Complex Variables and Elliptic Equations An International Journal ISSN: 1747-6933 (Print) 1747-6941 (Online) Journal homepage: http://www.tandfonline.com/loi/gcov20 A note on uniqueness boundary of holomorphic mappings Van Thu Ninh & Ngoc Khanh Nguyen To cite this article: Van Thu Ninh & Ngoc Khanh Nguyen (2016): A note on uniqueness boundary of holomorphic mappings, Complex Variables and Elliptic Equations, DOI: 10.1080/17476933.2016.1225202 To link to this article: http://dx.doi.org/10.1080/17476933.2016.1225202 Published online: 22 Aug 2016 Submit your article to this journal Article views: View related articles View Crossmark data Full Terms & Conditions of access and use can be found at http://www.tandfonline.com/action/journalInformation?journalCode=gcov20 Download by: [Northern Illinois University] Date: 29 August 2016, At: 20:01 COMPLEX VARIABLES AND ELLIPTIC EQUATIONS, 2016 http://dx.doi.org/10.1080/17476933.2016.1225202 A note on uniqueness boundary of holomorphic mappings Van Thu Ninh and Ngoc Khanh Nguyen Department of Mathematics, Vietnam National University at Hanoi, Hanoi, Vietnam ABSTRACT ARTICLE HISTORY In this paper, we prove Huang et al.’s conjecture stated that if f is a holomorphic function on + := {z ∈ C : |z| < 1, Im(z) > 0} with C ∞ -smooth extension up to ( − 1, 1) such that f maps ( − 1, 1) into a cone C := {z ∈ C : |Im(z)| ≤ C|Re(z)|}, for some positive number C, and f vanishes to infinite order at 0, then f vanishes identically In addition, some regularity properties of the Riemann mapping functions on the boundary and an example concerning Huang et al.’s conjecture are also given Received May 2016 Accepted 12 August 2016 COMMUNICATED BY Y Antipov KEYWORDS Riemann mapping function; infinite order; analytic cusp AMS SUBJECT CLASSIFICATIONS Primary: 32H12; Secondary: 32A10 Introduction Let be a domain in Rn with a ∈ ∂ A continuous function f : infinite order at a if, for every N ∈ N, lim x→a → C vanishes to f (x) = |x − a|N In 1991, Lakner [1] proved the following result Theorem 1: [1] Suppose that f is a holomorphic function on the upper half disc + := {z ∈ C : |z| < 1, Im(z) > 0} that extends continuously to the diameter ( − 1, 1), such that the extension maps ( − 1, 1) to a cone C := {z ∈ C : |Im(z)| ≤ C|Re(z)|} for some C > If f |(−1,1) has an isolated zero at the origin, then f vanishes to finite order at It is known that the function √ f (z) = exp ( − eiπ/4 / z) is holomorphic on + , extends C ∞ -smoothly to + , and vanishes to infinite order at (see [1]) Hence, this example shows that the condition that f maps ( − 1, 1) to a cone C is essential Baouendi and Rothschild [2] obtained the following result in which the condition f |(−1,1) has an isolated zero at is not necessary CONTACT Van Thu Ninh thunv@vnu.edu.vn © 2016 Informa UK Limited, trading as Taylor & Francis Group V T NINH AND N K NGUYEN Theorem 2: [2] Let f be a holomorphic function defined on the upper half disc + that extends continuously to the diameter ( − 1, 1) Assume that Ref (x) ≥ for every x = Re(z) ∈ ( − 1, 1) Then f has the boundary unique continuation property in the sense that if f vanishes to infinite order at 0, then f ≡ Huang et al [3], posted the following conjecture Conjecture: [Huang et al.’s conjecture] Let + be the upper half disc in C Assume that f is a holomorphic function on + with continuous extension up to ( − 1, 1) such that f maps ( − 1, 1) into C , for some positive C If f vanishes to infinite order at 0, then f vanishes identically