Analyticity on translates of a Jordan curve

Let C be real-analytic simple closed curve in the complex plane which is symmetric with respect to the real axis. Let r>0 be such that C+ir misses C-ir. We prove that if a continuous function f extends holomorphically from C+it for each t in [-r,r] then f is holomorphic.

A positive answer was obtained for real analytic functions by M. Agranovsky and the author [AG] and independently by L. Ehrenpreis [E] and for continuous functions by A. Tumanov [T1].
To answer the question above one passes in both [AG] and [T1] to an associated problem in C 2 . In [AG] the authors use semi-quadrics Λ a,ρ = {(z, w): (z − a)(w − a) = ρ 2 , 0 < |z − a| < ρ} which are attached to Σ = {(z, z): z ∈ C} along the circles {(z, z): z ∈ b∆(a, ρ)}. They use the property of Λ a,ρ that a continuous function on b∆(a, ρ) extends holomorphically through if and only if the function F defined on bΛ a,ρ = {(z, z): z ∈ b∆(a, ρ)} by F (z, z) = f (z) (z ∈ b∆(a, ρ)) has a bounded continuous extension to Λ a,ρ ∪bΛ a,ρ which is holomorphic on Λ a,ρ . Thus, when studying holomorphic extensions of a function f from circles in the plane one defines F (z, z) = f (z) in a region in Σ and then studies bounded holomorphic extensions of F from bΛ a,ρ through Λ a,ρ [AG,G3].
Tumanov then uses methods of CR theory to show that F does not depend on the second variable which means that f is holomorphic on IntS. He also discovers that this is actually a finite strip problem. Very recently Tumanov [T2] studied a similar problem for a family of circles with centers sliding along a smooth curve and with smoothly changing radii. He used semi-quadrics. He obtained the result by using a classical argument of H. Lewy about holomorphic extensions of CR functions. In particular, he found a very simple proof of the theorem on the strip.
In the present paper we generalize the result of Tumanov [T1] from vertical translates of circles to vertical translates of real-analytic simple closed curves which are symmetric with respect to the real axis.

The main results
Our first result is about analyticity on vertical translates of curves which are symmetric with respect to the real axis.
Theorem 2.1 Let Ω be a domain in C which is symmetric with respect to the real axis and whose boundary is a real analytic simple closed curve. Let r > 0 and let f be a continuous function on K = ∪{bΩ + it, −r ≤ t ≤ r} such that for each t, −r ≤ t ≤ r, the function f |(bΩ + it) has a continuous extension to Ω + it which is holomorphic on Ω + it. Suppose that (Ω − ir) ∩ (Ω + ir) = ∅. Then f is holomorphic on IntK.
We will deduce Theorem 2.1 from a more general result below which involves vertical translates of general domains and their images under conjugation.
Theorem 2.2 Let λ: [α, β] → IR be a continuous function and let a, b, c, d be real numbers such that D + ia, D * + ic are both contained in {t + is: s < λ(t), α ≤ t ≤ β} and such that D + ib, D * + id are both contained in {t + is: s > λ(t), α ≤ t ≤ β}. Let and let f be a continuous function on Q 1 ∪ Q 2 such that for each t, a ≤ t ≤ b, the function f |(bD + it) has a continuous extension to D + it which is holomorphic on D + it, (2.1) for each s, c ≤ s ≤ d, the function f |(bD * + is) has a continuous extension to D * + is which is holomorphic on D * + is. (

