Lectures on Meromorphic Functions

By W.K. Hayman

Notes by K.N. Gowrisankaran

and K. Muralidhara Rao

Contents

1 Basic Theory 1 1.1 ...... 1 1.2 The Poisson-Jensen Formula ...... 1 1.3 The Characteristic Function ...... 5 1.4 Some Inequalities ...... 11 1.5 ...... 13 1.6 The Ahlfors-Shimizu Characteristic: ...... 15 1.7 Functions in the plane ...... 23 1.8 Representation of a function...... 29 1.9 Convergence of Weierstrass products ...... 33

2 Nevanlinna’s Second Fundamental Theorem 41 2.1 ...... 41 2.2 Estimation of the error term ...... 44 2.3 ...... 52 2.4 Applications ...... 54 2.5 Picard values of meromorphic...... 61 2.6 Elimination of N(r, f ) ...... 68 2.7 Consequences ...... 71

3 Univalent Functions 73 3.1 Schlicht functions ...... 73 3.2 Asymptotic behaviour ...... 95

iii

Part I Basic Theory

1.1 1 We shall develop in this course Nevanlinna’s theory of meromorphic functions. This theory has proved a tool of unparallelled precision for the study of the roots of equations f (z) = a, f (1)(z) = b, etc. whether single or multiple and their relative frequency. Basic to this study is the Nevanlinna Characteristic T(r) which is an indication of the growth of the function f (z). We shall see in Theorem 2 that for every a T(r) is, apart from a bounded term, the sum of two components m(r, a) + N(r, a) of which the second measures the number of roots of the equation f (z) = a in |z| < r and the ﬁrst the average closeness of f (z) to a on |z| = r. The second fundamental theorem shows that in general the second term dominates and many applications giving well beyond Picard’s theorem result.

1.2 The Poisson-Jensen Formula We shall start with Poisson-Jensen formula which plays a fundamental role in our study.

Theorem 1. If f (z) is meromorphic in |z| ≤ R and has there zeros ap i and poles bν and if ζ = re , f (ζ) , 0, then for 0 ≤ r ≤ R we have 2π iφ 2 2 1 log |(Re )|(R − r )dr R(ζ − aµ) log f (reiθ) = + log 2π R2 − 2Rr cos (φ − θ) + r2 R2 − a ζ Z Xµ µ 0

1 2 Basic Theory

R(ζ − b ) (1.1) − log ν 2 Xν R − bνζ

2 Proof. Let f (z) , 0, in |z| ≤ 1. Then, since we can deﬁne an analytic branch of log f (z) in |z|≤ 1, we have by the residue theorem

1 dz log f (z) = log f (0) 2πi Z z |z|=1

By change of variable,

2π 1 log f (eiφ)dφ = log f (0) 2π Z 0

and now taking real part on both sides

2π 1 log | f (eiφ)|dφ = log | f (0)| 2π Z 0

For any ζ with |ζ| < 1, we eﬀect the conformal transformation w = z − ζ dz for the integral log f (z) . This in turn becomes, z 1 − ζz |z|R=1 1 dw log ϕ(w) = log f (ζ) where ϕ(w) = f {z(w)} 2πi Z w |w|=1

so that ϕ(0) = f (ζ). Substituting in the integral z = eiφ and taking real part we get,

2π 1 log | f (eiφ)| (1 − r2) dφ = log | f (ζ)|, ζ = reiθ 2π Z 1 − 2r cos(φ − θ) + r2 0 Basic Theory 3

Now for the function f (z) with poles bν and zeros aµ none of them being on |z| = 1, let us deﬁne

(z − b ) π ν (1 − b z) ψ(z) = f (z) ν (z − aµ) π (1 − aµz)

On |z| = 1, |ψ(z)| = | f (z)| and the function has no zeros or poles in |z|≤ 1. 3 By the above result,

2π 1 (1 − r2) log |ψ(eiθ)| dφ = log |ψ(ζ)| 2π Z 1 − 2r cos(φ − θ) + r2 0 ζ = reiθ, r < 1

Substitution for ψ gives the theorem for R = 1. In the case when there are poles and zeros on the circumference of the unit circle we proceed as follows. We have only to show that if f (z) has no zeros or poles in |z| < 1, but has poles and zeros on |z| = 1, then

