Relations of the Spaces $${\varvec{A^p(\varOmega )}}$$ A p ( Ω ) and $${\varvec{C^p(\partial \varOmega )}}$$ C p ( ∂ Ω )

Relations of the Spaces $${\varvec{A^p(\varOmega )}}$$ A p ( Ω ) and... Results Math (2018) 73:86 Online First c 2018 The Author(s) Results in Mathematics https://doi.org/10.1007/s00025-018-0842-5 p p Relations of the Spaces A (Ω) and C (∂Ω) N. Georgakopoulos, V. Mastrantonis, and V. Nestoridis Abstract. Let Ω be a Jordan domain in C, J an open arc of ∂Ω and φ : D → Ω a Riemann map from the open unit disk D onto Ω.Un- der certain assumptions on φ we prove that if a holomorphic function f ∈ H(Ω) extends continuously on Ω ∪ J and p ∈{1, 2,... } ∪ {∞},then (l) the following equivalence holds: the derivatives f , 1 ≤ l ≤ p, l ∈ N, ex- tend continuously on Ω ∪ J if and only if the function f | has continuous derivatives on J with respect to the position of orders l, 1 ≤ l ≤ p, l ∈ N. Moreover, we show that for the relevant function spaces, the topology induced by the l-derivatives on Ω, 0 ≤ l ≤ p, l ∈ N, coincides with the topology induced by the same derivatives taken with respect to the posi- tion on J. Mathematics Subject Classification. Primary 30E99; Secondary 30H50. Keywords. Riemann map, Poisson Kernel, Jordan curve, smoothness on the boundary. 1. Introduction In this paper we investigate the relationship between the continuous extend- ability of the derivatives of a function f ∈ A(Ω), for some Jordan domain Ω, and the differentiability of the map t → f (γ(t)) for some parametrization γ of ∂Ω. Here, A(Ω) is the collection of all complex functions holomorphic on Ω and continuous on Ω. Specifically, it is well known that the first p derivatives of a function f ∈ A(D), D being the unit disk in C, continuously extend over it D if and only if the map t → f (e )is p times continuously differentiable [6]. We generalize this for functions that are holomorphic on the unit disk but now continuously extend over an open arc of T = ∂D and prove an analogous equiv- alence for functions defined on Jordan domains that have sufficiently smooth Riemann maps. 0123456789().: V,-vol 86 Page 2 of 13 N. Georgakopoulos et al. Results Math The spaces A (Ω), p ∈{0, 1, 2,...}∪{∞}, consist of all holomorphic (l) functions f in Ω whose derivatives f , l ∈{0, 1,...},l ≤ p, extend con- tinuously on Ω. It is well known that for any f in the disc algebra A(D), p it p f ∈ A (D) if and only if the map t → f (e )is C smooth. In other words, p p A (D)= A(D) ∩ C (T) both as sets and as topological spaces. Additionally, if it f ∈ A(D)and g(t)= f (e ),t ∈ R, the equation that relates the continuous ex- dg it  it tension of f on T and the derivative of g is as expected, i.e. (t)= ie f (e ). dt To prove this, one can use the Poisson representation, to recover the values it of f in the disk from its boundary values, i.e. f (re )=(g ∗ P )(t) where P r r denotes the Poisson kernel, differentiate both sides with respect to t and let r → 1 . A detailed proof can be found in [6]. In this paper we prove analogous results for functions f ∈ A(D)whose derivatives continuously extend on an open arc of the unit circle but not nec- essarily the entire circle. Moreover, using Riemann’s mapping theorem, we can drop our initial assumption f ∈ A(D) and instead assume that it only extends continuously over the specific arc we are interested in. Precisely, if f : D → C is holomorphic and continuously extends over an open arc J ⊆ T, then its first p it derivatives continuously extend over that arc if and only if the map t → f (e ) p it is in C (I), where I =(a, b) is an interval in R with J = {e : t ∈ I}. This motivates a more general definition of the spaces A . In Sect. 3 we consider functions f holomorphic on a Jordan domain Ω and continuous on Ω ∪ J , for some open arc J of ∂Ω. We prove that for any (l) p ∈{1, 2,... }∪{∞}, the derivatives f , 0 ≤ l ≤ p, continuously extend over Ω ∪ J if and only if the continuous extension of f on J is p times continuously differentiable on J with respect to the position [2]. To do this, we place a smoothness assumption for the Riemann map φ : D → Ω from the open unit −1 disk D onto Ω. The condition is that (φ ) has a continuous extension on −1 Ω ∪ J and that (φ ) (z) =0 on Ω ∪ J . Professor Paul Gauthier informed us that results in the same direction have already been proven in a paper of Steve Bell and Laszlo Lempert entitled “A C Schwarz Reflection principle in one and several complex variables” which appeared in J. Differential Geometry 32(1990)899–915. However, they do not cover our result Theorem 5. He also added that section 8 of a paper of Steven R. Bell and Steven G. Krantz entitled “Smoothness to the boundary of conformal maps”, which appeared in Rocky Mountain Journal of Mathematics, vol. 17, no 1, winter 1987, shows that our result is the best possible in a certain sense. 2. Extendability Over an Open Arc of the Unit Circle For 0 ≤ p ≤ +∞, A (D) denotes the space of holomorphic functions on D whose derivatives of order l ∈ N, 0 ≤ l ≤ p, extend continuously over D.Itis topologized via the semi-norms: p p Relations of the Spaces A (Ω)and C (∂Ω) Page 3 of 13 86 (l) (l) |f | =sup |f (z)| =sup |f (z)|, 0 ≤ l ≤ p, l ∈ N. z∈T z∈D The following theorem is well known. A detailed proof can be found in [6], Theorem 1. For all f ∈ A(D) the following equivalence holds: f ∈ A (D) if it and only if the map g(t)= f (e ), t ∈ R,is p times continuously differentiable. In that case: dg it  it (t)= ie f (e ). (1) dt We now generalize this on functions that are holomorphic on D and con- tinuously extend over an open arc J of the unit circle. We prove that for any such function f and p ∈{1, 2,...}∪{∞} the first p derivatives of f continu- it ously extend over D∪J if and only if the map t → f (e )is p times continuously it differentiable in I = {t ∈ [a, a +2π]: e ∈ J } for a suitable a ∈ R. Denote by A (D, J ) the space of holomorphic functions whose first p derivatives contin- uously extend over D ∪ J and let C (J ) be the class of functions f : J → C, it such that the map t → f (e ), t ∈ I,is p times continuously differentiable. The p p aim is to show the equality A (D, J)= A(D, J ) ∩ C (J ). For simplicity, we it iθ take J = {e :0 <t< 1} throughout this section. For any z = re ∈ C we denote by P (t)or P (t) the Poisson kernel [1]. z r Proposition 1. If u :[0, 1] → R is a continuous function then: A(z)= u(t)P (t)dt (2) 2π is well defined in C\J and C harmonic. Proof. To see that A(z) is well defined in C\J observe that: 1 1 −it 1 1 1+ e z A(z)= u(t)P (t)dt = Re u(t) dt (3) −it 2π 2π 1 − e z 0 0 −it i(t+2kπ) it and 1 − e z =0 ⇔ z = e ,k ∈ Z.For afixed z ∈ C\{e :0 ≤ t ≤ 1} −it we have δ = dist(1, {ze :0 ≤ t ≤ 1}) > 0 and: −it 1+ e z 1+ |z| u(t) ≤ sup |u(t)| < + ∞ (4) −it 1 − e z δ t∈[0,1] for all t ∈ [0, 2π]. Since the quantity on the right hand side is integrable we deduce that A(z) is indeed well defined in C\J . In order to prove that A is C harmonic it suffices to show that g(z)= −it 1 1+e z u(t) dt is holomorphic in C\J , since A is the real part of g accord- −it 2π 0 1−e z ingto(3). Note that for z = z : 0 86 Page 4 of 13 N. Georgakopoulos et al. Results Math −it g(z) − g(z ) 1 2e = u(t) dt (5) −it −it z − z 2π (1 − e z)(1 − e z ) 0 0 and hence for z sufficiently close to z : 1 −it g(z) − g(z ) 1 2e − u(t) dt (6) −it 2 z − z 2π (1 − e z ) 0 0 1 −2it 1 2e (z − z ) = u(t) dt (7) −it −it 2 2π (1 − e z)(1 − e z ) 1 2 z→z ≤ sup |u(t)|2 |z − z | −− −→ 0. (8) 2π δ t∈[0,1] 0 Therefore g is holomorphic and the proof is complete. Lemma 1. Let u :[0, 1] → R be a continuous function and define A(z) as in iθ Proposition 1. For all z = re ∈ C\J : dA 1 −2r(1 − r )sin(θ − t) iθ (re )= u(t) dt. (9) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) Thus, for θ ∈ (1, 2π) and r =1: dA iθ (e )=0. (10) dθ iθ 1 1 Proof. Since A(re )= u(t)P (θ − t)dt = (f ∗ P )and P is dif- r r r 2π 0 2π iθ ferentiable in respect to θ we have that A(re ) is differentiable in respect to θ, dA 1 d(f ∗ P ) 1 dP r r iθ iθ iθ (re )= (re )= f ∗ (re )=⇒ (11) dθ 2π dθ 2π dθ dA 1 −2r(1 − r )sin(θ − t) iθ (re )= u(t) dt. (12) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) (10) is derived from (9) substituting r =1 and θ ∈ (1, 2π). Because every continuous function u :[0, 1] → C can be considered a 2π-periodic function u : R → C such that v(x)= u(x)for x ∈ [0, 1] + 2πZ and v(x) = 0 otherwise, one can expect that such a function when convolved with the Poisson kernel would retain the nice properties. More specifically, it will uniformly converge to 0 and to u(x) on the compact subsets of the respective open arcs. We prove this in Propositions 2 and 3. Proposition 2. If A(z) is as before, then for all θ ∈ (1, 2π) and l ∈ N: d A iθ lim (re )=0. (13) r→1 dθ The convergence is uniform in the compact subsets of (1, 2π). p p Relations of the Spaces A (Ω)and C (∂Ω) Page 5 of 13 86 Proof. We start with l = 1. It suffices to prove that for all [θ ,θ ] ⊂ (1, 2π)the 1 2 convergence is uniform. Observe that 0 <θ − t<θ − t< 2π for all t ∈ [0, 1] 1 2 and therefore cos(θ − t) ≤ max{cos(θ − 1), cos(θ )} = M< 1, ∀t ∈ [0, 1] 1 2 2 2 and ∀θ ∈ [θ ,θ ]. This implies that 1 + r − 2r cos(θ − t) ≥ 1+ r − 2rM = 1 2 (1 − r) +2r(1 − M ) for all t ∈ [0, 1] and θ ∈ [θ ,θ ]. Thus, for all θ ∈ [θ ,θ ] 1 2 1 2 and 0 <r < 1: dA 1 −2r(1 − r )sin(θ − t) iθ (re ) = u(t) dt (14) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) 1 2r(1 − r ) ≤ u (15) 2 2 2π ((1 − r) +2r(1 − M )) r→1 dA iθ and hence sup (re ) − −−− → 0 since the right hand side of (15)con- dθ θ∈[θ ,θ ] 1 2 verges to0as r → 1 . Note that no matter how many times we differentiate P in respect to θ we 2 k l m will have a finite sum of fractions with numerator c(1 − r ) cos(θ) sin(θ) for c =0,k,l,m ∈ N and denominator a power of (1 + r − 2r cos(θ − t)). So for any l ≥ 2 the same arguments apply. Denote by C ([0, 1]) the class of functions u :[0, 1] → C that are p times (l) continuously differentiable in (0, 1) and u continuously extend on [0, 1] for all 0 ≤ l ≤ p, l ∈ N. Proposition 3. Let p ∈{0, 1,...}∪{∞}, u :[0, 1] → R of class C ([0, 1]) and define A(z) as in Proposition 1. For all θ ∈ (0, 1) and 0 ≤ l ≤ p, l ∈ N, d A (l) lim (z)= u (θ ). (16) iθ 0 dθ z→e |z|<1 The convergence is uniform in the compact subsets of (0, 1). Proof. By a theorem of Borel (see [5]), we can find a function q : R → R of ∞ l l l l class C (R) such that q (0) = u (0) and q (1) = u (1) for 0 ≤ l ≤ p, l ∈ N. Thus, the function: u(t),t ∈ [0, 1] g(t)= q(t),t ∈ (1, 2π) is of class C (T). Define: 2π G(z)= g(t)P (t)dt. (17) 2π It is well known that uniformly for all θ ∈ R, 0 ≤ l ≤ p, l ∈ N: d G (l) lim (z)= g (θ) (18) iθ dθ z→e |z|<1 86 Page 6 of 13 N. Georgakopoulos et al. Results Math Let 1 2π 1 1 A(z)= u(t)P (t)dt and B(z)= q(t)P (t)dt. (19) z z 2π 2π 0 1 Note that G(z)= A(z)+ B(z) for all |z| < 1 and hence for 0 ≤ l ≤ p, l ∈ N: l l l d G d A d B (z)= (z)+ (z). (20) l l l dθ dθ dθ l l d G iθ (l) d B iθ From (18) Proposition 2 we have that (re ) → u (θ)and (re ) → 0 l l dθ dθ as r → 1 , uniformly in the compact subsets of (0, 1). As a result, l l l d A d G d B (l) iθ lim (z) = lim (z) − (z) = u (e ) (21) l l l iθ iθ 0 0 dθ dθ dθ z→e z→e |z|<1 |z|<1 while the convergence is uniform in the compact subsets of (0, 1). Remark 1. By linearity, Propositions 1, 2, 3 and Lemma 1 hold for complex functions f = u + iv where u = Ref and v = Imf , since we can apply them to the real and imaginary part separately. We now adapt the proof of Theorem 1 for functions whose derivatives only extend over an open arc of the unit circle. it Theorem 2. Let p ∈{1, 2,...}∪{∞}, f ∈ A(D) and g(t)= f (e ),t ∈ (0, 1). (l) The following are equivalent: f continuously extends on D ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if g is p times continuously differentiable in (0, 1).In that case, for all t ∈ (0, 1) dg it  it (t)= ie f (e ). (22) dt Proof. We prove it by induction on p.For p =1, let t ∈ (0, 1) and t <t <t 0 1 0 2 (l) such that [t ,t ] ⊂ (0, 1). Assuming that f continuously extends over D ∪ J 1 2 it it  it for all 0 ≤ l ≤ p, l ∈ N,let f (t)= f (re ), h(t)= ie f (e ) and: df it  it h (t)= (t)= ire f (re ) (23) dt for all t ∈ (0, 1) and 0 <r < 1. We have h → h as r → 1 uniformly in [t ,t ], r 1 2 it since f is continuous in D ∪ J . Note that f (t ) → f (e ) and hence from a r 0 well-known theorem f → hdt + c, for a some c ∈ C,as r → 1, while the convergence is uniform in [t ,t ]. Additionally, f → g,as r → 1, uniformly 1 2 r it  it in [t ,t ] and therefore g (t)= h(t)= ie f (e ),t ∈ (t ,t ). However, t was 1 2 1 2 0 arbitrary thus, g ∈ C ((0, 1)) and (22) holds. For the converse let g ∈ C ((0, 1)). We now use the Poisson representa- tion: 2π it f (z)= f (e )P (t)dt. (24) 2π 0 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 7 of 13 86 for |z| < 1. Define: t 2π+t 2 1 1 1 it A(z)= g(t)P (t)dt and B(z)= f (e )P (t)dt. (25) z z 2π 2π t t 1 2 Note that f (z)= A(z)+ B(z) for all |z| < 1. Consequently, df dA dB it it it (re )= (re )+ (re ) (26) dt dt dt for all 0 <r < 1and t ∈ R. Since [t ,t ] ⊂ (0, 1), Propositions 2 and 3 imply 1 2 that: dA dg it lim (re )= (t) (27) r→1 dt dt and dB it lim (re ) = 0 (28) r→1 dt uniformly for t ∈ [t ,t ]. Combining (26), (27) and (28) we get: 1 2 1 df 1 dg it it lim f (re ) = lim (re )= (t) (29) it it − − r→1 r→1 ire dt ie dt uniformly for t ∈ [t ,t ]. Since t was arbitrarily chosen in (0, 1) we deduce that 1 2 0 f extends continuously on D ∪ J . To complete the induction, let us assume (l) that the theorem holds for some p ≥ 1. If f continuously extends on D ∪ J (l) for all 0 ≤ l ≤ p +1,l ∈ N it follows that (f ) continuously extends on D ∪ J for all 0 ≤ l ≤ p, l ∈ N. By the induction hypothesis the map t → it p f (e ) belongs in the class C ((0, 1)) and since, by the case of p =1, g (t)= it  it  p p+1 ie f (e )wehave g ∈ C ((0, 1)) and hence g ∈ C ((0, 1)). For the converse, p+1  it  it p if g ∈ C ((0, 1)) it follows from (22) that g (t)= ie f (e ) ∈ C ((0, 1)) it p and therefore, the map t → f (e ) is of class C ((0, 1)). By the induction (l) hypothesis, (f ) continuously extends on D ∪ J , for all 0 ≤ l ≤ p, l ∈ N and (l) hence f continuously extends on D ∪ J , for all 0 ≤ l ≤ p +1,l ∈ N. The case of p = ∞ follows easily. Using Riemann’s mapping theorem [1] we can drop the assumption of continuity over D. Indeed, we can just have f continuously extend over the open arc we are dealing with, i.e. f ∈ A(D, J ). Theorem 3. Let p ∈{1, 2,...}∪{∞}, f : D ∪ J → C continuous on D ∪ J and it holomorphic in D and g(t)= f (e ),t ∈ (0, 1). The following are equivalent: (l) f continuously extends over D ∪ J , for all 0 ≤ l ≤ p, l ∈ N,ifand only if g is p times continuously differentiable in (0, 1). In that case: dg it  it (t)= ie f (e ) (30) dt for all t ∈ (0, 1). 