Hartogs-type extension for tube-like domains in C2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb C^2$$\end{document}

In this paper we consider the Hartogs-type extension problem for unbounded domains in C2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb C^2$$\end{document}. An easy necessary condition for a domain to be of Hartogs-type is that there is no a closed (in C2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb C^2$$\end{document}) complex variety of codimension one in the domain which is given by a holomorphic function smooth up to the boundary. The question is, how far this necessary condition is from the sufficient one? To show how complicated this question is, we give a class of tube-like domains which contain a complex line in the boundary which are either of Hartogs-type or not, depending on how the complex line is positioned with respect to the domain.


Introduction
In this paper we consider a Hartogs-type extension problem for unbounded domains in C 2 . The following classical holomorphic extension result of Hartogs [11] in 1906 is considered by most researchers as a formal beginning of Complex Analysis of several variables. After some basic definitions (in Sect. 3), we consider the case of tube-like domains in C 2 (see Fig. 3) where the projection of the domain along a 2-dimensional real plane is bounded in a transversal real plane. Here, we formulate the following conjecture: Conjecture for tube-like domains in C 2 . Let ⊂ C 2 be a tube-like domain with connected C 1 boundary. Then is a Hartogs-type domain if there is no complex line contained in , see Fig. 1.
The conjecture is still open. It should be mentioned that for non-tube-like domains, the non-existence of complex varieties in the closure of the domain is not a suitable sufficient condition. This will be considered in a forthcoming paper.
The main theorems of this paper are proved in Sects. 4 and 5. We consider tubelike domains along the complex line {0} × C ⊂ C 2 with smooth (C ∞ ) boundary M = b , see Fig. 2. Moreover, the line {0} × C is contained entirely in the boundary b . For each w ∈ C, let w be the cross-section of by C × {w}, and M w = b w . Finally let η : C −→ R be a continuous function such that for every w ∈ C, e iη(w) is the unit normal vector to b w at (0, w) pointing inside w .
In the theorems below, depending how the domain is positioned with respect to the line {0} × C, in one case the domain is of Hartogs-type in the other case not.
Here we formulate non-invariant versions of the main theorems which are easier to visualize. Invariant versions can be found in Theorems 4.6 and 5.4.
In Sect. 4, we give a sufficient condition for a domain not to be of Hartogs type: Then there is a C ∞ smooth CR function on M that cannot be continuously extended to a holomorphic function in , i.e., is not a Hartogs-type domain.
In the theorem, in particular, if the function η(w) is bounded on C, then the harmonic function χ(w) is constant. Consequently, the normal vector e iη(w) is contained in an open half-plane. Very roughly (and imprecisely) speaking, the domain is lying on "one side" of the complex line {0} × C.  The above theorem is an improvement (and generalization) of an example by the authors in [3], where a domain G ⊂ C 2 was constructed which contained two complex lines in its closure G (actually the lines were "mostly" contained in G, only touching the boundary at some points). Also a continuous CR function on the boundary bG was constructed (actually the function was smooth except a small portion of the boundary) that could not be holomorphically extended to G.
In Sect. 5 we give a sufficient condition for a domain containing a complex line as above to be of Hartogs-type, namely Theorem 2 (Corollary 5.2) Let ⊂ C 2 be a domain with smooth (C ∞ ) boundary M = b containing the line L, L = {0} × C ⊂ M. With the notation as above, assume additionally that M w is a smooth Jordan curve for every w ∈ C, and that If there does not exist an entire harmonic function χ : C −→ R such that then every smooth CR function on M continuously extends to a holomorphic function in , i.e., is a Hartogs-type domain.
Again, if the function η(w) is bounded on C, then non-existence of a harmonic function satisfying (2) means that the normal vector e iη(w) rotates more than the angle π when w varies. Very roughly (and imprecisely) speaking, the domain "goes around" the complex line {0} × C.
In the above theorems we see that the difference between the sufficient condition and the necessary condition is very small. The theorems also show that it will be hard to formulate a necessary and sufficient condition for a domain to be of Hartogs-type.

