Journal of Mathematical Physics, Analysis, Geometry 2013, vol. 9, No. 3, pp. 304–315 On Analytic and Subharmonic Functions in Unit Disc Growing Near a Part of the Boundary S.Ju. Favorov and L.D. Radchenko V.N. Karazin Kharkiv National University 4 Svobody Sq., Kharkiv 61077, Ukraine E-mail: [email protected] [email protected] Received January 17, 2012, revised December 20, 2012 In the paper there was found an analog of the Blaschke condition for analytic and subharmonic functions in the unit disc, which grow at most as a given function ϕ near some subset of the boundary. Key words: analytic function, subharmonic function, Riesz measure, Blaschke condition. Mathematics Subject Classification 2010: 30D50, 31A05. 1. Introduction It is well known that the zeroes {z } of any bounded analytic function in the n unit disk D satisfy the Blaschke condition (cid:88) (1−|z |) < ∞. (1) n There have been a number of papers published where there can be found the analogous conditions for various classes of unbounded analytic and subharmonic functions (see, for example, [6–9] ). In [5], there was studied the case of analytic functions in D of an exponential growth near a finite subset E ⊂ ∂D. In [1], the case of an arbitrary compact subset E ⊂ ∂D was considered. Namely, there were considered subharmonic functions v in D such that v(z) ≤ dist(z,E)−q, z ∈ D, (2) 1 for some 0 < q < ∞. The Riesz measures (generalized Laplacians) µ = (cid:52)v 2π are proven to satisfy the condition (cid:90) (1−|λ|)(dist(λ,E))(q−α)+dµ(λ) < ∞ (3) D (cid:176)c S.Ju. Favorov and L.D. Radchenko, 2013 On Analytic and Subharmonic Functions in Unit Disc for α such that (cid:90)2 m{s ∈ ∂D : dist(s,E) < t}dt < ∞. (4) tα+1 0 Here x = max{0,x}, and m is the normalized Lebesgue measure on ∂D, that is, + m(∂D) = 1. If m(E) > 0, then (3) and (4) are valid for any α < 0. If (4) is valid with some α > 0 and q ≤ α, then µ satisfies the Blaschke condition for bounded from above in D subharmonic functions (cid:90) (1−|λ|)dµ(λ) < ∞. (5) It was also proved in [1] that (3) is not valid for the subharmonic function v (z) = dist(z,E)−q in the case of divergent integral in (4). 0 If f(z) is an analytic function, then log|f(z)| is a subharmonic function with (cid:80) the Riesz measure k δ , where δ are the unit masses in the zeros {z } of n n zn zn n f(z), and k are the multiplicities of the zeros. Hence, if this is the case, (2) has n the form |f(z)| ≤ expdist(z,E)−q, (6) while (3) has the form (cid:88) (1−|z |)dist(z ,E)(q−α)+ < ∞. (7) n n zn Note that condition (6) seems to be too restrictive to be applied to the op- erator theory [1]. It is natural to consider subharmonic (and analytic) functions such that v(z) (cid:54) ϕ(dist(z,E)), (8) where ϕ(t) is the monotonically decreasing on R+ function, ϕ(t) → +∞ as t → +0. It is clear that for any subharmonic function v in D, which grows as dist(z,E) → 0, condition (8) is valid for ϕ(t) = sup{v(z) : dist(z,E) ≥ t}. The Main Results To formulate our results we need some notations. Let ρ(z) = dist(z,E), F(t) = F (t) = m{ζ ∈ ∂D : ρ(ζ) < t}. E Consider (cid:90)2 I(ϕ,E) := ϕ(s)dF(s). 