A PROPER SUBCLASS OF MACLANE ’ S CLASS

The MacLane’s class of analytic functions is the class of nonconstant analytic functions in the unit disk that have asymptotic values at a dense subset of the unit circle. In this paper, we define a subclass of consisting of those functions that have asymptotic values at a dense subset of the unit circle reached along rectifiable asymptotic paths. We also show that the class is a proper subclass of by constructing a function f ∈ that admits no asymptotic paths of finite length.


Preliminaries.
In all what follows, f is a nonconstant analytic function on the unit disk.
Definition 1 [2].We say that a simple curve Γ : z(t), 0 ≤ t < 1 is a boundary path ending at ζ if |z(t)| → 1 as t → 1 and if Γ ∩ C = {ζ}.The number a is called an asymptotic value associated with ζ if there is a boundary path Γ such that f (z(t)) → a as t → 1 and Γ ∩ C = {ζ}.In that case, we call Γ an asymptotic path.Definition 2. We define the set A(f ) to be the set of all points ζ at which f has an asymptotic value.In particular, we denote by A a the set of all points ζ associated with the asymptotic value a, and by A ∞ the set of all points ζ associated with the asymptotic value ∞.We also define the set A R to be the set of all points on the unit circle at which f has asymptotic values reached along rectifiable asymptotic paths.
Definition 3 [3].If A(f ) is a dense subset of C, we say that f ∈ Ꮽ, the MacLane class of analytic functions and we define the set to be the subset of Ꮽ for which A R is a dense subset of the unit circle C.
Definition 4 [1].Let H ⊂ D be a relatively closed subset of D. We say that H is an Arakelyan set or We say that H is a set of tangential approximations (by analytic functions of H) provided that, for each function g continuous on H and analytic on the interior H 0 of H, and for each positive continuous function (t), 0 < t < 1, there is an analytic function f on D such that, for all z ∈ H, Note that when H 0 = φ, the function g is only required to be continuous on H.
2. Main theorem. is a proper subset of Ꮽ.
Proof.The strategy is to construct a function f ∈ Ꮽ by approximating a function g on an Arakelyan set H, with H 0 = φ, using Lemma 1.
The set H is the union of (a) a sequence of circles converging to the circumference C, each having small equally spaced gaps in it, and (b) the boundary paths that snake through the gaps.The gaps in a circle have a total length that approaches zero quickly as the circles approach the circumference.Also, the gaps on consecutive circles are rotated enough so that the asymptotic paths approaching a point of the circumference that pass through the circles only in the gaps (of most of the circles) have infinite length.
The set H so constructed turns out to be a set of tangential approximation.We define a continuous function g on the set H as follows.On each circle (minus its gaps), the function is constant, and on consecutive circles, it has values 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3,... .Along the asymptotic paths of the set H, the function approaches infinity (uniformly) as the modulus approaches 1. Arakelyan's theorem allows us to extend the function g into an analytic function f with the desired property.The set H and the function g and, consequently, f are constructed in such a way that any asymptotic path along which f converges must be either funneled only through the gaps of the circle (if the limit is other than 0, 1, 2, or 3), or must eventually only hit at most one of four circles (hence, pass through the gaps most of the time).In either cases, the asymptotic path must be of infinite length because it is trying to avoid the circles minus the gaps on which the function keeps alternating between the four finite values.More specifically, we start with a sequence of circles {C j n } converging to the circumference C each having small equally spaced gaps so that the circle minus the gaps form the first set of arcs {γ j n,k }, (n = 2, 3,..., k = 1, 2,...,n, j = 0, 1, 2, 3).The arcs {γ j n,k } will be positioned so that if any asymptotic path were to avoid a 'good number' of them, that asymptotic path would have to be of infinite length.The second set of arcs {Γ p\q }, (q = 1, 2, 3,... and p = 1,...,q − 1) consists of boundary paths ending at a dense subset of C. The set H is the union of the sets {γ j n,k } and {Γ p\q }.First, we show that H ∈ K(D) then we define a function g on H with the property that g → ∞ as |z| → 1 along the boundary paths Γ p\q , while the function g takes four different constant values on subarcs {γ j n,k } of four consecutive concentric circles C j n , j = 0, 1, 2, 3. Finally, we show that g and H satisfy the conditions of Lemma 1 and that, for an appropriate choice of (t) > 0, the function f in the conclusion of the lemma has the desired property: f ∈ Ꮽ − .

Construction of the set H.
We only take the values of p and q that are relatively prime ((p, q) = 1).
