Ak = B(z, 2 a)\ B(z, 2 a). Then, w Ka w dH( ) = dH( ) Ω ( , w)α H( ( , ( , w)) ∩ ∂Ω) Ω ( , w)α H( ( , ( , w)) ∩ ∂Ω) ∂ \B(z,a) d z B z d z = ∂ ∩Ak d z B z d z ¹ Õk 0 ¹ K K a 1 H(B(z, 2k+1a) ∩ ∂Ω) 1 a 1 ≈ · . 2−kα ≈ , k α k Ω α α = (2 a) H(B(z, 2 a) ∩ ∂ ) a = a Õk 0 Õk 0 Ω where Ka ≔ ⌈log2 diam(∂ )/a⌉− 1. Lemma 5.2. Fix x ∈ Ω and let δ = dist(x,∂Ω). Suppose that a ∈ (δ, 2 diam(Ω)) and α ∈ (0, ∞). Then, d(x, z)α dH(z) . aα . Ω H(B(z, d(x, z)) ∩ ∂Ω) ¹∂ ∩B(x,a) Proof. Let x ∈ Ω and δ = dist(x,∂Ω). Then, there is x˜ ∈ ∂Ω with d(x, x˜) < min{a, 3δ/2} such that B(x, a) ⊂ B(x˜, 2a). On one hand, we can estimate d(x, z) ∈ [δ, 7δ/2) whenever z ∈ ∂Ω ∩ B(x˜, 2δ). On the other hand, for every z ∈ ∂Ω \ B(x˜, 2δ), we have 3δ 3 d(z, x)− d(z, x˜) ≤ d(x, x˜) < < d(z, x˜), 2 4 which yields that 1 7 d(z, x˜)≤ d(z, x) ≤ d(z, x˜) . 4 4 Consequently, d(x, z)α dH(z) Ω H(B(z, d(x, z)) ∩ ∂Ω) ¹∂ ∩B(x,a) δ α d(z, x˜)α . dH(z) + dH(z). Ω H(B(x˜, δ) ∩ ∂Ω) Ω H(B(x˜, d(z, x˜)) ∩ ∂Ω) ¹∂ ∩B(x˜,2δ ) ¹∂ ∩B(x˜,2a)\B(x˜,2δ ) k ka We can nd ka,kδ ∈ Z such that δ < 2 δ ≤ 2δ and a ≤ 2 < 2a. For the sake of brevity, let k+1 k Ak = B(x˜, 2 )\ B(x˜, 2 ), k ∈ Z. Then, d(z, x˜)α dH(z) Ω H(B(x˜, d(z, x˜)) ∩ ∂Ω) ¹∂ ∩B(x˜,2a)\B(x˜,2δ ) k k a 2kα a . dH(z) . 2kα . Ω , , k ∂Ω = ∂ ∩Ak H(B(z d(x 2 )) ∩ ) = kÕkδ ¹ kÕkδ TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 27

Altogether, d(x, z)α dH(z) . δ α + aα . aα . Ω H(B(z, d(x, z)) ∩ ∂Ω) ¹∂ ∩B(x,a) Denition 5.3 (cf. [15, Section 9.1]). A domain Ω ⊂ X is called a John domain with John constant c J ∈ (0, 1] and John center at y ∈ Ω if every x ∈ Ω can be joined to y by a rectiable curve γ : [0,lγ ] → Ω parametrized by arc-length such that γ (0) = x, γ (lγ ) = y, and

(5.4) dist(γ (t), X \ Ω)≥ c J t for all t ∈ [0,lγ ]. Suppose that Ω is a John domainwith compactclosure. Then, we can invoke the Ascoli theorem (cf. [40, p. 169]) to see that every z ∈ ∂Ω can also be joined to the John center by a rectiable curve such that (5.4) is satised. 1 Ω ∂Ω > R Lemma 5.5. Let u ∈ Lloc( ), z ∈ , and c 0. Assume that there is a ∈ such that

lim |u − a| dµ = 0. R→0 KB(z,R)∩Ω ∞ Ω ∞ For every sequence of points {xk }k=1 ⊂ with xk → z as k →∞ and every sequence of radii {rk }k=1 such that rk → 0 as k →∞ and rk ≥ cd(xk , z) for all k = 1, 2,..., we then have

lim |u − a| dµ = 0. k→∞ K B(xk,rk )∩Ω

Proof. Using the doubling property of µ⌊∂Ω, we can estimate

µ B(z,rk + d(xk , z)) ∩ Ω |u − a| dµ ≤ |u − a| dµ K , Ω µ B(x ,r ) ∩ Ω K , + , Ω B(xk rk )∩ k k B(z rk d(xk z))∩ . |u −a| dµ → 0 as k →∞. K B(z,(1+1/c)rk )∩Ω 1,p Theorem 5.6. Let Ω be a John domain whose closure is compact. Suppose that u ∈ Pq (Ω) for some 0 < q < p, where p ≥ 1 and p > θ. Then, the trace Tu dened by (1.2), whose existence was 1−θ /p Ω = 1−θ /p Ω established in Proposition 4.11, lies in Bp (∂ ) Bp,p (∂ ). + p Ω Moreover, kTuk 1−θ /p Ω ≤ C(kukLp (Ω) kgkLp (Ω)), where g ∈ L ( ) is a q-PI gradient of u. Bp (∂ ) The restriction p > θ is not an artifact of the proof of the theorem above. In fact, we will show in Section 7 below that any Lθ (∂Ω) boundary data can be extended to an N 1,θ (Ω) function. In Ω 1,θ Ω = 1,θ Ω particular, if admits a θ-Poincaré inequality with some 1 ≤ θ, then N ( ) P1 ( ), which shows that there is no chance to obtain any kind of smoothness of the trace in case p = θ > 1. The case of p = θ = 1 for Ω admitting 1-Poincaré inequality was covered in [32, 34], where it was 1,1 Ω = 1,1 Ω Ω 1 Ω proven that the trace class of N ( ) P1 ( ) and of BV ( ) is L (∂ ), which again does not allow for any Besov-type smoothness of the trace function. The proof below would fail at one step if we allowed q = p. Namely, we will use the non- centered Hardy–Littlewood maximal operator M to dene an auxilliary function h ≔ (Mgq )1/q. If q < p, then khkLp ≈ kgkLp , but we would have only a weak estimate khkLp,∞ . kgkLp in case q = p. However, if Ω is a uniform domain, then usage of the maximal function can be circumvented, see Theorem 5.13 below. 1,p Ω p Ω Proof. Let u ∈ Pq ( ) with a q-PI gradient g ∈ L ( ) and a dilation factor λ ≥ 1 on the right-hand side of (2.12) be given. Let δ = dist{a,∂Ω} > 0, where a ∈ Ω denotes the John center of Ω. Fora 28 LUKÁŠ MALÝ

xed pair of points y, z ∈ ∂Ω with d(y, z) ≤ δ, let γy and γz be the arc-length parametrized curves that connect the points y and z, respectively, to the John center a ∈ Ω and satisfy (5.4). = , c J k = c J = , ,... Let tk d(y z)(1 − 2λ ) and rk 2λ tk for k 1 2 . Next, we dene a chain of balls {Bk }k ∈Z by setting