86 Page 8 of 13 N. Georgakopoulos et al. Results Math Proof. We prove it by induction on p.For p = 1, the only if part is proven like Theorem 2; we also obtain (30). For the converse, let t ∈ (0, 1), t <t <t 0 1 0 2 it such that [t ,t ] ⊆ (0, 1) and J = {e : t <t<t }. Set V = {|z| < 1 2 1 2 1,z =0: ∈ J }. It is easily verified that V is simply connected and hence |z| there is a conformal map φ : D → V which extends to homeomorphism over the closures φ : D → V , by the Osgood–Carath´ eodory theorem [4]. Since it it it 1 2 φ(T)= {e : t ≤ t ≤ t }∪{re :0 ≤ r ≤ 1}∪{re :0 ≤ r ≤ 1} and 1 2 it −1 it {e : t ≤ t ≤ t } is connected we deduce that φ ({e : t ≤ t ≤ t }) is closed 1 2 1 2 −1 it and connected in T. Without loss of generality, assume that φ ({e : t ≤ it t ≤ t })= {e :0 ≤ t ≤ 1}. Consequently, f ◦ φ : D → C is continuous on D and holomorphic in D that is, of class A(D). Using the Reflection Principle, we it deduce that φ conformally extends on an open G ⊃ V ∪{e : t ≤ t ≤ t } thus, 1 2 it ∞ t → φ(e ),t ∈ (t ,t ) is of class C ((t ,t )) with non-vanishing derivative (see 1 2 1 2 1 it ∞ [1] pp. 233–235). Since g ∈ C ((t ,t )) and t → φ(e )isin C ((0, 1)) we have 1 2 it 1 that the map t → f (φ(e )) is in C ((0, 1)). By Theorem 2 the derivative (f ◦ φ) continuously extends on D ∪ J and hence f continuously extends in −1   −1 −1 D ∪ J , because f =(f ◦ φ) ◦ φ and f =(f ◦ φ) ◦ φ · (φ ) . Our choice of t was arbitrary and therefore f continuously extends on D ∪ J . To complete the induction, we follow the proof of Theorem 2. The case of p = ∞ follows easily. In other words, Theorem 3 states that for any open arc J of T and p p p ∈{0, 1, 2,...}∪{∞}, A (D, J)= A(D, J ) ∩ C (J ) holds. Remark 2. In all Propositions and Lemmas of this section, [0, 1] can be re- placed by any interval [a, b] with 0 ≤ a<b<a +2π. 3. Jordan Domains Motivated by Theorem 3 we now give a more general definition of the spaces p p A .Let Ω be a Jordan domain and J an open arc of ∂Ω. Denote by A (Ω, J ) (l) the collection of all functions f holomorphic on Ω such that f continuously p p extends on Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N. Note that A (Ω, ∂Ω)= A (Ω). If γ : I → J is a parametrization of J , denote by C (J ) the class of all functions f : G → C defined on a varying set G, J ⊂ G, such that f ◦ γ : I → C is p times continuously differentiable. Moreover, if φ : D → Ω is a Riemann map, p it denote by C (J ) the class C (J ) where γ(t)= φ(e ) . Since any two Riemann maps differ by an automorphism of the unit disk it is easily verified that the spaces C (J ) do not depend on the chosen Riemann map. Next, we consider differentiability on J with respect to the position [2]. Definition 1. Let J be a Jordan arc and f : J → C. We define the derivative of f on z ∈ J by: df f (z) − f (z ) (z ) = lim (31) dz z→z ,z∈J z − z 0 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 9 of 13 86 if this limit exists and is a complex number. df In order to go one step further , we consider the derivative on J of dz Definition 1 and we take its derivative on J with respect to the position. df Definition 2. A function f : J → C belongs to the class C (J)if (z)exists dz p−1 d f and is continuous for z ∈ J . Inductively, suppose that is well defined on p−1 dz J for some p =2,..., +∞, we say that f is of class C (J)if p−1 d f d f d( ) p−1 dz (z)= (z) (32) dz dz exists and is continuous on J . Remark 3. In [2] the following fact is proven. If γ : I → J is a C regular (with non vanishing derivative) parametrization of a Jordan arc J, n ∈{1, 2,... }∪ {∞}, a function f is of class C (J ) if and only if g(t)=(f ◦ γ)(t),t ∈ I,isof class C (I),p ∈{1, 2,...}∪{∞},p ≤ n. Additionally, dg df (t)= (γ(t)) · γ (t). (33) dt dz In this section we prove that given a Jordan domain Ω, φ : D → Ω a −1 1 −1 Riemann map such that φ ∈ A (Ω, J ) and (φ ) (z) =0,z ∈ Ω ∪ J , f ∈ (l) A(Ω, J)and p ∈{1, 2,... }∪{∞} the following are equivalent: f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if f | is of class C (J ). p p That is A (Ω, J)= A(Ω, J ) ∩ C (J ). Let us first prove a straightforward fact concerning the parametrization of an arc J of ∂Ω induced by a Riemann map φ. Theorem 4. Let Ω be a Jordan domain, φ : D → Ω a Riemann map and J an −1 n open arc of ∂Ω such that φ is of class A (Ω, J ) for some n ∈{1, 2,... }∪ −1 {∞} and (φ ) (z) =0,z ∈ Ω ∪ J . The following holds for any p ∈{1, 2,... }∪ p p {∞},p ≤ n: A (Ω, J)= A(Ω, J ) ∩ C (J ). In that case, if f ∈ A (Ω, J ) and it g(t)=(f ◦ φ)(e ): dg it  it  it (t)= ie f (φ(e ))φ (e ) (34) dt is for all t ∈{s ∈ R : φ(e ) ∈ J }. On the right hand side of (34) f denotes the continuous extension of f (z)from Ω to Ω ∪ J . Before proceeding to the proof, let us note that the choice of the Riemann map is irrelevant. Suppose that Φ, Ψ are two Riemann maps and J an open arc −1 n −1 of ∂Ω such that Φ is of class A (Ω, J ),n ∈{1, 2,...}∪{∞}, and (Φ ) (z) = −1 n −1 0 for all z ∈ Ω ∪J . Then, Ψ is also of class A (Ω, J ) and (Ψ ) is non zero in −1 Ω ∪J . To see this, observe that Φ ◦Ψ is an automorphism of the unit disk and z−a hence there are a ∈ D and c ∈ T such that Ψ = Φ◦φ , where φ (z)= c ,z ∈ a a 1−az D. Note that φ is holomorphic in D(0, ) ⊃ D with non vanishing derivative a 86 Page 10 of 13 N. Georgakopoulos et al. Results Math (l) thus, Ψ can be extended over D for all 0 ≤ l ≤ n, l ∈ N. Additionally, 1−|a| −1  −1  −1 (Ψ ) (z)= · (Φ ) (z) = 0 for all z ∈ Ω ∪ J.Moreover, Φ being −1 2 (1+aΦ (z)) n n −1 of class A (Ω, J ) is equivalent to Φ being of class A (D, I), I = φ (J ). This is −1 easy to see given that (Φ ) (z) =0 for z ∈ Ω ∪J . In other words, if a Riemann map induces a regular n times continuously differentiable parametrization of J then, the same holds for any other Riemann map. −1 is Proof. (Theorem 4) Set I = φ (J)and I = {s ∈ R : φ(e ) ∈ J }. Given an f ∈ A(Ω, J ) one can easily verify that f is of class A (Ω, J ) if and only if f ◦ φ is of class A (D, I), since φ is non zero in D ∪ I and p ≤ n. By Theorem 3, p it f ◦ φ is of class A (D, I) if and only if the map t → (f ◦ φ)(e ),t ∈ I, is p times continuously differentiable which by definition is equivalent to f being it of class C (J ). To see that (34) holds, set g (t)= f (φ(re )),t ∈ R, 0 < r< 1. Note that g → g as r → 1 uniformly in the compact subsets of dg r it  it  it I. Additionally, (t)= ire f (φ(re ))φ (e ) which uniformly converges to dt it  it  it − ie f (φ(e ))φ (e )as r → 1 in the compact subsets of I . From a well known theorem of calculus (34) follows. In the next theorem we prove that for any function f ∈ A(Ω, J ) the first p derivatives of f continuously extend over Ω ∪ J if and only if f | is of class C (J ), given that J is a smooth enough open arc of the Jordan domain Ω. −1 1 −1 That is, if φ is a Riemann map and φ is of class A (Ω, J ) with (φ ) (z) = p p 0,z ∈ Ω ∪ J then A (Ω, J)= A(Ω, J ) ∩ C (J ) for all p ∈{1, 2,...}∪{∞}. Theorem 5. Let Ω be a Jordan domain, φ : D → Ω a Riemann map and J an −1 1 −1 open arc of ∂Ω such that φ is of class A (Ω, J ) with (φ ) (z) =0,z ∈ Ω∪J . (l) Given an f ∈ A(Ω, J ) and p ∈{1, 2,...