Basic definitions and notation
CR manifolds Let M be a real embedded submanifold of class C 1 in C n . Let T p M be the real tangent space to M at p. We can consider T p M as a real vector subspace of the complex space T p C n . We say that M is a Cauchy- This constant is called CR dimension of M and denoted by dim CR M. If dim CR M = 0, then M is called a totally real manifold.
CR functions By CR functions we mean functions f : M −→ C of class C 1 that satisfy the tangential Cauchy-Riemann equations, L f = 0 for any L = a 1 ∂ ∂z 1 + · · · + a n ∂ ∂z n tangent to M.
Obviously, the restriction f = F| M of a holomorphic function F : U −→ C, M ⊂ U , to M is CR, but in general, not all CR functions arise as restrictions of holomorphic functions.
Smoothness By smooth (manifold or function) we shall mean of class C ∞ , unless otherwise specified. Usually the regularity conditions can be weakened, for instance, to consider CR functions in the distributional sense (e.g., continuous CR functions), however, it is not our intention here to formulate and prove results under the weakest regularity hypothesis. In what follows, by we will denote a domain (open, connected) in C 2 (n = 2) with connected boundary b and also C 2 \ is connected. Usually we assume that the boundary b is smooth, unless otherwise specified.
Tubes and tube-like domains Let P be a 2-dimensional real subspace in C 2 and π : C 2 −→ P be a projection, an R-linear mapping. By a tube (see Fig. 3) we will mean T = π −1 (U ), where U ⊂ P is a bounded domain with smooth boundary. A tube-like domain is an unbounded domain ⊂ C 2 with smooth boundary such that π( ) ⊂ P is bounded.
If π −1 (0) is a totally real plane, we say that the tube or tube-like domain is along a totally real plane. If π −1 (0) is a complex line, we say that the tube or tube-like domain is along a complex line.    In general, the whole intersection M w = M ∩ (C × {w}) does not have to be a smooth curve. However, since the real tangent planes T (0,w) (M) and T (0,w) (C × {w}) intersect transversally (as the former contains {0} × C), a neighborhood of 0 in M w is a smooth simple (open) arc and, for some δ > 0, the intersection D(0, δ)∩(M w ∪ w ) is diffeomorphic to a half-disc {(x, y) : x ≥ 0, x 2 + y 2 < 1}. Due to this, there is a unique unit normal vector n(w) to M w at 0 that points inside w . Letting w vary, the above transversality argument also implies that the function is continuous. We omit details of the proof.
Finally (by monodromy) there is a continuous function η : C −→ R such that (η(·) is unique up to addition of a constant multiple of 2π ; we fix one choice.) Assuming the above set-up and notation, suppose further that − π 2 < η(w) < π 2 for all w ∈ C. Then there is a C ∞ -smooth CR function ϕ : M −→ C that does not extend continuously to a holomorphic function in .
Remark 4.2 Actually in the above theorem, the domain need not be tube-like. However, the line L = {0} × C is contained in the boundary.

Proof of Theorem 4.1
Special holomorphic function For the proof we need to construct a special holomorphic function. Denote by P the semi-infinite strip (see Fig. 5), and by S the enlarged semi-infinite strip  Fig. 6. Under the map z −→ 1/z, angles are preserved and half-lines and intervals are mapped to circular arcs.

Lemma 4.3
There is a function f : C −→ C such that • f is holomorphic in C\{0} with an essential singularity at 0, • for every n ≥ 0 we have lim C\H z→0 f (n) (z) = 0.
For construction and proof, see the Appendix.
Define F :

Lemma 4.4 The function F is smooth on
Postponing for a moment the proof of the lemma, define now ϕ : ϕ satisfies trivially the CR condition at the points of {0} × C, and so is a CR function on M. Suppose ϕ can be continued to , i.e., there is a function  which implies that f has removable singularity at 0, contrary to its construction. Thus an extension is impossible. The proof of the theorem will be complete when Lemma 4.4 will be proved.

