Meromorphic extensions from small families of circles and holomorphic extensions from spheres

Let B be the open unit ball in C^2 and let a, b, c be three points in C^2 which do not lie in a complex line, such that the complex line through a and b meets B and such thatis different from 1 if one of the points a, b is in B and the other in the complement of B and such that at least one of the numbers,is different from 1. We prove that if a continuous function f on the sphere bB extends holomorphically into B along each complex line which passes through one of the points a, b, c then f extends holomorphically through B. This generalizes recent work of L.Baracco who proved such a result in the case when the points a, b, c are contained in B. The proof is different from the one of Baracco and uses the following one variable result which we also prove in the paper and which in the real analytic case follows from the work of M.Agranovsky: Let D be the open unit disc in C. Given a in D let C(a) be the family of all circles in D obtained as the images of circles centered at the origin under an automorphism of D that maps the origin to a. Given distinct points a, b in D and a positive integer n, a continuous function f on the closed unit disc extends meromorphically from every circle T in either C(a) or C(b) through the disc bounded by T with the only pole at the center of T of degree not exceeding n if and only if f is of the form f(z) = g_0(z)+g_1(z)\bar z +...+ g_n(z)\bar z^n where the functions g_0, g_1, ..., g_n are holomorphic on D.


The main results
Denote by ∆ the open unit disc in C . If a ∈ C and r > 0 write ∆(a, r) = {ζ ∈ C: |ζ − a| < r}. Given α ∈ ∆ the Moebius map maps the circles in ∆ centered at the origin to the circles α − Rζ 1 − αRζ : ζ ∈ b∆ , 0 < R < 1 which are called the circles with the hyperbolic center α and we denote this family of circles by C α . In particular, C 0 is the family of all circles in ∆ centered at the origin. If Γ is a circle we denote by c(Γ) its center, moreover, if Γ ⊂ ∆ we denote by h(Γ) its hyperbolic center, that is, the unique point such that Γ ∈ C h(Γ) . In the case when α ∈ b∆ then we denote by C α the family of all circles in ∆ which pass through α.
We say that a continuous function f on a circle Γ extends holomorphically (meromorphically) from Γ if it extends holomorphically (meromorphically) through the disc bounded by Γ.
If a function f on ∆ has the form f (z) = a 0 (z) + a 1 (z)z + · · · + a n (z)z n (z ∈ ∆) (1.1) where a i , 0 ≤ i ≤ n, are holomorphic functions on ∆ then we say that f is a polyanalytic function on ∆ (of order n ≤ 1 if a n ≡ 0). Polyanalytic functions of order zero are holomorphic functions. If f is a polyanalytic function of order ≤ n on ∆ then for each circle Γ ⊂ ∆, the function z → (z − c(Γ)) n f (z) extends holomorphically from Γ; in other words, f has a meromorphic extension from Γ (to the disc bounded by Γ) with the only pole at the center of Γ, which is of degree ≤ n. Indeed, if z ∈ b∆(a, r) then z = a + r 2 /(z − a) so (z − a) n f (z) = (z − a) n a 0 (z) + a 1 (z) a + r 2 z − a + · · · + a n (z) a + r 2 z − a n provides the necessary holomorphic extension through ∆(a, r).
We begin with a one-variable result.
In [A2] M. Agranovsky obtained, in a real analytic case, a characterization of polyanalytic functions in terms of meromorphic extendibility from various families of circles. In the special case when f is real-analytic Theorem 1.1 follows from his work. In the case when n = 0, Theorem 1.1 follows from the result of A. Tumanov when α, β ∈ b∆ [T2] and reduces to the results of the author when α ∈ ∆, β ∈ b∆ [G2] and when α, β ∈ ∆ [G3].
Given a ∈ C 2 we denote by L(a) the family of all complex lines passing through a Given b ∈ C 2 , b = a, we denote by Λ(a, b) the complex line passing through a and b. We denote by B the open unit ball in C 2 . Using Theorem 1.1 we prove THEOREM 1.2 Let a, b be two points in C 2 such that Λ(a, b) meets B. Suppose that if one of the points a, b is in B and the other in C 2 \ B then < a|b > = 1. Assume that a continuous function f on bB extends holomorphically into B along every complex line L ∈ L(a) ∪ L(b). Then for any c ∈ Λ(a, b) ∩ B, the function f extends holomorphically into B along any complex line L ∈ L(c).