B0 = B(z, 3d(y, z)), B−k = B(γy (tk ),rk ), and Bk = B(γz (tk ),rk ), for all k = 1, 2,.... = , = ∞ For the sake of simpler notation, let r0 3d(y z) and r−k rk . The subchains {Bk }k=1 and ∞ {B−k }k=1 consist of ballsof bounded overlap where the upper bound on the number of overlapping ≔ c J balls depends on the John constant c J . Moreover, abbreviating α 2 − 2λ , we have Bk+1 ⊂ αBk , which follows from the triangle inequality. Indeed, if x ∈ Bk+1, then c J d x,γ (t ) ≤ d x,γ (t + ) + d γ (t + ),γ (t ) ≤ r + + t − t + = r + + r = r 2 − . z k z k 1 z k 1 z k k 1 k k 1 k 1 k k 2λ The triangle inequality also implies for all k ≥ 1 and x ∈ αλBk that

c J t dist(γ (t ),∂Ω) d(γ (t ), z) d(γ (t ), x) + d(x, z) αr d(x, z) r = k ≤ k ≤ k ≤ k ≤ k + . k 2λ 2λ 2λ 2λ 2 2λ Conversely, 2λ d(x, z) ≤ d(x,γ (t )) + d(γ (t ), z)≤ αλr + t = αλ + r . k k k k c k J Hence, d(x, z) 2d(x, z) (5.7) ≤ rk ≤ if x ∈ αλBk . λ(α + 2/c J ) c J

We can also use the triangle inequality to show that if x ∈ αλBk , k ≥ 1, then

dist(x,∂Ω)≥ dist(γ (tk ),∂Ω)− αλrk

c J c J d(x, z) ≥ c J tk − αλrk = (2λ − αλ)rk = rk ≥ ≕ β d(x, z) 2 2λ(α + 2/c J ) and conversely,

2 c J α dist(x,∂Ω)≤ d(x, z) ≤ λ α + r = + 1 t ≤ 2d(y, z) . c k 2 k J An analogous argument can be carried out for the balls Bk with k ≤−1. Consequently, the chains of inated balls {αλBk }k ≥1 and {αλB−k }k ≥1 are contained in “John carrots” ∞ (5.8) αλBk ⊂ x ∈ Ω : βd(x, z) ≤ dist(x,∂Ω) ≤ 2d(z,y) ≕ Cz,y and k=1 ∞Ø αλB−k ⊂ x ∈ Ω : βd(x,y) ≤ dist(x,∂Ω)≤ 2d(y, z) ≕ Cy,z = kØ1 Lemma 5.5 yields that

Tu(y) = lim u dµ and Tu(z) = lim u dµ. k→∞ K k→∞ K B−k ∩Ω Bk ∩Ω TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 29

Since p > θ, we can nd ε > 0 such that p −θ −ε > 0. Using a chaining argument and the doubling property of µ similarly as before, we obtain that 1/q . . q (5.9) |Tu(y)− Tu(z)| |u − uαBk | dµ αrk g dµ . KΩ KΩ Z ∩αBk Z ∩αλBk kÕ∈ kÕ∈ Note that αr0 = C(c J )· r1. Let 1/q q h(x) = sup g dµ ≕ Mqg(x), x ∈ Ω, KΩ B ∋x ∩B be a non-centered maximal function of the Hardy–Littlewood type in Ω for g. Since µ⌊Ω is dou- q q bling, the maximal operator Mq maps L (Ω) → weak-L (Ω), cf. [9, Theorem III.2.1], as well as ∞ ∞ L (Ω) → L (Ω). Hence, khkLp (Ω) ≈ kgkLp (Ω) by the Marcinkiewicz interpolation theorem (with a slightly more involved, yet fairly straightforward, reasoning in case q < 1). Moreover, 1/q 1/q gq dµ ≤ inf h(x) ≤ hq dµ KΩ x ∈Ω∩αλB K ∩αλB0 0 αλB1 q q 1/q since αλB1 ⊂ Ω ∩ B0 ⊂ Ω ∩ αλB0. For the sake of brevity, let g¯ = (g + h ) , in which case kg¯kLp (Ω) ≈ kgkLp (Ω). Then, (5.9) and the Hölder inequality yield that (5.10) 1/q 1/p + + |Tu(y)− Tu(z)| . r g¯q dµ ≤ r 1−(θ ε)/pr (θ ε)/p g¯p dµ k K k k K Z αλBk Z αλBk k ∈Õ\{0} k ∈Õ\{0} ′ 1/p r θ +ε 1/p . r (p−θ −ε)/(p−1) k g¯p dµ . k µ(B ) Z Z k αλBk k ∈Õ\{0} k ∈Õ\{0} ¹ As the radii rk form a geometric sequence, we can compute the sum 1/p′ ′ (p−θ −ε)/(p−1) , (p−θ −ε)/(p−1) 1/p = , 1−(θ +ε)/p , rk ≈ d(y z) d(y z) k ∈Z\{0} Õ where the constants depend on p, θ, ε, and c J . It follows from (5.7) that d(γz (tk ), z)≈ rk for k ≥ 1 and hence µ(Bk ) = µ(B(γz (tk ),rk )) ≈ µ(B(z,rk ) ∩ Ω) by the doubling condition of µ⌊∂Ω. Similarly, µ(B−k ) = µ(B(γy (tk ),rk )) ≈ µ(B(y,rk ) ∩ Ω). Consequently, (2.3) yields that r θ 1 r θ 1 k . and −k . , k ≥ 1 . µ(Bk ) H(B(z,rk ) ∩ ∂Ω) µ(B−k ) H(B(y,r−k ) ∩ ∂Ω) ∞ ∞ As each of the subchains of balls {αλBk }k=1 and {αλB−k }k=1 has bounded overlap, we have ∞ r θ +ε ∞ r εg¯(x)p k g¯(x)p dµ(x) . k dµ(x) ( ) H( ( , ) ∩ ∂Ω) = µ Bk αλBk = αλBk B z rk Õk 1 ¹ Õk 1 ¹ d(x, z)εg¯(x)p . dµ(x), H(B(z, d(x, z)) ∩ ∂Ω) ¹Cz,y where Cz,y is the “carrot” (5.8) whose tip is at z ∈ ∂Ω. The trace Tu lies in Lp (∂Ω) by Theorem 4.4. It suces to estimate the Besov seminorm consid- Ω2 ≔ Ω Ω ering only pairs of points (y, z) ∈ ∂ δ {(y, z) ∈ ∂ × ∂ : d(y, z) < δ }, where δ > 0 is the 30 LUKÁŠ MALÝ distance of the John center of Ω from ∂Ω. For each such couple of points, we will apply (5.10) and the subsequent estimates. Hence |Tu(y)− Tu(z)|p dH 2(y, z) − ∂Ω2 d(y, z)p θ H(B(z, d(y, z)) ∩ ∂Ω) º δ d(y, z)p−θ −ε d(x, z)εg¯(x)p (I) . dµ(x)dH 2(y, z) p−θ Ω ∂Ω2 d(y, z) H(B(z, d(y, z)) ∩ ∂Ω) C H(B(z, d(x, z)) ∩ ∂ ) º δ ¹ z,y d(y, z)p−θ −ε d(x,y)εg¯(x)p (II) + dµ(x)dH 2(y, z) p−θ Ω ∂Ω2 d(y, z) H(B(y, d(y, z)) ∩ ∂Ω) C H(B(y, d(x,y)) ∩ ∂ ) º δ ¹ y,z