}∪{∞} we have that: f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if f | is of class C (J ). p p That is, A (Ω, J)= A(Ω, J ) ∩ C (J ). In that case: d f | (l) J f (z)= (z) (35) dz for all z ∈ J and 0 ≤ l ≤ p, l ∈ N. (l) (l) On the left hand side of (35), f (z) denotes the continuous extension of f on Ω ∪ J , while on the right hand side the differentiation is with respect to the position. is Proof. Let p =1 and f ∈ A(Ω, J ). Denote by I = {s ∈ R : φ(e ) ∈ J }. By Theorem 4 we have that f ∈ A (Ω, J ) if and only if the map g(t)= dg it it  it f (φ(e )),t ∈ I is continuously differentiable. Additionally, (t)= ie φ (e ) dt it f (φ(e )) and: dg df | dφ| df | it it it it it  it it J T J (e )= (φ(e )) · (e ) · ie = ie φ (e ) (φ(e )) (36) dt dz dz dz p p Relations of the Spaces A (Ω)and C (∂Ω) Page 11 of 13 86 it  it Since ie φ (e ) is non zero we have that g is continuously differentiable if and 1 1 only if f | is of class C (J ). Consequently, f is of class A (Ω, J ) if and only if g 1 1 is of class C (I) if and only if f | is of class C (J ). Equation (35) follows from Eqs. (34) and (36). For the induction step, assume that p> 1. We have that p  p−1 f ∈ A (Ω, J ) if and only if f ∈ A (Ω, J ) which by the induction hypothesis p−1 p is equivalent to f | ∈ C (J ) and that is equivalent to f | ∈ C (J ), by (35) J J for l =1. Moreover, (l−1) (l−1) d f | d df | d J J   (l−1) (l) (z)= (z) = (f (z)) = (f ) (z)= f (z) l (l−1) (l−1) dz dz dz dz (37) for all z ∈ J and 1 ≤ l ≤ p, l ∈ N. Remark 4. Note that if Ω is a Jordan domain, J an analytic arc of ∂Ω and −1 φ : D → Ω a Riemann map we know from [1] (p. 235) that φ has a conformal extension over an open set G ⊃ Ω ∪ J . Consequently, if J is an analytic arc p p we immediately have A (Ω, J)= A(Ω, J ) ∩ C (J ) for all p ∈{1, 2,...}∪{∞}. In addition, if we have γ a C regular parametrization of J, n ∈{1, 2,...}∪ {∞} we have a triple equivalence as derived from Remark 3 and Theorem 5. Theorem 6. Let Ω be a Jordan domain, φ a Riemann map and J an open −1 1 −1 arc of ∂Ω such that φ is of class A (Ω, J ) and (φ ) (z) =0,z ∈ Ω ∪ J , γ a C regular parametrization of J, n ∈{1, 2,...}∪{∞},f ∈ A(Ω, J ) and p{1, 2,...}∪{∞},p ≤ n. The following are equivalent: (l) 1. f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N. That is, f is of class A (Ω, J ). 2. f | is of class C (J ). 3. g(t)=(f ◦ γ)(t),t ∈ I is p times continuously differentiable. That is f is of class C (J ). p p p In other words, A (Ω, J)= A(Ω, J ) ∩ C (J)= A(Ω, J ) ∩ C (Ω, J ). In that case: dg df | (t)= (γ(t))γ (t)= f (γ(t))γ (t) (38) dt dz for all t ∈ I. Remark 5. We notice that the assumption on Theorem 5 that the Riemann −1 −1 map φ : D → Ω is such that φ ∈ A(Ω, J ) with (φ ) (z) =0 for z ∈ Ω ∪ J is in fact equivalent to say that the parametrization of J induced by φ is C regular. We topologize A (Ω) by the semi-norms (l) (l) |f | =sup |f (z)| =sup |f (z)|, 0 ≤ l ≤ p, l ∈ N (39) z∈∂Ω z∈Ω 86 Page 12 of 13 N. Georgakopoulos et al. Results Math and C (∂Ω) by the semi-norms d g |g| =sup | (z)|, 0 ≤ l ≤ p, l ∈ N (40) dz z∈∂Ω see also [2]. By Eq. (35) it easily seen that the topology on A (Ω)and A(Ω) ∩ p p C (∂Ω) induced by C (∂Ω) coincide. Summarizing: Corollary 1. Let Ω be a Jordan domain and φ : D → Ω a Riemann map of class A (D) such that φ (z) =0 for all z ∈ D. The following equivalence holds for all f ∈ A(Ω) and p ∈{1, 2,...}∪{∞}: f ∈ A (Ω) if and only if f | is ∂Ω p times continuously differentiable with respect to the position. In this case we have: d f (l) (z)= f (z) (41) dz (l) For all z ∈ ∂Ω and 0 ≤ l ≤ p, l ∈ N. On the right hand side of (41) f (z) (l) denotes the continuous extension of f from Ω to Ω ∪ J . We also have, p p A (Ω)= A(Ω) ∩ C (∂Ω) as topological spaces. Remark 6. There is a natural way to topologize the spaces A (Ω, J ) for any Jordan domain Ω,openarc J ⊂ ∂Ω and p ∈{1, 2,... }∪{∞}. Then, the p p equality A (Ω, J)= A(Ω, J ) ∩ C (J ) holds taking into account the topologies of the spaces as well. However, we will not deal with the topological properties of the spaces A (Ω, J ) in this paper. One can easily extend the results of this section for the complement of a Jordan domain Ω given that the functions considered vanish at infinity. That is, if A (C\Ω) is the class of all functions f with lim f (z) = 0 and holo- z→∞ morphic in C\Ω such that their first p derivatives extend over C\Ω for all ˆ ˆ 0 ≤ l ≤ p, l ∈ N, then A (C\Ω)= A (C\Ω) ∩ C (∂Ω) under similar assump- tions to that of Theorem 5.For J an open arc of ∂Ω we define A (C\Ω, J ) similarly. Remark 7. The previous results can be generalized to the case of finitely con- nected domains bounded by a finite set of disjoint Jordan curves. For this we use the Laurent Decomposition [3]. We will take a moment to sketch the proof. Let Ω be a bounded domain whose boundary consists of a finite number of disjoint Jordan curves. If V ,V ,..., 0 1 ˆ ˆ V are the connected components of C\Ω, ∞∈ V and Ω = C\V , ..., n−1 0 0 0 Ω = C\V , then for every f which is holomorphic in Ω we know that n−1 n−1 there exist functions f ,...,f which are holomorphic in Ω ,...,Ω re- 0 n−1 0 n−1 spectively such that f = f + f + ··· + f and lim f (z)=0,j ∈ 0 1 n−1 z→∞ j {1,...,n − 1} [3]. Let φ : D → Ω ,j ∈{0, 1,... ,n − 1}, be the respective con- j j formal maps. Without loss of generality, take J an open arc of ∂Ω such that −1 φ is of class A (Ω ,J ) with non vanishing derivative. Let also f be a holo- morphic function in Ω and continuous in Ω∪J such that f = f +f +···+f , 0 1 n−1 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 13 of 13 86 f holomorphic in Ω , is the extended Laurent decomposition of f . Since f is j j j holomorphic in a neighborhood of ∂Ω for all j = 0, we have that f continu- 0 0 ously extends over Ω ∪ J since f continuously extends over Ω ∪ J . Similarly, (l) f continuously extends over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N, if and only if f p p is of class A (Ω ,J ). By Theorem 5, f is of class A (Ω ,J ) if and only if f | 0 0 0 p p is of class C (J ), which happens if and only if f | is of class C (J ). Open Access. This article is distributed under the terms of the Creative Com- mons Attribution 4.0 International License (http://creativecommons.org/licenses/ by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. References [1] Ahlfors, L.V.: Complex Analysis, 3rd edn. Mc Graw-Hill Book Co, New York (1978) [2] Bolkas, E., Nestoridis, V., Panagiotis, C., Papadimitrakis, M.: One sided extend- ability and p-continuous analytic capacities, submitted. arXiv:1606.05443 [3] Costakis, G., Nestoridis, V., Papadoperakis, I.: Universal Laurent series. Proc. Edinb. Math.Soc.(2) 48(3), 571–583 (2005) [4] Koosis, P.: An Introduction of H Spaces. Cambridge University Press, Cam- bridge (1998) [5] Krantz, S., Parks, H.: A Primer of Real Analytic Functions, 2nd edn. Springer Birkh¨ auser, Boston (2002) p p [6] Mastrantonis, V.: Relations of the spaces A (Ω)and C (∂Ω). arXiv:1611.02791 N. Georgakopoulos, V. Mastrantonis and V. Nestoridis Department of Mathematics National and Kapodistrian University of Athens Panepistimioupolis 157 84 Athens Greece e-mail: nikolaosgmath@gmail.com V. Mastrantonis e-mail: blasismas@hotmail.com V. Nestoridis e-mail: vnestor@math.uoa.gr Received: August 24, 2017. Accepted: March 13, 2018. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Results in Mathematics Springer Journals