From circles to general curves
In this section we describe the idea how to pass from circles to general curves. Let Ω be a domain bounded by a real-analytic simple closed curve which is symmetric with respect to the real axis. With no loss of generality assume that Ω contains the origin. Let f be a continuous function on ∪{bΩ + it: t ∈ IR} such that for each t ∈ IR, the function ζ → f (ζ + it) (ζ ∈ bΩ) has a continuous extension to Ω which is holomorphic on Ω.
Semi-quadrics are related to circles so they cannot be used to study the analyticity of functions on a family of translates of a given curve that is not a circle. We look again at the way how Tumanov [T1] adds the extra variable in the case of the circles. An important point in his setting is that on b∆ the conjugation z → z extends to the map z → 1/z which carries ∆ \ {0} biholomorphically onto C \ ∆. This is not the case for general curves so for domains Ω more general than a disc it seems difficult to work with the manifold {(ζ + it, ζ): ζ ∈ Ω, t ∈ IR} used in [T1] when Ω is a disc. However, since the reflection (1.1) takes place only in the second variable the idea is to replace the manifold {(ζ + it, ζ): ζ ∈ Ω, t ∈ IR} with a manifold that is attached to the cylinder {(z, w): |w| = 1}. To do this we take the conformal map Φ: Ω → ∆ that satisfies Φ(ζ) = Φ(ζ) (ζ ∈ Ω), Φ(0) = 0, notice that Φ extends to a diffeomorphism Φ from Ω to ∆ and define the smooth manifold N 1 by We define a continuous function F on N 1 by letting, for each t ∈ IR, the function ζ → F (ζ + it, Φ(ζ)) (ζ ∈ Ω) be the holomorphic extension of ζ → f (ζ + it) (ζ ∈ bΩ) through Ω.
If ζ ∈ bΩ then ζ ∈ bΩ and if t ∈ IR then which makes possible to extend F continuously to a new geometric object, the smooth manifold by using the equality (3.1) for ζ ∈ Ω \ {0} as a definition. Thus we get a continuous CR function F on N 1 ∪ N 2 , the union of two manifolds with the common boundary We then show that the classical argument of H. Lewy which Tumanov used with semiquadrics works also in the present situation. This helps us to prove that F depends only on the first variable which implies that f is holomorphic. In fact, our main result, Theorem 2.2, is somewhat more general than the one just described. Its proof, although technically a bit complicated, uses essentially the idea above.

The manifold N
We now begin with the proof of Theorem 2.2. With no loss of generality we assume that 0 ∈ D and that the imaginary axis intersects bD transversely.
Let Φ: D → ∆ be a conformal map such that Φ(0) = 0. Since bD is real-analytic the map Φ extends to a biholomorphic map from a neighbourhood of D to a neighbourhood of ∆. Define Ψ: The map Ψ maps D * conformally onto ∆ and extends to a biholomorphic map from a neighbourhood of D * to a neighbourhood of ∆. Define where p and q are real functions. Then The function θ is well defined and continuous on C since w = 1/w (w ∈ b∆). The function θ is invariant with respect to w → 1/w, the reflection across b∆. Note that θ|∆ is smooth as it extends to a harmonic function in a neighbourhood of ∆. Similarly, which shows that we obtain N by taking the graph {(θ(w), w): w ∈ C} of θ in IR × C = IR × {0} × C and then making the union of all translates of this graph in the extra perpendicular direction (i, 0), that is, we see that N is the union of manifolds N 1 and N 2 with boundary which meet along the common boundary The complement of the graph of θ in IR×C has two components: {(t, w): t > θ(w), w ∈ C} and {(t, w): t < θ(w), w ∈ C}, which, by (4.1) implies that C 2 \ N has two components

Intersecting N with complex lines
We will apply the reasoning of H. Lewy about holomorphic extensions of CR functions. To this end, we look first at the intersections of N with complex lines L(z) = {(z, w): w ∈ C}. We shall use the fact that since bD is real-analytic and compact there are at most finitely many points ζ ∈ bD such that the tangent line to bD at ζ is parallel to the imaginary axis.
For z ∈ S writeẼ Since θ is invariant with respect to the reflection across b∆ it follows that we get Λ 2 (τ ) by reflecting Λ 1 (τ ) ⊂ ∆ across b∆. So it is enough to study Λ 1 (τ ). Clearly If τ ∈ [α, β] is such that τ + iIR meets bD transversely, as happens for all but finitely many τ , then Λ 1 (τ ) consists of finitely many pairwise disjoint closed arcs with endpoints on b∆ but otherwise contained in ∆ which meet b∆ transversely. By transversality and by the fact that Φ extends across bD as a conformal map, these arcs change smoothly with τ as long as τ + iIR meets bD transversely. If τ ∈ (α, β) is such that τ + iIR does not meet bD transversely then Λ(τ ) consists of a finite number of arcs with endpoints on b∆ and pairwise disjoint interiors plus a possible finite set on b∆. There may be points on b∆ which are common endpoints of two (but not more than two ) of these arcs. Since we get Λ 2 (τ ) by reflecting Λ 1 (τ ) across b∆ it follows that if τ = 0 and if τ + iIR is transverse to bD then Λ(τ ) consists of finitely many pairwise disjoint simple closed curves, symmetric with respect to b∆. If τ = 0 and τ + iIR does not meet bD transversely then Λ(τ ) consists of finitely many pairwise disjoint simple closed curves, symmetric with respect to b∆ plus a possible finite subset of b∆. There may be points on b∆ that are common points of two, but not more than two of these curves. Except for these points, the curves are pairwise disjoint. Clearly Λ(α) and Λ(β) are finite sets.
Since iIR meets bD transversely Λ 1 (0) consists of finitely many pairwise disjoint closed arcs with endpoints on b∆ but otherwise contained in ∆ which meet b∆ transversely. One of these arcs passes through the origin so its image under the reflection across b∆ passes through infinity. Thus, Λ(0) ∪ {∞} consists of finitely many pairwise disjoint simple closed curves on the Riemann sphere one of which contains infinity.
A similar proposition holds for α < τ < 0 and for P 2 in the place of P 1 .