L. Baracco [B2] proved recently a conjecture of M. Agranovsky [A1]: If a, b, c ∈ B do not lie on a complex line then L(a) ∪ L(b) ∪ L(c) is a test family for holomorphic extendibility for C(bB), that is, if f ∈ C(bB) extends holomorphically into B along each complex line L ∈ L(a) ∪ L(b) ∪ L(c) where the points a, b, c are in B and do not lie on a complex line then f extends holomorphically through B. Another attempt to prove the conjecture was presented in [A3] by M.Agranovsky who later found that the proof in [A3] is incomplete. Our proof of Theorem 1.2 provides a new, different proof of the result of Baracco. Indeed, Theorem 1.2 implies that f extends into B along every complex line that meets B and so f extends holomorphically through B [AV, S]. Since in our Theorem 1.2 the points a, b do not have to lie in B we get a more general result: COROLLARY 1.3 Let a, b, c be points in C 2 which do not lie on a complex line, such that Λ(a, b) meets B and such that if one of the points a, b lies in B, the other in C 2 \ B then < a|b > = 1 and such that at least one of the numbers < a|c >, < b|c > is different from 1. Then L(a) ∪ L(b) ∪ L(c) is a test family for holomorphic extendibility for C(bB).
Remark For functions f in C ∞ (bB), L(a) ∪ L(b) above is a test family for holomorphic extendibility, that is, the complex lines through two points suffice [G3]. This is no more true for functions in C k (bB) [G3]. Corollary 1.3 implies that if N + 1 points in the open unit ball of C N do not lie in a (N − 1)dimensional complex plane then the complex lines passing through these points form a test family for holomorphic extendibility for continuous functions on the unit sphere in C N . It is known that the complex lines through two points in the unit ball suffice for functions of class C ∞ for any dimension N [G3].

Poles at the hyperbolic centers
We begin by a simple. but important observation of M. Agranovsky [A2]: PROPOSITION 2.1 Let Γ ⊂ ∆ be a circle bounding the open disc D which is not centered at 0. The rational extension of z → ϕ(z) = 1 − |z| 2 from Γ has one zero in D, a single zero at h(Γ), and one pole in D, a simple pole at c(Γ).
Proof. If Γ = b∆(a, r) then z ∈ Γ implies that z = a + r 2 /(z − a) so is a rational function which has one pole a in D which proves the second statement. To prove the first statement, write where α ∈ ∆, α = 0, and 0 < R < 1. If ζ ∈ b∆ then which has one zero at ζ = 0 and one zero at ζ = ∞ which implies that the rational extension of ϕ from Γ has one zero at z = α and the other zero at z = 1/α which proves the first statement, Remark If Γ ⊂ ∆ meets b∆, that is, if Γ ∈ C α with α ∈ b∆ then the rational extension of ϕ from Γ has one pole in D, a single pole at c(Γ), no zero in D and double zero at α. To see this, assume with no loss of generality that α = 1 and let Γ = {(1 − r) + rζ: ζ ∈ b∆} where 0 < r < 1. Then, for ζ ∈ b∆, which has a double zero at ζ = 1. This, in particular, implies that if Γ ∈ C α the function ζ → (z − α) 2 /(1 − |z| 2 ), defined an Γ \ {α} extends continuously to Γ.
Given a continuous function f on ∆ and n ∈ IN, set Given γ ∈ ∆ consider the following two conditions (H) If α ∈ ∆ and Γ ∈ C α then by the preceding discussion (z − c(Γ)) n f (z) extends holomorphically from Γ if and only if (z − α) n F (z) extends holomorphically from Γ. Moreover, if α ∈ b∆ and Γ ∈ C α then (z − c(Γ)) n f (z) extends holomorphically from Γ if and only if (z − α) 2n F (z) extends holomorphically from Γ. This gives LEMMA 2.2 Let f be a continuous function on ∆ and let n ∈ IN. Let Then for each γ ∈ ∆, (H) and (C) are equivalent.
In the same way we show that if α, β ∈ b∆ and if (C) holds for γ = α, γ = β, then, using Theorem 1.1 in the case (iii) above we see that (C) holds for every γ ∈ ∆.
All this shows that, after a rotation if necessary, it is enough to prove Theorem 1.1 in the cases when α = t, 0 < t < 1, and β = 0, (2.2) Suppose that we have done this. Then we have also proved THEOREM 2.3 Let F be a continuous function on ∆ such that z → (1 − |z| 2 ) n F (z) is continuous on ∆. Let α, β ∈ ∆, α = β and assume that (H) holds for γ = α and γ = β. Then where f is polyanalytic of degree ≤ n. In particular, (H) holds for every γ ∈ ∆.
In this section we deduce Theorem 1.2 from Theorem 2.3. Along the way we describe, for a given complex line Λ in C 2 that meets B , those continuous functions on bB that extend holomorphically into B along each complex line L ∈ L(c), c ∈ Λ ∩ B. We already know that each such function which is of class C ∞ necessarily extends holomorphically through B.
So let a, b be two points in C 2 such that Λ(a, b) meets B, and suppose that if one of the points is in B, the other in C 2 \ B then < a|b > = 1. Assume that f ∈ C(bB) extends holomorphically into B along every complex line L ∈ L(a) ∪ L(b).
