5ru.
m,k P,q
Consequently we can find positive integers p and q such that if z is in
Vp q and p(z,C) < l/p, then p(DJ,f(z)) > 4ru. Let n denote the index
of the disk D_r:which has the same center as E and radius rn = 2ru. So
P(¥,f(z)) > rn if z is in VP,q and P(z,C) < 1/p. Due to the choice of
E, there exists a sequence of elements 2 in Vm,k which approaches f
and a corresponding sequence of elements f(E) which approaches a point
a E 3;. So for an infinite set of positive numbers, there exist points
‘2 in V , for l/(3s) < p(§,C) < l/s such that D—[: and {f(2)} are not
disjoint. Thus 5* e 0L? F for all t and is therefore also in
S=t n,m,k,s
n,m,k'
N . Then b the definition of
ow suppose f e Fn,m,k and En,p,q y
E — ' int. Since is in F , it is in
n’p’q, CVp, and DD. are diSJO f n,m,k
F for an infinite number of 5's. From the definition 0f
n,m,k,s
F . D— re not dis'oint and so C is
n,m,k,s’ it follows that CVm,k and n a j Vm,k
nOt Equal to CV
P,q
We will now show that E Suppose on the contrary
W is a-porous .
. , _ 'th
there « and E Which is not 0 P0rous W1
eXists a point in Fn,m,k n,p,q
43
respect to G. Then the angle Vﬁ. close to its vertex is covered by a
,k
union of angles V" for n in F and E . So by the definition
P,q n,m,k n,p,q
of E at points z in V ‘which are sufficiently close to the point,
n:P:q m,k
the values of f(z) are at a distance Zrn from.D;l Therefore CV k and
m
3
D; are disjoint and the point is not in F k and E . Thus
3 3 3 3
F E is porous on C and E is a-porous.
n,m,k n n,p,q W
This theorem is closely related to the Collingwood Maximality
Theorem (Collingwood, 3), which states that for an arbitrary single-
valued function f(z) defined in D and any Stolz angle Atwith vertex at
(9 93(f,() = C(f,() except for a set of first category. It is also re-
lated to Theorem 14 which states that for an arbitrary single-valued
function f(z) defined in D the outer angular cluster set CA = 9A ex-
cept for a set of measure zero.
Theorem.22: If f(z) is an arbitrary function, then EUU is of G60 type
and of 0-porosity for some a, (Yanagihara, l, Theorem.l, p.424)
The proof of this theorem is quite similar to that of Theorem 21.
Theorem.23: If f(z) is an arbitrary function, then EUV is of Gﬁa type
M—_
and of o-porosity' (0 ). (Yanagihara, 1, Theorem 2, p.425)
Yanagihara (1, Theorem 4, p.426) has shown that there exists a
bounded holomorphic function f(z) for which EUV is of measure 2n. For
example, we pick an inscribed disk U(l)‘= {21 IZ ' Pl‘< 1 ’ Pi for
0 < P‘< 1. Then there exists a constant b such that an are A = iz-
reie :9 = bﬁ/l-i:i is contained in U(l). In addition we choose tn
44
00
such that 0 < tn < 1, is strictly increasing to l and 2V1 - t * oo-
57
M-Topology for Continuous Functions
Suppose f is a continuous complex-valued function defined in D.
Then we let T(p) denote the set of all Jordan arcs contained in DlJ{p}
and having one endpoint at p, and let Gf(p) = {Ct(f,p) :t e T(p)}. In
order to define the metric M, we choose two nonempty closed sets A and
B in W and set M(A,B) = max(sug p and such that for each
point z on any rectilinear segment [sj’sj+l] the condition d(f(Z),Kn) > /
l/Zn is satisfied. Some subset of the union of segments Sj’sj+1] con-
stitutes an element of T(p). Since Cs(f,P)n Ct (f:P)=:CS(f’p)n Kn: ¢’
n
p is an ambiguous point of f.
Corollary: Let f be a continuous function in D and E be the set of
points p for which Gf(p) is not compact in the M-topology. Then E is a
countable set. (Belna and Lappan, 1, Corollary 1, p.212)
This corollary follows immediately from Theorem 33 and the
Bagemihl Ambiguous Point Theorem (Theorem 4).