(1) Description of the arcs γ j n,k , n = 2, 3,...; j = 0, 1, 2, 3; k = 1, 2,...,n.We start with a sequence of circles {C 0 n }, n = 2, 3,..., centered at the origin, each of radius r 0 n = 1 − 1/n, and we let D n , n = 2, 3,... be the annulus D n = {r 0 n ≤ |z| < r 0 n+1 }.Let D 1 be the disc {z : |z| < r 0 2 }.Within each annulus D n , we define three concentric circles C j n , (j = 0, 1, 2, 3) centered at the origin, of radius r j n = r 0 n + (j/(4n(n + 1))), where j = 1, 2, 3. So, the three circles are equally spaced consecutively between C 0 n and C 0 n+1 .Note that here and in all that follows, j is used as an index rather than an exponent in r j n .On every circle C j n , (n = 2, 3,...; j = 0, 1, 2, 3) we arrange n (equally spaced) arcs {γ j n,k }, (k = 1, 2,...,n) of equal length so that the gaps consist of n open arcs σ j n,k , each of length πr j n /2 n , such that, for j = 0 or 2, the arcs σ j n,k , k = 1, 2,...,n, all have their midpoint at the point r j n e 2kπ i/n , whereas for j = 1 or 3, the arcs σ j n,k , k = 1, 2,...,n, have their midpoint at r j n e (2k+1)π i/n (a rotation of angle π/n from the previous case).The arcs σ j n,k , which are the gaps on the circles, are rotated enough so that a boundary path funneling through a 'good number' of them would be of infinite length.Note that the length of In the future, we refer to the arcs γ because of the arrangement of the arcs γ n and σ n .Consider a curve J n ⊂ D n such that Suppose, in addition, that J n ∩ γ j n,k = φ for at most one value of j ∈ {0, 1, 2, 3}.Therefore, J n crosses a pair of circles C j n and C j n+1 for some value of j ∈ {0, 1, 2, 3} at points of some arcs σ n and σ n+1 .By the previous remark, the length of such a path is (By doing so, we have made sure that the gaps on consecutive circles are rotated enough so that asymptotic paths approaching a point of the circumference that pass through the circles only in the gaps (of 'most' of the circles) have infinite length.) (2) Description of the boundary paths Γ p/q .For q = 1, 2,..., and p = 1, 2,...,q −1 and (p, q) = 1, let S = e 2πip/q . (8) Note that S is a dense subset of C. We define a sequence of disjoint boundary paths {Γ p/q }, where p and q are as mentioned above, and such that Γ p/q ∩{z : |z| = r } consists of exactly one point for all r satisfying r 0 q ≤ r < 1. (Γ p/q is not defined for r < r 0 q .)We need to construct the arcs Γ p/q ∩ C j n for j ∈ {0, 1, 2, 3}, and n = q, q + 1,..., so that if p/q = p /q .In the special case of Γ 1 , we make for j ∈ {0, 2}, r j n e πi/2 n+1 e π i/n for j ∈ {1, 3}. (10) For q > 1, we make for j ∈ {0, 2} r j n e (2πi p/q /n)+(pπ i/2 n+1 q) e π i/n for j ∈ {1, 3}, where is the greatest integer function.The first part of the argument, 2πi np/q /n, is the center of the 'gap' σ j n,k , while the second part of the argument, pπ i/2 n+1 q, determines the distance from the point Γ p/q ∩ C j n to the midpoint of the arc σ j n,k ; it ensures that the point of intersection is still within σ j n,k .Such paths Γ p/q must end at the dense subset S of C since, for any p and q as described above, r j n → 1 and e (2πi np/q /n)+(pπ i/2 n+1 q) → e 2πip/q as n → ∞. ( Note that two different paths intersect the circle C 0 n at two different points since if p/q = p /q , then r j n e (2πi np/q /n)+(pπ i/2 n+1 q) = r j n e (2πi np /q /n)+(p πi/2 n+1 q ) . ( We define to be the polygonal arc that begins at z(0) = r 0 q e (2πip/q)+(pπ i/2 q+1 q) .