Continuity of an integral
We shall need Proposition 6.1 Let z 0 ∈ S and suppose that G is a continuous function on a neighbourhood ofẼ(z 0 ) in N . Then the function defined in a neighbourhood of z 0 in S, is continuous at z 0 .
Proof. We prove the continuity of The proof for Θ 2 (z) = Λ 2 (z) G(z, w)dw will be analogous; note that Θ = Θ 1 + Θ 2 since Λ 1 (z) and Λ 2 (z) meet in a finite set. Recall that Φ and Φ ′ extend holomorphically into a neighbourhood of D, so the continuity of Θ 1 depends on how K(t) = D ∩ (t + iIR) changes with t near t 0 = ℜz 0 . Assume for a moment that α < t 0 < β. There are at most finitely many t, α ≤ t ≤ β such that t + iIR does not intersect bD transversely. Thus there is an η > 0 such that t +iIR intersects bD transversely for every t, 0 < |t −t 0 | < η. In particular, for each t, t 0 − η < t < η, K(t) is a finite collection of pairwise disjoint closed segments with endpoints varying continously with t. Since bD ∩ (t 0 + iIR) is a finite set and since bD is compact it follows that each of these endpoints has a limit as t ր t 0 . As t ր t 0 some segments may degenerate into points in the limit K(t 0 −0) and some pairs of segments may get a common endpoint. Clearly Since the set bD ∩ (t 0 + iIR) is finite it follows that K(t 0 − 0) ⊂ K(t 0 ) is a finite set. Thus, lim z→z 0 ,ℜz≤ℜz 0 Θ(z) = Θ(z 0 ). Similarly we show that lim z→z 0 ,ℜz≥ℜz 0 Θ(z) = Θ(z 0 ) which proves that Θ is continuous at z 0 . The same (one sided) reasoning applies if z 0 ∈ bS. The proof is complete.

The manifold M and the function F
We now define a submanifold of N that is more closely related to our problem. Write Suppose that f is a continuous function on Q 1 ∪ Q 2 which satisfies (2.1) and (2.2). For each t, a ≤ t ≤ b, let g t be a continuous extension of ξ → f (ξ + it) (ξ ∈ bD) to D which is holomorphic on D and for each s, c ≤ s ≤ d, let h s be the continuous extension and the function H on M 2 by In particular, on the part of bM 1 contained in bN 1 we have and on the part of bM 2 contained in bN 2 we have Suppose that (z, w) ∈ M 1 ∩ M 2 . Then there are ξ ∈ bD, ζ ∈ bD * and t, s, a ≤ t ≤ b, c ≤ s ≤ d, such that (ξ + it, Φ(ξ)) = (z, w) = (ζ + is, 1/Ψ(ζ)) = (ζ + is, Ψ(ζ)) = (ζ + is, Φ(ζ)) which implies that ζ = ξ and ξ is a well defined continuous function on M 1 ∪M 2 which is holomorphic on each holomorphic leaf of M 1 and on each holomorphic leaf of M 2 . Our aim is to show that F depends only on z which will imply that f is holomorphic on IntQ 1 ∪ IntQ 2 .

