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A counterexample to the Arakelyan Conjecture

A ``self--similar'' example is constructed that shows that a conjecture of N. U. Arakelyan on the order of decrease of deficiencies of an entire function of finite order is not true.


Introduction
Let f be an entire function and δ(a, f ) denotes the Nevanlinna deficiency of f at the point a ∈ C. Standard references are [8,9]. (No knowedge of Nevanlinna theory is necessary to understand this paper. We really deal with a problem of potential theory.) Since δ(a, f ) ≥ 0 and the deficiency relation of Nevanlinna states that a∈C δ(a, f ) ≤ 2, it follows that the set of deficient values, that is, {a : δ(a, f ) > 0}, is at most countable. We denote the sequence of deficiencies by {δ n }. In 1966 Arakelyan [2] (see also [8] or [7]) constructed the first example of an entire function of finite order having infinitely many deficient values. In this example the deficiencies satisfy ∞ n=1 1 log(1/δ n ) < ∞, (1.1) and he conjectured that (1.1) is true for every entire function of finite order. Another method of constructing such examples was proposed in [3], but the function in [3] also satisfies (1.1).
For meromorphic functions of finite order Weitsman [11] proved and this is known to be best possible [9,4]. The only known improvement of (1.2) for entire functions is due to Lewis and Wu [10]: where ǫ 0 is an absolute constant. In fact, the value ǫ 0 = 2 −264 is given in [10].
In this note we will give a construction that produces an entire function of finite order having infinitely many deficiencies δ n with the property δ n ≥ c −n , (1.4) where c > 1 is a constant. Thus Arakelyan's conjecture (1.1) fails.
Of course a substantial gap still remains between the theorem of Lewis and Wu and our example. It is natural to ask whether is true with arbitrary ǫ > 0 for entire functions of finite order.
It is more or less well known that the problem of estimating deficiencies for entire functions of finite order is equivalent to a problem of potential theory. Namely, the following statements are equivalent: A. Given any ρ > 1/2 and a sequence of complex numbers a n , there exists an entire function f of order ρ with the property δ(a n , f ) ≥ cδ n with some constant c > 0.
B. There exist a bounded subharmonic function u in the annulus A = {z : 1 < |z| < 2} and disjoint open sets E n ⊂ A, 1 ≤ n < ∞ with the following properties: We indicate briefly how to prove the equivalence. To prove A→B we take a sequence of Pólya peaks [9, p. 101] r k for log M (r, f ) and consider the sequence of subharmonic functions This sequence is precompact in an appropriate topology and we may take a subsequence that converges to a subharmonic function u. If f has deficient values then u satisfies (i) and (ii). See [1,6] for details.
To prove B→A we apply the construction from [3] that involves an extension of u to a subharmonic function in C with the property of self-similarity: u(2z) = ku(z), k = const > 0, approximation of u by the logarithm of modulus of an entire function g and performing a quasi-conformal modification on the function g that produces the entire function f satisfying A. It is also plausible that Arakelyan's original method could be applied directly as soon as a subharmonic function with the properties (i) and (ii) is constructed.
Remark. The above-mentioned paper of Lewis and Wu contains also the solution of a problem of Littlewood on the upper estimate of mean spherical derivative of a polynomial. The connection between the two problems seems somewhat obscure. An example that gives a lower estimate in the Littlewood's problem was constructed in [5] using some self-similar sets arising in the iteration theory of polynomials. It is interesting that the example we are going to construct now also has the property of self-similarity. Instead of iteration of a polynomial here the crucial role is played by a semigroup of Möbius transformations of the plane.