As in [G3] We use Moebius transforms to show that it is enough to prove the statement of Theorem 1.2 in the special case when Λ(a, b) is the z-axis and in one of the following two cases: As in [G3] or [A1] we now use the Fourier series decomposition and averaging to reduce the problem in C 2 to a series of one variable problems. For each z ∈ ∆, write the Fourier series Clearly the coefficients c n (z) = 1 are continous on ∆ and if n < 0 they extend continuously to ∆ with zero boundary values. If z 0 ∈ C and if f extends holomorphically into B along each complex line passing through (z 0 , 0) then in the sum (3.2) the same holds for each term: a continuous function on bB. Converse also holds: If each term w n c n (z) in (3.2) extends holomorphically into B along each complex line through (z 0 , 0) then the same holds for f . This is so since f is uniformly continuous on bB and so the family of functions e iθ → f (z, e iθ 1 − |z| 2 ), z ∈ ∆ is uniformly equicontinuous on b∆. The proof of Fejer's theorem [H] shows in such a case f (z, e iθ 1 − |z| 2 ) is the limit of Cezaro means of the Fourier series (3.1) which is uniform with respect to z ∈ ∆. Notice that w n c n (z) extends holomorphically into B along each Λ ∈ L((t, 0)) if and only if (z − t) n c n (z) extends holomorphically from each Γ ∈ C t in the case when 0 ≤ t ≤ 1 and from each Γ ∈ C 1/t in the case when 1 < t < ∞ [G3].
We will now apply Theorem 2.3 to each c n . If n ≤ 0 then c n is continuous on ∆ and vanishes identically on b∆ if n < 0. If n ≥ 0 then we know that (1 − |z| 2 ) n/2 c n (z) extends continuously through ∆ and so the same holds for (1 − |z| 2 ) n c n (z).
Assume now that f extends holomorphically into B along each L ∈ L(a) ∪ L(b) where a, b are as in (i). If n < 0 this implies that c n extends holomorphically from each circle belonging to C τ 1 ∪ C τ 2 where 0 ≤ τ 1 < τ 2 ≤ 1 and since c n extends continuously to ∆ it follows by the main results of [G2, G3] that c n is holomorphic on ∆. Since it is continuous on ∆ and vanishes identically on b∆ it follows that it vanishes identically on ∆. Now, let n > 0. Now again, we have that (z − τ j ) n c n (z) extends holomorphically from each Γ ∈ C τ j , j = 1, 2, which, by Theorem 2.3 implies that where the function g n is polyanalytic of order ≤ n on ∆ and consequently, for every γ ∈ ∆, (z − γ) n c n (z) extends holomorphically from each Γ ∈ C γ which implies that for each n > 0 and for each γ ∈ ∆ the function w n c n (z) extends into B holomorphically along each L ∈ L((γ, 0)) and hence, by the preceding discussion, the same holds for f . Similar reasoning applies in the case (ii). This completes the proof of Theorem 1.2.
One should mention that the idea of multiplying c n with (1 − |z| 2 ) n to achieve the regularity at the boundary and thus shifting the poles from the hyperbolic centers to the centers is due to M. Agranovsky [A1, A2].
Remark On bB we have |w| 2 = 1 − |z| 2 so it follows that, for n > 0, It is easy to check directly that for each k, 0 ≤ k ≤ n, and for each complex line L ∈ L((γ, 0)) where γ ∈ ∆, different from the z−axis, the function extends holomorphically from each circle Γ ∈ C γ . In fact, the holomorphic extension of z from Γ has the pole of order 1 at c(Γ) and the same holds for 1 − |z| 2 . On the other hand, the rational extension of 1 − |z| 2 from Γ has the only zero at γ.
By the preceding reasoning we proved be its Fourier series. The following are equivalent: whwre c n0 , c n1 , · · · c nn are holomorphic functions on ∆.
Obviously, by Theorem 1.2, for either (i) or (ii) to hold it is enough that it holds for complex lines through two points of ∆ × {0}.
Remark When we want to construct examples we must take into account that the functions (z, w) → w n c n (z) must be continuous on bB. For instance, the following standard example is of this sort: Remark We present another example, due to M. Agranovsky. Given α, β ∈ ∆, α = β, the function is continuous on bB and extends holomorphically into B along any complex line meeting ∆ × {0} as well as along any complex line in L((t 1 , 0)) ∪ L((t 2 , 0)), yet does not extend holomorphically through B. Obviously, finitely many points on the z axis outside B are also possible.