Integrals of CR functions on M
Denote by π 1 the projection π 1 (z, w) = z. With no loss of generality assume that λ(0) = 0. Our assumptions imply that there is an η > 0 such that if Recall that for z ∈ Σ 1 the set E(z) is the boundary of Y (z) = P 1 ∩ L(z) in L(z) and for z ∈ Σ 2 the set E(z) is the boundary of Y (z) = P 2 ∩ L(z) in L(z). Let A j = ∪{Y (z): z ∈ Σ j } = π 1 (Σ j ) ∩ P j (j = 1, 2).
For each j = 1, 2, A j is an open subset of Σ j × C whose relative boundary is N ∩ (Σ j × C). Using Proposition 5.1 we see that the complement of A j in Σ j × C is connected, j = 1, 2.
We shall prove that the function F extends holomorphically into A 1 and into A 2 . We begin to follow the reasoning of H. Lewy. In [L] this was done for smooth functions on smooth manifolds and for more general, including continuous, functions on smooth manifolds this was done in [R]. We cannot refer to these results directly since in our case the manifold is not smooth but consists of two smooth pieces. However, these two pieces are both foliated by analytic discs which simplifies the situation. We provide the details to make the proof self contained.
Let ∆(u, r) be contained in either Σ 1 or Σ 2 and assume that Θ is a continuous function on π −1 1 (∆(u, r)) ∩ N which is holomorphic on each holomorphic leaf. The function is, as we know, well defined and by Proposition 6.1 it is continuous on ∆(u, r). Recall that there are at most finitely many real values τ such that τ + iIR is not transversal to bD. So, if we want to prove that Q is holomorphic on ∆(u, r) it is enough to prove that Q is holomorphic in a neighbourhood of each z 0 ∈ ∆(u, r) such that z 0 + iIR intersects bD transversely. Let z 0 be such a point. Let ρ > 0 be such that for each z ∈ ∆(z 0 , ρ), z + iIR meets bD transversely. Passing to a smaller ρ if necessary we may assume that there is a γ > 0 such that whenever U is a closed disc contained in ∆(z 0 , ρ) of radius not exceeding γ, the circle bU + it, t ∈ IR, either misses bD or meets bD in one point or in two points.
Let U be such a disc. By transversality, there is a positive integer ν such that for each z ∈ U the set L(z)∩N = L(z)∩M consists of ν pairwise disjoint simple closed curves, each being the union of two smooth arcs with endpoints on {z} × b∆, one having its interior contained in {z} × ∆, and the other having its interior contained in {z} × (C \ ∆), which change smoothly with z. So π −1 is an open subset of N which has ν components whose closures are pairwise disjoint; the boundary of this set is π −1 1 (bU ) ∩ N = ∪{L(z) ∩ N : z ∈ bU }. Let Ω be one of these components. Write T = bN 1 = bN 2 . The set Ω consists of three pairwise disjoint parts: the domains Ω 1 = Ω ∩ IntN 1 , Ω 2 = Ω ∩ IntN 2 and the two dimensional surface Ω ∩ T . For each z ∈ bU , L(z) ∩ bΩ is a simple closed curve so bΩ is a torus and Ω is a solid torus in N . We now want to show that bΩ Θ(z, w)dz ∧ dw = 0. (8.1) Note first that Ω ∩ T is the common part of bΩ 1 and bΩ 2 so to prove (8.1) it is enough to prove that Consider (8.2). The properties of U imply that Ω 1 can be written as the union of a continous family of pairwise disjoint analytic discs and bΩ 1 is the union of their pairwise disjoint boundaries These analytic discs, if nonempty, are of two sorts: either their boundaries are smooth simple closed curves which meet Ω ∩ T in at most one point (which happens if U ⊂ D + it), or their boundaries are simple closed curves consisting of two arcs, one contained in IntN 1 and the other contained in T (which happens if U meets D + it but is not contained in D + it). Recall that N 1 = {(Υ(w) + it, w): w ∈ ∆, t ∈ IR} where the conformal map Υ = Φ −1 extends to a biholomorphic map from R∆ for some R > 1 to a neighbourhood of D. Define the function ρ by ρ(z, w) = (1/i)(z − Υ(w)) and notice that ρ is real on N 1 .
where the integral on the right vanishes by the by the Cauchy theorem since the function (z, w) → Θ(z, w)/Υ ′ (w) is continuous on A t and holomorphic on A t . This proves that the integral on the left in (8.4) vanishes. We repeat the reasoning for N 2 to get (8.3). This proves (8.1). Thus, bU bΛ(z) Θ(z, w)dw dz = 0 for every disc U ⊂ ∆(z 0 , ρ), which, by the Morera theorem, implies that the function Q is holomorphic on D(z 0 , ρ). This proves that Q is holomorphic on Σ 1 ∪ Σ 2 .