0 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 305 S.Ju. Favorov and L.D. Radchenko Note that F(t) is continuous on [0,2] and has a discontinuity at t = 0 if and only if m(E) > 0. Theorem 1. Let E be a closed subset of ∂D, v be a subharmonic function in D, v (cid:54)≡ −∞, and v(z) satisfy (8). If I(ϕ,E) < ∞, then the Riesz measure µ of the function v(z) satisfies (5). E x a m p l e 1. Let E = {ζ ,...,ζ }, ϕ(t) = t−1log−α(1/t), α > 1. 1 k It is readily seen that F(t) = 2kt/(2π) + o(1) as t → 1. The assumption of Theorem 1 is fulfilled and the Riesz measure of any subharmonic function v(z) ≤ ρ−1(z)log−α(1/ρ(z)) satisfies the Blaschke condition (5). Theorem 2. Let E, v and µ satisfy the assumption of Theorem 1, ϕ: R+ → R+ be an absolutely continuous nonnegative monotonically decreasing function, ϕ(t) → +∞ as t → +0, and ψ(t) be an absolutely continuous monotonically increasing function on [0,2] such that ψ(0) = 0. If 1(cid:90)/25 ψ(50y)ϕ(cid:48)(y)F(y)dy > −∞, (9) 0 then (cid:90) ψ(ρ(λ))(1−|λ|)dµ(λ) < ∞. (10) D R e m a r k 1. In the case m(E) > 0, we always get I(ϕ,E) = ∞ and thus the assumption of Theorem 1 is not satisfied. Further, (9) takes the form 1(cid:90)/25 ψ(50y)ϕ(cid:48)(y)dy > −∞. 0 In the case E = ∂D, the result coincides with that obtained in [8]. E x a m p l e 2. Let E = {ζ ,...,ζ }, ϕ(t) = e1/t. The assumption of 1 k Theorem 2 is satisfied with ψ(t) = e−c/t for c > 50. Hence, if v is a subharmonic function v(z) such that v(z) ≤ exp(1/ρ(z)), then for c > 50 the integral (cid:90) exp(−c/ρ(z))(1−|λ|)dµ(λ) converges. The following theorem shows the accuracy of Theorem 2. 306 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 On Analytic and Subharmonic Functions in Unit Disc Theorem 3. Let E, ϕ, ψ be the same as above and, in addition, ϕ(1/t) be convex with respect to logt. If (cid:90)2 ψ(y)ϕ(cid:48)(y)F(y)dy = −∞, (11) 0 then for the Riesz measure µ of the function v (z) = ϕ(ρ(z)) we get 0 0 (cid:90) ψ(ρ(λ))(1−|λ|)dµ (λ) = +∞. 0 Now consider the case v(z) = log|f(z)| with the analytic function f(z). Theorem 1(cid:48). Let E be a closed subset of ∂D, ϕ(t) be an absolutely continuous nonnegative monotonically decreasing function on R+, ϕ(t) → +∞ as t → +0, I(ϕ,E) < ∞, f be an analytic function in D with zeros z . If n |f(z)| (cid:54) exp(ϕ(ρ(z)), then its zeros satisfy the Blaschke condition (1). Theorem 2(cid:48). Let E, f, ϕ be the same as in Theorem 1(cid:48), but the condition I(ϕ,E) < ∞ for some absolutely continuous monotonically increasing function ψ on [0,2] such that ψ(0) = 0 be replaced by (9). Then (cid:88) ψ(ρ(z ,E))(1−|z |) < ∞. n n zn R e m a r k 2. For ϕ(t) = t−q, ψ(t) = ts, Theorems 1–3, 1(cid:48), 2(cid:48) were proved earlier in [1]. 2. Demonstrations We begin with the lemmas. Lemma 1. Let ν be a nonnegative finite measure on X, g(x) be a Borel function on X, ϕ(t) be a Borel function on R. Then (cid:90) (cid:90)∞ (ϕ◦g)(x)dν(x) = ϕ(y)dH(y), X −∞ where H(y) = ν{x : g(x) < y}. The proof of this lemma can be easily reduced to the case ν(X) = 1 which is well known (see, for example, [2, formula (15.3.1)]). Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 307 S.Ju. Favorov and L.D. Radchenko Lemma 2. For each z ∈ D and τ ∈ [0,1] we have ρ(z) (cid:54) 2ρ(τz). P r o o f. Consider ζ = z/|z| and z(cid:48) ∈ [0,ζ] such that ρ(z(cid:48)) = dist(E,[0,ζ]). The cases z(cid:48) = 0 and z ∈ [0,z(cid:48)] are trivial. Suppose z ∈ (z(cid:48),ζ). We have ρ(z) (cid:54) ρ(z(cid:48))+|z(cid:48)−z| (cid:54) ρ(z(cid:48))+|z(cid:48)−ζ|. Consider ζ(cid:48) ∈ E such that ρ(z(cid:48)) = |z(cid:48) −ζ(cid:48)| and the triangle with vertexes on z(cid:48), ζ, ζ(cid:48). The angle in z(cid:48) is right, and the angle in ζ is greater than π/4. Thus, |z(cid:48)−ζ| (cid:54) |z(cid:48)−ζ(cid:48)| and therefore ρ(z) (cid:54) 2ρ(z(cid:48)) ≤ 2ρ(τz). Note that the result of the lemma was used in [1, p.41] without proof. P r o o f of Theorem 1. First, suppose v(0) = 0. Let I(ϕ,E) < ∞. Using Lemma 1 with the measure m(ζ) on ∂D and taking into account the equalities F(y) ≡ 0 for y < 0 and F(y) ≡ 1 for y > 2, we get (cid:90) (cid:90)∞ (cid:90)2 (ϕ◦ρ)(ζ)dm(ζ) = ϕ(y)dF(y) = ϕ(y)dF(y) = I(ϕ,E) < ∞. (12) ∂D 0 0 First prove that there exists a harmonic majorant for v in D. Using the properties of the Poisson integral (cid:90) 1−|z|2 U(z) = ϕ(ρ(ζ))dm(ζ), |ζ −z|2 ∂D we obtain limU(z) = ϕ(ρ(ζ)), ζ ∈ ∂D\E. z→ζ Hence lim (v(z)−U(z)) (cid:54) 0 for ζ ∈ ∂D\E and lim U(z) = +∞. z→ζ z→ζ∈E Let Ω , t ∈ (0,1), be the connected component of the set {z ∈ D : ρ(z) > t} t such that 0 ∈ Ω . Put t Γ = {z ∈ D : ρ(z) = t}, E = {ζ ∈ ∂D : ρ(z) < t}, Ec = ∂D\E . t t t t Since E is a finite union of disjoint open sets, it is seen that Ec is a finite union t t of disjoint closed sets. Moreover, ∂Ω ⊂ Ec ∪Γ . Note that for the connected t t t {z ∈ D : ρ(z) > t} we have ∂Ω = Ec∪Γ . t t t Forz ∈ Γ , takeζ(cid:48) ∈ E such that|z−ζ(cid:48)| = t. Putγ = {ζ ∈ ∂D : |ζ−ζ(cid:48)| (cid:54) t}. t z Clearly, γ ⊂ E . Since the function ϕ(t) is positive and ϕ(ρ(ζ)) ≥ ϕ(t) on E , z t t we get (cid:90) 1−|z|2 U(z) ≥ ϕ(ρ(ζ))dm(ζ) ≥ ϕ(t)ω(z,γ ,D), |ζ −z|2 z γz 308 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 On Analytic and Subharmonic Functions in Unit Disc where (cid:90) 1−|λ|2 ω(λ,γ ,D) = dm(ζ) z |ζ −λ|2 γz is the harmonic measure at the point λ = z with respect to the arc γ (see, for z example, [3, §4.3]). On the other hand, the harmonic measure at any pointλ ∈ D with respect to the arc γ equals β/(2π), where β = β(λ) is the length of the arc z of the unit circle formed by the strait lines passing through λ and the ends of γ z (see [10, chapter I, §5]). By direct calculations, β(z) = π −2arcsin(t/2). Since 0 < t ≤ 1, we obtain ω(z,γ ,D) ≥ 1/3 and z 1 v(z) U(z) (cid:62) ϕ(t)ω(z,γ ,D) (cid:62) ϕ(t) (cid:62) , z ∈ Γ . z t 3 3 Therefore, lim [v(z)−3U(z)] ≤ 0, ζ ∈ Γ , lim [v(z)−U(z)] ≤ 0, ζ ∈ Ec. z→ζ t z→ζ t Using the maximum principle, we get v(z) (cid:54) 3U(z) for all z ∈ Ω . t Furthermore, by the Green formula [3, Theorem 4.5.4], we have (cid:90) v(z) = u (z)− G (z,λ)dµ(λ), (13) t Ωt Ωt where