4. Towards the beginning of the proof of Theorem 1.1 We have already seen that it is enough to prove Theorem 1.1 in the cases (2.2), (2.3) and (2.4). We first look at (2.2). In this case we have to prove THEOREM 4.1 Suppose that 0 < t < 1 and let n ∈ IN ∪ {0}. Suppose that f is a continuous function on ∆ such that for each circle Γ ∈ C 0 ∪ C t , the function f |Γ extends meromorphically through the disc bounded by Γ with the only pole of order ≤ n at the center of Γ. Then f is of the form The proof of Theorem 4.1 will be our major task. We will provide a detailed proof. At the end we will indicate how to modify the proof to treat (2.3) and (2.4) As in [G2, G3] we shall use semiquadrics introduced in [AG] and [G1] to transform the problem into a problem about holomorphic extensions of CR functions on surfaces consisting of these semiquadrics. The principal idea to apply the reasoning of H. Lewy and H. Rossi in this context is due to A. Tumanov and is described in [T2], and, for holomorphic extensions applied in [G2, G3]. There, the final step was to apply the Liouville theorem to conclude that the main function is constant in the second variable. The proof is considerably more complicated in the case of meromorphic extensions, where we apply the generalized Liouville theorem to conclude that the function is a polynomial in the second variable. In [G2, G3] we were dealing with holomorphic extensions to semiquadrics where the maximum principle assured continuous dependence of extensions on the parameter. Now we will be dealing with meromorphic extensions where we do not have the maximum principle so we will need the following preliminary LEMMA 4.1 Let I ⊂ IR be an interval. Let n ∈ IN and let (ζ, t) → Φ(ζ, t) be a continuous function on b∆×I such that for each t ∈ I the function ζ → Φ(ζ, t) extends holomorphically through ∆ \ {0} and has a pole at 0 of degree ≤ n. Denote the extension byΦ. Theñ where the functions d j are continuous on I, Θ is continuous on ∆ × I, and for each t ∈ I, ζ → Θ(ζ, t) is holomorphic on ∆ .
Proof. The function (ζ, t) → Ψ(ζ, t) = ζ n Φ(ζ, t) is continuous on b∆ × I and for each t, ζ → Ψ(ζ, t) extends holomorphically through ∆. Denote this extension byΨ. We havẽ is continuous on I and where for each t ∈ I, the function ζ → Θ(ζ, t) is holomorphic on ∆. The function is continuous on b∆ × I and so by the maximum principle it follows that it is continuous on ∆ × I. This completes the proof.

The geometry of semiquadrics, Part 1
Given a ∈ C and r > 0 let be the semiquadric associated with the circle b∆(a, r).
The crucial property of these semiquadrics which will connect our problem with a problem in CR geometry is the following: A continuous function g extends meromorphically from a circle b∆(a, r) with the pole of degree ≤ n at a if and only if the function G, defined on bΛ(a, r) by G(ζ, ζ) = g(ζ) (ζ ∈ b∆(a, r)) extends holomorphically through Λ(a, r) and has a pole of degree ≤ n at the point at infinity. In fact, if we denote by the letterg the holomorphic extension of g through ∆(a, r) \ {a} the functioñ provides the necessary holomorphic extension of G through Λ(a, r). In our case, we have two families of circles; where ρ(T ) = (T − t)(T − 1/t) [G3] and consequently we will deal with two families of semiquadrics: the first family {Λ(0, R), 0 < R ≤ 1} and the second family {Λ(T, (T − t)(T − 1/t)): 0 ≤ T < t}. Our function F (ζ, ζ) = f (ζ) (ζ ∈ ∆) extends holomorphicall to each semiquadric of each family and these extensions will then define a CR function on a hypersurface in C 2 that we study and describe in detail.
In the first family the semiquadrics are pairwise disjoint. Let L be the union of their closures in C × C; In [C \ {0}] × C , the set L is a smooth hypersurface with piecewise smooth boundary consisting of Λ(0, 1) and {(ζ, ζ): ζ ∈ ∆ \ {0}.
Assume now that ℑz = 0. Denote by λ z the arc on the circle passing through t, 1/t and z (and consequently passing through 1/z) which joins z and 1/z and does not contain t and 1/t. Let T (z) be such that z ∈ b∆(T (z), ρ(T (z))). For the semiquadric from the second family the intersection (z, w(T )) ∈ Λ(T, ρ(T )) exists if and only if 0 ≤ T < T (z). As T increases from 0 to T (z), w(T ) moves along λ z from 1/z to z [G3]. Now, let z = η, 0 < η < t. Let v(R) be such that (η, v(R)) ∈ Λ(0, R) and let w(T ) be such that (η, w(T )) ∈ Λ(T, ρ(T )). It is easy too see that as R increases from R(η) = η to 1, v(R) ∈ IR increases from η to 1/η. As T increases from 0 to η, w(T ) ∈ IR increases from η to 1/η. As T increases from 0 to η, w(T ) increases from 1/η to ∞. As T increases from η to T (η) then w(T ) ∈ IR increases from −∞ to η.