1 dz log f (z) = log f (0) 2πi Z z |z|=1

For if f (z) has zeros and poles in |z| < 1 we can consider ψ(z), in place of f (z). Further we can assume that there is only one zero (the case of pole being treated in the same manner) on |z| = 1. For the case when f (z) has a finite number (it can have at most only a finite number) of zeros (poles) can be treated similarly. Let therefore z = a, |a| = 1 be a zero of f (z) on |z| = 1. Let P be the point z = a and consider a circle of radius ρ < 1 about P, ρ being small. Consider the contour SQR (fig.) Inside it f (z) has no zeros or poles. Hence by the residue theorem, log f (z) dz = log f (0). Thus it is enough to prove that log f (z) dz tendsR to zero as ρ tends to zero. QRR 4 Basic Theory

Let z = a be zero of order k. Then f (z) = (z − a)kλ(z), λ(a) , 0, in 4 a certain neighbourhood of a; and we can assume the choice of ρ such that this expansion is valid within and on the circle of radius ρ about P. dz dz dz log f (z) = k log(z − a) + log λ(z) Z z Z z Z z QR QR Since λ(z) remains bounded the second integral tends to zero. So we dz have only to prove that log(z − a) tends to zero as ρ → 0, Now, z QRR

dz log |(z − a)| log(z − a) ≤ max Z z ( |z| ) QR

1 πρ ≤ [log(1/ρ) + 0(1)]πρ → 0 if ρ ≤ 2 This proves the result, in the case when the function f (z) has zeros or poles on the unit circle. In case R , 1, we consider the function f (Rz) instead of f (z) and arrive at the result. Hence the theorem is proved completely. Corollary. In the special case when ζ = 0 we get the Jensen’s formula

2π 1 aµ b (1.2) log | f (0)| = log | f (Reiφ)|dφ + log − log ν 2π R R Z Xµ Xν 0

the summation ranging over poles and zeros of f (z) in |z|≤ R. The above formula does not hold if zero is a pole or a zero of f (z). If f (0) = 0 or ∞ Basic Theory 5

λ and f (z) is not identically constant then f (z) = Cλz +···+··· . Consider f (z)/zg. This has neither zero nor pole at zero. Hence we get,

2π iφ 1 f (Re ) |aµ| |b | log |C | = log dφ + log − log ν λ λ 2π Z R X R X R 0 2π 1 |aµ| |b | = log | f (Reiφ)|dφ + log − log ν − λ log R 2π Z R R 0 X X where sums are taken over zeros and poles of f (z) in 0 ≤ |z|≤ R. 5

1.3 The Characteristic Function Set for x, real and positive, log+ x = log x if x > 1, log+ x = 0 if x ≤ 1, Then clearly log x = log+ x − log+(1/x). So

R R 1 log dn(t) = log | | Z t ν bν 0 X given on integrating by parts,

R R R dt n(t) log + n(t) = log(R/|bν|) t 0 Z t ν 0 X 6 Basic Theory

The ﬁrst term is zero, in consequence of the fact n(t) = 0, near zero. We write n(r, f ) for the number of poles of f (z) in |z| ≤ r, so that n(r, 1/ f ) is equal to the number of zeros of f (z) in |z| ≤ r. We deﬁne N(r, f ) to be

r dt (3) n(t, f ) Z t 0 r dt 6 If f (0) = ∞ we deﬁne N(r, f ) = [n(t, f ) − n(0, f )] + n(0, f ) log r. t R0 Then the formula (1.2) becomes, for f (0) , 0, ∞

2π 2π 1 1 1 log | f (0)| = log+ | f (reiθ)|dθ − log+ dθ + N(r. f ) 2π Z 2π Z | f (reiθ)| 0 0 −N(r, 1/ f ).