Relations of the Spaces $${\varvec{A^p(\varOmega )}}$$ A p ( Ω ) and $${\varvec{C^p(\partial \varOmega )}}$$ C p ( ∂ Ω )

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Results Math (2018) 73:86 Online First c 2018 The Author(s) Results in Mathematics https://doi.org/10.1007/s00025-018-0842-5 p p Relations of the Spaces A (Ω) and C (∂Ω) N. Georgakopoulos, V. Mastrantonis, and V. Nestoridis Abstract. Let Ω be a Jordan domain in C, J an open arc of ∂Ω and φ : D → Ω a Riemann map from the open unit disk D onto Ω.Un- der certain assumptions on φ we prove that if a holomorphic function f ∈ H(Ω) extends continuously on Ω ∪ J and p ∈{1, 2,... } ∪ {∞},then (l) the following equivalence holds: the derivatives f , 1 ≤ l ≤ p, l ∈ N, ex- tend continuously on Ω ∪ J if and only if the function f | has continuous derivatives on J with respect to the position of orders l, 1 ≤ l ≤ p, l ∈ N. Moreover, we show that for the relevant function spaces, the topology induced by the l-derivatives on Ω, 0 ≤ l ≤ p, l ∈ N, coincides with the topology induced by the same derivatives taken with respect to the posi- tion on J. Mathematics Subject Classification. Primary 30E99; Secondary 30H50. Keywords. Riemann map, Poisson Kernel, Jordan curve, smoothness on the boundary. 1. Introduction In this paper we investigate the relationship between the continuous extend- ability of the derivatives of a function f ∈ A(Ω), for some Jordan domain Ω, and the differentiability of the map t → f (γ(t)) for some parametrization γ of ∂Ω. Here, A(Ω) is the collection of all complex functions holomorphic on Ω and continuous on Ω. Specifically, it is well known that the first p derivatives of a function f ∈ A(D), D being the unit disk in C, continuously extend over it D if and only if the map t → f (e )is p times continuously differentiable [6]. We generalize this for functions that are holomorphic on the unit disk but now continuously extend over an open arc of T = ∂D and prove an analogous equiv- alence for functions defined on Jordan domains that have sufficiently smooth Riemann maps. 0123456789().: V,-vol 86 Page 2 of 13 N. Georgakopoulos et al. Results Math The spaces A (Ω), p ∈{0, 1, 2,...}∪{∞}, consist of all holomorphic (l) functions f in Ω whose derivatives f , l ∈{0, 1,...},l ≤ p, extend con- tinuously on Ω. It is well known that for any f in the disc algebra A(D), p it p f ∈ A (D) if and only if the map t → f (e )is C smooth. In other words, p p A (D)= A(D) ∩ C (T) both as sets and as topological spaces. Additionally, if it f ∈ A(D)and g(t)= f (e ),t ∈ R, the equation that relates the continuous ex- dg it  it tension of f on T and the derivative of g is as expected, i.e. (t)= ie f (e ). dt To prove this, one can use the Poisson representation, to recover the values it of f in the disk from its boundary values, i.e. f (re )=(g ∗ P )(t) where P r r denotes the Poisson kernel, differentiate both sides with respect to t and let r → 1 . A detailed proof can be found in [6]. In this paper we prove analogous results for functions f ∈ A(D)whose derivatives continuously extend on an open arc of the unit circle but not nec- essarily the entire circle. Moreover, using Riemann’s mapping theorem, we can drop our initial assumption f ∈ A(D) and instead assume that it only extends continuously over the specific arc we are interested in. Precisely, if f : D → C is holomorphic and continuously extends over an open arc J ⊆ T, then its first p it derivatives continuously extend over that arc if and only if the map t → f (e ) p it is in C (I), where I =(a, b) is an interval in R with J = {e : t ∈ I}. This motivates a more general definition of the spaces A . In Sect. 3 we consider functions f holomorphic on a Jordan domain Ω and continuous on Ω ∪ J , for some open arc J of ∂Ω. We prove that for any (l) p ∈{1, 2,... }∪{∞}, the derivatives f , 0 ≤ l ≤ p, continuously extend over Ω ∪ J if and only if the continuous extension of f on J is p times continuously differentiable on J with respect to the position [2]. To do this, we place a smoothness assumption for the Riemann map φ : D → Ω from the open unit −1 disk D onto Ω. The condition is that (φ ) has a continuous extension on −1 Ω ∪ J and that (φ ) (z) =0 on Ω ∪ J . Professor Paul Gauthier informed us that results in the same direction have already been proven in a paper of Steve Bell and Laszlo Lempert entitled “A C Schwarz Reflection principle in one and several complex variables” which appeared in J. Differential Geometry 32(1990)899–915. However, they do not cover our result Theorem 5. He also added that section 8 of a paper of Steven R. Bell and Steven G. Krantz entitled “Smoothness to the boundary of conformal maps”, which appeared in Rocky Mountain Journal of Mathematics, vol. 17, no 1, winter 1987, shows that our result is the best possible in a certain sense. 2. Extendability Over an Open Arc of the Unit Circle For 0 ≤ p ≤ +∞, A (D) denotes the space of holomorphic functions on D whose derivatives of order l ∈ N, 0 ≤ l ≤ p, extend continuously over D.Itis topologized via the semi-norms: p p Relations of the Spaces A (Ω)and C (∂Ω) Page 3 of 13 86 (l) (l) |f | =sup |f (z)| =sup |f (z)|, 0 ≤ l ≤ p, l ∈ N. z∈T z∈D The following theorem is well known. A detailed proof can be found in [6], Theorem 1. For all f ∈ A(D) the following equivalence holds: f ∈ A (D) if it and only if the map g(t)= f (e ), t ∈ R,is p times continuously differentiable. In that case: dg it  it (t)= ie f (e ). (1) dt We now generalize this on functions that are holomorphic on D and con- tinuously extend over an open arc J of the unit circle. We prove that for any such function f and p ∈{1, 2,...}∪{∞} the first p derivatives of f continu- it ously extend over D∪J if and only if the map t → f (e )is p times continuously it differentiable in I = {t ∈ [a, a +2π]: e ∈ J } for a suitable a ∈ R. Denote by A (D, J ) the space of holomorphic functions whose first p derivatives contin- uously extend over D ∪ J and let C (J ) be the class of functions f : J → C, it such that the map t → f (e ), t ∈ I,is p times continuously differentiable. The p p aim is to show the equality A (D, J)= A(D, J ) ∩ C (J ). For simplicity, we it iθ take J = {e :0 <t< 1} throughout this section. For any z = re ∈ C we denote by P (t)or P (t) the Poisson kernel [1]. z r Proposition 1. If u :[0, 1] → R is a continuous function then: A(z)= u(t)P (t)dt (2) 2π is well defined in C\J and C harmonic. Proof. To see that A(z) is well defined in C\J observe that: 1 1 −it 1 1 1+ e z A(z)= u(t)P (t)dt = Re u(t) dt (3) −it 2π 2π 1 − e z 0 0 −it i(t+2kπ) it and 1 − e z =0 ⇔ z = e ,k ∈ Z.For afixed z ∈ C\{e :0 ≤ t ≤ 1} −it we have δ = dist(1, {ze :0 ≤ t ≤ 1}) > 0 and: −it 1+ e z 1+ |z| u(t) ≤ sup |u(t)| < + ∞ (4) −it 1 − e z δ t∈[0,1] for all t ∈ [0, 2π]. Since the quantity on the right hand side is integrable we deduce that A(z) is indeed well defined in C\J . In order to prove that A is C harmonic it suffices to show that g(z)= −it 1 1+e z u(t) dt is holomorphic in C\J , since A is the real part of g accord- −it 2π 0 1−e z ingto(3). Note that for z = z : 0 86 Page 4 of 13 N. Georgakopoulos et al. Results Math −it g(z) − g(z ) 1 2e = u(t) dt (5) −it −it z − z 2π (1 − e z)(1 − e z ) 0 0 and hence for z sufficiently close to z : 1 −it g(z) − g(z ) 1 2e − u(t) dt (6) −it 2 z − z 2π (1 − e z ) 0 0 1 −2it 1 2e (z − z ) = u(t) dt (7) −it −it 2 2π (1 − e z)(1 − e z ) 1 2 z→z ≤ sup |u(t)|2 |z − z | −− −→ 0. (8) 2π δ t∈[0,1] 0 Therefore g is holomorphic and the proof is complete. Lemma 1. Let u :[0, 1] → R be a continuous function and define A(z) as in iθ Proposition 1. For all z = re ∈ C\J : dA 1 −2r(1 − r )sin(θ − t) iθ (re )= u(t) dt. (9) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) Thus, for θ ∈ (1, 2π) and r =1: dA iθ (e )=0. (10) dθ iθ 1 1 Proof. Since A(re )= u(t)P (θ − t)dt = (f ∗ P )and P is dif- r r r 2π 0 2π iθ ferentiable in respect to θ we have that A(re ) is differentiable in respect to θ, dA 1 d(f ∗ P ) 1 dP r r iθ iθ iθ (re )= (re )= f ∗ (re )=⇒ (11) dθ 2π dθ 2π dθ dA 1 −2r(1 − r )sin(θ − t) iθ (re )= u(t) dt. (12) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) (10) is derived from (9) substituting r =1 and θ ∈ (1, 2π). Because every continuous function u :[0, 1] → C can be considered a 2π-periodic function u : R → C such that v(x)= u(x)for x ∈ [0, 1] + 2πZ and v(x) = 0 otherwise, one can expect that such a function when convolved with the Poisson kernel would retain the nice properties. More specifically, it will uniformly converge to 0 and to u(x) on the compact subsets of the respective open arcs. We prove this in Propositions 2 and 3. Proposition 2. If A(z) is as before, then for all θ ∈ (1, 2π) and l ∈ N: d A iθ lim (re )=0. (13) r→1 dθ The convergence is uniform in the compact subsets of (1, 2π). p p Relations of the Spaces A (Ω)and C (∂Ω) Page 5 of 13 86 Proof. We start with l = 1. It suffices to prove that for all [θ ,θ ] ⊂ (1, 2π)the 1 2 convergence is uniform. Observe that 0 <θ − t<θ − t< 2π for all t ∈ [0, 1] 1 2 and therefore cos(θ − t) ≤ max{cos(θ − 1), cos(θ )} = M< 1, ∀t ∈ [0, 1] 1 2 2 2 and ∀θ ∈ [θ ,θ ]. This implies that 1 + r − 2r cos(θ − t) ≥ 1+ r − 2rM = 1 2 (1 − r) +2r(1 − M ) for all t ∈ [0, 1] and θ ∈ [θ ,θ ]. Thus, for all θ ∈ [θ ,θ ] 1 2 1 2 and 0 <r < 1: dA 1 −2r(1 − r )sin(θ − t) iθ (re ) = u(t) dt (14) 2 2 dθ 2π (1 + r − 2r cos(θ − t)) 1 2r(1 − r ) ≤ u (15) 2 2 2π ((1 − r) +2r(1 − M )) r→1 dA iθ and hence sup (re ) − −−− → 0 since the right hand side of (15)con- dθ θ∈[θ ,θ ] 1 2 verges to0as r → 1 . Note that no matter how many times we differentiate P in respect to θ we 2 k l m will have a finite sum of fractions with numerator c(1 − r ) cos(θ) sin(θ) for c =0,k,l,m ∈ N and denominator a power of (1 + r − 2r cos(θ − t)). So for any l ≥ 2 the same arguments apply. Denote by C ([0, 1]) the class of functions u :[0, 1] → C that are p times (l) continuously differentiable in (0, 1) and u continuously extend on [0, 1] for all 0 ≤ l ≤ p, l ∈ N. Proposition 3. Let p ∈{0, 1,...}∪{∞}, u :[0, 1] → R of class C ([0, 1]) and define A(z) as in Proposition 1. For all θ ∈ (0, 1) and 0 ≤ l ≤ p, l ∈ N, d A (l) lim (z)= u (θ ). (16) iθ 0 dθ z→e |z|<1 The convergence is uniform in the compact subsets of (0, 1). Proof. By a theorem of Borel (see [5]), we can find a function q : R → R of ∞ l l l l class C (R) such that q (0) = u (0) and q (1) = u (1) for 0 ≤ l ≤ p, l ∈ N. Thus, the function: u(t),t ∈ [0, 1] g(t)= q(t),t ∈ (1, 2π) is of class C (T). Define: 2π G(z)= g(t)P (t)dt. (17) 2π It is well known that uniformly for all θ ∈ R, 0 ≤ l ≤ p, l ∈ N: d G (l) lim (z)= g (θ) (18) iθ dθ z→e |z|<1 86 Page 6 of 13 N. Georgakopoulos et al. Results Math Let 1 2π 1 1 A(z)= u(t)P (t)dt and B(z)= q(t)P (t)dt. (19) z z 2π 2π 0 1 Note that G(z)= A(z)+ B(z) for all |z| < 1 and hence for 0 ≤ l ≤ p, l ∈ N: l l l d G d A d B (z)= (z)+ (z). (20) l l l dθ dθ dθ l l d G iθ (l) d B iθ From (18) Proposition 2 we have that (re ) → u (θ)and (re ) → 0 l l dθ dθ as r → 1 , uniformly in the compact subsets of (0, 1). As a result, l l l d A d G d B (l) iθ lim (z) = lim (z) − (z) = u (e ) (21) l l l iθ iθ 0 0 dθ dθ dθ z→e z→e |z|<1 |z|<1 while the convergence is uniform in the compact subsets of (0, 1). Remark 1. By linearity, Propositions 1, 2, 3 and Lemma 1 hold for complex functions f = u + iv where u = Ref and v = Imf , since we can apply them to the real and imaginary part separately. We now adapt the proof of Theorem 1 for functions whose derivatives only extend over an open arc of the unit circle. it Theorem 2. Let p ∈{1, 2,...}∪{∞}, f ∈ A(D) and g(t)= f (e ),t ∈ (0, 1). (l) The following are equivalent: f continuously extends on D ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if g is p times continuously differentiable in (0, 1).In that case, for all t ∈ (0, 1) dg it  it (t)= ie f (e ). (22) dt Proof. We prove it by induction on p.For p =1, let t ∈ (0, 1) and t <t <t 0 1 0 2 (l) such that [t ,t ] ⊂ (0, 1). Assuming that f continuously extends over D ∪ J 1 2 it it  it for all 0 ≤ l ≤ p, l ∈ N,let f (t)= f (re ), h(t)= ie f (e ) and: df it  it h (t)= (t)= ire f (re ) (23) dt for all t ∈ (0, 1) and 0 <r < 1. We have h → h as r → 1 uniformly in [t ,t ], r 1 2 it since f is continuous in D ∪ J . Note that f (t ) → f (e ) and hence from a r 0 well-known theorem f → hdt + c, for a some c ∈ C,as r → 1, while the convergence is uniform in [t ,t ]. Additionally, f → g,as r → 1, uniformly 1 2 r it  it in [t ,t ] and therefore g (t)= h(t)= ie f (e ),t ∈ (t ,t ). However, t was 1 2 1 2 0 arbitrary thus, g ∈ C ((0, 1)) and (22) holds. For the converse let g ∈ C ((0, 1)). We now use the Poisson representa- tion: 2π it f (z)= f (e )P (t)dt. (24) 2π 0 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 7 of 13 86 for |z| < 1. Define: t 2π+t 2 1 1 1 it A(z)= g(t)P (t)dt and B(z)= f (e )P (t)dt. (25) z z 2π 2π t t 1 2 Note that f (z)= A(z)+ B(z) for all |z| < 1. Consequently, df dA dB it it it (re )= (re )+ (re ) (26) dt dt dt for all 0 <r < 1and t ∈ R. Since [t ,t ] ⊂ (0, 1), Propositions 2 and 3 imply 1 2 that: dA dg it lim (re )= (t) (27) r→1 dt dt and dB it lim (re ) = 0 (28) r→1 dt uniformly for t ∈ [t ,t ]. Combining (26), (27) and (28) we get: 1 2 1 df 1 dg it it lim f (re ) = lim (re )= (t) (29) it it − − r→1 r→1 ire dt ie dt uniformly for t ∈ [t ,t ]. Since t was arbitrarily chosen in (0, 1) we deduce that 1 2 0 f extends continuously on D ∪ J . To complete the induction, let us assume (l) that the theorem holds for some p ≥ 1. If f continuously extends on D ∪ J (l) for all 0 ≤ l ≤ p +1,l ∈ N it follows that (f ) continuously extends on D ∪ J for all 0 ≤ l ≤ p, l ∈ N. By the induction hypothesis the map t → it p f (e ) belongs in the class C ((0, 1)) and since, by the case of p =1, g (t)= it  it  p p+1 ie f (e )wehave g ∈ C ((0, 1)) and hence g ∈ C ((0, 1)). For the converse, p+1  it  it p if g ∈ C ((0, 1)) it follows from (22) that g (t)= ie f (e ) ∈ C ((0, 1)) it p and therefore, the map t → f (e ) is of class C ((0, 1)). By the induction (l) hypothesis, (f ) continuously extends on D ∪ J , for all 0 ≤ l ≤ p, l ∈ N and (l) hence f continuously extends on D ∪ J , for all 0 ≤ l ≤ p +1,l ∈ N. The case of p = ∞ follows easily. Using Riemann’s mapping theorem [1] we can drop the assumption of continuity over D. Indeed, we can just have f continuously extend over the open arc we are dealing with, i.e. f ∈ A(D, J ). Theorem 3. Let p ∈{1, 2,...}∪{∞}, f : D ∪ J → C continuous on D ∪ J and it holomorphic in D and g(t)= f (e ),t ∈ (0, 1). The following are equivalent: (l) f continuously extends over D ∪ J , for all 0 ≤ l ≤ p, l ∈ N,ifand only if g is p times continuously differentiable in (0, 1). In that case: dg it  it (t)= ie f (e ) (30) dt for all t ∈ (0, 1). 