Denote by S + 0,0 the square S + 0,0 = {z : |ℜz| ≤ 3 10 , |ℑz − 1| ≤ 3 10 } and set S + n,k = γ n,k (S + 0,0 ). It is easy to see that the squares S + n,k are disjoint. Consider the domain The boundary ∂D 0 consists of the real axis, boundaries of the squares and horizontal line l 0 = {z : ℑz = 4/3}. The domain D 0 is Γ-invariant and the transformation z → z/2 maps D 0 onto The boundary of D 1 consists of the real axis, boundaries of some squares, and the horizontal line l 1 = {z : ℑz = 2/3} ⊂ D 0 . Let u be the harmonic function in D 0 that solves the Dirichlet problem This Dirichlet problem has a unique solution. So we conclude from translation invariance that It follows that the function u has a positive minimum M −1 < 1 on the line l 1 ⊂ D 0 . Comparing u(z) and u(2z) on ∂D 1 and using the maximum principle, we conclude that u(2z) ≤ M u(z), z ∈ D 1 , which is equivalent to It follows from (2.6) and (2.7) that Now we are going to extend u to the strip S + = {z : 0 < ℑz < 4 3 }, that is, to define u in the squares. We start by defining u in S + 0,0 . The normal derivative (in the direction of the outward normal to the boundary of the square) of u has positive infimum on ∂S + 0,0 ; it tends to +∞ as we approach a corner of the square. Denote by G > 0 the Green function for S + 0,0 with the pole at the point i = √ −1. It is clear that the normal derivative of G on the boundary of the square is bounded (it tends to zero as we approach a corner). Set where t > 0. If t is small enough we obtain a subharmonic extension of u into S + 0,0 , because the jump of the normal derivative will be positive as we cross the boundary of the square from inside. Fix such t, and extend u to the remaining squares by the formula u(γ n,k (z)) = M −n u(z), z ∈ S + 0,0 . (2.9) It follows from (2.8) that the normal derivative always has a positive jump as we cross the boundary of S + n,k , so the extended function is subharmonic in S + . Now consider the smaller squares for some β > 0 and all n and k. Now we are going to extend u to the strip S = S + ∪ S − where To do this we repeat the above construction starting with the square S − 0,0 = {z : |ℜz − 1 2 | ≤ 3 10 , |ℑz + 1| ≤ 3 10 } and using the same semigroup Γ. We obtain the squares S − n,k = γ n,k (S − 0,0 ) and K − n,k = γ n,k (K − 0,0 ), where K − 0,0 = {z : |ℜz − 1 2 | ≤ 2 7 , |ℑz + 1| ≤ 2 7 }, and the function u 1 subharmonic in S − that satisfies the inequality similar to (2.10): with some β 1 > 0 and M 1 > 1. Extend u to S = S + ∪ S − by setting u(z) = u 1 (z), z ∈ S − and u(x) = 0, x ∈ R. The extended function u is continuous in S. We will prove that it is subharmonic in S.
Consider the strips Π n = {z : |ℑz| < 4 3 2 −n }. Define the functions v n in the following way: v n (z) = u(z), z ∈ S\Π n ; v n are continuous in S and harmonic in Π n . Then v n (x) > 0, x ∈ R, and it follows from the maximum principle (applied to Π + n = {z : 0 < ℑz < 4 3 2 −n }) that v n ≥ u in S. We conclude that v n are subharmonic because the sub-mean value property holds in every point of S. Furthermore, it is evident that v n → u uniformly in S as n → ∞, so u is subharmonic in S.
Denote K + n = k K + n,k and K − n = k K − n,k and remark that each vertical line ℜz = x 0 intersects K + n ∪ K − n . Indeed the projection of K + n onto the real axis is It is clear that the union of these two sets is the whole real axis. Now set E n = k (S + n,k ∪ S − n,k ) and K n = K + n ∪ K − n . Then each E n is a union of some components of the set {z ∈ S : u(z) < 0} and for every vertical line l = {z : ℜz = x 0 } the length of intersection l ∩ K n is at least b −n for some b > 1. So we have in view of (2.10) and (2.11): l∩En u(x 0 + iy) dy ≤ l∩Kn u(x 0 + iy) dy ≤ −c −n with some constant c > 1.