We deﬁne,

(1.4) T(r, f ) = N(r, f ) + m(r, f )

where,

2π 1 (1.5) m(r, f ) = log+ | f (reiθ)|dθ 2π Z 0 Again (1.2) takes the form, for f (0) , 0, ∞

(1.2′) T(r, f ) = T(r, 1/ f ) + log | f (0)|

λ If f (z) ∼ CλZ near z = 0, where λ , 0, then we obtain T(r, f ) = T(r, 1/ f ) + log |Cλ|. In future such modiﬁcations will be taken for granted. The function T(r, f ) is called the Characteristic Function of f (z). This is the Nevanlinna characteristic function. Theorem 2. First fundamental theorem. Basic Theory 7

For any finite complex a, T(r, f ) = T[r, 1/( f − a)] + log | f (0) − a| + ǫ(a) where |ǫ(a)|≤ log+ |a| + log 2. + + + Proof. We note that log |z1 + z2|≤ log z1| + log |z2| + log 2 + + + and log |z1 − z2|≥ log |z1|− log |z2|− log 2. Whence, 7 log+ | f (z) − a|− log+ | f (z)|≤ log 2 + log+ |a| Integrating we get, − log 2 − log+ |a| + m(r, f − a) ≤ m(r, f ) ≤ log 2 + log+ |a| + m(r, f − a) Since f and f − a have the same poles, N(r, f ) = N(r, f − a). Therefore, T(r, f − a) − log+ |a|− log 2 ≤ T(r, f ) ≤ log 2 + log+ |a| + T(r, f − a) That is, |T(r, f ) − T(r, f − a)|≤ log 2 + log+ |a| T(r, f ) + T(r, f − a) + ǫ(a), where |ǫ(a)|≤ log 2 + log+ |a| From (1.2′) we have, 1 T(r, f ) = T r, + log | f (0) − a| + ǫ(a) f − a! where |ǫ(a)|≤ log 2 + log+ |a|. Hence the theorem is proved. 1 1 If we write m(r, a), N(r, a) for m r, , N r, , then m(r, a) f − a! f − a! represents the average degree of approximation of f (z) to the value a on the circle |z| = r and N(r, a) the term involving the numbers of zeros of f (z) − a. Their sum can be regarded as the total affinity of f (z) for the value a and we see than apart from a bounded term the total affinity for every value of a. However, the relative size of the two terms m, N remains in doubt. We shall see in the second fundamental theorem that in general it is N(r, a) that is the larger component. The value of a for which this is not the case will be called exceptional. 8 Basic Theory

Let us now consider some examples 8 1. Let f (z) = P(z)/Q(z), P(z) and Q(z) being polynomials of degrees m and n respectively, and prime to each other, that is have no common roots. Then for large r, | f (z)|∼ Crm−n. So if m is greater than n, m(r, f ) = (m − n) log r + 0(1) and since

log+ x = 0, for 0 < x ≤ 1, m(r, 1/ f ) = 0 (1.6) and m(r, 1/( f − a)) = 0.

for large r and fixed a. Since f (z) = ∞ has n roots in the open plane, n(t, f ) = n, for t > t0 and N(r, f ) = n log r + 0(1). Again T(r, f ) = m(r, f ) + N(r, f ) which is equal to m log r + 0(1) for a finite again by (1.6) and theorem 2, N(r, 1/( f −a)) = m log r+0(1), 1 a finite. Thus N r, /T(r, f ) → 1 as r →∞. In this case " f − a! # a = ∞ is the only exceptional value. Similar conclusions follow if m is less than n by taking 1/ f in place of f in the above discussion. In this case a = 0 = f (∞) is the only exceptional value. If m is equal to n, f = c + 0(z−λ) (say). Or writing, f − c ∼ b(z−λ), we see that m(r, 1/( f − c)) = λ log r + 0(1)

m(r, 1/( f − a)) = 0(1), when a , c m(r, f ) = 0(1), N(r, f ) = n log r + 0(1)

Thus T(r, f ) = n log r + 0(1) and,

N(r, a) = n log r + 0(1), a , c N(r, c) = (n − λ) log r + 0(1).

Thus in any case

T(r, f ) = (Max . m, n) log r + 0(1) i.e., T(r, f ) ∼ (Max . m, n) log r. Basic Theory 9

9 Thus in the case of a rational function there is always only one exceptional value, viz., f (∞)

2. Let f (z) = ez. In this case the value of m(r, f ) is,

2π m(r, f ) = 1/2π log+ er cos θdθ Z 0 π/2 = 1/2π cos θ dθ Z −π/2 = r/π

because cos θ ≤ 0 in π/2 ≤ θ ≤ π, −3π/2 ≤ θ ≤ −π/2 and so log+ er cos θ is equal to zero.