Proof of Lemma 4.4 It is enough to show that
Let Then α 0 > 0 and − π 2 + α 0 ≤ η(w) ≤ π 2 − α 0 , |w| ≤ R. This means that the smallest unoriented angle between the tangent line to M w at 0 (for |w| ≤ R) and the positive real axis is α 0 , see Fig. 7. For a fixed α ∈ (0, α 0 ) there is a ρ 0 > 0 such that the circular sector Denote these three sets as Set (1)  Theorem 4.1 can be reformulated in a more invariant way, namely, instead of the inequalities − π 2 < η(w) < π 2 we use a condition that involves an entire harmonic function. We have the following Then there is a smooth CR function on M that cannot be continuously extended to a holomorphic function in .
Before we give a proof of the Corollary, we take a closer look at the notion of a vector pointing inside a domain. Let G ⊂ C be a domain with smooth boundary bG and let v be a vector with the initial point p ∈ bG. We say that v is pointing inside the domain if it lies in the open half-space determined by the tangent hyperplane to bG at p and containing the unit normal vector to bG at p pointing inside the domain (see Fig. 8). Obviously, this notion can be generalized for higher dimensions.
Proof of Corollary 4.5 Let λ : C −→ R be a harmonic conjugate of χ . We use now the biholomorphic map K of C 2 onto itself, to reduce the general case to that of Theorem 4.1. Namely, the hypersurface M 1 := K (M) and domain 1 := K ( ) satisfy the assumptions of Theorem 4.1 due to condition (5), so there is a CR function ϕ 1 ∈ C ∞ (M 1 ) which cannot be continuously extended to a holomorphic function on 1 It is clear then that ϕ cannot be continuously extended to a holomorphic function in .
Observe that condition (5) can be given the following invariant form ( ) Let be a domain with smooth boundary M = b and let V be a closed complex variety contained in M. There exists a nowhere vanishing holomorphic vector field Since χ is harmonic, this implies condition (5).
Conversely, if (5) holds, let λ be an entire harmonic function such that λ + iχ is holomorphic. Let Therefore, Corollary 4.5 can be reformulated as Then there is a smooth CR function on M that cannot be continuously extended to a holomorphic function in .

Hartogs-type domains with a complex line in the boundary
The main goal of this section is to investigate whether the condition (5) is necessary for failure of the Hartogs extension phenomenon in the class of domains containing a complex line in its boundary M = b . The following theorem and its corollary suggest that it is "close" to being such, but there is a gap. The class of domains considered in this section is very similar to those from Sect. 4. We need to make an additional assumption that the tubes are shrinking at infinity, because otherwise there might be other obstacles to extension. We use the notation and meanings of η = η(w), w , M w as in the previous section.
The proof of the theorem is given in the rest of this section. As an immediate consequence of the theorem, we obtain Corollary 5.2 Let be a domain as in the above theorem. Assume additionally that there does not exist an entire harmonic function χ : C −→ R such that Then every smooth CR function on M continuously extends to a holomorphic function in , i.e., is a Hartogs-type domain. Remark 5.3 The discrepancy between the necessary condition (7) and (5) is small: the first says χ − η ∞ ≤ π 2 , the second As in the previous section, the above corollary can be formulated more invariantly. Let G ⊂ C be a domain with smooth boundary bG and let v be a vector with the initial point p ∈ bG. We say that v is pointing into the closed side of the domain if v = 0 and it lies in the closed half-space determined by the tangent hyperplane to bG at p and the unit normal vector to bG at p pointing inside the domain (see Fig. 9). Obviously, this notion can be generalized to higher dimensions.  Here we give an example of a class of domains, described in an easy geometric way (illustrated in Fig. 10), which satisfy the conditions of the theorem.
Then every entire harmonic function χ such that η − χ ∞ ≤ π 2 would have to be bounded, and so constant, and clearly such constant does not exist. Hence Corollary 5.2 applies.