Situation is entirely different if z = η where either −1 < η < 0 or t < η < 1. In this case we have "folding": when R increases from R(η) to 1, v(R) moves from η to 1/η and when T increases from 0 to T (η) then w(T ) moves from 1/η to η.
For each z ∈ ∆, ℑz = 0, J z ∪ λ z is a simple closed curve bounding a domain that we denote by D z . If ℑz > 0 then D z is contained in the lower halfplane and if ℑz < 0 then D z is contained in the upper halfplane. When z ∈ ∆ approaches a point a ∈ b∆, ℑa = 0, then the domains D z shrink to the point {a}. When z, ℑz > 0, approaches a point η, 0 < η < t, then D z become larger and larger and in the limit they become the lower halfplane. When z, Imz < 0, approaches a point η, 0 < η < t, then D z become larger and larger and in the limit they become the upper halfplane. When z, ℑz > 0, approaches a point η ∈ (−1, 0) ∪ (t, 1) then the domains D z become thinner and thinner and in the limit they shrink to the segment with endpoints η and 1/η. The same takes place for ℑz < 0.

Our CR function on L ∪ N
Let n ∈ IN ∪ {0}, 0 < t < 1 and suppose that f is a continuous function on ∆ that satisfies the assumptions of Theorem 4.1.

Define
F is a continuous function on {(ζ, ζ): ζ ∈ ∆}. The assumptions together with the continuity of F and Lemma 4.1 imply that the function F extends from L∩Σ = {(ζ, ζ): ζ ∈ ∆\{0}} continuously to L such that the extension Φ is holomorphic on each fiber Λ(0, R) of L with a pole of degree ≤ n at infinity. Thus, Φ is a CR function. Similarly, the function F extends from N ∩Σ = {(ζ, ζ): ζ ∈ ∆\{t}} continuously to N so that the extension Ψ is holomorphic on each fiber Λ(T, ρ(T )) \ {(t, 1/t)}, 0 ≤ T < t, with a pole of degree ≤ n at infinity. Thus, Ψ is a CR function. The functions Φ and Ψ coincide on {(ζ, ζ): ζ ∈ ∆ \ {0, t}} ∪ Λ(0, 1), the common boundary of L and N . Since in π −1 ∆ \ [(−1, 0] ∪ [t, 1)] there are no other common points of L and N we can define the function F on M where which extends the original F and is equal to Φ on L and to Ψ on N . The function F so obtained is continuous on M and holomorphic on each fiber of M so it is a CR function on M .
Our final goal will be to show that on M we have F (z, w) = a 0 (z) + a 1 (z)w + · · · + a n (z)w n where a j , 0 ≤ j ≤ n, are holomorphic functions on ∆ which will then imply that f (z) = F (z, z) = a 0 (z) + a 1 (z)z + · · · + a n (z)z n on ∆, which is the statement of Theorem 4.1.
In exactly the same way as in [G2, G3], following an idea of A.Tumanov [T2], we now use an argument of H. Lewy [L], extended by H. Rossi [R], to show that the function F that is CR on M extends holomorphically through {z} × D z for each z ∈ ∆, ℑz = 0 and that the extension so obtained is holomorphic in z. In this way we obtain a function F that is holomorphic on Ω + and on Ω − and which extends continuously to z∈∆,ℑz>0 {z} × bD z , a part of boundary of Ω + , and to z∈∆,ℑz<0 {z} × bD z , a part of boundary of Ω − .
Before we proceed, notice that our function F is well defined on (0, t) × IR.
We will now show that the continuity of F on M implies that We will also show that our assumptions imply that for each η, 0 < η < t there is a constant c(η) such that |F (η, ζ)| ≤ c(η)(1 + |ζ|) n (ζ ∈ C). (8.3) Suppose for a moment that we have done this.

Polynomial in the second variable
The fact that for each η, 0 < η < t, the function w → F (η, w) is a polynomial of degree ≤ n implies, for instance, that for each such η, For each η, 0 < η < t, and for each w 0 , ℑw 0 < 0, there is a neighbourhood U ⊂ C of η and a neighbourhood W of w 0 , such that for each k, ∂ k F ∂w k is continuous on {z ∈ U, ℑz ≥ 0} × W (a consequence of expressing the derivatives with the Cauchy integral formula) and holomorphic on {z ∈ U : ℑz > 0} × W . By the preceding discussion, for each k ≥ n + 1, ∂ k F ∂w k vanishes identically on {z ∈ U, ℑz = 0} × W , so it follows that it vanishes identically on {z ∈ U, ℑz > 0} × W , an open subset of Ω + , and since it is holomorphic on Ω + , it follows that it vanishes identically on Ω + as Ω + is connected. Thus, (9.1) In the same way we prove that (9.1) holds for all z ∈ Ω − . Recall that for each z ∈ ∆, ℑz = 0, D z is connected and hence (9.1) implies that for each such z, there are numbers a 0 (z), · · · , a n (z) such that F (z, w) = a 0 (z) + a 1 (z)w + · · · + a n (z)w n (w ∈ D z ).