86 Page 8 of 13 N. Georgakopoulos et al. Results Math Proof. We prove it by induction on p.For p = 1, the only if part is proven like Theorem 2; we also obtain (30). For the converse, let t ∈ (0, 1), t <t <t 0 1 0 2 it such that [t ,t ] ⊆ (0, 1) and J = {e : t <t<t }. Set V = {|z| < 1 2 1 2 1,z =0: ∈ J }. It is easily verified that V is simply connected and hence |z| there is a conformal map φ : D → V which extends to homeomorphism over the closures φ : D → V , by the Osgood–Carath´ eodory theorem [4]. Since it it it 1 2 φ(T)= {e : t ≤ t ≤ t }∪{re :0 ≤ r ≤ 1}∪{re :0 ≤ r ≤ 1} and 1 2 it −1 it {e : t ≤ t ≤ t } is connected we deduce that φ ({e : t ≤ t ≤ t }) is closed 1 2 1 2 −1 it and connected in T. Without loss of generality, assume that φ ({e : t ≤ it t ≤ t })= {e :0 ≤ t ≤ 1}. Consequently, f ◦ φ : D → C is continuous on D and holomorphic in D that is, of class A(D). Using the Reflection Principle, we it deduce that φ conformally extends on an open G ⊃ V ∪{e : t ≤ t ≤ t } thus, 1 2 it ∞ t → φ(e ),t ∈ (t ,t ) is of class C ((t ,t )) with non-vanishing derivative (see 1 2 1 2 1 it ∞ [1] pp. 233–235). Since g ∈ C ((t ,t )) and t → φ(e )isin C ((0, 1)) we have 1 2 it 1 that the map t → f (φ(e )) is in C ((0, 1)). By Theorem 2 the derivative (f ◦ φ) continuously extends on D ∪ J and hence f continuously extends in −1   −1 −1 D ∪ J , because f =(f ◦ φ) ◦ φ and f =(f ◦ φ) ◦ φ · (φ ) . Our choice of t was arbitrary and therefore f continuously extends on D ∪ J . To complete the induction, we follow the proof of Theorem 2. The case of p = ∞ follows easily. In other words, Theorem 3 states that for any open arc J of T and p p p ∈{0, 1, 2,...}∪{∞}, A (D, J)= A(D, J ) ∩ C (J ) holds. Remark 2. In all Propositions and Lemmas of this section, [0, 1] can be re- placed by any interval [a, b] with 0 ≤ a<b<a +2π. 3. Jordan Domains Motivated by Theorem 3 we now give a more general definition of the spaces p p A .Let Ω be a Jordan domain and J an open arc of ∂Ω. Denote by A (Ω, J ) (l) the collection of all functions f holomorphic on Ω such that f continuously p p extends on Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N. Note that A (Ω, ∂Ω)= A (Ω). If γ : I → J is a parametrization of J , denote by C (J ) the class of all functions f : G → C defined on a varying set G, J ⊂ G, such that f ◦ γ : I → C is p times continuously differentiable. Moreover, if φ : D → Ω is a Riemann map, p it denote by C (J ) the class C (J ) where γ(t)= φ(e ) . Since any two Riemann maps differ by an automorphism of the unit disk it is easily verified that the spaces C (J ) do not depend on the chosen Riemann map. Next, we consider differentiability on J with respect to the position [2]. Definition 1. Let J be a Jordan arc and f : J → C. We define the derivative of f on z ∈ J by: df f (z) − f (z ) (z ) = lim (31) dz z→z ,z∈J z − z 0 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 9 of 13 86 if this limit exists and is a complex number. df In order to go one step further , we consider the derivative on J of dz Definition 1 and we take its derivative on J with respect to the position. df Definition 2. A function f : J → C belongs to the class C (J)if (z)exists dz p−1 d f and is continuous for z ∈ J . Inductively, suppose that is well defined on p−1 dz J for some p =2,..., +∞, we say that f is of class C (J)if p−1 d f d f d( ) p−1 dz (z)= (z) (32) dz dz exists and is continuous on J . Remark 3. In [2] the following fact is proven. If γ : I → J is a C regular (with non vanishing derivative) parametrization of a Jordan arc J, n ∈{1, 2,... }∪ {∞}, a function f is of class C (J ) if and only if g(t)=(f ◦ γ)(t),t ∈ I,isof class C (I),p ∈{1, 2,...}∪{∞},p ≤ n. Additionally, dg df (t)= (γ(t)) · γ (t). (33) dt dz In this section we prove that given a Jordan domain Ω, φ : D → Ω a −1 1 −1 Riemann map such that φ ∈ A (Ω, J ) and (φ ) (z) =0,z ∈ Ω ∪ J , f ∈ (l) A(Ω, J)and p ∈{1, 2,... }∪{∞} the following are equivalent: f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if f | is of class C (J ). p p That is A (Ω, J)= A(Ω, J ) ∩ C (J ). Let us first prove a straightforward fact concerning the parametrization of an arc J of ∂Ω induced by a Riemann map φ. Theorem 4. Let Ω be a Jordan domain, φ : D → Ω a Riemann map and J an −1 n open arc of ∂Ω such that φ is of class A (Ω, J ) for some n ∈{1, 2,... }∪ −1 {∞} and (φ ) (z) =0,z ∈ Ω ∪ J . The following holds for any p ∈{1, 2,... }∪ p p {∞},p ≤ n: A (Ω, J)= A(Ω, J ) ∩ C (J ). In that case, if f ∈ A (Ω, J ) and it g(t)=(f ◦ φ)(e ): dg it  it  it (t)= ie f (φ(e ))φ (e ) (34) dt is for all t ∈{s ∈ R : φ(e ) ∈ J }. On the right hand side of (34) f denotes the continuous extension of f (z)from Ω to Ω ∪ J . Before proceeding to the proof, let us note that the choice of the Riemann map is irrelevant. Suppose that Φ, Ψ are two Riemann maps and J an open arc −1 n −1 of ∂Ω such that Φ is of class A (Ω, J ),n ∈{1, 2,...}∪{∞}, and (Φ ) (z) = −1 n −1 0 for all z ∈ Ω ∪J . Then, Ψ is also of class A (Ω, J ) and (Ψ ) is non zero in −1 Ω ∪J . To see this, observe that Φ ◦Ψ is an automorphism of the unit disk and z−a hence there are a ∈ D and c ∈ T such that Ψ = Φ◦φ , where φ (z)= c ,z ∈ a a 1−az D. Note that φ is holomorphic in D(0, ) ⊃ D with non vanishing derivative a 86 Page 10 of 13 N. Georgakopoulos et al. Results Math (l) thus, Ψ can be extended over D for all 0 ≤ l ≤ n, l ∈ N. Additionally, 1−|a| −1  −1  −1 (Ψ ) (z)= · (Φ ) (z) = 0 for all z ∈ Ω ∪ J.Moreover, Φ being −1 2 (1+aΦ (z)) n n −1 of class A (Ω, J ) is equivalent to Φ being of class A (D, I), I = φ (J ). This is −1 easy to see given that (Φ ) (z) =0 for z ∈ Ω ∪J . In other words, if a Riemann map induces a regular n times continuously differentiable parametrization of J then, the same holds for any other Riemann map. −1 is Proof. (Theorem 4) Set I = φ (J)and I = {s ∈ R : φ(e ) ∈ J }. Given an f ∈ A(Ω, J ) one can easily verify that f is of class A (Ω, J ) if and only if f ◦ φ is of class A (D, I), since φ is non zero in D ∪ I and p ≤ n. By Theorem 3, p it f ◦ φ is of class A (D, I) if and only if the map t → (f ◦ φ)(e ),t ∈ I, is p times continuously differentiable which by definition is equivalent to f being it of class C (J ). To see that (34) holds, set g (t)= f (φ(re )),t ∈ R, 0 < r< 1. Note that g → g as r → 1 uniformly in the compact subsets of dg r it  it  it I. Additionally, (t)= ire f (φ(re ))φ (e ) which uniformly converges to dt it  it  it − ie f (φ(e ))φ (e )as r → 1 in the compact subsets of I . From a well known theorem of calculus (34) follows. In the next theorem we prove that for any function f ∈ A(Ω, J ) the first p derivatives of f continuously extend over Ω ∪ J if and only if f | is of class C (J ), given that J is a smooth enough open arc of the Jordan domain Ω. −1 1 −1 That is, if φ is a Riemann map and φ is of class A (Ω, J ) with (φ ) (z) = p p 0,z ∈ Ω ∪ J then A (Ω, J)= A(Ω, J ) ∩ C (J ) for all p ∈{1, 2,...}∪{∞}. Theorem 5. Let Ω be a Jordan domain, φ : D → Ω a Riemann map and J an −1 1 −1 open arc of ∂Ω such that φ is of class A (Ω, J ) with (φ ) (z) =0,z ∈ Ω∪J . (l) Given an f ∈ A(Ω, J ) and p ∈{1, 2,...