Let us now examine the rst of the two integrals. The integration domain is contained in the set (x,y, z) ∈ Ω × ∂Ω × ∂Ω : βd(x, z) ≤ dist(x,∂Ω) ≤ 2d(y, z) < δ . By the Fubini theorem, we change the order of integration and then extend the integration domain so that I ≤ ··· dH(y)dH(z)dµ(x) . Ω Ω Ω Ω Ω Ω ¹{x ∈ :dist(x,∂ )<δ } ¹{z ∈∂ ∩B(x,dist(x,∂ )/β } ¹{y ∈∂ \B(z,dist(x,∂ )/2)} We apply Lemma 5.1 for the innermost integral and then we proceed by Lemma 5.2, which yields d(x, z)ε 1 I ≤ g¯(x)p · dH(z) dµ(x) Ω Ω Ω H(B(z, d(x, z)) ∩ ∂Ω) dist(x,∂Ω)ε ¹ δ ¹∂ ∩B(x,dist(x,∂ )/c) Ω ε p dist(x,∂ ) p ≤ g¯(x) dµ(x) ≤ kg¯k p Ω , Ω dist(x,∂Ω)ε L ( ) ¹ δ where Ωδ = {x ∈ Ω : dist(x,∂Ω) < δ } is the δ-strip in Ω near the boundary. Integral (II) can be processed analogously as (I) by mere ipping order of integration with respect to y and z. As noted after the theorem’s statement above, the Hardy–Littlewoodmaximal function we used in the proof fails to be in Lp (Ω) if we allow q = p. However, we will obtain the desired critical Besov smoothness of the trace under the additional assumption that Ω is a uniform domain.

Denition 5.11. A domain Ω ⊂ X is called uniform if there is a constant cU ∈ (0, 1] such that every pair of distinct points x,y ∈ Ω can be connected by a rectiable curve γ : [0,lγ ] → Ω = = −1 , parametrized by arc-length such that γ (0) x, γ (lγ ) y, lγ ≤ cU d(x y), and

(5.12) dist(γ (t), X \ Ω) ≥ cU min{t,lγ − t} for all t ∈ [0,lγ ]. Similarly as for John domains, one can use the Ascoli theorem (cf. [40, p. 169]) to see that if Ω is uniform with compact closure, then every pair of distinct boundary points x,y ∈ ∂Ω can be −1 , connected by a curve γ of length lγ ≤ cU d(x y) such that (5.12) is satised. Ω 1,p Ω Theorem 5.13. Let be a uniform domain whose closure is compact. Suppose that u ∈ Pp ( ) for p Ω 1−θ /p Ω = 1−θ /p Ω some p ≥ 1 while θ < p. Then, the trace Tu ∈ L (∂ ) lies in Bp (∂ ) Bp,p (∂ ). Moreover, p Ω kTuk Û 1−θ /p Ω ≤ CkgkLp (Ω), where g ∈ L ( ) is a p-PI gradient of u. Bp (∂ ) Sketch of proof. The proof can be led along the same lines as for Theorem 5.6 with a slightly modied construction of chains of balls connecting pairs of boundary points, using the fact that Ω is uniform, which will yield that a Hardy–Littlewood-type maximal function is no longer needed. TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 31

For a xed couple of points y, z ∈ ∂Ω, let γ : [0,lγ ] → Ω be a curve that connects the two points, i.e., γ (0) = z and γ (lγ ) = y, and satises the conditions from the denition of a uniform domain. Let lγ c k c t t = 1 − U and r = U k , k = 0, 1, 2,.... k 2 2λ k 2λ Next, we dene a chain of balls {Bk }k ∈Z by setting

Bk = B(γ (tk ),rk ) and B−k = B(γ (lγ − tk ),rk ), for all k = 0, 1, 2,....

With this choice of balls, we obtain that the chains of inated balls (including αλB0) are contained in the carrots ∞ αλBk ⊂ x ∈ Ω : βd(x, z) ≤ dist(x,∂Ω) ≤ 2d(z,y) ≕ Cz,y and k=0 ∞Ø αλB−k ⊂ x ∈ Ω : βd(x,y) ≤ dist(x,∂Ω)≤ 2d(y, z) ≕ Cy,z . = kØ0 Hence αλB0 does not need any special treatment via the maximal function and one can proceed as in the proof of Theorem 5.6. Corollary 5.14. Under the hypothesis of Theorems 5.6 or 5.13 and assuming that H is Ahlfors codimension-θ regular, we have that

kTuk Ûα ∂Ω ≤ CkgkLp (Ω) for every 0 < α ≤ 1 − θ/p, q ∈ [p, ∞], and Bpα ,q ( )

inf kTu − akLp∗,p (∂Ω) ≤ inf kTu − akLp∗ (∂Ω) ≤ CkgkLp (Ω), a ∈R a ∈R = p(s−θ ) ∗ = p(s−θ ) where pα s−p(1−α ) and p s−p . Proof. Both estimates follow from the embeddings in Theorem 3.23 and Corollary 3.18.

1−ϑ /p Ω 6. Bounded linear extension from the Besov class Bp,p (∂ ) to the Newtonian class N 1,p (Ω) In this section, we will nd a linear extension operator that maps the Besov class B1−ϑ /p (∂Ω) into the class of locally Lipschitz Newtonian functions N 1,p (Ω). The extension will be dened using a Whitney-type cover of Ω and the corresponding partition of unity. For the extension theorem, we will assume that the relation between µ⌊Ω and H⌊∂Ω satises both (2.2) and (2.3) with 0 < ϑ ≤ θ ≤ p. In fact, it suces to assume that (2.3) is localized, i.e., µ(B(z,r) ∩ Ω) H(B(z,r) ∩ ∂Ω)≤ C(z) r θ holds for H-a.e. z ∈ ∂Ω and for all r < R(z), where C(z) > 0 and R(z) > 0. Such a localization would aect the proofs of Lemmata 6.6, 6.12, and 6.13, making the argument somewhat obscured by technical details, whence their formulation and proof will use the non-localized form (2.3). Let us rst establish the Whitney covering of an open set Ω as in [21, Section 4.1], see also [5]. Proposition 6.1 ([21, Proposition 4.1.15]). Let Ω ( X be open. Then there exists a countable collection WΩ = {B(pj,i ,rj,i ) = Bj,i } of balls in Ω so that = Ω • j,i Bj,i , • χ ≤ C, Ðj,i B(pj,i,2rj,i ) • 2j−1 < r ≤ 2j for all i, and Í j,i 32 LUKÁŠ MALÝ

= 1 , Ω • rj,i 8 dist(pj,i X \ ), where C > 0 depends only on the doubling constant of the measure µ.