Notice that if C = 1, then Re[f (x)] ≥ for every x ∈ ( − 1, 1) Therefore, it follows from Theorem that this conjecture is true in case C ≤ In this paper, we prove the following theorem which ensures that Huang et al.’s conjecture is true for the case that f is C ∞ -smooth up to the boundary Theorem 3: Suppose that f ∈ Hol( + )∩C ∞ ( + ) and f (−1, 1) ⊂ ∞ := C\iR ∪{0} Suppose also that the set of zeros of f |(−1,1) is discrete and its limit point is If f vanishes to infinite order at 0, then f ≡ Remark 1: The theorem is still true if is replaced by any domain (C\L) ∪ {0}, where L is a line in the complex plane passing through the origin Throughout this paper, we assume that f is holomorphic in + , C ∞ -smooth up to ( − 1, 1) The set of zeros of f on ( − 1, 1) is discrete and its limit point is Each zero of f on ( − 1, 1)\{0} is of finite order Let {rn } ⊂ R+ be a decreasing sequence such that all zeros of f in + lie in ∪n γrn , where γr := {z ∈ C : |z| = r, Im(z) ≥ 0} is an upper semicircle with radius r > Denoted by κ(n) the number of zeros of f on γrn ∩ + , counting multiplicities, by κ(n) ˜ the number of zeros of f on γrn ∩ ( − 1, 1), counting multiplicities, and by κ (n) the number of zeros of f on γrn ∩ ( − 1, 1) without counting multiplicities In addition, denote An := {reiθ : rn+1 < r < rn , ≤ θ ≤ π} Let us recall that the index of a piecewise smooth curve γ : [a, b] → C∗ with respect to is the real number Ind(γ ) := Re 2πi γ dz = Re z 2πi b a γ (t) dt γ (t) Notice that Ind(γ ) < 1/2 if γ ⊂ ∞ \{0} and Ind(γ ) < if γ ⊂ C∗ \{iy : y > 0} For more properties of indices we refer the reader to [1] Remark 2: Let f be a holomorphic function as in Theorem Suppose that f vanishes to infinite order at b ∈ (−1, 1)\{0} with the maximum modulus among zeros of f on (−1, 1) Noting that b is an isolated zero of f Thus, by [1, Lemma 2] (or Lemma in Section 2), there exists a sequence of upper semi-circles {γn } centred on b and of radii n such that + n → and Ind(f ◦ γn ) → +∞ as n → ∞ Moreover, one can choose a sequence of upper semi-circles {γn } centred on −b and of radii n such that n → 0+ and {Ind(f ◦ γn )} is bounded from blow Now, fix r, r > with r < |b| < r and with |r − r | small enough We can construct a closed path γ = γr + ( − γr ) + ( − γn ) + ( − γn ) + 4j=1 ln,j , where ln,1 = [−r, −b − n ], ln,2 = [−b + n , −r ], ln,3 = [r , b − n ], and ln,4 = [b + n , r] Then, applying Cauchy’s theorem to f /f , one obtains that COMPLEX VARIABLES AND ELLIPTIC EQUATIONS = Re 2πi γ f (z) dz f (z) = Ind(f ◦ γr ) − Ind(f ◦ γr ) + Ind(f ◦ lj ) − Ind(f ◦ γn ) − Ind(f ◦ γn ) j=1 By the above argument, one has Ind(f ◦ γn ) + Ind(f ◦ γn ) → +∞ Moreover, since f ◦ ln,j ⊂ ∞ , for ≤ j ≤ 4, one has Ind(f ◦ ln,j ) < j=1 This is a contradiction since |Ind(f ◦ γr ) − Ind(f ◦ γr )| is finite Therefore, without loss of generality, we may assume that f vanishes to infinite order only at Remark 3: In Theorem 3, the condition that f ∈ C ∞ ( + ) is the only technical condition but is important for showing the existence of left and right hand limits I(rn± ) (see the notation I(r) below) Remark 4: [Notations] Throughout the paper, taking the risk of confusion we employ the following notations (i) C := {z ∈ C : |Im(z)| ≤ C|Re(z)|}, C > 0; (ii) ∞ := {z ∈ C : Re(z) = 0} ∪ {0}; + := {z ∈ C : |z| < 1, Im(z) > 0}; (iii) (iii) R+ := {x ∈ R : x > 0}; (iv) I(r) := Ind(f ◦ γr ) (r > 0), where γr := {z ∈ C : |z| = r, Im(z) ≥ 0} Let f , g : A → C be functions defined on A ⊂ C with ∈ A such that limz→0 f (z) = limz→0 g(z) = We write (v) f ∼ g at on A if limz→0 f (z)/g(z) = 1; (vi) f ≈ g at on A if there is C > such that 1/C|g(z)| ≤ |f (z)| ≤ C|g(z)| for all z ∈ A This paper is organized as follows In Section 2, we give several lemmas needed later and then we prove the Theorem Some regularity properties of the Riemann mapping functions on the boundary and an example are given in Section Finite order vanishing of boundary values of holomorphic functions In this section, we shall prove some technical lemmas which generalize Lakner’s lemmas (cf [1]) The proofs follow very closely the proofs of [1] In fact, M Lakner proved his lemmas for the case when f has no zeros in ( − 1, 1), except the origin In this paper, we consider the case when the origin is not an isolated zero of the restricted function f |(−1,1) Therefore, in order to prove these lemmas, we shall modify Lakner’s method for the general case First of all, we have the following lemma 4 V T NINH AND N K NGUYEN Lemma 1: Suppose that f ∈ Hol( + )∩C ∞ ( + ) and f (−1, 1) ⊂ := C\iR+ Suppose, also that the set of zeros of f |(−1,1) is discrete, its limit point is 0, and each zero of f on ( − 1, 1)\{0} is of finite order Then, we have ˜ (i) I(rn+ ) − I(rn− ) = κ(n) + κ(n) ; (ii) |I(r) − I(r )| < 2, rn+1 < r, r < rn Proof: (i) For each n we can write f in the form f (z) = (z − α1 )l1 · · · (z − αm )lm ϕ(z), where α1 , , αm are all zeros of f on γrn and ϕ is a continuous function without zeros in An−1 ∪ An ∪ γrn and holomorphic in the interior Therefore, we have m lj ϕ (z) f (z) = , + f (z) z − αj ϕ(z) j=1 and thus m lj Ind(γr − αj ) + Ind(ϕ ◦ γr ) I(r) = j=1 Fix n ∈ N, The limits limr→rn+ Ind(γr − a) and limr→rn− Ind(γr − a) are calculated at the point a ∈ + with |a| = rn as follows: lim Ind(γr − a) = 3/4 and lim Ind(γr − a) = −1/4 r→rn+ r→rn− Moreover, it is also calculated that lim Ind(γr − b) = 1/2 and lim Ind(γr − b) = r→rn+ r→rn− for b ∈ γrn ∩ ( − 1, 1) Therefore, one obtains ˜ + Ind(ϕ ◦ γrn ), I(rn+ ) = lim IndI(r) = 3κ(n)/4 + κ(n)/2 I(rn− ) r→rn+ = lim IndI(r) = −κ(n)/4 + Ind(ϕ ◦ γrn ) r→rn− Hence, we conclude that I(rn+ ) − I(rn− ) = κ(n) + κ(n)/2 ˜ (ii) Fix rn+1 < r < r < rn and construct a closed path γ = γr + ( − γr ) + l + + l − , where l + = [r , r] and l − = [−r, −r ] Employing Cauchy’s theorem to f /f we obtain = Re 2πi γ f dz = I(r) + Ind(f ◦ l + ) − I(r ) + Ind(f ◦ l − ) f Since, the paths f ◦ l ± ⊂ < , |Ind(f ◦ l ± )| < This implies that |I(r) − I(r )| COMPLEX VARIABLES AND ELLIPTIC EQUATIONS Lemma 2: Suppose that f ∈ Hol( + ) ∩ C ∞ ( + ) and f ( − 1, 1) ⊂ := C\iR+ ∪ {0} Suppose also that the set of zeros of f |(−1,1) is discrete, its limit point is 0, and each zero of f on ( − 1, 1)\{0} is of finite order Then, I(r) is bounded from above if one of the following conditions is satisfied: (i) is the cone (C\L) ∪ {0}, where L is a line passing through the origin (ii) Each zero of f on ( − 1, 1) is of order at least ∞ ∞ ˜ − ∞ (iii) n=1 κ(n) + n=1 κ(n) n=1 κ (n) = +∞ (iv) is a half-plane {z ∈ C : Re(az) ≥ 0} for some a ∈ C∗ Proof: In the case when the origin is an isolated zero of f , one easily see that I(r) is bounded Therefore, we consider the case when the set of zeros of f in + is a sequence of points converging to the origin Let {rn } ⊂ R+ be a decreasing sequence such that all zeros of f in + lie in ∪γrn Now, + + l− , for > small enough we consider the closed path γn = γrn − + ( − γrn+1 + ) + ln, n, + − where ln, = [rn+1 + , rn − ] and ln, = [−rn + , −rn+1 − ] Then, employing Cauchy’s