Some properties of holomorphic and plurisubharmonic functions
We proceed now with the proof of Theorem 5.1, which will take a couple of steps. First we prove some lemmas.  there is a neighborhood V of (z, w) such that V is bounded from above and S := [C×(C\E)]\W . Then W is a pseudoconvex set and S is a pseudoconcave, relatively closed subset of C × (C\E), whose fibers S w are R + -cones, S w = {0}, for every w ∈ C\E.
Proof Representing F(z, w) by a Laurent series in z, z = 0, where a n , b n : C −→ C are entire functions, we obtain that Thus E is closed. If E is not discrete, all b n ≡ 0 and F has entire extension to C 2 , a contradiction.
We denote (z, w) := sup{ t (z, w) : 0 < t ≤ 1}. Then Lemma 6.2 implies that W is relatively pseudoconvex in (C\{0})×(C\E). Since the latter set is pseudoconvex, so is W , and S is pseudoconcave by definition. The fibers S w are cones because it follows immediately from the definition of S , namely if there is no neighborhood V of (z, w) ∈ S such that V is bounded then the same property has the point (tz, w) for t > 0 (Fig. 11).
Proof Denote The function A is a continuous branch of arg(z) on the set and so pluriharmonic on its domain of definition. In particular, for fixed z ∈ C\{0} the function A = A (z, w) is independent of w. Define Because S w being a cone, we can write We note that for each w ∈ C\E, the cone S w is closed and the set S w ∩ S 1 is compact, see Fig. 11. Consequently the maximum and minimum actually exist and for each w ∈ C\E there are z max (w) ∈ S 1 and z min (w) ∈ S 1 such that and Now we prove that χ 0 = χ 0 (w) is an upper-semi-continuous (usc) in C\E; the proof that μ 0 is a lower-semi-continuous (lsc) is analogous. Let's calculate lim sup w→w 0 χ 0 (w), namely take any sequence w n converging to w 0 . Without any loss of generality, we can assume that z max (w n ) converges to z max 0 (w 0 ). If we see, using (10) and (11), that which proves that χ 0 is upper-semi-continuous.
To show that χ 0 is subharmonic in C\E, we prove the following Proof of the Sublemma If χ is not subharmonic, then there is a z 0 ∈ U and R > 0 and a harmonic function F on the disc D(z 0 , R) such that Therefore, there exists ε > 0 such that The function ψ(z) = φ(z) + ε|z − z 0 | 2 is upper semicontinuous on D(z 0 , R) and thus assumes its maximum at a point a which must lie on the inside of D(z 0 , R) since ψ < 0 on the boundary of this disc and ψ(z 0 ) > 0. We therefore have Since ψ(z) ≤ ψ(a), we have Since a belongs to the inside of D(z 0 , R), there is an r > 0, such that (12) holds as well for z ∈ D(a, r ). Also observe that we can write where (z) is an affine linear function of z and hence is harmonic. We now let h(z) = (z) − F(z). Clearly h is harmonic, and from the definitions we have Note that α satisfies the desired estimate in view of (12) and note that α(a) = 0 in view of its definition. Thus h satisfies the properties stated in the sublemma.
End of the proof of Lemma 5.6 Now coming back to the proof of the lemma, we show that χ 0 is subharmonic in C\E. Suppose not. Then, using the Sublemma, there is a point w 0 ∈ C\E and a smooth subharmonic function α(w) in {|w − w 0 | < r }, r > 0, such that for some ε > 0 and for |w − w 0 | < r .
Let now Since U is a neighborhood of (z 0 , w 0 ) and ϕ(·, ·) a plurisubharmonic function in U , we obtained a contradiction with local maximum principle for pseudoconcave sets, due essentially to Wermer [26] (or see [25], Theorem 2.1(iv) for the exact form used above). Thus χ 0 is subharmonic in C\E.
A symmetrical reflection of the above proof shows that μ 0 defined in (11) is superharmonic in C\E. In addition, assumption (9) implies Then (χ 0 − μ 0 )(w) ≤ π , w ∈ C\E, is uniformly bounded from the above (on C\E) subharmonic function, which can be extended to C to be subharmonic at points of E also, and so is constant, say χ 0 − μ 0 = c 0 . As χ 0 = μ 0 + c 0 is both sub-and super-harmonic, it is harmonic. Since χ 0 is continuous at points of E, inequalities (13) imply that χ 0 is locally bounded (above and below) in neighborhoods of points of E, and so has entire harmonic extension χ : C −→ R satisfying |χ(w) − η(w)| ≤ π 2 , w ∈ C. The lemma is proved.

