On the convolution of convex 2-gons

We study the convolution of functions of the form \[ f_\alpha (z) := \dfrac{\left( \frac{1 + z}{1 - z} \right)^\alpha - 1}{2 \alpha}, \] which map the open unit disk of the complex plane onto polygons of 2 edges when $\alpha\in(0,1)$. We extend results by Cima by studying limits of convolutions of finitely many $f_\alpha$ and by considering the convolution of arbitrary unbounded convex mappings. The analysis for the latter is based on the notion of angle at infinity, which provides an estimate for the growth at infinity and determines whether the convolution is bounded or not. A generalization to an arbitrary number of factors shows that the convolution of $n$ randomly chosen unbounded convex mappings has a probability of $1/n!$ of remaining unbounded. We also extend Cima's analysis on the coefficients of the functions $f_\alpha$ by providing precise asymptotic behavior for all $\alpha$.


Introduction
As usual, let D and T be the open unit disk and the unit circle in the complex plane, respectively.If f and g are two holomorphic functions in D, we define their convolution as Moreover, it can be checked that where {a n } n≥0 and {b n } n≥0 denote the sequences of Taylor coefficients of f and g, so the convolution is also known as Hadamard product in the bibliography.The convolution of functions satisfying certain geometric properties has been extensively studied (see for instance [8] or Chapter 14 of [4] for some general references).For this paper, it is especially relevant the following result, which was originally conjectured by Pólya and Schoenberg [5, Appendix II] and was proved by Ruscheweyh and Sheil-Small [7].
Theorem A (Ruscheweyh, Sheil-Small).Let f and g belong to the class C of univalent functions in D whose ranges are convex sets.Then, f * g ∈ C as well.
In this paper we will focus on the mappings , 0 < α < 1.
The mappings f α are univalent and f α (D) corresponds to the convex 2-gon with vertex at − 1 2α and edges with directions e απ 2 i and e − απ 2 i .If we allow α to assume the value 1, it is clear that so f 1 turns out to be the identity element for the convolution in the subspace of analytic functions vanishing at the origin.On the other hand, taking limits when α → 0 we get that f 0 (z) := lim which is again one-to-one and f 0 (D) = {z ∈ C : | Im z| < π 4 }.
Much of the motivation of the present paper stems from Cima's paper [2] studying the convolution f α * f β .We will collect several of the results therein and bring attention to his comment on the technical difficulties faced in the analysis.The paper is organized as follows.In Section 2, we extend the results in [2] by giving precise asymptotic behavior for the coefficients of the function f α .We determine when f α * f β is bounded or not for all choices of the parameters, and study all possible limits of the convolution of finitely many mappings f α .In Section 3, we recall the notion of angle at infinity and use it to establish bounds for the growth of general unbounded convex mappings.From this, we find the probability that the convolution of finitely many randomly chosen unbounded convex mappings is bounded.
(iii) For every α ∈ (0, 1), there exists a constant C = C(α) such that Observe that, in particular, this last inequality yields that f α * f β is bounded if α+β < 1, as was explicitly mentioned in [2, Theorem 2].Our aim is to complement this theorem by considering the case α + β ≥ 1, showing in fact that a lower bound of the same type as in (2) is also true.
Proof.On the one hand, we trivially have lim On the other hand, observe that c n (α The Euler-Gauss formula for the Gamma function (see for instance [3, p. 255 Let ε > 0 and n 0 be such that and by letting ε → 0 we find that which implies (4).