Analyticity of the coefficients
Our extended function F is well defined on both where ∆ + = {ζ ∈ ∆: ℑz > 0} is the upper half of ∆ and ∆ − is the lower half of ∆.
For each z ∈ ∆, ℑz = 0, denote by I z the segment from z to 1/z (the straight part of bD z ) and by J z the circular arc part of bD z . The sets I z and J z meet at the points z and 1/z. For −1 < x < 0 we define I x = J x = (1/x, x). For z = x ∈ (−1, 0) denote the extension by ψ(x, w) (w ∈ (1/x, x). Here ψ(x, w) is the value of the extension of original F to the semiquadric Λ(T, ρ(T )) of the second family which passes through (x, w), at the point (x, w).
We need the following LEMMA 10.1 Let p m (w) = a m0 + a m1 w + · · · + a mn w n be a sequence of polynomials. Let w mi , 1 ≤ i ≤ n+1, be sequences of points in C, converging to distinct points w 1 , · · · , w n+1 , respectively. Suppose that for each i, 1 ≤ i ≤ n + 1, the sequence p m (w mi ) converges. Then there are α 0 , · · · , α n such that a mi converges to α i for each i, 0 ≤ i ≤ n, and therefore, the sequence p m converges, uniformly on compacta, to the polynomial p(w) = α 0 + α 1 w + · · · + α n w n . In particular, p m (w mi ) converge to p(w i ), 1 ≤ i ≤ n + 1, for any sequences w mi converging to w i , 1 ≤ i ≤ n + 1.
Proof. Since w 1 , · · · , w n+1 are distinct the Vandermonde matrix is nonsingular and consequently for all sufficiently large m the matrices V (w m1 , · · · , w m,n+1 ) =    1, w m1 , · · · , w n m1 · · · 1, w m,n+1 , · · · w n m,n+1    are nonsingular and as m → ∞, converge to the nonsingular matrix V (w 1 , · · · , w n ). Consequently, for m sufficiently large, the matrices V (w m1 , · · · , w m,n+1 ) −1 are well defined and, as m → ∞, they converge to V (w 1 , · · · , w n+1 ) −1 . Since and since both factors on the right converge it follows that the columns on the left converge. This completes the proof. ∼ Now, fix x, − 1 < x < 0, and fix distinct points y 1 , · · · y n+1 ∈ (1/x, x). Choose a sequence z n , ℑz n > 0, converging to x and observe that by the nature of bD z n for each i, 1 ≤ i ≤ n +1, there is a sequence u mi ∈ I z m , m ∈ IN that converges to y i , and a sequence w mi ∈ J z m that converges to y i . By (10.1) F (z m , u mi ) converges to ϕ(x, y i ) and by (10.2), F (z m , w mi ) converges to ψ(x, y i ). Now apply Lemma 10.1 to the sequence of polynomials w → p m (w) = F (z m , w) = a 0 (z m ) + a 1 (z m )w + · · · + a n (z m )w n to see that there are numbers α i = lim m→∞ a i (z m ), 0 ≤ i ≤ n and that for each i, 1 ≤ i ≤ n + 1, ϕ(x, y i ) = α 0 + α 1 y i + · · · + α n y n i = ψ(x, y i ). In particular, since y j were arbitrary, it follows that ϕ(x, y) ≡ ψ(x, y) (1/x < y < x).

Removable singularities of the coefficients at 0 and at t
In the preceding section we proved that where the functions a j , 0 ≤ j ≤ n, are holomorphic on ∆ \ {0, t}. Recall that our f has the property that for each Γ ∈ C 0 ∪ C t the function z → (z − c(Γ)) n f (z) extends holomorphically from Γ. (11.1) PROPOSITION 11.1 Suppose that f is as above and suppose that (11.1) holds. Then the functions a j , 0 ≤ j ≤ n, extend holomorphically through ∆.