Holomorphic extensions of F and the completion of the proof
We continue to follow the reasoning of H. Lewy. For each z ∈ Σ 1 ∪ Σ 2 and each w ∈ C \ Λ(z) (that is, for each w such that (z, w) ∈ N ) define For a fixed z the function w → Ξ(z, w) is holomorphic on C \ Λ(z). Now, fix z 0 ∈ Σ 1 and w ∈ C \ Λ(z 0 ). We show that z → Ξ(z, w) is holomorphic in a neighbourhood of z 0 . There is a ρ > 0 such that ∆(z 0 , ρ) ⊂ Σ 1 and (z, w) ∈ N (z ∈ ∆(z 0 , ρ)) so the function (z, ζ) → F (z, ζ)/(ζ − w) is continuous on ∪{L(z) ∩ N : z ∈ ∆(z 0 , ρ)} and holomorphic on each holomorphic leaf. By the reasoning in Section 8 it follows that z → Ξ(z, w) is holomorphic on ∆(z 0 , ρ). This shows that Ξ is holomorphic on π −1 1 (Σ 1 ) \ N . Fix a large w. We know that z → Ξ(z, w) is continuous on Σ 1 ∪ [α + i(−δ, δ)] and holomorphic on Σ. Since Λ(α) is a finite set it follows that Ξ(z, w) approaches 0 as z approaches a point of α + i(−δ, δ). It follows that Ξ(z, w) ≡ 0 (z ∈ Σ 1 ). Since this holds for all sufficiently large w, the connectedness of ( It follows, in particular, that for all z ∈ Σ 1 the function w → F (z, w), defined on E(z) = bY (z), has a continuous extension into Y (z) ∪ bY (z) which is holomorphic on Y (z). The same reasoning shows this for all z ∈ Σ 2 . Recall that Λ(0) ∪ {∞} consists of finitely many pairwise disjoint simple closed curves on the Riemann sphere one of which passes through infinity. By transversality, Λ(t) changes continuously with t near 0 and contains ∞ only if t = 0. As t ր 0 the open sets Y (t) and their oriented boundaries converge to a domain Y − and its oriented boundary bY − which, as a set, coincides with Λ(0). As t ց 0 the open sets Y (t) and their oriented boundaries Λ(t) converge to a domain Y + and its oriented boundary bY + which, as a set, coincides with Λ(0). We have Y − ∪ Λ(0) ∪ Y + = C. Since the function F is bounded and continuous on M it follows that as t ր 0, the holomorphic extensions of F |M ∩ L(t + iλ (t) to a bounded entire function on L(0), which, by the Liouville theorem, must be constant. This implies that F |M ∩ L(0) is a constant. In the same way we show that F |M ∩ L(is) is constant for each s, −η < s < η. It follows that there is an r > 0 such that on r∆ the holomorphic extensions of all functions f |(it+bD) to it+D, for all t such that r∆ ⊂ it+D, coincide, and that on r∆ the holomorphic extensions of all functions f |(it+bD * ) to it+D * , for all t such that r∆ ⊂ it + D * , coincide. This implies first that f is holomorphic on ∪{it + bD: r∆ ⊂ it + D, a < t < b} ∩ IntS, and then, by translating bD further along iIR, that f is holomorphic on ∪{it + bD: a < t < b} ∩ IntS = IntQ 1 . Similarly, we show that f is holomorphic on IntQ 2 . The proof is complete.

Remarks
It remains an open problem to prove Theorem 2.1 without the assumption that Ω is symmetric with respect to the real axis. It is known that one cannot drop the assumption that (Ω − ir) ∩ (Ω + ir) = ∅. In fact the function f (z) = z 2 /z (z = 0) 0 (z = 0) is continuous on C and extends holomorphically from each circle which either surrounds the origin or contains the origin, yet f is not holomorphic.