End of the proof of Theorem 5.1
Proof Consider a smooth CR function f : M −→ C that cannot be continuously extended to a holomorphic function in . We can decompose is proper. Consequently, all leaves of the foliation (C\{0}) × C by the complex lines {z}×C, z = 0, intersect compactly = M ∪ , and the intersection is empty for large |z|. To extend the function F − we apply the foliation method derived in [7], p. 561, and applied in [3], Sec. 3, p. 121. First we extend F − smoothly (C ∞ ) to a function F − on with support close to b ∩ [(C\{0}) × C], then define the (0, 1)-form Obviously ω is a smooth ∂-closed form on (C\{0}) × C. Next we solve the ∂-problem in such a way that u vanishes outside because the fibers {z} × C do not intersect for |z| large; see details in [7], pp. 561-563. Therefore the function F = F − − u has the properties that F (C\{0})×C is holomorphic and F C 2 \ is continuous. Observe that F is not entire, for otherwise F + − F would be a holomorphic extension of f to , a contradiction.
Thus F satisfies the assumption of Lemma 5.5. Let E, S be sets with properties stated there. Let n(w), w ∈ C, denote the unit normal vector at 0 ∈ M w to M w , pointing inside the Jordan domain w . Then (by monodromy) there is a continuous function η : C −→ R such that n(w) = e iη(w) . We will show now that this function η satisfies condition (9), which is equivalent to Consider (z 0 , w 0 ), z 0 = 0, with (z 0 , e iη(w 0 ) ) > π 2 . By smoothness of M, there is 1 ≥ t 0 > 0, ε > 0, such that Since F(z, w) is continuous on C 2 \ , it is bounded on Q, so z 0 / ∈ S w 0 . We are allowed now to apply Lemma 5.6, and so obtain a harmonic function χ : C −→ R with |χ(w) − η(w)| ≤ π 2 for all w ∈ C, which completes the proof of the theorem.
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Appendix
The Appendix has two parts: proof of Lemma 4.3 and proof of some properties of supremum of plurisubharmonic functions.

Proof of Lemma 4.3
In this section we will use the sets P and S that were defined in (3) and (4). For the proof of Lemma 4.3 we need the following fact Lemma 6.1 There is an entire function g : C −→ C (g ≡ 0) such that for every k, N = 0, 1, 2, . . ., we have Proof of Lemma 6.1 We split the proof into few steps.
It is not hard to see thatg(ζ ) is holomorphic. By repeated integration by parts (note the compact support of ψ), we havẽ Fix any A > 0. For Re ζ ≤ A we have and so which establishes the Assertion. It should be mentioned that the same argument handles derivatives ofg.

Approximation of a branch of
Consider now a holomorphic function h 0 : C\[2π, +∞) −→ C that is a continuous branch of the multi-valued function i √ ζ − 2π defined as follows: Then for ζ ∈ C\[2π, ∞), Observe that E is closed in C and its complement in C is connected and locally connected, as C\E = P ∪ {∞}. Therefore h 0 E can be uniformly approximated (on E) by entire functions (Theorem 3 in Brown et al. [5], see also Arakeljan [1]). In particular, there is an entire function It follows that and so |ζ | ≤ 4|h(ζ )| if ζ ∈ C\P and |ζ | ≥ 4π.

Definition of the function g
With A = 1 in the Assertion, let g(ζ ) =g(h(ζ )), ζ ∈ C. Then g is an entire function. Furthermore, using (14) and (15), for any N = 1, 2, . . . the estimate which proves Lemma 6.1 for the function g, i.e., when k = 0. It remains to prove for derivatives of g.