Since the radii of the balls are suciently small, we have that 2Bj,i ⊂ Ω for every Bj,i ∈ WΩ. By the boundedness of Ω there is a largest exponent j that occurs in the cover; we denote this j exponent by j0. Note that 2 0 is comparable to diam(Ω). Note also that no ball in level j intersects a ball in level j +2. This follows by the reverse triangle j+4 j+3 j+3 j−1 j inequality d(pj,i ,pj+2,k ) ≥ 2 − 2 = 2 and the bounds on the radii: 2 < rj,i ≤ 2 and j+1 j+2 2 < rj+2,k ≤ 2 . As in [21, Theorem 4.1.21], there is a Lipschitz partition of unity {φj,i } subordinate to the Whit- ney decomposition WΩ, that is, j,i φj,i ≡ χΩ and 0 ≤ φj,i ≤ χ2Bj,i for every ball Bj,i ∈ WΩ while φj,i is C/rj,i -Lipschitz continuous. Í 6.2. Extending boundary data via Whitney-type partition of unity. Given f ∈ L1(∂Ω), we will construct a function F : Ω → R whose trace will (under certain hypothesis, cf. Lemmata 6.12 and 6.13) be the original function f on ∂Ω. Moreover, we will show that F ∈ N 1,p (Ω) provided 1−ϑ /p Ω that f ∈ Bp,p (∂ ). Consider the center of the Whitney ball pj,i ∈ Ω and choose a closest point qj,i ∈ ∂Ω. Dene = ( , ) ∩ ∂Ω = ( ) H( ) Uj,i : B qj,i rj,i . We set aj,i : U f y d y . Then, set ⨏ j,i F(x) = Ef (x) ≔ aj,iφj,i (x), x ∈ Ω. , Õj i We willshow that E : Lp (∂Ω) → Lp (Ω) isboundedforeveryp ≥ 1 in Lemma 6.3. In Proposition 6.8, 1−ϑ /p Ω 1,p Ω Ω we will see that E : Bp,p (∂ ) → N ( ) ∩ Liploc( ) provided that p ≥ max{1, ϑ }. Finally, Lemmata 6.12 and 6.13 show that Ef has a trace on ∂Ω in the sense of (1.2) and the trace coincides = p Ω 1−ϑ /p Ω with the given function f if either ϑ θ and f ∈ L (∂ ) with any p ≥ 1, or f ∈ Bp,p (∂ ) and p > θ. Let us rst establish the Lp -estimates for F = Ef . Lemma 6.3. Let p ≥ 1. Then, ϑ /p kF kLp (Ω) . diam(Ω) k f kLp (∂Ω).

Proof. We rst consider a xed ball Bℓ,m from the Whitney cover. Recall that if 2Bj,i ∩ Bℓ,m , ∅, ℓ ∗ ≔ , 6 ∂Ω ∗ then |j − | ≤ 1. Thus, Uj,i ⊂ Uℓ,m B(qℓ,m 2 rℓ,m ) ∩ , whence H(Uj,i )≈H(Uℓ,m). Then, p |F(x)|p dµ(x) = f (y)dH(y)φ (x) dµ(x) K j,i ¹Bℓ,m ¹Bℓ,m j,i s.t. Uj,i 2B , ∩ÕBℓ, ,∅ j i m p ≤ |f (y)| dH(y) φ (x) dµ(x) K j,i ¹Bℓ,m j,i s.t. Uj,i 2Bj,i ∩ÕBℓ,m ,∅ p . |f (y)| dH(y) φj,i (x) dµ(x) K ∗ Bℓ,m U , ¹ ℓ,m Õj i p = |f (y)| dH(y) dµ(x). B KU ∗ ¹ ℓ,m ℓ,m TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 33

From the construction of the Whitney balls, the doubling condition of µ⌊Ω, and (2.2), we deduce that Ω . ϑ (6.4) µ(Bℓ,m )≈ µ(B(qℓ,m,rℓ,m ) ∩ ) rℓ,m H(Uℓ,m). Hence, p . p . ϑ p (6.5) |F(x)| dµ(x) µ(Bℓ,m ) |f (y)| dH(y) rℓ,m |f (y)| dH(y). B KU ∗ U ∗ ¹ ℓ,m ℓ,m ¹ ℓ,m ℓ Ω = Recall that rℓ,m ≈ 2 . As ℓ,m Bℓ,m and the balls have uniformly bounded overlap, we have j Ð 0 p . ϑ p . ℓ ϑ p |F | dµ rℓ,m |f | dH 2 |f | dH Ω ∗ ∗ ℓ, Uℓ, ℓ= m Uℓ, ¹ Õm ¹ m Õ−∞ Õ ¹ m j0 . ℓ ϑ p Ω ϑ p 2 |f | dH ≈ diam( ) k f k p ∂Ω . Ω L ( ) ℓ= ∂ Õ−∞ ¹ We will use the extension constructed in this section to nd a nonlinear bounded extension from Lp (∂Ω) to N 1,p (Ω) in the subsequent section in case p = ϑ = θ ≥ 1. There, we will need the following estimates for the integral of the function F and its gradient (Lemma 6.11 below) on boundary shells of Ω. Lemma 6.6. Let z ∈ ∂Ω, r > 0, and 0 < ρ < diam(Ω)/2. Set Ω(ρ) = {x ∈ Ω : dist(x, X \ Ω) < ρ}. Then, |F |p dµ . min{r, ρ}ϑ |f |p dH . Ω 8 Ω ¹B(z,r )∩ (ρ) ¹B(z,2 r )∩∂ Proof. Since B(z,r) ∩ Ω(ρ) = B(z,r) ∩ Ω(min{r, ρ}), we do not lose any generality by assuming that ρ ≤ r. Let ℓρ be the greatest value of ℓ ∈ Z for which there exists a ball Bℓ,m that intersects Ω(ρ) ∩ B(z,r). For each ℓ ≤ ℓρ , we dene I(ℓ) to be the collection of all m ∈ N for which Bℓ,m ∩ Ω(ρ) ∩ B(z,r) is non-empty. Then by (6.5),

ℓρ ℓρ p p . ϑ p |F | dµ ≤ |F | dµ rℓ,m |f | dH. Ω ∗ B(z,r )∩ (ρ) ℓ= Bℓ,m ℓ= Uℓ, ¹ Õ−∞ mÕ∈I(ℓ) ¹ Õ−∞ mÕ∈I(ℓ) ¹ m The triangle inequality yields that

d(z,qℓ,m )≤ d(z,pℓ,m ) + d(pℓ,m ,qℓ,m )≤ 2d(z,pℓ,m ) ≤ 2(r + rℓ,m ), where Bℓ,m = B(pℓ,m ,rℓ,m ) and Uℓ,m = B(qℓ,m,rℓ,m ) ∩ ∂Ω with qℓ,m ∈ ∂Ω being a boundary point = , Ω , + 1 lying closest to pℓ,m . Moreover, 8rℓ,m dist(pℓ,m X \ ) ≤ d(pℓ,m z) ≤ r rℓ,m . Hence, rℓ,m ≤ 7r. , 16 , 16 + 1 ∗ , 8 Consequently, d(z qℓ,m )≤ 7 r and Uℓ,m ⊂ B(z ( 7 7 )r). Thus, Uℓ,m ⊂ B(z 2 r) and