Theorem 23: Suppose f is a continuous function in D and p 6 C- If {tn}
is a directed sequence of arcs in T(p) such that Ctn(f,p) = KH and if K
is a Continuum such that M(Kn’K) —) 0 but K é Gf(p), then there exists
a directed sequence of arcs {8k }in T(p) and 6 > 0 such that for each
is between tnk
integer k > 0 there exists an integer nk > 0 such that sk
a d ]] Iemma ] p.88)
n t and d C f p),K) > e. (Lappan, a ’
nk'l'l ( Sk( ,
59
2592:: We will prove this theorem by assuming that it is false and then
showing that we obtain a contradiction. If this theorem is false, then
for each positive integer k there exists an integer Nk such that for
each 6 > 0 which is sufficiently small, n > Nk implies that all of the
sets tnr1{z E D: 'z- pl < 5} lie in the same component of {z e D:
d(f(z),K) < l/k, 'z-—p| < 5}. Therefore, for each n > N and each 6 > 0
k
there exists a Jordan arc qn leading from a point of tn to a point of
tn+l such that qnc:{z e D: 'z- p' < 5 and d(f(z),K) < l/k}. So we may
choose a subsequence {tnk} of {tn} such that nk > Nk for each positive
integer R. Then for each k there exists a Jordan arc pk leading from a
point on t to a oint on t such that p c:{z E D: Iz- p < l/k and
”k p “H1 k |
d(f(z),K) < l/k}, and the portion té of tnk between the terminal point
0f pk_1 and the starting point of pk satisfies the relationship
M(f(t£,K) < l/k. Without loss of generality we may assume that pk
meets tn and t in exactly one point each. Then letting t be the
n
k k+1
I
Jordan arc obtained by splicing together all of the arcs tk and pk, we
have Ct(f,P) = K contradicting the hypothesis K é Gf(p).
IEEQEEE ii: Suppose f is a continuous function in D and p is a point
in C such that Gf(p) is not compact in the M-topology. Then there exist
directed sequences {tn} and {Sn} of arcs in T(p), e > 0 and a continuum
K suCh that if Kn = Ctn(f,p) and Ln = Csn(f,p), then for each n > 0
M(Kn’K) < l/n, d(Ln’K) > 6 and the arc sn is between tn and tn+l'
(Lappan, 11, Lemma 2, p.89)
' d‘tions
ﬂﬁwf: Let{t%l}be a sequence of arcs in T(P) satisfying the con 1
re not
Ctn(f:P) = Kn and M(Kn’K) < l/n Where K é Gf(p). If the arcs a
60
mutually disjoint, they can be shortened individually so that an infi-
nite subset of the shortened arcs are mutually disjoint. If this was
not true, there would exist an arc t e T(p) where t is contained in the
union of the tn's and Ct(f,p) = K which contradicts the assumption on K.
Now we can choose a directed subsequence of the tn's. In addition we
can select an appropriate continuum K since Gf(p) is not compact. So
the conclusion of this theorem follows from Theorem 34.
Light Interior Functions
A function f from D into W is called a light interior function if
f is a continuous open map which does not take any continuum into a
Single point. It has been shown that f has a factorization f = goh
where h is a homeomorphism of the unit disk onto itself or onto the
finite complex plane and g is a nonconstant meromorphic function.
Let A(f) denote the set of all eie for which there exists an as-
ymptotic path of f in D which includes e19 in its end and let Ap(f)
denote the set of all ei6 for which there exists an asymptotic path of
f in D which ends at the point eie. For any homeomorphism h of D onto
D, we define B(h) to be the set of all e16 for which there exists an as-
ymptotic path of h in D with end E and ei6 is contained in the interior
of E.