We must now show that each a j has a removable singularity at the origin. To do this, we use the fact that from each circle b∆(λ, ρ(λ)) ∈ C t the function f extends holomorphically through ∆(λ, ρ(λ)) \ {λ} with a pole at λ. On b∆(λ, ρ(λ)) we have so our requirement will be that must have no pole at the origin. This is the same as to say that has a zero of order ≥ n at 0 for each λ, 0 ≤ λ < t. Requiring this we should then conclude that h 0 has a zero of order ≥ n at the origin h 1 has a zero of order ≥ n − 1 at the origin · · · h n−1 has a zero of order ≥ 1 at the origin which is the same as to conclude that each of the functions H 0 = h 0 , H 1 = zh 1 , · · · , H n = z n h n has a zero of order ≥ n at the origin. Thus, denoting we have to show that if has, for each λ, zero of order ≥ n at the origin, then the same hods for each H j . Suppose that (11.3) has a zero at the origin for each λ. This implies that We now use the fact that is a large set. This is so by the continuity: when λ → 0, λ − ρ(λ) 2 /λ is very large and when λ is near t, ρ(λ) is near 0 so λ − ρ(λ) 2 /λ is near t.. Thus, (11.4) implies that H 0 (0) = · · · = H n (0) = 0 so H 0 (z) = zG 0 (z), · · · , H n (z) = zG n (z) with G i holomorphic. If (11.3), which now becomes z G 0 (z) + G 1 (z)ϕ(z, λ) + · · · + G n (z)ϕ(z, λ) n , has a zero of order ≥ n at the origin then G 0 (z) + G 1 (z)ϕ(z, λ) + · · · + G n (z)ϕ(z, λ) n has a zero of order ≥ n − 1 at the origin. In particular, and so, after n steps we deduce that each H j has zero of order ≥ n at the origin. This completes the proof of Theorem 4.1 and thus proves Theorem 1.1 in the case when β = 0 and α ∈ ∆ \ {0},assuming that we have proved (8.1), (8.2) and (8.3).
12. Proving (8.1), (8.2) and (8.3), Part 1 We first look at (8.1), that is, at the function F on Ω + . Write Clearly M + is a part of bΩ + . We have We know that the function F is well defined and continuous on M + . We also know that the function F is continuous on z∈∆, ℑz>0 and that the extension is holomorphic on Ω + . We now want to show that F extends continuously to Ω + ∪ M + ∪ Θ + where The set Θ + is a disjoint union of halfplanes attached to M + along z∈(0,t) {z} × IR and is contained in bΩ + . In other words, we want to show that our function, holomorphic on Ω + , extends continuously to which is (8.1) that we want to prove. This will imply that for each η ∈ (0, t) the function ζ → F (η, ζ) is continuous on {ℑζ ≤ 0} and holomorphic on {ℑζ < 0}. We will also show that it satisfies an estimate of the form |F (η, ζ)| ≤ M η (1 + |ζ|) n (ℑζ ≤ 0) (12.1) which will, after proving the inequality also for ℑζ ≥ 0, give (8.3). To get an estimate of the form (12.1) we first prove the following LEMMA 12.1 Given η, 0 < η < t there are an open disc U centered at η and a constant M = M (U ) such that Remark By the maximum priciple the estimate (12.2) implies that and which, after proving the desired continuous extendibility, gives (12.1).
With no loss of generality assume that 0 ∈ E z for all z ∈ U, ℑz > 0. We understand that E z = IP for z ∈ U ∩ IR. The domains E z change continously with z in the sense that bE z change continuously with z in Frechet's sense [Ts,p.383] which, by a theorem of Courant [Ts,p.383] implies that if for each z ∈ U, ℑz ≥ 0, ψ z is the conformal map from ∆ to E z , ψ z (0) = 0, ψ ′ z (0) > 0, these maps change continuously with z, uniformly on ∆. It follows that if we set Ψ(z, w) = (z, ψ z (w)) then Ψ maps {z ∈ U, ℑz ≥ 0} × ∆ homeomorphically onto z∈U, ℑz≥0 {z}×E z . Recall that the function G is continuous on z∈U ℑz≥0 {z}×bE z and is also continuous on z∈U, ℑz>0 {z} × E z and holomorphic on z∈U, ℑz>0 {z} × E z .
Let J = G • Ψ. The function J is continuous on {z ∈ U, ℑz > 0} × ∆ and holomorphic on each {z} × ∆, z ∈ U, ℑz > 0, and is also continuous on {z ∈ U, ℑz ≥ 0} × b∆. By the continuity it follows that for each z ∈ (0, t) the function J extends holomorphically through {z} × ∆ and if we define the extensionJ so that on each {z} × ∆ it is the holomorphic extension of J from {z} × b∆, then so extended function will be, by the continuity of J on {z ∈ U, ℑz ≥ 0} × b∆ and by the maximum principle, continuous on z∈U, ℑz≥0 {z} × ∆. It follows thatJ • Ψ −1 provides the necessary extension of G to z∈U, ℑz≥0 {z} × E z .
This completes the proof of (8.1), and "half" of (8.3). In the same way we prove (8.2) and the "other half" of (8.3). Theorem 4.1 is proved. This proves Theorem 1.1 in the case (2.2).