Arbitrary number of convolutions.
In this section we study the convolution of an arbitrary number of 2-gons, including the limiting case f 0 .For α ∈ D \ {0}, we define the function f α as the rotation of f |α| by e i arg α , i.e., We are interested in the set of all possible limits of finite convolutions of 2-gons given by According to the lemma below, we may assume that all of the elements of the sequence {α n } n≥1 , except perhaps α 1 , are positive.Lemma 2.5.
uniformly for all compact subset K of D. Since g 2 (α) = α, we have that the sequence of partial products n j=1 α j n≥1 is convergent.Let p be its limit.At this point, we split the proof by cases.(i) If p = 0, we are going to show that f (z) = z.Without loss of generality, we might assume that |α n | ≥ 1 2 for all n ≥ 1.Because of Lemma B, we have that and thus it is clear that f (z) = z, which is obviously the limit of "convolutioning" f 1 2 , with itself infinitely many times.(ii) On the other hand, if p = 0 we have that the infinite product converges.By continuity, we have that n≥1 |α n | = |p|, and therefore the product λ := n≥1 e i arg αn is also convergent.Observe that |λ| = 1, and consider the sequence β 1 := λ|α 1 | and β n := |α n |, n ≥ 2. We are going to show that Indeed, observe that the k-th Taylor coefficient of f is given by which is trivially equal to the k-th coefficient of lim n→∞ f β 1 * . . .* f βn because of the definition of the sequence {β n } n≥1 .Therefore, it is clear that the identity (5) holds.
It may seem from this proof that {α n } n≥1 and {β n } n≥1 can give rise to the same limit as long as lim n→∞ n j=1 α j = lim n→∞ n j=1 β j , but this is not true.For instance, if {α n } n≥1 ⊂ (0, 1) is a sequence such that its infinite product is convergent, and consider the sequence {β n } n≥1 defined as A straightforward computation shows that the function ) is a supermultiplicative function.In other words, for all α, β ∈ (0, 1).Then, we have that and hence the convergence of the product j≥1 β j yields that The following result highlights the effects of having a vanishing Taylor coefficient in L.
Two cases are possible: (i) If there exists a subsequence {α n j } j≥1 such that α n j ≤ 1  2 for all j, then the monotonocity of the polynomials {g k (α)} k≥2 with respect to α implies that and hence f must be the identity.(ii) If there exists N ≥ 2 such that α n > 1  2 for all n ≥ N , we will show that lim n→∞ n j=N α j = 0, and then Lemma B yields that f (z) ≡ z.
In order to get a contradiction, assume that n≥1 α n is convergent.In other words, that k 0 > 2.Then, the product n≥1 k 0 −1 k 0 +1−2αn is also convergent.However, Lemma B implies that so the product on the left-hand side is bounded from below by the multiplication of two convergent and positive products, which is not possible.Thus, the partial products of the sequence {α n } n≥1 converge to zero, which means that f is the identity.
The following analogue of Theorem 2.4 can also be drawn. .
Proof.(i) If B > 1, then there exists N big enough such that the partial sum N n=1 (1 − α n ) is greater than 1 as well.Thus, we have that (ii) Since the sum is convergent observe α n < 1  2 for, at most, a finite number of indexes.Because |1−L(α)| = O(1−α) in 1  2 , 0 , note that, in particular, we have that the product n≥1 L(α n ) is convergent.As a consequence of the uniform convergence of {G k (α)} k≥1 , we have that there exists k 0 ≥ 0 such that and therefore (6) holds.(iii) It is similar to the previous case.
Although finite convolutions of elements of {f α } 0≤α≤1 do not belong to L, the latter theorem can be also applied for this case.Nevertheless, these finite convolutions may be approximated by functions in L, as it is implicitly shown in the following proposition.
Proposition 2.8.The functions f 0 and f 1 do not belong to L, but they belong to the closure of L. In particular, L is not closed.
Proof.Since the coefficients of f α , α ∈ D, have moduli strictly less than 1, it is clear that f 1 ∈ L. On the other hand, f 0 does not belong to L because of Proposition 2.6.Now, consider the quantities and the sequence of functions Observe that g k (α n,j ) ≤ g k (α n,j+1 ) < 1 for all n, j ≥ 1 and k ≥ 2, so the sequence of k-th Taylor coefficients of the functions {F j } j≥1 is convergent.Hence, the limit F = lim j→∞ F j exists.It remains to show that F ≡ f 1 , but this a consequence of the extremal properties of f 1 .Indeed, it is a well-known fact that and that equality is attained if and only if f (z) ≡ λf 1 (λz) for some unimodular constant λ.It is clear that F ∈ C and, by the construction of {F j } j≥1 , its second Taylor coefficient is equal to 1. Therefore F must coincide with f 1 .
Once we know that f 1 ∈ L, it is easy to see that f 0 ∈ L (for example, it can be checked that G j := f 1/j * F j ∈ L converges to f 0 when j → ∞) and consequently we have finished the proof.
Unfortunately, a complete characterization of L seems to be out of reach.However, we can provide the following necessary condition.Proof.Take f ∈ L different than the identity.Consider a sequence of functions {F j } j≥1 ⊂ L such that F = lim j→∞ F j .Since f (z) ≡ z, we might assume that F j (z) ≡ z for all j.Let {α n,j } n≥1 (all of them positive except, possibly, the first one) such that F j = lim n→∞ f α 1,j * . . .* f α n,j .Note that we have that (7) g since the elementary estimate g k (α) ≥ α k−1 holds for every 0 < α < 1 and k ≥ 2.Moreover, Lemma B implies that and for all j, m ≥ 1.Consequently, there is no odd number k such that lim If k is an even number such that the k-th Taylor coefficient of f is zero, it follows from ( 7) and ( 8) that f must be an odd function.