1322£§m_29: Suppose f is a light interior function with factorization
f = g°h - If A(g) is dense on C, then A(f)LJB(h) is dense on 0.. Fur-
thermore, if Ap(g) and Ap(h) are dense on C, then Ap(f)LJB(h) is dense
on C. (3. Mathews, 1, Theorem, p.79)
61
Proof: We will prove this theorem by assuming that it is false and show
that we have a contradiction. Let the arc (w1,11,2)CC - A(f) be arbi-
trary and [91,92]C(\y1,\y2) with 0 < 92 - 91<2Tr. Let I‘1 and 1‘2 be Jordan
i6 16
arcs inD ending ate 1 and e 2 respectively with 1‘1“ 1‘2 = {0}. Then h
maps the domain A bounded by 1‘1 UI‘2 and the arc [91,62] onto a domain A1
in D.
Then there exist a point ela e Cr1(h,9 )0 Cr2(h,92) and sequences
. . . id .
{Zn} and {zn} in 1‘1 and 1‘2 respectively With h(zn) —-> e and h(zn) —>
.a .
e1 . Let A be a Jordan are at eLe which passes consecutively through
the points h(zl), h(zi), h(zz), h(zlfl),.... According to Collingwood
and Cartwright (Lemma 1, p.93), either [61,62]CCA(h-1,CL) or [62,61+2«n]
C CA(h-l,a). Therefore, either (61,92)CB(h) or (62,91+2TT)CB(h) and
Case (ii) [ZS-10 C]3Cr1(h,91) uCr2(h,92), with a proper inclusion.
Then E = [510 CJ-[Cr1(h,91) Ucr2(h,92)] is a nonempty open subarc
of C. Let eia be in bothEand A(g). Then e1C1 is in the end of an
asymptotic path A of g. But C(h-1,a)C[91,62] so that h-1(A) is an
asymptotic path of f whose end intersects [91,62]. Therefore, [91,62] 0
A(f) 7‘ ¢, a contradiction.
Consequently both cases lead to contradictions. Since ($1,412) was
arbitrary A(f) UB(h) is dense on C. The second part of the theorem is
proved s imi 1 ar ly .
62
Locally Univalent Functions
Any function f(z) meromorphic in D is called locally univalent if
f(z) has at most simple poles and f'(z) # 0. The function has Koebe
grgs if there exist curves JnCZD such that for some a < B < a + 2H and
some constant c which is possibly x
(i) Jn intersects the radii arg z =u and arg z = B for each n,
(ii) IzI-—+ l for z e Jn as n-+ m,
(iii) |f(z)- c| < e for z e Jn as n-—+ m.
For any set C, the boundary of G is denoted by BC.
Theorem 31: Let f(z) be a meromorphic locally univalent function with-
out Koebe arcs. Then f(z) has three distinct asymptotic values on each
arc of C. (McMillan and Pommerenke, Theorem, p.31)
Erggf: Suppose that there exists an arc A of C on which there is at
most one asymptotic value. So we may assume without loss of generality
that f(z) has no finite asymptotic value on A. Let d(z) denote the
radius of the largest disk around f(z) having no branch points on the
Riemann image surface F. Since f is locally univalent there is a boun-
dary point on the periphery of this disk. Seidel and Walsh (p.133) have
shown that d(z) <;(1 - Izl2)'f'(z)' for 'zl < 1. There exists a se-
quence {Zn} converging to some interior point g of A such that f'(z) is
bounded. Consequently d(zn)-+ 0. Assume f(zn)-—+ c where c is pos—
sibly m. Let Pn be the pre-image of the segment on F from f(zn) to the
nearest boundary point bn' Thus f(z)-—+ bn for z e Pn as '2' —+~1.
Since there are no Koebe arcs, Pn ends at a point, say Cn' lf(z)-
f(znﬂ < d(zn)-+ 0, f(zn)-—+ c and zn-+ g for z 6 P11 as n-—+ w.
63
Then Cn—y g because there are no Koebe arcs on which f(z) --> c.
Therefore, f(z) has the finite asymptotic value bn at gm 6 A.