The proof of Theorem 1.1 in the cases (2.3) and (2.4) is almost the same. The case (2.3) is the limiting case of (2.2) when t tends to 1. For z ∈ ∆, z ∈ IR, the domain D z is now bounded by I z , the segment joining z and 1/z, and by λ z , the arc of the circle passing thhrough z, 1/z and 1, with endpoints z and 1/z, which does not contain 1. Folding occurs only on the interval (−1, 0) and there is only the singularity at the origin to remove in proving the analyticity of the coefficients a 0 , · · · , a n on ∆. The case (2.4) is even simpler. The domain D z is now bounded by λ z and µ z where µ z is the arc on the circle passing through z, 1/z and −1, with endpoints z and 1/z, which does not contain −1. There is no folding and no singularity to remove. Theorem 1.1 is proved.

Remarks and open questions
Remark In Theorem 1.1 the continuity of f at the boundary is essential as shown by the example f (z) = z(z − 1/2) 1 − |z| 2 (z ∈ ∆) of a function which extends holomorphically from every circle Γ ∈ C 0 ∪ C 1/2 yet it is not holomorphic on ∆.
Remark Note that if f is continuous on ∆ and of the form (1.1) then a j , 0 ≤ j ≤ n, need not be continuous on ∆. To see this, let g be a bounded holomorphic function on ∆ that does not extend continuously to ∆. Then f (z) = (1 − |z| 2 )g(z) = g(z) − [zg(z)]z extends continuously through ∆ yet the functions g and zg do not.
Example 14.1 The function g(z, w) = |w| 2 ((z, w) ∈ bB) extends holomorphically into B along each complex line passing through the origin and along each complex line parallel to one of the coordinate axes that meets B . Let M be the Moebius transform in C 2 which maps the origin to the point a = (1/2, 1/2) [Ru]. M maps the the complex lines through the origin to L(a), the complex lines parallel to the z−axis to L(b) and the complex lines parallel to the w−axis to L(c) where b, c are contained in C 2 \ B and satisfy < a|b >=< a|c >= 1, < b|c > = 1. The function f = g • M −1 is real analytic on bB and extends into B along each complex line L ∈ L(a) ∪ L(b) ∪ L(c) yet it does not extend holomorphically into B, and so, by the main result of [G3], a is the only point in B such that f extends holomorphically into B along each complex line through this point.
This example shows that in Theorem 1.2 and Corollary 1.3 one cannot drop the assumption that < a|b > = 1 if one of the points a, b is in B and the other in C 2 \ B. It is easy to see that the points a, b, c do not lie on the same complex line so the example shows also that there are triples a, b, c of points in C 2 , not lying on the same complex line, such that L(a) ∪ L(b) ∪ L(c) is not a test family for holomorphic extendibility for C ∞ (bB), This is a function of class C k on bB which extends holomorphically into B along any complex line which meets {0} × ∆. In particular, it extends holomorphically into B along every complex line that is parallel to the z−axis. Let M be the Moebius map in C 2 that maps the origin to the point (1/2, 0). The function f = g • M −1 extends holomorphically into B along any complex line that meets L ∩ B where L = {(z, w): z = 1/2} and along each complex line passing through (2, 0) yet it does not extend holomorphically through B. Choosing a, b ∈ L ∩ B and c = (2, 0) we have < a|b >=< a|c >= 1 which shows that in Corollary 1.3 we cannot drop the assumption that one of the numbers < a|b >, < a|c > is different from 0.
If E is a complex line that misses B then given a finite set {a 1 , a 2 , · · · , a n } ⊂ E there is a real analytic function on bB which extends holomorphically into B along every L ∈ L(a 1 ) ∪ L(a 2 ) ∪ · · · ∪ L(a n ), yet it does not extend holomorphically through B [KM,G3]. The situation in the case when E is tangent to bB is unclear. If E meets B the situation is different for functions of class C ∞ : the lines through two points a 1 , a 2 suffice provided that < a 1 |a 2 > = 1 in the case that one of the points a 1 , a 2 is in B and the other in C 2 \ B [G3]. The situation is different for functions of class C k : Given numbers t 1 , t 2 , · · · , t n , 1 < t 1 < t 2 < · · · < t n < ∞ the function is a function of class C k on bB which extends holomorphically into B along each complex line which meets ∆ × {0} as well as along each complex line L ∈ L((t 1 , 0)) ∪ L((t 2 , 0)) ∪ · · · ∪ L((t n , 0)) yet does not extend holomorphically through B. We conclude with the following open QUESTION Suppose that a, b, c ∈ C 2 do not lie on the same complex line and assume that Λ(a, b) ∩ B = Λ(b, c) ∩ B = Λ(a, c) ∩ B = ∅. Is L(a) ∪ L(b) ∪ L(c) a test family for holomorphic extendibility for C(bB)? If not, is this family a test family for holomorphic extendibility for C ∞ (bB)?