ℓρ |F |p dµ . 2ℓϑ |f |p dH ≈ ρϑ |f |p dH, Ω 8 Ω 8 Ω B(z,r )∩ (ρ) ℓ= B(z,2 r )∩∂ B(z,2 r )∩∂ ¹ Õ−∞ ¹ ¹ ℓ ℓ where we used that rℓ,m ≈ 2 and 2 ρ ≈ ρ, which follow from the construction of WΩ. By choosing any r > diam Ω in the preceding lemma, we obtain the following corollary. Corollary 6.7. With the notation of Lemma 6.6, we have

|F |p dµ . ρϑ |f |p dH. Ω Ω ¹ (ρ) ¹∂ 34 LUKÁŠ MALÝ

Since Lip F is an upper gradient of locally Lipschitz functions, cf. [3, Proposition 1.14], the following proposition provides us with the desired Lp -normboundforanuppergradientof F = Ef . Proposition 6.8. Given Ω ⊂ X and p ≥ max{1, ϑ }, then 1−ϑ /p Ω k Lip F kLp (Ω) . k f k 1−ϑ /p Ω whenever f ∈ Bp,p (∂ ). Bp,p (∂ )

Proof. Fix a ball Bℓ,m ∈ WΩ, and x a point x ∈ Bℓ,m. Then, for all y ∈ Bℓ,m,

Thus, we want to bound terms of the form |aj,i − aℓ,m |. Note also that if 2Bj,i ∩ Bℓ,m , ∅, then |j − ℓ| ≤ 1. Therefore,

|a − a | = f (z)dH(z)− f (w)dH(w) j,i ℓ,m K K Uj,i Uℓ,m

(6.9) ≤ |f (z)− f (w)| dH(w)dH(z) . |f (z)− f (w)| dH(w)dH(z), K K K ∗ K ∗ Uj,i Uℓ,m Uℓ,m Uℓ,m ∗ ≔ , 6 ∂Ω where Uℓ,m B(qℓ,m 2 rℓ,m )∩ denotes the expanded subset of the boundary. Recall that H is ∗ Ω . doubling by (2.3) and by the assumption that µ⌊ is doubling, and that is why H(Uℓ,m) H(Uℓ,m) ∗ . and H(Uℓ,m) H(Uj,i ), which was used in (6.9). The above estimates together with the bounded overlap of the Whitney balls yield the following inequality: |F(y)− F(x)| 1 (6.10) Lip F(x) = lim sup . |f (z)− f (w)| dH(w)dH(z) y→x d(y, x) rℓ,m K ∗ K ∗ Uℓ,m Uℓ,m p p whenever x ∈ Bℓ,m. Therefore, k Lip F k p ≤ ℓ, k Lip F k p and hence (6.10) and (6.4) L (Ω) m L (Bℓ,m ) yield that Í ( ) p p µ Bℓ,m w w (Lip F) dµ . p |f (z)− f ( )| dH( )dH(z) Ω K ∗ K ∗ ℓ, r Uℓ, Uℓ, ¹ Õm ℓ,m m m H(Uℓ, ) . m |f (z)− f (w)|p dH(w)dH(z) p−ϑ K ∗ K ∗ ℓ, r Uℓ, Uℓ, Õm ℓ,m m m j 0 1 . |f (z)− f (w)|p dH(w)dH(z) p−ϑ ∗ K ∗ ℓ= m r Uℓ, Uℓ, Õ−∞ Õ ℓ,m ¹ m m j 0 1 . w p w , +ℓ |f (z)− f ( )| dH( )dH(z) 7 p−ϑ Ω K 7+ℓ ℓ= (2 ) ∂ B(z,2 ) Õ−∞ ¹ TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 35

∗ ℓ where the last inequality follows from the uniformly bounded overlap of the balls Uℓ,m for each . + Without loss of generality, we may choose R = 2j0 7 in the denition of the Besov norm (3.5), in which case R ≈ diam(Ω). Finally,

j0 1 w p w +ℓ |f (z)− f ( )| dH( )dH(z) 7 p−ϑ Ω K 7+ℓ ℓ= (2 ) ∂ B(z,2 ) Õ−∞ ¹ R 1 dt w p w = . ≈ |f (z)− f ( )| dH( )dH(z) k f k Û1−ϑ /p ∂Ω = tp−ϑ Ω K t Bp,p ( ) ¹t 0 ¹∂ B(z,t) Next, we obtain a localized estimate for the Lp -norm of the p-weak upper gradient Lip F on the layer Ω(ρ) if the boundary function f Lipschitz is continuous. The Lipschitz constant of f on ∂Ω will be denoted by LIP(f ,∂Ω), i.e., we dene |f (x)− f (y)| LIP(f ,∂Ω) = sup . x,y ∈∂Ω:x,y d(x,y) Lemma 6.11. For 0 < ρ < diam(Ω)/2, set Ω(ρ) = {x ∈ Ω : dist(x, X \ Ω) < ρ}. If f is Lipschitz continuous on ∂Ω, then

(Lip F(x))p dµ(x) . µ(Ω(ρ)) LIP(f ,∂Ω)p . Ω ¹ (ρ) Proof. It follows from (6.10) that Lip F(x) . LIP(f ,∂Ω) for all x ∈ Ω. 1−ϑ /p Ω So far, we have seen that if f ∈ Bp,p (∂ ), the corresponding extension F is in the Newtonian space N 1,p (Ω). This extension is linear by construction. We will now prove that the trace of F 1−ϑ /p Ω = returns the original function f , i.e., T ◦ E is the identity function on Bp,p (∂ ) whenever ϑ θ or p > θ. For the sake of clarity, let us explicitly point out that the following lemma shows that the N 1,p extension has a well-dened trace even though no Poincaré inequality for Ω is assumed. = p ∂Ω Lemma 6.12. Assume that ϑ θ. Let f ∈ Lloc( ), p ≥ 1. Then, p = ∂Ω lim+ |Ef (x)− f (z)| dµ(x) 0 for H-a.e. z ∈ . r →0 KB(z,r )∩Ω In particular, the trace T (Ef )(z) exists and equals f (z) for H-a.e. z ∈ ∂Ω.

Proof. Since H⌊∂Ω is doubling, we know that H-a.e. z ∈ ∂Ω is a Lebesgue point of f . Let z ∈ ∂Ω be such a point. Let fz (w) = f (w)− f (z), w ∈ ∂Ω. Note that Efz (x) = Ef (x)− f (z) for every x ∈ ∂Ω by linearity of the extension operator E. From Lemma 6.6 applied to Efz with r = ρ, we obtain that

α Ω Lemma 6.13. Let f ∈ Bp,p (∂ ) with α ∈ (0, 1) and p ≥ 1 such that αp > θ − ϑ. Then,

p = ∂Ω lim+ |Ef (x)− f (z)| dµ(x) 0 for H-a.e. z ∈ . r →0 KB(z,r )∩Ω Ω 1−ϑ /p Ω In particular, the trace T (Ef )(z) exists and equals f (z) for H-a.e. z ∈ ∂ if f ∈ Bp,p (∂ ) and 0 < ϑ ≤ θ < p. Proof. By Lemma 3.16, there is g ∈ Lp (∂Ω) and E ⊂ ∂Ω with H(E) = 0 such that |f (z)− f (w)| ≤ α d(z, w) (g(z) + g(w)) for all z, w ∈ ∂Ω \ E. As H⌊∂Ω is doubling, the Hardy–Littlewood maximal operator M : L1(∂Ω) → weak-L1(∂Ω) is bounded by [9, Theorem III.2.1]. Pick z ∈ ∂Ω \ E such that g(z) < ∞ and Mgp (z) < ∞, which is satised for H-a.e. z ∈ ∂Ω as Mgp ∈ weak-L1(∂Ω). Then, following the steps of proof of Lemma 6.12 to its ultimate displayed formula, we arrive at

|Ef (x)− f (z)|p dµ(x) . r ϑ −θ |f (w)− f (z)|p dH(w) KB(z,r )∩Ω KB(z,r )∩∂Ω . r ϑ −θ d(z, w)αp (g(z)p + g(w)p )dH(w) KB(z,r )∩∂Ω + ≤ r ϑ −θ αp (g(z)p + Mgp (z)), which approaches 0 as r → 0.