Since ez has no poles, N(r, f ) = 0. Consequently T(r, f ) is equal to m(r, f ) which in turn is equal to r/π. We employ the notation m(r, a) = m(r, 1/( f − a)) for ﬁnite a and m(r, ∞) = m(r, f ), and similarly with the functions n, N, and T. We have |ez − a| ≥ ||ez| − |a|| = |er cos θ − |a|| if z = reiθ. If a , 0, we have for large r e r < |a| < er. Therefore |a| = er cos α for 0 ≤ α ≤ π. Thus 2π 1 2m(r, a) = log+ dθ |ez − a| R0

2 1 ≤ log+ dθ Z |er cos θ − er cos α| 0 2 1 1 = 2π log+ + 2 log+ dθ er cos α Z |er(cos θ−cos α) − 1| 0

If cos θ − cos α ≥ 0, we have |er(cos θ−cos α) − 1| 10

r2 = r(cos θ − cos α) + (cos θ − cos α)2 + ···≥ r(cos θ − cos α) 2 10 Basic Theory

≥ (cos θ − cos α) if r ≥ 1.

1 r(cos θ−cos α) r(cos θ−cos α) 1 If 2 < |e − 1| and cos α − cos θ > 0 we have e > 2 or 0 < r(cos α − cos θ) < 1. Since 1 − e−x = xe−x we have

|er(cos θ−cos α) − 1| = 1 − e−r(cos α−cos θ) ≥ r(cos α − cos θ)e−r(cos α−cos θ) r(cos α − cos θ) cos α − cos θ ≥ ≥ 2 2

1 2 ≤ 2 log+ + 2 log+ 2dθ + 2 log+ dθ. |a| Z Z | cos θ − cos α| E [0,π]−E π 1 2 ≤ 2 log+ + 4π log 2 + 2 log dθ |a| Z | cos θ − cos α| 0

π 2 11 Further log dθ is a continuous function of α in | cos θ − cos α| R0 0 ≤ α ≤ π and hence is bounded. Thus

m(r, a) = 0(1) Basic Theory 11

Hence

T(r, a) = T(r, f ) + 0(1) = r/π + 0(1). i.e. N(r, a) + m(v, a) = r/π + 0(1). i.e. N(r, a) = r/π + 0(1).

N(r, a) This shows that → 1. T(r)

1.4 Some Inequalities We have already seen that

+ + + log z1 + z2 log z1 + log |z2| + log 2

More generally,

ν=n + + + + + log zν ≤ log |n max |zν| ≤ log n log | max |zν| Xν=1

ν=n + + ≤ log n + log |zν| Xν=1 Hence, N n=N ≤ + m r, fn m(r, fn) log N  Xn=1  Xn=1   N   Now F = fn(z) has poles only where the fn(z) have poles and n=1 the multiplicityP of such pole is at most the maximal multiplicity of the poles of fn(z) which is not greater than the sum of the multiplicities. This gives

N N(r, F) ≤ N(r, fn) Xn=1 N N ≤ + and T r, fn T(r, fn) log N  Xn=1  Xn=1     12 Basic Theory

Again if |a1a2 ... an|≤ 1 12 n + + log |a1a2 ... an| = 0 ≤ log |aν| X1 + If |a1a2 ... an| > 1, log |a1a2 ... an| = log |a1a2 ... an|, which is in turn is equal to sum of log |ai| i = 1 to m, and hence less than or equal to n + log |ai|. i=1 P Therefore we get

If a, b, c, d are constants with ad − bc , 0,

T(r, (a f + b)/(c f + d)) = T(r, f ) + 0(1)

For,

T(r, (a f + b)/(c f + d)) = T(r, a/c + bc − ad)/(c f + d)c) = T(r, k/( f + k′)) + 0(1) and by the ﬁrst fundamental theorem,

= T(r, ( f + k′)/k) + 0(1) ≤ T(r, f + k′) + 0(1) ≤ T(r, f ) + 0(1)