theorem to f /f one has = Re 2πi γn f + − dz = −I(rn+1 + ) + Ind(f ◦ ln, ) + I(rn − ) + Ind(f ◦ ln, ) f Letting → 0+ , one obtains that + ) + s(n) = 0, I(rn− ) − I(rn+1 + ) + Ind(f ◦ l − ) where s(n) := lim →0+ Ind(f ◦ ln, n, From this and Lemma (i), we have + κ(n) + κ(n)/2 ˜ = I(rn+ ) − I(rn+1 ) + s(n) (1) Summing the first N relations (1) one gets + )= I(r1+ ) − I(rN+1 N κ(n) + n=1 N N κ(n) ˜ − n=1 s(n) (2) n=1 N Now, fix N ≥ We shall prove that N n=1 s(n) ≤ n=1 κ (n) for the case f ( − 1, 1) ⊂ N N + C\iR and n=1 s(n) ≤ 1/2 n=1 κ (n) for the case f ( − 1, 1) ⊂ (C\L) ∪ {0}, where L is a line passing through the origin Indeed, suppose that there exist i, j ∈ N∗ with i < j such that f (ri ) = f (rj ) = and f (rk ) = for every i < k < j Then, if f ( − 1, 1) ⊂ C\iR+ , then j−1 + Ind(f ◦ lk, ) = lim Ind(f ◦ [rj + , ri − ]) ≤ lim →0+ k=i →0+ V T NINH AND N K NGUYEN Since, f ◦ [rj , ri ] ⊂ C\iR+ Similarly, if f ( − 1, 1) ⊂ (C\L) ∪ {0}, then j−1 + Ind(f ◦ lk, ) = lim Ind(f ◦ [rj + , ri − ]) ≤ lim →0+ →0+ k=i since f ◦ [rj , ri ] is contained in a half-plane Furthermore, one can consider the sequence of points {−rj } instead of {rj } and one thus obtains similar estimates Hence, these estimates yield our assertions ∞ ∞ Therefore, if ∞ ˜ − ∞ n=1 κ(n) + n=1 κ(n) n=1 κ (n) = +∞, then n=1 κ(n) + ∞ ∞ + κ(n) ˜ − s(n) = +∞, and thus by (2) one has I(r ) → −∞ Consequently, n n=1 n=1 by Lemma (ii), it follows that I(r) is bounded from above This proves the assertion for the case (iii) + ˜ ≥2 N For the case (ii), since N n=1 κ(n) n=1 κ (n), it follows from (2) that either {I(rn )} + is bounded or I(rn ) → −∞ as n → ∞ So by Lemma 2, I(r) is bounded from above Next, if is the infinite cone (C\L) ∪ {0}, where L is a line passing through the origin, then N N N 1 s(n) ≤ κ (n) ≤ κ(n) ˜ 2 n=1 n=1 n=1 Therefore, by (2) and Lemma we also have I(r) bounded from above, and hence, i/ is proved N Finally, if is a half-plane, then N n=1 s(n) ≤ 1/2 n=1 κ (n) for any N ≥ Hence, (2) implies that I(r) is bounded from above Thus, this proves iv/ The following lemma is a generalization of Lemma in [1] Lemma 3: Suppose that f ∈ Hol( + ) ∩ C ∞ ( + ) and f ( − 1, 1) ⊂ := C\iR+ Suppose also that the set of zeros of f |(−1,1) is discrete and its limit point is If f vanishes to infinite order only at 0, then I(t) dt = +∞ lim sup t r→0+ ln (1/r) r Proof: Without loss of generality, we may assume that there exists a sequence of zeros of f converging to the origin Let {rn } ⊂ R+ be a sequence with rn → 0+ , such that all zeros of f lie in ∪γrn Denote An := {reit : ≤ t ≤ π, rn+1 < r < rn } Then, there exists a holomorphic function (z) := un (z) + ivn (z) such that f (z) = e (z) on each An One can see that un (z) = ln |f (z)| on An Hence, we have I(r) := Ind(f ◦ γr ) 1 f (z) dz = Re (z)dz = Re 2πi γr f (z) 2πi γr 1 (reiπ ) − (r) = (reiπ ) − (r) = Re 2πi 2π π d π ∂ 1 = (r, t)dt = (r, θ)dθ 2π dt 2π ∂θ π ∂ = r un (r, θ)dθ 2π ∂r (3) COMPLEX VARIABLES AND ELLIPTIC EQUATIONS By Lemma 1, I(r) is piecewise continuous on (0, 1], and thus it is integrable on [r, 1] Let us denote J(r) := 1 ln 1/r r I(t) dt t Then, (3) yields that J(r) = 1 ln 1/r = lim →0+ = lim →0+ + I(t) dt t ⎡ r ⎣ ln 1/r ln 1/r rn0 e− r π t = lim →0+ + π rn0 e− r I(t) dt + t r 0, ˆ ∀p ∈ B ˆ which is C 0,α up to Then, by Corollary 2, one can find a biholomorphism ψ : + → B, + 0,α the boundary, such that (f ◦ ψ) ∈ Hol( ) ∩ C ( + ), and such that (f ◦ ψ) vanishes to infinite order at Therefore, Theorem tells us that f ≡ Now, we assume that f ∈ Hol( + ) ∩ C ∞ ( + ) and f vanishes to infinite order at Then, it follows from Corollary that there exists a sequence {pj } ⊂ + such that pj → and Re f (pj ) = In addition, Daghighi and Krantz [12] recently proved that either f ≡ or there is a sequence in + , converging to 0, along which Im(z)/Re(z) is unbounded It is a natural question whether there exists a subdomain Bˆ with ∈ ∂ Bˆ such that ˆ However, the following example points out that one cannot Re(f (z)) = for all z ∈ B remove the condition f ( − 1, 1) ⊂ ∞ in Theorem and there is no such a subdomain ˆ Bˆ ⊂ + with ∈ ∂ Bˆ and Re(f (z)) = on B In order to introduce the example, we need the following well-known result (with possible repetitions) satisfies Theorem 5: [See Theorem 9.1.5 in [13]] If {aj }∞ j=1 ⊂ ∞ (1 − |aj |) < +∞ j=1 which has zero set and no aj = 0, then there is a bounded holomorphic function on consisting precisely of the aj ’s, counted according to their multiplicities Specifically, the infinite product ∞ j=1 −aj Ba (z) |aj | j converges uniformly on compact subsets of to a bounded holomorphic function B(z), z − aj The zeros of B are precisely the aj ’s, counted according to their where Baj (z) = − a¯ j z multiplicities Example 2: Denote by H := {z ∈ C : Im(z) > 0} the upper half-plane and consider the sequence {am,n }m∈Z,n∈N∗ ⊂ H given by am,n = m3 ∈ H − in3 12 V T NINH AND N K NGUYEN Let ψ : H → be the biholomorphism defined by ψ(z) = z−i z+i and let us denote αm,n := ψ(am,n ) = − n3 − im3 − m6 − n6 − 2im3 = + n3 + im3 (1 + n3 )2 + m6 Then, a computation shows that (m6 + n6 + 2n3 + 1)2 − (m6 + n6 − 1)2 − 4m6 (m6 + n6 + 2n3 + 1)2 6 4(m + n + n3 )(n3 + 1) − 4m6 = (m6 + n6 + 2n3 + 1)2 4n (m6 + n6 + n3 ) + 4(n6 + n3 ) = (m6 + n6 + 2n3 + 1)2 4(n3 + 1) < m + n6 + 2n3 + < n +1 − |αm,n |2 = Hence, it yields that (1 − |αm,n |) < +∞ m,n Now, we define the holomorphic function G : H → C by ∞ G(w) = n=1 Then the function F := G ◦ ψ : H → the function ∞ −αm,n Bα (z) |αm,n | m,n m=−∞ satisfies |F(z)| ≤ for every z ∈ H Therefore, f := e iπ/4 − e √z F(z) is holomorphic in H, continuous in H and it vanishes to infinite order at z = Moreover, the origin is only the isolated zero of f |∂H Acknowledgements This work was completed when the first author was visiting the Vietnam Institute for Advanced Study in Mathematics (VIASM) He would like to thank the VIASM for financial support and hospitality Disclosure statement No potential conflict of interest was reported by the authors COMPLEX VARIABLES AND ELLIPTIC EQUATIONS 13 Funding This research is funded by the Vietnam National University, Hanoi (VNU) under project number [QG.16.07] References [1] Lakner M Finite order vanishing of boundary values of holomorphic mappings Proc Am Math Soc 1991;112:521–527 [2] Baouendi MS, Rothschild LP Unique continuation and a Schwarz reflection principle for analytic sets Commun Part Differ Equ 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[9,10] One says that has an analytic cusp at if the boundary of at has two regular analytic curves such that the opening angle of at vanishes It is shown in [9] that, after a change of variable... Krantz SG A note on a conjecture concerning boundary uniqueness Complex Var Elliptic Equ 2016;60:945–950 [12] Daghighi A, Krantz S Erratum: a note on a conjecture concerning boundary uniqueness. .. Mathematics, Vietnam National University at Hanoi, Hanoi, Vietnam ABSTRACT ARTICLE HISTORY In this paper, we prove Huang et al.’s conjecture stated that if f is a holomorphic function on + :=