SES OF UNIQUENESS FOR DIRICHLE YPE SPACES KARIM KELLAY Abstract. We study the uniqueness sets on the unit circle for weighted Dirichlet spaces.. Introduction Let D be the open unit disk in the complex plane and let = D be the unit circle. For µ a positive Borel measure on the unit circle, the Dirichlet-type space D(µ) consists of all analytic functions f defined on D such that D µ (f) := D ξ (f)dµ(ξ) <. where D ξ (f) is the local Dirichlet integral of f at ξ given by f(e it ) f(ξ) 2 dt D ξ (f) := e it ξ 2 2π. he space D(µ) is endowed with the norm f 2 µ := f 2 H 2 + D µ(f). Note that if dµ(e it ) = dt/2π, the normalized arc measure on, then the space D(µ) coincides with the classical space of functions with finite Dirichlet integral. hese spaces were introduced by Richter [] and by Aleman []. he space D(µ) were studied in [,, 2, 3, 4, 5, 6]. Let D h (µ) the harmonic version of D(µ) given by We define the capacity C µ of a set E by D h (µ) := { f L 2 () : D µ (f) < }. C µ (E) := inf { f 2 µ : f D h (µ) and f a.e. on a neighborhood of E }, see [4, 5]. We say that the proprety hold C µ quasi everywhere (C µ q.e) if it holds everywhere outside a set of zero C µ capacity. We have C µ (E) := inf { f 2 µ : f D h (µ) and f C µ q.e on E }. 2000 Mathematics Subject Classification. primary 30H05; secondary 3A25, 3C5. Key words and phrases. weighted Dirichlet spaces, capacity, uniqueness set. his work was partially supported by ANR Dynop.
2 KELLAY see [7, heorem 4.2]. Since D(µ) H 2, every function f D(µ) has non tangential limits f almost everywhere on. We have more, the radial limit of f D h (µ) given by f (ζ) := lim f(rζ) exists C µ quasi everywhere [4, heorem 2..9]. r Let E be a subset of. he set E is said to be a uniqueness set for D(µ) if, for each f D(µ) such that f = 0 on E, we have f = 0. In order to state our main result, we introduce some notion. Given E, we write E for the Lebesgue measure of E. Let 0 α <, we call the mesure µ is α regular if for all open set O, all open arc and γ (0, ), we have µ(o) c(α)o α, \ dµ(ζ) d(ζ, γ) 2 c(α, γ) +α. he Lebesgue mesure dµ(ζ) = dζ/2π is 0 regular, if dµ(ζ) = ζ α dζ, the mesure µ is α regular for α (0, ) and dµ(ζ) = d(ζ, E) α dζ is also α regular where E is a Cantor set see [6]. heorem. Let µ be a α regular measure. Let E be a Borel subset of of Lebesgue measure zero. We assume that there exists a family of pairwise disjoint open arcs (I n ) of such that E I n. Suppose n C µ (E I n ) =, then E is a uniqueness set for D(µ). n he case of Dirichlet space, dµ = dm, was obtained by Khavin and Mazya [9], see also [4, 3, 8]. In [0], we give the generalization of their result in the Dirichlet spaces D s, 0 < s, which consists of all analytic functions f on the open disc D such that D s (f) = f(z) f(w) 2 dz z w +s 2π 2. Proof dw 2π. Let I be an open arcs of and f be a function. We set f(z) f(w) 2 dz D I,µ (f) := I I z w 2 2π dµ(w) and m I(f) := I We begin with the following key lemma I f(ξ)dξ Lemma 2. Let µ be a α regular measure. Suppose that 0 < γ <. Let E and f D(µ) be such that f E = 0. hen, for any open arc I with I γπ m I (f) 2 D I,µ(f) C µ (E I), where the implied constants depend only on α and γ.