(15) (Observe that the arc Γ p/q starts on the arc σ 0 q,p whose midpoint is of argument 2πip/q.)In the annulus D n , for n = q, q+1,... the arc joins r 0 n e (2πi np/q /n)+(pπ i/2 n+1 q) on C 0 n successively to r 1 n e (2πi np/q /n)+(pπ i/2 n+1 q) e π i/n on C 1 n , to r 2 n e (2πi np/q /n)+(pπ i/2 n+1 q) on C 2 n , to r 3 n e (2πi np/q /n)+(pπ i/2 n+1 q) e π i/n on C 3 n , and finally to r 1 n+1 e (2πi np/q /n)+(pπ i/2 n+1 q) on C 0 n+1 .Observe the rotation provided by the e π i/n factor.From the definition of Γ p/q and from the fact that Γ p/q ∩C j n = Γ p /q ∩C j n in case p/q = p /q , the paths are disjoint.As an illustration, we take the path Γ 4/5 .Its initial point is the point r 0 5 e 2πi(4/5)+4πi/2 6 6 on the arc σ 0 4,5 ⊂ C 0 5 .The path Γ 4/5 ends at the point of C of the argument 2πi(4/5).The second circle that Γ 4/5 crosses at a point on an arc σ n whose midpoint is of argument 2πi(4/5) (more informally, the gap at 2πi(4/5)), is the circle C 0 10 .The kth such circle is the circle C 0 5k .(Note that this is a justification for using the greatest integer function.In fact, the solution for n in the equation e 2πi n4/5 /n = e 2πi(4/5) is n = 5k.)Now, consider the paths Γ 1/3 and Γ 1/4 .They both intersect C 0 8 at different points of the same arc σ 0 8,2 .In fact, Γ 1/3 ∩ C 0 8 = r 0 8 e (2πi 8(1/3) /8)+(π i/2 9 3) = r 0 8 e (π i/2)+(π i/2 9 3) (16) while Γ 1/4 ∩ C 0 8 = r 0 8 e (2πi 8(1/4) /8)+(π i/2 9 3) = r 0 8 e (π i/2)+(π i/2 9 4) . ( In general, there might be more than two paths Γ p/q intersecting a circle C 0 n at different points of the same arc σ j n,k .However, there are finitely many (n 2 ) such paths since q ≤ n and p < q.Note that the arcs {γ n } and {σ n } have been arranged so that any such path Γ p/q is of infinite length since, Γ p/q ∩ D n contains an arc joining a point of σ j n,k to a point of σ j+1 n,t , j = 0, 1, 2, similar to the arcs J n in a previous remark.Finally, define the set H to be the following disjoint union: The set H is relatively closed because every arc γ j n,k is closed and Γ p\q are closed arcs in D n .

Proof of H ∈ K(D).
Let z 0 ∈ D − H.We need to find a path Γ 0 ∈ D − H that connects z 0 to C. Note that z 0 ∈ γ n for any n, and z 0 ∈ Γ p/q for any Γ p/q ⊂ H. Choose n so that z 0 ∈ D n .Since D − H is open in D, we can construct a path Γ 0 in D − H that first joins z 0 to some point z 0 ∈ σ n − H for some σ n ⊂ D. Observe that there might be more than one path Γ p/q crossing the same arc σ n .Let Γ h /m ⊂ H be the boundary path in H with the property that Γ h/m ∩ σ n is the closest to z 0 on σ n .From z 0 , we make the path Γ 0 ⊂ D − H follows Γ h/m so closely in D n − H that Γ 0 intersects no path Γ p/q ⊂ H, and so that the distance between Γ 0 and Γ h/m in the annulus D n approaches 0 as n → ∞.Thus, Γ 0 connects z 0 to the boundary without intersecting H, and it ends at the point e 2π i(h/m) as desired.
Observe that g → ∞ as |z| → 1 along Γ p/q .Note that H has no interior and g is continuous on H.
Construction of a function f ∈ Ꮽ − .Since g and H satisfy the conditions of Lemma 1, there corresponds to every positive continuous function (t) : 0 < t < 1, some analytic function f on D with the property that f (z)− g(z) < d(z, C) for all z ∈ H. (20) Let f denote the function corresponding to 0 (t) = 1/9 for all t.Since g → ∞ as |z| → 1 when z ∈ Γ p/q , (21) then f → ∞ as |z| → 1 when z ∈ Γ p/q (22) as well, and since the paths Γ p/q end at a dense subset S of C, it follows that ∞ is reached as an asymptotic value at a dense subset of the unit circle, that is A ∞ is a dense subset of C, so that f ∈ Ꮽ, the MacLane class.
Since the function g has values that differ by one on the arcs γ n of the different circles C j n , j ∈ {0, 1, 2, 3}, the function f has values that differ at least by 1−2(1/8) = 3/4 on the arcs γ n of the different circles C j n , j ∈ {0, 1, 2, 3}.Therefore, if n is sufficiently large, no asymptotic path can cross the arcs γ n on more than one of the four circles C j n , j ∈ {0, 1, 2, 3}; hence, by a previous remark, Finally, since the regions D n are disjoint and it follows that, for any asymptotic path Γ , In other words, no asymptotic path is rectifiable, and so f ∈ .
), j = 0, 1, 2, 3, in D n as the arcs γ n (σ n ) provided there is no ambiguity.The distance from a point of σ n on C j n to a point of σ n+1 on C