Now suppose there are no asymptotic values on the arc A except 0
and 00. From the preceding paragraph it follows that 0 and 00 are asymp-
totic values on a dense subset of A. Let a e A be a point at which
there is the asymptotic value 0. Hence there is a path P ending at a
such that f(z) —-> 0 as z—> a for z e P. Let GO») denote the component
of [z: lf(z)| < X, A > 0} that contains the part of P near a. Then
C ﬂaGOQCA for small positive )\ because there are no Koebe arcs on
which f(z) —-> 0. For such a value of A, G(X) does not contain any as-
ymptotic path for values 7‘ 0, but it does contain the path P on which
f(z) —> 0. Since the Riemann image surface F does not contain branch
points it follows that f(z) maps G(l) onto a copy of the universal
covering surface of {O < le < A]. This construction can be performed
infinitely often to obtain disjoint domains GkCD that are mapped by
f(z) onto the universal covering surface of [0 < lwl *

l5 'zol/p < 1 for n suffi-
ciently large, a contradiction.
Theorem lg: Let f(z) be a normal function in D that omits the value w
If
which is either finite or infinite and let A be an open subarc of C.
the set of Fatou points of f(z) on A is of measure zero, then A contains
a Fatou point of f(z) at which the corresponding angular limit of f(z)
is w. (Bagemihl, 1, Theorem 1, p.3)
If f(z) were bounded in some neigh-
Proof: Assume w is 00. Let g e A.
borhood of Q, then by a simple extension of Fatou's Theorem, the set of
Fatou points of f(z) on A would be of positive measure, which is con—
trary to the hypothesis. So f(z) is unbounded in every neighborhood of
g. Hence there exists a number 5 > 0 such that the region H = D n {z
’2 ' Q! < 5} satisfies the conditions that H0 C C A and f(z) is un-
Consequently there exists a sequence of points {Zn} in D
(M <...<
bounded in H.
l 2
such that zn -+§ and Mn = 'f(zn) 1 ->oo as n+oo where l < M
For 11 any positive integer, let Vn be the open set of points
M
n <. . . .
Let Rn denote the component of Vn that
in D for which 1f(z)l > Mn ~ 1.
contains zn [f(z)! = Mn - 1 at all boundary points of Rn that lie in
D By the maximum principle, Emma is non-empty. Suppose the diameter
__ miE
Let rn - 26erlzl. Since f(z)
of Rn does not tend to zero as n->oo.
l and there exists a Koebe se-
omits oo in D b a um tion lim
y 38 P ’ n+oo
Quence of arcs along which f(z) +00, a contradiction of Theorem 15.
rn
Thus there exists a natural number N such that RNCH.
86
We want to show that f(z) is unbounded in G1 = RN' Let G? be the
smallest simply connected region containing G1 and z ? ¢(w) be a func-
tion that maps D conformally onto G*. The set B* = G%FWC is non-empty.
1
We denote by Bf the set of all points of B* that are accessible from CY.
111 _ lim ill . . . .
let ¢*(e )‘- r_+_1 ¢(re ) for every u for which the limit ex1sts. By
Fatou's Theorem this limit exists at almost every point of C. The set
E1 = (sin: '¢*(elu)' = l} is a Borel set and hence measurable. In addi-
iu
tion B? = {¢*(elu) :e GEE We want to show that the function g(w) =
l]'
f(¢(w)) is unbounded in D. Assume not.
Suppose m(E1) > 0. Let EO denote the Borel set of positive measure
which is the subset of E1 consisting of all the points for which g(w)
possesses a radial limit and B: be the image of EO under the mapping
2 = ¢(w). From an extension of Lowner's Theorem (Tsuji, l, p.322), B:
is a measurable subset of Bf with m(B:) > 0. Let goeng. Then there
exists a path in G? terminating at £0, and this path is the image under
Z = ¢(w) of a path in D that terminates at a point e1u°~~umd¢l = Jmew>d¢l<¢> ->folnm(¢(u)>d¢l<¢(u>>
because Hm(w(u)) is continuous on [0,1] and ¢1(w(u)) is continuous and
of bounded variation on [0,1].