SES OF UNIQUENESS 3 Proof. We assume that I = (e iθ, e iθ ) with θ < γπ/2. Let = (e 2iθ/(+γ), e 2iθ/(+γ) ) and I γ = (e iθγ, e iθγ ) with θ γ = 3+γ θ. Note that I I 2(+γ) γ. Let φ be a positive function on such that supp φ = I γ, φ = on I and φ(z) φ(w) z w/ for z, w. Let f be such that { f(e f(e it it ), e it I, ) = f(e i 3θ t 2 ), e it \I. hen by a change of variable, we get see [0]. Now, we consider the function D I,µ (f) D,µ ( f) () F (z) = φ(z) f(z), z. m (f) Hence F 0, F E I = C µ q.e. herefore, We claim that C µ (E I) F 2 µ. (2) F 2 µ D I,µ(f) m I (f) 2. (3) he Lemma follows from (2) and (3). Now, we proof the claim (3). We have F 2 µ = F (ζ) 2 dζ 2π + m (f) 2 + + F (ζ) F (ξ) 2 dζ dµ(ξ) ζ ξ 2 2π 2π 2π + F (ζ) F (ξ) 2 dζ ζ ξ 2 2π m (f) f(ξ) 2 dζ dµ(ξ) ζ ξ 2 2π 2π 2 dζ m (f) f(ζ) m (f) 2 m (f) 2 ζ \ ξ I γ ξ \ ζ I γ m (f) f(ζ) 2 dζ ζ ξ 2 2π dµ(ξ) 2π dµ(ξ) 2π A = 2πm (f) + B 2 4π + C 2 4π 2 m (f) + D 2 4π 2 m (f). 2 (4) Since a measure µ is α regular, we have for l.s.c positive function g(ξ)dξ α g(ξ)dµ(ξ). (5)
4 KELLAY Indeed g(ξ)dµ(ξ) = µ({ξ : g(ξ) > t})dt {ξ : g(ξ) > t} α dt α {ξ : g(ξ) > t}dt α g(ξ)dξ. Hence by () and (5) A := m (f) f(ζ) 2 dζ f(ξ) f(ζ) 2 dξdζ f(ξ) f(ζ) 2 dξdµ(ζ) α If (z, w), then we write f(ξ) f(ζ) 2 ξ z 2 dξdµ(ζ) D I,µ ( f). (6) F (z) F (w) = ( φ(z) f(z) f(w) ) + (φ(z) φ(w)) f(w) m (f) m (f) m (f) m (f) f(z) f(w) + z w m m (f) f(w). So, by () again, B := F (ζ) F (ξ) 2 ζ w 2 f(ζ) f(ξ) 2 m (f) 2 ζ ξ 2 ( + m (f) 2 f(η) f(w)dη) 2dµ(ξ)dζ 4 f(η) f(ξ) 2 dηdµ(ξ) D I,µ(f) m (f) 2 η ξ 2 m (f). (7) 2
Now, Finally, by (5) C := D := SES OF UNIQUENESS 5 m (f) ζ \ ξ Iγ f(ξ) 2 ζ ξ 2 dζ m ζ \ d(ζ, I γ ) 2 (f) f(ξ) 2 dµ(ξ) ξ I γ f(η) f(ξ) 2 dηdµ(ξ) 2 I γ f(η) f(ξ) 2 D η ξ 2 I,µ (f). (8) ξ \ ξ \ 2+α m (f) ζ Iγ f(ζ) 2 ζ ξ 2 dµ(ξ) m d(ξ, I γ ) 2 (f) f(ζ) 2 dζ ζ I γ f(η) f(ζ) 2 dηdζ I γ f(η) f(ζ) 2 dηdµ(ζ) 2 I γ f(η) f(ζ) 2 η ζ 2 dζdµ(ζ) D I,µ (f). (9) By (6), (7), (8), and (8) we get (3) and the proof is complete. Proof of heorem. Since E = 0, we can assume that sup n γπ with γ (0, ). Let f D(µ) be such that f E = 0. We set l = n I n. By Lemma 2 and ensen inequality log f(ξ)dξ S In n = n n n ( cdin,µ(f) ) C µ (E I n ) l C µ (E I n ) + l n ( c C µ (E I n ) + l log l C µ (E I n ) + l log ( c l D µ(f) ( cdin,µ(f) ) log ) D In,µ(f) n ) =. By Fatou heorem we obtain f = 0 and the proof is complete.
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