Similarly we get T(r, (a f + b)/(c f + d)) ≥ T(r, f ) + 0(1). Basic Theory 13

1.5 13 2π Theorem 3 (H. Cartan). We have T(r, f ) = (1/2π) N(r, eiθ)dθ + R0 log+ | f (0)| that is, T(r, f ) apart from a constant term is the average of N(r, a), for a on the circle |a| = 1. For the proof we require the following lemma. Lemma 1. If a is complex,

2π 1 log |eiθ − a|dθ = log |a| = log+ |a|, |a|≥ 1 2π Z 0 If

|a| < 1 write, |a − eiθ| = |a||1 − eiθ/a| = |a||1 − e−iθ/a| = |a||eiθ − i/a| 1 and we get by the ﬁrst part, since > 1, a

2π 1 1 log |eiθ − a|dθ = log |a| + log = 0 = log+ |a|. 2π Z a 0

To prove the theorem consider now Jensen’s formula applied to f (z)−eiθ,

2π 1 log | f (0) − eiθ| = log | f (reiφ) − eiθ|dφ + N(r, f ) − N(r, 1/( f − eiθ)) 2π Z 0 14 Basic Theory

since N(r, f ) = N(r, f − eiθ) ( f and f − eiθ have the same poles). Now integrate both sides with respect to θ and invert the order of integration in the ﬁrst term in the right hand side which we can do by Fubini’s + 14 Theorem, because the integrand is bounded above by log [| f (reiφ)| + 1] and this is positive, integrable and independent of θ.

2π 2π 2π 1 1 1 log | f (0) − eiθ|dθ = dθ log | f (reiφ) − eiθ|dφ 2π Z 2π Z 2π Z 0 0 0 2π 2π 1 1 + N(r, f )dθ − N(r, eiθ)dθ 2π Z 2π Z 0 0 and by the previous lemma,

2π 2π 1 1 log+ | f (0)| = log+ | f (eiφ)|dφ + N(r, f ) − N(r, eiθ)dθ 2π Z 2π Z 0 0 2π 1 = T(r, f ) − N(r, eiθ)dθ 2π Z 0 which gives the result.

Corollary 1. T(r, f ) is an increasing convex function of log r. It is also easily seen that m(r, f ) need not be either an increasing or a convex function of log r (e.g. by considering rational functions). In fact from theorem 3,

2π r 2π d(T(r)) d 1 dt 1 = dθ n(t, eiθ) = n(r, eiθ)dθ d log r d log r 2π Z Z t 2π Z 0 0 0 and the right hand side is non-negative and non-decreasing with r.

1 2π Corollary 2. m(r, eiθ)dθ ≤ log 2. 2π R0 Basic Theory 15

In fact by theorem 2,

1 m(r, eiθ) + N(r, eiθ) = T r, f − eiθ ! = T(r, f ) − log | f (0) − eiθ)|+ < log 2 > where <ǫ > denotes any quantity that is less than ǫ in modulus. 15 Integrating,

2π 2π 1 1 m(r, eiθ)dθ + N(r, eiθ)dθ 2π Z 2π Z 0 0 = T(r, f ) − log+ | f (0)|+ < log 2 > by the lemma. So

2π 1 m(r, eiθ)dθ + T(r, f ) − log+ | f (0)| 2π Z 0 = T(r, f )+ < log 2 > − log+ | f (0)| which is the required result. i.e.

2π 1 m(r, eiθ)dθ =< log 2 > 2π Z 0

1.6 The Ahlfors-Shimizu Characteristic: We have deﬁned the Nevalinna characteristic function of a meromorphic function f (z). We now proceed to deﬁne the characteristic function after Ahlfors [1] and Shimizu [1]. Prior to that let us prove the following lemma.