7. Extension theorem for Lp data on a boundary of codimension p In this section, we assume that ∂Ω is Ahlfors codimension-p regular, i.e., both (2.2) and (2.3) are satised and ϑ = θ = p ≥ 1. Given an Lp -function on ∂Ω, we will construct its N 1,p extension in Ω using the linear extension operator for Besov boundary data. However, the mapping f ∈ Lp (∂Ω) 7→ Ext f ∈ N 1,p (Ω) will be nonlinear, which is to be expected in view of [38, 39]. Instead of constructing the extension using a Whitney decomposition of Ω, we will set up a sequence of layers inside Ω whose widths depend not only on their distance from X \ Ω, but also on the function itself. Using a partition of unity subordinate to these layers, we will glue together N 1,p extensions of Lipschitz approximations of the boundary data. The core idea of such a construction can be traced back to Gagliardo [10] who discussed extending L1(Rn−1) functions 1,1 n to W (R+), and the construction was successfully generalized later in [34, 46]. p ∂Ω ∞ We start by approximating f in L ( ) by a sequence of Lipschitz functions {fk }k=1 such that 2−k k fk+1 − fk kLp (∂Ω) ≤ 2 k f kLp (∂Ω). Note that this rate of convergence of fk to f also ensures that fk → f pointwise H-a.e. in ∂Ω. For technical reasons, we choose f1 ≡ 0. ∞ Next, we choose a decreasing sequence of real numbers {ρk }k=1 such that: • ρ1 ≤ diam(Ω)/2; • 0 < ρk+1 ≤ ρk /2; Ω • k ρk LIP(fk+1,∂ ) ≤ Ck f kLp (∂Ω). These will now be used to dene layers in Ω. Let Í ρk − dist(x, X \ Ω) ψk (x) = max 0, min 1, , x ∈ Ω. ρ − ρ + k k 1 Recall that Ω(ρ) denotes the shell {x ∈ Ω : dist(x, X \ Ω) < ρ}. The sequence of functions {ψk−1 −ψk : k = 2, 3,...} serves as a partition of unity in Ω(ρ2) subordinate to the system of layers given by {Ω(ρk−1)\ Ω(ρk+1) : k = 2, 3,...}. TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 37

0 Since Lipschitz continuous functions lie in the Besov class Bp,p , we can apply the linear exten- 0 Ω 1,p Ω sion operator E : Bp,p (∂ ) → N ( ), whose properties were established in Section 6, to dene the extension of f ∈ Lp (∂Ω) by extending its Lipschitz approximations in layers, i.e.,

∞ ∞ (7.1) F(x) = Ext f (x) := ψk−1(x)− ψk (x) Efk (x) = ψk (x) Efk+1(x)− Efk (x) , x ∈ Ω. k=2 k=1 Õ Õ The following result shows that the above extension is in the class N 1,p (Ω) with appropriate norm bounds since Lip F is an upper gradient of F, which is locally Lipschitz by construction.

Proposition 7.2. Given f ∈ Lp (∂Ω), the extension dened by (7.1) satises

p kF kLp (Ω) . diam(Ω) k f kLp (∂Ω) and 1/p k Lip F kLp (Ω) . (1 + H(∂Ω) )k f kLp (∂Ω) .

Proof. Corollary 6.7 allows us to obtain the desired Lp estimate for F. Since the extension operator for Besov boundary data E is linear, we have that Efk+1 − Efk = E(fk+1 − fk ). Therefore,

∞ ∞

p p p kF kL (Ω) ≤ kψk E(fk+1 − fk )kL (Ω) ≤ kE(fk+1 − fk )kL (Ω(ρk )) k=1 k=1 Õ∞ Õ p p p . + p Ω . p Ω . Ω p Ω . ρk k fk 1 − fk kL (∂ ) ρ1 k f kL (∂ ) diam( ) k f kL (∂ ) = Õk 1 In order to obtain the Lp estimate for Lip F, we rst apply the product rule for locally Lipschitz functions, which yields that

∞ Lip F = |E(fk+1 − fk )| Lipψk + ψk Lip(E(fk+1 − fk )) k=1 Õ∞ |E(f + − f )|χΩ k 1 k (ρk ) + ≤ χΩ(ρk ) Lip(E(fk+1 − fk )) = ρk − ρk+1 Õk 1 Thus,

∞ E(fk+1 − fk ) k k p Ω ≤ + ( + − ) . Lip F L ( (ρ1)) Lip E fk 1 fk Lp (Ω(ρ )) ρk − ρk+1 p Ω k k=1 L ( (ρk )) Õ

It follows from Corollary 6.7 that

∞ ∞ E(fk+1 − fk ) ρk . k fk+1 − fk kLp (∂Ω) ρk − ρk+1 p Ω ρk − ρk+1 k=1 L ( (ρk )) k=1 Õ Õ ∞

. k fk+1 − fk kLp (∂Ω) . k f kLp (∂Ω) . = Õk 1 38 LUKÁŠ MALÝ

Next, we apply Lemma 6.11 to see that ∞ ∞ Ω 1/p Ω Lip E(fk+1 − fk ) p . ρk H(∂ ) LIP(fk+1 − fk ,∂ ) L (Ω(ρk )) k=1 k=1 Õ Õ ∞ 1/p . H(∂Ω) ρk LIP(fk+1,∂Ω) + LIP(fk ,∂Ω) = Õk 1 1/p . H(∂Ω) k f kLp (∂Ω), ∞ where we used the dening properties of {ρk }k=1 to obtain the ultimate inequality. Altogether, we 1/p p . + Ω p have shown that k Lip F kL (Ω(ρ1)) (1 H(∂ ) )k f kL (∂Ω). Finally, we will show that the trace of the extended function yields the original function back. Proposition 7.3. Let F ∈ N 1,p (Ω) be the extension of f ∈ Lp (∂Ω) as constructed in (7.1). Then,

lim |F − f (z)| dµ = 0 r →0 KB(z,r )∩Ω for H-a.e. z ∈ ∂Ω.