Properties of derivatives of g
It remains to observe that all derivatives of g have the same property in the slightly smaller set C\S. Fix N = 1, 2, . . . . If we denote Using the Cauchy formula we obtain, for |ζ 0 | ≥ 2 and ζ 0 ∈ C\S, Proof of Lemma 4.3 Define the function f : C −→ C by Then f C\{0} is holomorphic, and f has an essential singularity at 0. To see this, suppose that f has a removable singularity or a pole, then lim z→0 f (z) exists either as a finite number or as infinity. Then lim ζ →∞ g(ζ ) exists in the same sense and so f (z) is a polynomial, perhaps a constant one. This however is not possible by the way g was constructed.
To see that f (z) and f (n) (z) approach 0 as z → 0 within C\H we use the following fact that can be shown by induction: For every natural number n = 1, 2, . . . there are complex polynomials P 1,n , P 2,n , . . . , P n,n , all of degree ≤ 2n, such that for z = 0 Now, if z → 0 within C\H , then 1/z → ∞, 1/z ∈ C\S, and by Lemma 6.1, we have Thus f (n) (z) → 0 for z → 0, z ∈ C\H , what we wanted to prove. for some neighborhood V of (z, w) with V ⊂ U, V is bounded from above .
Then W is relatively pseudoconvex in U .

Comments
The lemma (which is true in C n ) is presumably well-known, but we could not find a reference. Note that the function does not have to be upper semi continuous. We base the proof on the well-known characterization of pseudoconvex domains in terms of Hartogs figures. Recall the definition of a (compact) Hartogs figure in C 2 . Let (z 0 , w 0 ) ∈ C 2 and a, b ∈ C 2 be two vectors linearly independent over C. Then by a Hartogs figure with center at (z 0 , w 0 ) and frame a, b we mean the compact set K = {(z 0 , w 0 ) + u a : u ∈ C, |u| ≤ 1} ∪ {(z 0 , w 0 ) + e iθ a + v b : θ ∈ R, v ∈ C, |v| ≤ 1}, and a filled Hartogs figure is the compact set K := {(z 0 , w 0 ) + u a + v b : u, v, ∈ C, |u|, |v| ≤ 1}.
We note that K is a bi-disc; the notation K is justified as this bi-disk is actually a polynomial hull of K . We use a characterization of relatively pseudoconvex sets given in the following Proof of Lemma 6.2 Consider any Hartogs figure K ⊂ W with K ⊂ U . It has some center (z 0 , w 0 ) and frame a, b. Since K ⊂ W , by definition K has a covering {V j } j∈J , by open neighborhoods, such that V j ⊂ W , j ∈ J , and there are finite constants M j such that Select a finite covering V j 1 , V j 2 , . . . , V j n of K , and let V := V j 1 ∪ V j 2 ∪ · · · ∪ V j n , M := max(M j 1 , M j 2 , . . . , M j n ).
Then we have Consider now another Hartogs figure K ε , also with center (z 0 , w 0 ), where ε > 0 is yet to be chosen, and the frame Clearly, if ε > 0 is small enough, Fix such ε. Since K ε ⊂ V , we obtain for every α ∈ A (note that α itself is upper semi continuous, so the maximum exists). Since K ε ⊂ U , α is a plurisubharmonic function on a neighborhood of K ε . Since K ε is a polynomial hull of K ε , or simply by looking at the way complex discs fill K ε , we conclude Observe finally that by the way it is constructed, the bi-disc K ε is a neighborhood of K , i.e., K ⊂ Int( K ε ). By (16), Int( K ε ) ⊂ W , and so K ⊂ W . The lemma is proved.
The finite covering argument used in the proof implies also the following Corollary 6.4 In the setting of Lemma 6.2, the function is uniformly bounded from the above on every compact subset of W .
Observe that this statement is not completely trivial as is not shown to be upper semi continuous.