Lemma 2 (Spencer). Suppose that D is a bounded domain in the complex plane, whose boundary is composed of a ﬁnite number of analytic 16 Basic Theory

curves, γ, and that G(R) is twice continuously diﬀerentiable on the set of values R assumed by | f (z)| in D and on γ, f (z) being regular in D and on γ. Then ∂ G(| f (z)|)ds = g(| f (z)|)| f ′(z)|2dx dy Z ∂n " γ D

16 where g(R) = G′′(R)+(1/R)G′(R) and we diﬀerentiate out of the domain ∂ along the normal in . ∂n Let us make use of Green’s formula which under the hypothesis of the lemma gives,

∂ ∂2 ∂2 G(| f |)ds = ∇2[G(| f |)]dx dy ∇2 = + Z ∂n " ∂x2 ∂y2 γ D

Suppose ﬁrst that f (z) has no zeros for z in D. Put u = log | f |, in order to facilitate the easy calculation of ∇2G(| f |). u is a harmonic function. Therefore, ∇2u = 0. Now | f | = eu, G(| f |) = G(eu). Calculating the partial derivatives with respect to x and y we see that,

∂2 ∂u 2 ∂2u [G(eu)] = euG′(eu) + e2uG′′(eu) + euG′(eu) ∂x2 ∂x ∂x2 ! n o and

∂2 ∂u 2 ∂2u G(eu) = euG′(eu) + e2uG′′(eu) + G′(eu)eu ∂y2 ∂y ! ∂y2 n o ∂u ∂u d And u = Rl. log f (z) and − i = (log f ) = f ′/ f . ∂y ∂y dz Hence on addition we get,

| f ′(z)|2 ∇2[G(eu)] = eu G′(eu) + euG′′(eu) | f (z)|2 since ∇2u = 0. Basic Theory 17

Writing | f (z)| = R,

∇2[G(R)] = | f ′(z)|2[G′(R) + RG′′(R)]/R = g(R)| f ′(z)|2 where g(R) = (1/R)G′(R) + G′′(R). 17 This establishes the result provided that f (z) , 0 for z in D. Other- wise, exclude the zeros (which are necessarily finite in number) of f (z) ∂ in D, by circles of small radii, and in these G(| f |) is bounded, since ∂n by hypothesis G(R) is continuously differentiable near R = 0. Hence the contribution to the left hand side of the formula tends to zero with the radii of the circles, and so the lemma is proved. = 1 + 2 Let us now specialise with G(R) 2 log(1 R ) and f (z) a meromorphic function in |z|, r with no poles on |z| = r. Exclude the poles in |z| < r by circles of radii ρ (small). Let us apply the lemma to the region D consisting of |z| < r with the poles excluded. We find in this case,

g(R) = (1/R)G′(R) + G′′(R) = 2/(1 + R2)2, and ∂ G(| f |dx = g(| f |)| f ′(z)|2dx dy. Z ∂n " γ D

γ consists of the circumference of |z| = r and the smaller circles. Near a pole z0 of order k, of f (z)

k | f (z)| ∼ |0|/λ |z − z0| = λ and 1 G(| f |) = log(1 + | f |2) = log | f | + 0(1). 2 that is,

G(R) = k log γλ + 0(1) Therefore, ∂ ∂ 1 [G(R)] = − k log + 0(1) ∂n ∂ρ ρ! 18 Basic Theory

∂ since in the lemma the derivative is along the normal directed out of ∂n the region and hence here into the circle (isolating z0). 18 ∂ [G(R)] = k/ρ + 0(1). ∂n Hence the contribution to the left hand side in the lemma is 2πρ[k/ρ + 0(1)], that is, 2πk + 0(ρ). Adding over all the circles and letting ρ tend to zero,

∂ 2 1 2πn(r, f ) + [log(1 + | f | ) 2 ]ds Z ∂n |z|=r 2| f ′(z)|2 = dx dy " [1 + | f (z)|2]2 |z|

2π 1 i 2 1 2 1 N(r, f ) + log(1 + | f (re )| ) 2 dθ − log(1 + | f (0)| ) 2 2π Z 0 r A(t)dt (1.7) = . Z t 0 The integration is justiﬁed since both sides are continuous and have equal derivatives except for the isolated values of r for which |z| = r Basic Theory 19