Proof. Let E0 be the collection of all z ∈ ∂Ω for which limk fk (z) = f (z), and for k ∈ N let Ek be the collection of all z ∈ ∂Ω for which TEfk (z) = fk (z) exists. Lemma 6.12 yields that Ω ∞ = ∞ H(∂ \ k=0 Ek ) 0. We dene also an auxiliary sequence {Fn }n=1 of functions approximating F by Ñ n ∞ Fn = (ψk−1 − ψk )Efk + (ψk−1 − ψk )Efn, n ∈ N. = = + Õk 2 kÕn 1 Since Fn = Efn in Ω(ρn), the trace of Fn exists on ∂Ω and coincides with TEfn = fn. ∞ N Fix a point z ∈ k=0 Ek and let ε > 0. Then, we can nd j ∈ such that |fk (z)− f (z)| < ε for every k ≥ j. Next, we choose k0 > j such that R := ρk satises: Ñ 0 • R LIP(fj ,∂Ω) < ε; • | − ( )| < < B(z,r )∩Ω Fj fj z dµ ε for every r R; ⨏ ∞ • ρ LIP(f + ,∂Ω) < ε. k=k0 k k 1 For every r ∈ (0, ρ + ) ⊂ (0, R/2), we can then estimate Í k0 1

|F − f (z)| dµ ≤ |F − Fj | dµ + |Fj − fj (z)| dµ + |fj (z)− f (z)| KB(z,r )∩Ω KB(z,r )∩Ω KB(z,r )∩Ω kF − Fj kLp (B(z,r )∩Ω) (7.4) ≤ |F − F | dµ + 2ε ≤ + 2ε. j 1/p KB(z,r )∩Ω µ(B(z,r) ∩ Ω)

For such r, choose kr > k0 such that ρkr +1 ≤ r < ρkr . Then, ∞ p Ω kF − Fj kL (B(z,r )∩ ) ≤ (ψk−1 − ψk )|E(fk − fj )| Lp (B(z,r )∩Ω) k=k Õr ∞ ∞ (7.5) ≤ |E(fk − fj )| p . min{r, ρk−1 }k fk − fj k p ( ( , 8 )∩∂Ω) L (B(z,r )∩Ω(ρk−1)) L B z 2 r k=k k=k Õr Õr = by Lemma 6.6. For k kr in thelast inequalityabove, we used the factthatwe have chosen r < ρkr Ω = Ω and hence B(z,r) ∩ (ρkr −1) B(z,r) ∩ (r). TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 39

8 Let us, for the sake of brevity, write Ur = B(z, 2 r) ∩ ∂Ω. As fk − fj is Lipschitz continuous and due to the choice of j, we have for k ≥ j that + 1/p k f − fj k p ( ) ≤ f − fj − (f (z)− fj (z)) p f (z)− fj (z) H(Ur ) k L Ur k k L (Ur ) k (7.6) . r H(U )1/p LIP(f − f ,U ) + 2εH( U )1/p . r k j r r p Observe that r H(Ur ) . µ(B(z,r) ∩ Ω) by (2.3) and the doubling condition for µ⌊Ω. Note that ∞ ρ ≈ ρ ≤ R. Combining this with (7.5) and (7.6) gives us that k=kr k−1 kr −1 ∞ Í 1/p kF − Fj kLp (B(z,r )∩Ω) . ρk−1µ(B(z,r) ∩ Ω) LIP(fk ,∂Ω) + LIP(fj ,∂Ω) k=kr Õ ∞ min{r, ρ } + 2εµ(B(z,r) ∩ Ω)1/p k−1 = r kÕkr ∞ 1/p . µ(B(z,r) ∩ Ω) ρk LIP(fk+1,∂Ω) + R LIP(fj ,∂Ω) + ε = k k0 Õ . µ(B(z,r) ∩ Ω)1/pε. Plugging this estimate into (7.4) completes the proof.

8. Counterexamples for the trace theorems In Sections 4–7, we witnessed a certain mismatch of the target space of the trace operator and the domain of the extension operators for domains whose boundary has unequal lower and upper codimension bounds, i.e., ϑ < θ in (2.2) and (2.3). We have seen that the trace operator maps 1,p Ω 1−θ /p Ω 1−ϑ /p Ω 1−θ /p Ω Pp ( ) into Bp,∞ (∂ ) but we were able to extend functions of Bp,p (∂ ) ( Bp,p (∂ ) to a Sobolev-type function in Ω with p-integrable gradient. The following example shows that the trace operator, in general, does not map into a Besov space of smoothness any higher than 1 − θ/p with summability of the pth power. Example 8.1. Let X = R2 be endowed with the Euclidean distance and 2-dimensional Lebesgue Ω = 2 measure. Let {(x1, x2) ∈ X : 0 < x2 < x1 < x1}. Since the boundary consists of two line Ω = 2 = 2 = = segments connected by a parabola, i.e., ∂ {(x1, x2) ∈ [0, 1] : x2 x1 or x1 1 or x2 0}, it is natural to choose H to be the 1-dimensional Hausdor measure on ∂Ω. Note that for every x ∈ Ω, there is R(x) > 0 such that µ(B(x,r)∩ Ω)≈ r 2 whenever r ∈ (0, R(z)). However, µ(B(0,r) ∩ Ω)≈ r 3 for every r ∈ (0, 2 diam Ω). Therefore, µ(B(z,r) ∩ Ω) µ(B(z,r) ∩ Ω) . H(B(z,r) ∩ ∂Ω) . , z ∈ ∂Ω,r ∈ (0, 2 diam Ω). r r 2 , = 3 , 1 Given p ∈ (2 3), let α p − 1 ∈ (0 2 ) and dene a function 1 u(x , x ) = , (x , x )∈ Ω. 1 2 xα log( e ) 1 2 1 x1 Since u is really a function of one variable only, it is easy to verify (e.g., by the mean value theorem) that g(x , x ) ≔ C u(x1,x2) is a Hajłasz gradient of u in Ω. Then, 1 2 α x1 2 1 x1 1 p 1 1 g(x) dx ≈ + dx2 dx1 ≈ e dx1 < ∞. Ω (α 1)p e p x log( )p 0 0 x log( ) 0 1 x1 ¹ ¹ ¹ 1 x1 ¹ 40 LUKÁŠ MALÝ

1,p Ω 1,p Ω 1−2/p Ω Hence, u ∈ M ( ) ⊂ Pp ( ). Theorem 4.4 yields that Tu ∈ Bp,∞ (∂ ). If we restrict our attention to the horizontal line segment ∂Ω, we see that

1 dt (Tu)q dH = , tαq log( e )q ¹[0,1]×{0} ¹0 t which is convergent if and only if q ≤ 1/α. In particular, Tu < Lq (∂Ω) for any q > 1/α. From 1−2/p+ε Ω 1/(α −ε) Ω Remark 3.19, we can deduce that Bp,∞ (∂ ) ⊂ weak-L (∂ ) for any ε ≥ 0. Consequently, 1−2/p+ε Ω q Ω Tu < Bp,∞ (∂ ) for any ε > 0 since otherwise we would have Tu ∈ L (∂ ) for some q > 1/α. 1−θ /p Ω 1−θ /p Ω Note that Tu in the example above actually lies in Bp,p (∂ ) ( Bp,∞ (∂ ). In Theorem 5.6, 1,p Ω 1−θ /p Ω 1−θ /p Ω the trace operator was shown to map M ( ) into Bp,p (∂ ) rather than Bp,∞ (∂ ) provided Ω 1−θ /p Ω that is a John domain. However, it is currently unclear whether the target space is Bp,p (∂ ) also for general non-John domains. Considering the example above, let us now argue that the trace operator there is not surjective, 1−θ /p Ω 1−θ /p Ω = = be it for the target space Bp,∞ (∂ ), orfor Bp,p (∂ ). Recall that ϑ 1 < θ 2. The measure H restricted to the vertical segment ∂ΩV ≔ {1} × [0, 1] ⊂ ∂Ω, is codimension 1 regular with respect 1−1/p Ω ( 1−2/p Ω 1,p Ω to µ⌊Ω. Hence, Tu|∂ΩV ∈ Bp,∞ (∂ V ) Bp,∞ (∂ V ) whenever u ∈ Pp ( ). The following example shows that the issue with T not being surjective can arise when ϑ < θ even if Ω is a uniform domain that admits a 1-Poincaré inequality.