u is the least harmonic majorant of v in Ω . We have u (z) (cid:54) 3U(z) for t t t z ∈ Ω . The Green function G (z,λ) in Ω is equal to t Ωt t G (z,λ) = log1/(|z−λ|)−h (z,λ), Ωt t where h (z,λ) is the harmonic function in Ω with the boundary values log1/|z− t t λ|. By [1, proof of Theorem 1], G (0,λ) ≥ (1−|λ|)/6 for λ ∈ Ω with k = 25 > Ωt kt 6π+3. Consequently, (cid:90) (cid:90) (1−|λ|)dµ(λ) (cid:54) 6 G (0,λ)dµ(λ) = 6u (0) Ωt t Ωkt Ωt (cid:90) (cid:54) 18U(0) = 18 ϕ(ρ(ζ))dm(ζ) = 18I(ϕ,E). ∂D The later inequality is valid for all t > 0, hence we obtain the statement of the theorem for the case v(0) = 0. If v(0) > 0, consider the function v(z)−v(0) instead of v(z). v(z)−v(0) If −∞ < v(0) < 0, consider the function v (z) = ϕ(2) . We have 1 ϕ(2)−v(0) 1−v(0)/ϕ(ρ(z)) v (z) (cid:54) ϕ(ρ(z)) (cid:54) ϕ(ρ(z)). 1 1−v(0)/ϕ(2) Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 309 S.Ju. Favorov and L.D. Radchenko Notice that the Riesz measure of the function v (z) coincides with the Riesz 1 measure of the function v(z) up to a constant depending only on v(0). Hence the Blaschke condition (5) for v implies the same condition for v. 1 If v(0) = −∞, consider the harmonic function h(z) in the disk {|z| < 1/2} such that h(z) = v(z) for |z| = 1/2 and put (cid:40) max(v(z),h(z)),|z| < 1 v (z) = 2 1 v(z),|z| (cid:62) 1. 2 Clearly, v (z) (cid:54) max v(z) (cid:54) ϕ(1/2) for |z| ≤ 1/2 and v (0) (cid:54)= −∞. 1 1 |z|=1/2 Inaddition, v (z)issubharmonicinD(see[3, Theorem2.4.5])andtherestriction 1 of its Reisz measure µ to the set {z ∈ D : |z| > 1} is equal to µ. Applying the 1 2 proved statement to the function ϕ (z) = max{ϕ(z),ϕ(1/2)}, we obtain 1 (cid:90) (1−|λ|)dµ (λ) < ∞. 1 D Therefore the integral in (5) is also finite. Theorem 1 is proved. P r o o f of Theorem 2. Arguing as above, we can consider only the case v(0) = 0. Let Ω , Γ , E , Ec, γ , ω(λ,γ ,D) be the same as in the proof of Theorem 1. t t t t z z For z ∈ D, put (cid:90) (cid:90) 1−|z|2 1−|z|2 V (z) = ϕ(ρ(ζ))dm(ζ)+ϕ(t) dm(ζ). (14) t |ζ −z|2 |ζ −z|2 Etc Et Since γ ⊂ E , we get z t 1 V (z) (cid:62) ϕ(t)ω(z,γ ,D) (cid:62) ϕ(t). t z 3 Therefore, limsupv(z) (cid:54) lim3V (z) = 3V (ζ), ζ ∈ ∂Ω . t t t z→ζ z→ζ Using the maximum principle, we get v(z) (cid:54) 3V (z) for all z ∈ Ω , in particular, t t v(0) (cid:54) 3V (0). Clearly, we have t (cid:90) (cid:90) (cid:90) V (0) = V (ζ)dm(ζ) = ϕ(t)dm(ζ)+ ϕ(ρ(ζ))dm(ζ) t t ∂D Et (cid:90) Etc = ϕ(t)F(t)+ ϕ(ρ(ζ))dm(ζ). Ec t 310 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 On Analytic and Subharmonic Functions in Unit Disc Using Lemma 1 with g(ζ) = ρ(ζ) and taking into account the equality H(y) = m{ζ : ρ(ζ) < y}−m{ζ : ρ(ζ) (cid:54) t} = F(y)−F(t), we get (cid:90) (cid:90)2 ϕ(t)F(t)+ ϕ(ρ(ζ))dm(ζ) = ϕ(t)F(t)+ ϕ(y)dF(y). Ec t t Integrating by parts, we get (cid:90)2 V (0) = ϕ(t)F(t)+ϕ(2)F(2)−ϕ(t)F(t)− ϕ(cid:48)(y)F(y)dy t t (cid:90)2 = ϕ(2)− ϕ(cid:48)(y)F(y)dy. (15) t Therefore, using the Green formula (13) and the estimate u (z) ≤ 3V (z) of the t t least harmonic majorant u in Ω , we get t t   (cid:90) (cid:90)2 G (0,λ)dµ(λ) = u (0) (cid:54) 3V (0) = 3ϕ(2)− ϕ(cid:48)(y)F(y)dy. (16) Ωt t t Ωt t By [1, proof of Theorem 1], G (0,λ) ≥ (1−|λ|)/6 for λ ∈ Ω with k = 25 > Ωt kt 6π+3. Therefore,   (cid:90) (cid:90)2 (1−|λ|)dµ(λ) (cid:54) 18ϕ(2)− ϕ(cid:48)(y)F(y)dy (17) Ωkt t if kt ∈ (0,1). In particular, the measure (1−|λ|)dµ(λ) of the set {λ ∈ D : ρ(λ) (cid:62) ε} is finite for each ε > 0. Applying Lemma 1 with the function g = ρ and taking into account that ρ(λ) ≤ 2, we get (cid:90) (cid:90)2 ψ(ρ(λ))(1−|λ|)dµ(λ) = ψ(t)dG(t) {λ∈D:ρ(λ)(cid:62)ε} ε Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 311 S.Ju. Favorov and L.D. Radchenko (cid:82) with G(t) = (1−|λ|)dµ(λ). We have {λ∈D:ε(cid:54)ρ(λ)<t}   (cid:90)2 (cid:90)2 (cid:90)   ψ(t)dG(t) = − ψ(t)d (1−|λ|)dµ(λ) ε ε {λ:ρ(λ)>t}   (cid:90) (cid:90)2 (cid:90)   = ψ(ε) (1−|λ|)dµ(λ)+ ψ(cid:48)(t) (1−|λ|)dµ(λ)dt. (18) {λ:ρ(λ)≥ε} ε {λ:ρ(λ)>t} We claim that under the condition   (cid:90)2 (cid:90)   ψ(cid:48)(t) (1−|λ|)dµ(λ)dt < ∞ 0 {λ:ρ(λ)>t} we have (cid:90) ψ(ε) (1−|λ|)dµ(λ) → 0 as ε → 0. (19) {λ:ρ(λ)≥ε} Indeed, for any η > 0 and sufficiently small positive δ < ε < 2,   (cid:90)ε (cid:90)   ψ(cid:48)(t) (1−|λ|)dµ(λ)dt (cid:54) η. δ {λ:ρ(λ)>t} Therefore, (cid:90) (cid:90)ε (cid:90) (ψ(ε)−ψ(δ)) (1−|λ|)dµ(λ) = ψ(cid:48)(t)dt (1−|λ|)dµ(λ) {λ:ρ(λ)≥ε} δ {λ:ρ(λ)≥ε}   (cid:90)ε (cid:90)   (cid:54) ψ(cid:48)(t) (1−|λ|)dµ(λ)dt (cid:54) η. δ {λ:ρ(λ)>t} Passing to the limit δ → 0, we obtain (cid:90) ψ(ε) (1−|λ|)dµ(λ) (cid:54) η, {λ:ρ(λ)≥ε} which proves (19). 312 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 On Analytic and Subharmonic Functions in Unit Disc By Lemma 2, ifρ(z) > t, thenρ(τz) > t/2 for allz ∈ D and0 < τ < 1. Hence the interval [0,z] belongs to the set {ζ : ρ(ζ) > t/2}, and {z : ρ(z) > t} ⊂ Ω . t/2 Therefore,     (cid:90)2 (cid:90) (cid:90)2 (cid:90)     ψ(cid:48)(t) (1−|λ|)dµ(λ)dt (cid:54) ψ(cid:48)(t) (1−|λ|)dµ(λ)dt. ε {ρ(λ)>t} ε Ωt/2 By (18), to prove the convergence of the integral (cid:90) ψ(ρ(λ))(1−|λ|)dµ(λ), D it is sufficient to show the convergence of the integral     (cid:90)2 (cid:90) (cid:90)1/k (cid:90)     ψ(cid:48)(t) (1−|λ|)dµ(λ)dt = 2k ψ(cid:48)(2kt) (1−|λ|)dµ(λ)dt. 0 Ω 0 Ω t/2 kt We have   (cid:90)1/k (cid:90)2 (cid:90)2 1 ψ(cid:48)(2kt) ϕ(cid:48)(y)F(y)dydt = ψ(2) ϕ(cid:48)(y)F(y)dy 2k 0 t 1/k (cid:90)2 (cid:90)1/k ψ(2kt) 1 −lim ϕ(cid:48)(y)F(y)dy+ ψ(2ky)ϕ(cid:48)(y)F(y)dy t→0 2k 2k t 0 (cid:90)1/k 1 (cid:62) const+ ψ(2ky)ϕ(cid:48)(y)F(y)dy. 2k 0 By the condition of the theorem, the last integral is finite. The proof is complete. P r o o f of Theorem 3. Note that the function −logρ(z) is subharmonic, (cid:181) (cid:182) 1 hence the function ϕ(ρ(z)) = ϕ is subharmonic as well. e−logρ(z) Using the Green formula (13) for the function v (z) in the domain Ω , we get 0 t (cid:90) ϕ(1) = v (0) = u0(0)− (log1/|λ|−h (0,λ))dµ (λ), (20) 0 t t 0 Ωt where u0(z) is the least harmonic majorant for v (z) in Ω , and h (0,λ) is the t 0 t t solution of Dirichlet problem in Ω with the boundary values log1/|λ|. Clearly, t Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 3 313