}∪{∞} we have that: f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N if and only if f | is of class C (J ). p p That is, A (Ω, J)= A(Ω, J ) ∩ C (J ). In that case: d f | (l) J f (z)= (z) (35) dz for all z ∈ J and 0 ≤ l ≤ p, l ∈ N. (l) (l) On the left hand side of (35), f (z) denotes the continuous extension of f on Ω ∪ J , while on the right hand side the differentiation is with respect to the position. is Proof. Let p =1 and f ∈ A(Ω, J ). Denote by I = {s ∈ R : φ(e ) ∈ J }. By Theorem 4 we have that f ∈ A (Ω, J ) if and only if the map g(t)= dg it it  it f (φ(e )),t ∈ I is continuously differentiable. Additionally, (t)= ie φ (e ) dt it f (φ(e )) and: dg df | dφ| df | it it it it it  it it J T J (e )= (φ(e )) · (e ) · ie = ie φ (e ) (φ(e )) (36) dt dz dz dz p p Relations of the Spaces A (Ω)and C (∂Ω) Page 11 of 13 86 it  it Since ie φ (e ) is non zero we have that g is continuously differentiable if and 1 1 only if f | is of class C (J ). Consequently, f is of class A (Ω, J ) if and only if g 1 1 is of class C (I) if and only if f | is of class C (J ). Equation (35) follows from Eqs. (34) and (36). For the induction step, assume that p> 1. We have that p  p−1 f ∈ A (Ω, J ) if and only if f ∈ A (Ω, J ) which by the induction hypothesis p−1 p is equivalent to f | ∈ C (J ) and that is equivalent to f | ∈ C (J ), by (35) J J for l =1. Moreover, (l−1) (l−1) d f | d df | d J J   (l−1) (l) (z)= (z) = (f (z)) = (f ) (z)= f (z) l (l−1) (l−1) dz dz dz dz (37) for all z ∈ J and 1 ≤ l ≤ p, l ∈ N. Remark 4. Note that if Ω is a Jordan domain, J an analytic arc of ∂Ω and −1 φ : D → Ω a Riemann map we know from [1] (p. 235) that φ has a conformal extension over an open set G ⊃ Ω ∪ J . Consequently, if J is an analytic arc p p we immediately have A (Ω, J)= A(Ω, J ) ∩ C (J ) for all p ∈{1, 2,...}∪{∞}. In addition, if we have γ a C regular parametrization of J, n ∈{1, 2,...}∪ {∞} we have a triple equivalence as derived from Remark 3 and Theorem 5. Theorem 6. Let Ω be a Jordan domain, φ a Riemann map and J an open −1 1 −1 arc of ∂Ω such that φ is of class A (Ω, J ) and (φ ) (z) =0,z ∈ Ω ∪ J , γ a C regular parametrization of J, n ∈{1, 2,...}∪{∞},f ∈ A(Ω, J ) and p{1, 2,...}∪{∞},p ≤ n. The following are equivalent: (l) 1. f continuously extend over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N. That is, f is of class A (Ω, J ). 2. f | is of class C (J ). 3. g(t)=(f ◦ γ)(t),t ∈ I is p times continuously differentiable. That is f is of class C (J ). p p p In other words, A (Ω, J)= A(Ω, J ) ∩ C (J)= A(Ω, J ) ∩ C (Ω, J ). In that case: dg df | (t)= (γ(t))γ (t)= f (γ(t))γ (t) (38) dt dz for all t ∈ I. Remark 5. We notice that the assumption on Theorem 5 that the Riemann −1 −1 map φ : D → Ω is such that φ ∈ A(Ω, J ) with (φ ) (z) =0 for z ∈ Ω ∪ J is in fact equivalent to say that the parametrization of J induced by φ is C regular. We topologize A (Ω) by the semi-norms (l) (l) |f | =sup |f (z)| =sup |f (z)|, 0 ≤ l ≤ p, l ∈ N (39) z∈∂Ω z∈Ω 86 Page 12 of 13 N. Georgakopoulos et al. Results Math and C (∂Ω) by the semi-norms d g |g| =sup | (z)|, 0 ≤ l ≤ p, l ∈ N (40) dz z∈∂Ω see also [2]. By Eq. (35) it easily seen that the topology on A (Ω)and A(Ω) ∩ p p C (∂Ω) induced by C (∂Ω) coincide. Summarizing: Corollary 1. Let Ω be a Jordan domain and φ : D → Ω a Riemann map of class A (D) such that φ (z) =0 for all z ∈ D. The following equivalence holds for all f ∈ A(Ω) and p ∈{1, 2,...}∪{∞}: f ∈ A (Ω) if and only if f | is ∂Ω p times continuously differentiable with respect to the position. In this case we have: d f (l) (z)= f (z) (41) dz (l) For all z ∈ ∂Ω and 0 ≤ l ≤ p, l ∈ N. On the right hand side of (41) f (z) (l) denotes the continuous extension of f from Ω to Ω ∪ J . We also have, p p A (Ω)= A(Ω) ∩ C (∂Ω) as topological spaces. Remark 6. There is a natural way to topologize the spaces A (Ω, J ) for any Jordan domain Ω,openarc J ⊂ ∂Ω and p ∈{1, 2,... }∪{∞}. Then, the p p equality A (Ω, J)= A(Ω, J ) ∩ C (J ) holds taking into account the topologies of the spaces as well. However, we will not deal with the topological properties of the spaces A (Ω, J ) in this paper. One can easily extend the results of this section for the complement of a Jordan domain Ω given that the functions considered vanish at infinity. That is, if A (C\Ω) is the class of all functions f with lim f (z) = 0 and holo- z→∞ morphic in C\Ω such that their first p derivatives extend over C\Ω for all ˆ ˆ 0 ≤ l ≤ p, l ∈ N, then A (C\Ω)= A (C\Ω) ∩ C (∂Ω) under similar assump- tions to that of Theorem 5.For J an open arc of ∂Ω we define A (C\Ω, J ) similarly. Remark 7. The previous results can be generalized to the case of finitely con- nected domains bounded by a finite set of disjoint Jordan curves. For this we use the Laurent Decomposition [3]. We will take a moment to sketch the proof. Let Ω be a bounded domain whose boundary consists of a finite number of disjoint Jordan curves. If V ,V ,..., 0 1 ˆ ˆ V are the connected components of C\Ω, ∞∈ V and Ω = C\V , ..., n−1 0 0 0 Ω = C\V , then for every f which is holomorphic in Ω we know that n−1 n−1 there exist functions f ,...,f which are holomorphic in Ω ,...,Ω re- 0 n−1 0 n−1 spectively such that f = f + f + ··· + f and lim f (z)=0,j ∈ 0 1 n−1 z→∞ j {1,...,n − 1} [3]. Let φ : D → Ω ,j ∈{0, 1,... ,n − 1}, be the respective con- j j formal maps. Without loss of generality, take J an open arc of ∂Ω such that −1 φ is of class A (Ω ,J ) with non vanishing derivative. Let also f be a holo- morphic function in Ω and continuous in Ω∪J such that f = f +f +···+f , 0 1 n−1 p p Relations of the Spaces A (Ω)and C (∂Ω) Page 13 of 13 86 f holomorphic in Ω , is the extended Laurent decomposition of f . Since f is j j j holomorphic in a neighborhood of ∂Ω for all j = 0, we have that f continu- 0 0 ously extends over Ω ∪ J since f continuously extends over Ω ∪ J . Similarly, (l) f continuously extends over Ω ∪ J for all 0 ≤ l ≤ p, l ∈ N, if and only if f p p is of class A (Ω ,J ). By Theorem 5, f is of class A (Ω ,J ) if and only if f | 0 0 0 p p is of class C (J ), which happens if and only if f | is of class C (J ). Open Access. This article is distributed under the terms of the Creative Com- mons Attribution 4.0 International License (http://creativecommons.org/licenses/ by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. References [1] Ahlfors, L.V.: Complex Analysis, 3rd edn. Mc Graw-Hill Book Co, New York (1978) [2] Bolkas, E., Nestoridis, V., Panagiotis, C., Papadimitrakis, M.: One sided extend- ability and p-continuous analytic capacities, submitted. arXiv:1606.05443 [3] Costakis, G., Nestoridis, V., Papadoperakis, I.: Universal Laurent series. Proc. Edinb. Math.Soc.(2) 48(3), 571–583 (2005) [4] Koosis, P.: An Introduction of H Spaces. Cambridge University Press, Cam- bridge (1998) [5] Krantz, S., Parks, H.: A Primer of Real Analytic Functions, 2nd edn. Springer Birkh¨ auser, Boston (2002) p p [6] Mastrantonis, V.: Relations of the spaces A (Ω)and C (∂Ω). arXiv:1611.02791 N. Georgakopoulos, V. Mastrantonis and V. Nestoridis Department of Mathematics National and Kapodistrian University of Athens Panepistimioupolis 157 84 Athens Greece e-mail: nikolaosgmath@gmail.com V. Mastrantonis e-mail: blasismas@hotmail.com V. Nestoridis e-mail: vnestor@math.uoa.gr Received: August 24, 2017. Accepted: March 13, 2018.

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Results in MathematicsSpringer Journals

Published: May 28, 2018

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