Example 8.2. Let X = R2 be endowed with the Euclidean distance and measure dµ(x) = |x | dx. Then, X supports a 1-Poincaré inequalityby [19, Example 1.6]. Let Ω = [0, 1]2. Since Ω isa uniform domain, it also supports a 1-Poincaré inequalityby [6, Theorem 4.4], whence N 1,p (Ω) = M1,p (Ω) = 1,p Ω Ω Pp ( ) for every p > 1. Suppose that ∂ is endowed with 1-dimensional Hausdor measure. = < = , = 3 = −α −1 1,p Ω Then, 1 ϑ θ 2. Given p ∈ (2 3), let α p − 1. Thus, u(x) |x | log(e/|x |) ∈ N ( ). By 1−2/p Ω Theorem 5.6,Tu ∈ Bp,p (∂ ), but it lacks any higher smoothness (for the exponent p) similarly as 1,p Ω 1−2/p Ω in Example 8.1. On the other hand,T : N ( ) → Bp,p (∂ ) is not surjective since the restriction Ω 1−1/p of the trace to any compact set K ⊂ ∂ \ {0} lies in Bp,p (K). 1,θ Ω Let us now also look at the assumptions of Theorem 4.17, where the trace of a Pθ ( ) function was established to be Lθ (∂Ω) provided that its θ-PI gradient lies in a weighted Lθ (Ω) space. The following example shows that the weight in the integrability condition for the PI gradient cannot be removed even in uniform domains with smooth boundary.

Example 8.3. Let Ω = B(0, 1) ⊂ R2 be endowed with Euclidean distance. Let the measure in Ω be given by dµ(x) = dist(x,∂Ω)dx. Let H be the 1-dimensional Hausdor measure on ∂Ω. Then, = = , 1 ϑ θ 2 in (2.2) and (2.3). Let ε ∈ (0 2 ) and dene e ε C u(x) = log , g(x) = , x ∈ Ω. dist(x,∂Ω) dist(x,∂Ω) log( e )1−ε dist(x,∂Ω) Then, g is a Hajłasz gradient of u provided that the constant C > 0 if suciently large. Moreover, ,∂Ω 1 2 = dist(x ) r(1 − r) g(x) dµ(x) e dx ≈ e dr < ∞. Ω Ω dist(x,∂Ω)2 log( )2−2ε (1 − r)2 log( )2−2ε ¹ ¹ dist(x,∂Ω) ¹0 1−r TRACEANDEXTENSIONTHEOREMSINMETRICSPACES 41

1,2 Ω 1,2 Ω Ω Hence, u ∈ M ( ) ⊂ P2 ( ). However, for every z ∈ ∂ , e ε e ε TRu(z) = log dµ(x) ≥ log →∞ as R → 0. K Ω dist(x,∂Ω) R B(z,R)∩

It follows from the denition of the trace (1.2) that necessarily Tu(z) = limR→0 TRu(z) for H-a.e. z ∈ ∂Ω. In this case, we would obtain that Tu ≡∞ on ∂Ω, which would violate (1.2).

Moreover, the weight condition of Theorem 4.17 is essentially sharp whenever θ > 1, which is a consequence of the following proposition. On the other hand, if ϑ = θ = 1 and Ω admits Ω 1 Ω Ω 1,1 Ω = 1,1 Ω a 1-Poincaré inequality, then T : BV ( ) → L (∂ ) by [32], where BV ( ) ⊃ N ( ) P1 ( ) denotes the space of functions of bounded variation (see [36]).

Proposition 8.4. Let ε ∈ (0, 1). Then, there are: • w ( , ∞)→[ , ∞) 1 dt < ∞ a decreasing function : 0 1 such that 0 t w(t) ; • a uniform domain Ω ⋐ X with a doubling measure µ⌊Ω that admits 1-Poincaré inequality; ∫ • a measure H on ∂Ω that satises both (2.2) and (2.3) with ϑ = θ > 1; and 1,θ 1−ε θ θ • a function u ∈ M (Ω) such that g˜ε (x) ≔ g(x)w(dist(x,∂Ω)) ∈ L (Ω), where g ∈ L (Ω) is a Hajłasz gradient (and hence a 1-PI gradient) of u ( ) ( )→∞ → such that u does not have a trace in the sense of (1.2). Moreover, B(z,R)∩Ω u x dµ x as R 0 for every z ∈ ∂Ω. ⨏

θ θ Recall that if w, Ω, µ, and H are as in the proposition above and g˜0 ∈ L (Ω), then Tu ∈ L (∂Ω) exists by Theorem 4.17.

Proof. Let ε > 0 be given. Let X = R2 be endowed with the Euclidean distance and set Ω = B(0, 1). n−1 Pick n ≥ 2/ε. Dene dµ(x) = (dist(x, X \ Ω) + χX \Ω(x))dx. By [23, Theorem 3.4 (i)], cf. [22, Lemmata 8.3 and 8.4], Ω admits a 1-Poincaré inequality. Let H be the 1-dimensional Hausdor measure on ∂Ω, in which case ϑ = θ = n. Let w(t) = max{1, log(e/t)1+ε/4}, t ∈ (0, ∞). = e ε/4 1,θ Ω = 1,θ Ω Then, the function u(x) log( dist(x,∂Ω) ) lies in N ( ) M ( ), but lacks a trace since limR→0 TRu(z) = ∞ for every z ∈ ∂Ω. Note that C g(x) = ,∂Ω e 1−ε/4 dist(x ) log( dist(x,∂Ω) ) is a Hajłasz gradient of u (provided that C > 0 is suciently large) and hence

e (1+ε/4)(1−ε) C log( dist(x,∂Ω) ) C g˜ = = ε ,∂Ω e 1−ε/4 ,∂Ω e ε/2+ε2/4 dist(x ) log( dist(x,∂Ω) ) dist(x ) log( dist(x,∂Ω) )

θ Let us now verify that g˜ε ∈ L (Ω). Recall that θ = n ≥ 2/ε. Ω n−1 n dist(x,∂ ) dx kg˜ε k n (Ω) ≈ + 2 L Ω dist(x,∂Ω)n log( e )nε/2 nε /4 ¹ dist(x,∂Ω) dx 1 r dr ≤ + ≈ + < ∞. Ω dist(x,∂Ω) log( e )1 ε/2 (1 − r) log( e )1 ε/2 ¹ dist(x,∂Ω) ¹0 1−r 42 LUKÁŠ MALÝ

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Department of Mathematical Sciences, University of Cincinnati, Cincinnati, OH 45221-0025, U.S.A. E-mail address: [email protected]