Theorem 12; Suppose f(z) is holomorphic and nonconstant in D. Let
A* = {lz-ll < r}, n be a positive integer and po(z), ..., pn-1(z), q(z)
be holomorphic functions in.Aﬂy Let A1= ner? D and
n-1
¢(z) = f(n)(z) + 2 p (z)f(m)(z) + q(z) for 2623.
m=o m
If ¢EEALA] and there exists a finite constant c such that ¢1(z) = ¢(z) -
q(z) # c, then fesA[AJ and A*(f), the set of finite asymptotic values,
is dense on car? C. (MacLane, 2, Theorem 1, p.275)
Proof: Sincezﬁ* can be replaced by a smaller disk contained in A? with
its center on C, it is sufficient to show that f possesses a finite
asymptotic value at one point on Aﬁflc.
133
Let E denote the subset of points g on C such that for each point
g there exists a neighborhood UQ = [iz-§| < r}r) D and a Jordan arc JC
such that ¢(z) maps UQ into the complement of JC. In a similar manner
E1 is defined using ¢1 = ¢(t)-—q(t) instead of ¢. We set E2 = E LJEl.
Suppose AﬁWW C contains points of E2. By shrinking.A* we may as-
sume.&*f) C is contained in E2. From a simple generalization of Fatou
(1) both ¢ and ¢1 have finite angular limits almost everywhere since
u(z) - ¢1(z) = q(z) has angular limits almost everywhere. Using the
notation in the proof of Lemma 1, we see that 8m has finite angular
limits almost everywhere. So f also has finite angular limits almost
everywhere on.A*F\C. From a theorem of Privalow (l, p.210) the asymp-
totic values assumed by f(z) on any interval Aﬂlﬁ C contained in E2 form
a set containing a closed set of positive harmonic measure. Consequently
this set must be infinite. If E2 is dense on.A*rW C, we are finished.
So we now assume that.A*{) C is contained in the complement of E2.
According to MacLane (1, Theorem 7, p.19) each asymptotic tract of ¢1
must end at a single point because ¢1 omits the value c. Suppose that
the asymptotic values of ¢1 are bounded by a finite constant M. We
choose two distinct points g1 and g2 on.A*() C at which ¢1 has asymptotic
values and join C1 and g2 by a curve F(Z‘A which is an asymptotic path
at both Cl and g2. Then ¢1 is bounded on T and we denote the bound by
B. Let G be the domain bounded by F and part of Aﬁf) C. If ¢1 is
bounded in G, then the arc §1§2 is contained in E2, a contradiction. So
we pick a value w such that w = ¢ (z ) for some 2 EEG and [W I >
o o 1 o o o
max(B,M). Then by the lifting argument of MacLane (1, p.13, Section 2),
¢1 has an asymptotic value a at some boundary point of G on C satisfying
the condition x);.a >"wo| > M,eacontradiction. Therefore, we now assume
134
that ¢1 has two asymptotic values along r, whose magnitudes are greater
than 2IcI, where c is the constant defined above. So 1¢1-c‘ Z 6 > 0 on
r. Since ¢1 omits fewer values in G than a Jordan are, there exists a
216 G such that |¢1(zl) - cl < 6. By the lifting argument of MacLane (l,
p.13, Section 2), there exists r1(: G ending at a point of A*f) C such
that ¢1(z) maps f1 one-to-one onto a linear segment. By Case (iii) (b)
of Lemma 1, f has a finite asymptotic value on F1.
Special cases of Theorem 19 show that for any positive integer n
if f(n) EA and f(n) 9‘ c, then feA and A*(f) is dense on C.
Theorem 29; Let fezA and f(z) # c, where c is some finite constant.
Then A: is dense on C. (MacLane, 2, Theorem 5, p.278)
2529:: We may assume without loss of generality that c==0. Let Y be an
arbitrary arc of C. First we suppose that there is an interior point
of Y such that Tim inflf(z)l = O. From MacLane (1, Theorem 11 and its
2
Corollary, pp.25 - g8) we can find a crosscut F of C from gle‘Y to gze‘y
with the properties that f has nonzero asymptotic values on F at both g1
and Q2 and Q is in the open are from £1 to C2. Let A be the domain
bounded by r and the open are from £1 to g2. |f(z)l 2 m > 0 on F. We
pick a point zo~~