Estimates for unbounded convex sets
In this section we deduce estimates for general unbounded convex sets.We recall that the angle at infinity of a convex set is the infimum of the angles of all sectors containing that set.It follows from convexity that an unbounded convex set contains a 2-gon of this amplitude if it is positive.Theorem 3.1.Let 0 < α < 1.Let g be a unbounded convex mapping with g(1) = ∞.If g(D) has an angle at infinity of πα, then there exists δ > 0 such that Proof.Without loss of generality, we may assume that Ω = g(D) contains a sector of the form S α := c + f α (D), that g(0) = c + f α (0) and that 0 lies outside the closure g(D).Thus, there exists an analytic function ϕ : It remains to analyze the behavior as z → 1 with z ∈ D \ D α , where D α := ϕ(D).Since g is convex, then its Schwarzian norm Sg is less than or equal to 2. Because the angle at infinity is positive it follows also that g(D) is a quasidisk on the Riemann sphere, hence Sg < 2 by Theorem 4 in [1].We conclude that the function On the other hand, it follows from subordination that if the angle at infinity of the unbounded convex region Ω = g(D) is πα, then for any β > α there exists An interesting issue is whether there exists a sector of minimal amplitude containing Ω.This will happen only if the vertices of the sectors of a minimizing sequence remain in a bounded region of the plane.For example, a parabolic region has zero angle at infinity but no parallel strip contains it.Simple modifications provide similar examples for positive angle at infinity (for example when Ω is the intersection of angles of decreasing amplitudes with aligned and divergent vertexes, like the one represented in Figure 1), so in principle the choice β = α is not allowed.
Example of convex set with vertex at infinity.
We state now an analogue of Theorem 2.4.Although it can be derived from this result and the fact that the convolution of convex mappings preserves subordination (see Theorem 4.1 in [7]), we present an alternative and self-contained proof.Theorem 3.2.Let g 1 , g 2 be unbounded convex mappings defined in D with angles πα, πβ at infinity, respectively.(i) If α + β < 1 then g 1 * g 2 is bounded.(ii) If α+β ≥ 1 then g 1 * g 2 is unbounded and the angle at infinity is at least π(α+β −1).
The angle at infinity is equal to π(α + β − 1) if both regions g 1 (D), g 2 (D) admit sectors of minimal amplitude containing them.
Proof.We may assume that both α, β < 1.After a rotation, we may assume that both regions contain the ray (0, ∞), with g 1 (1) = g 2 (1) = ∞ and 0 ∈ g 1 (T), g 2 (T) as shown in the Figure 2. Using the integral representation of the convolution, we have that If z → e it 0 = 1 then the integral remains bounded because the factors are each integrable and unbounded at different values of t.In other words, g 1 * g 2 can become unbounded only as z → 1.Because the convolution is continuous in the spherical metric, we may assume that z is real.For small δ > 0 fixed, the integral remains bounded as r → 1, hence it suffices lo analyze the behavior of as r → 1.For this we write g 1 = u 1 + iv 1 , g 2 = u 2 + iv 2 , and express The normalizations of g 1 , g 2 imply that u 1 ( 1+r 2 e it ), u 2 ( 2r 1+r e −it ) > 0 while v 1 ( 1+r 2 e it ) and v 2 ( 2r 1+r e −it ) have the sign of t.Furthermore, all four quantities u 1 , u 2 , v 1 , v 2 will be comparable in size because the domains contain and are contained in certain sectors.
For part (i) we use (10) for g 1 and g 2 with angles α ′ > α and β ′ > β and α ′ + β ′ < 1 to conclude that I(r) will stay bounded as r → 1.For part (ii), we see from the above remarks on the size and signs of the u j , v k that it suffices to analyze no only the real part of I(r), but just the integral J(r) = .

Proposition 2 . 9 .
Let f ∈ L.Then, one of the following statements is true: (i) All of the Taylor coefficients of f are different than 0. (ii) The function f is an odd function and all the odd Taylor coefficients are different than 0. (iii) f (z) ≡ z.

Figure 2 .
Figure 2. Example of g j (D) after the normalization.