+ 2 1 contains poles of the function f (z). Now log R ≤ log(1 + R ) 2 ≤ 1 + + 1 + 2 2 + 1 19 log R 2 log 2, if R ≥ 1 so that log(1 R ) ≤ log R < 2 log 2 > and 1 + + 2 2 1 = 1 + if R ≤ 1, 0 ≤ log(1 R ) ≤ 2 log 2 2 log 2 log R so that in all 1 + 1 + 2 2 + 1 + iθ 2 2 = cases log(1 R ) ≤ log R < 2 log 2 >. Hence log[1 | f (re )| ] 1 + + iθ < 2 log 2 > log | f (re )| substituting,

r 2π dt 1 1 A(t) = log+( f (reiθ)|dθ + N(r, f )+ < log 2 > − log+ | f (0)| Z t 2π Z 2 0 0 1 = m(r, f ) + N(r, f )+ < log 2 > − log+ | f (0)| 2 1 = T(r, f )+ < log 2 > − log+ | f (0)| 2 Now if we put, r dt T0(r, f ) = A(t) Z t 0 We get 1 T (r, f ) = T(r, f )+ < log 2 > − log+ | f (0)| 0 2

Deﬁnition. T0(r, f ) is called Ahlfors-Shimizu Characteristic function of f (z).

Interpretation of (1.7) 20 Basic Theory

Consider the Riemann Sphere of diameter one lying over the w-plane and touching it at w = 0. To every point w corresponds the point P(w) in which the line Nw (fig) intersects the sphere. From the figure NP = 2 1 cos θ = 1/(1 + R ) 2 , R = |w|. NP is called the chordal distance between w and ∞ and is denoted by [w, ∞]. 20 Then 2π 2π 1 iθ 2 1 1 1 log[1 + | f (re )| ] 2 dθ = log dθ 2π Z 2π Z [ f (z), ∞] 0 0 and so is the average of the chordal distances between f (z) and infinity, when z runs over |z| = r. The left hand side of the above equation is an alternative for m(r, f ) in the sense that it differs from m(r, f ) by less than 1 2 log 2. It is easy to see that if ds is an element of length in the plane and dσ the corresponding element on the Riemann-sphere then dσ = ds/(1 + R2). Hence if du dv is an element of area in the w-plane, then the corresponding element of area on the sphere is du dv/(1 + R2)2. If dx dy is an element of area in z-circle, |z| < r, its image in the w-plane is | f ′(z)|2dx dy, since the element of length is multiplied by | f ′(z)|. Thus the element of area corresponding to dx dy on the Riemann-w-sphere is precisely | f ′(z)|2dx dy/[1 + | f (z)|2]2. Therefore, we interpret πA(r) as the area on the Riemann-w-sphere of the image, with due count of multiplicity (the map w = f (z) may not be one-one, and more than one lement of arc x in the z-plane may go into the same element of arc x in the w-plane) of |z| < r by w = f (z). Since the area of the sphere is π, A(r) itself being the area of the image divided by the area of the sphere may be interpreted as the average number of roots in |z| < r of the equation f (z) = w, as w moves over the Riemann sphere.

21 1 We have seen above that A(r) = area of image of |z| < r by f (z) on the π Riemann Sphere. So that A(r) can be interpreted as the average value of the number n(r, a) of roots of f (z) = a, in |z| < r as a runs over the complex plane. Basic Theory 21

Let us now rotate the sphere. This corresponds to a one-one conformal map of the closed plane onto itself and in fact to a bilinear map of the type w′ = eiλ(1 + aw)/(w − a) (for the proof refer Carath´eodery, [1]). Further under this rotation A(r) will remain unaltered. Set ϕ(z) = eiλ[1 + a f (z)]/[ f (z) − a], and observe that [ϕ(z), ∞] = [ f (z), a], where w, a is the chordal distance between the points corresponding to w and a on the Riemann Sphere. In particular [w, ∞] = 2 1 {1 + |w| } 2 . If we apply our previous result to ϕ(z), we get

r 2π dt 1 1 1 A(t) = N(r,ϕ) + log dθ − log Z t 2π Z [ϕ(reiθ), ∞] [ϕ(0), ∞] 0 0 2π 1 1 1 = N(r, a) + log dθ − log . 2π Z [ f (reiθ), a] [ f (0), a] 0 Thus we have proved the theorem. Theorem. For every complex a and a = ∞

r 2π dt 1 1 1 A(t) = N(r, a) + log dθ − log . Z t 2π Z [ f (reiθ), a] [ f (0), a] 0 0 The new chordal distance is 22 | − | = 1 = w a [w, a] 1 1 1 + aw 2 2 [(1 + |a|2)(1 + |w|2)] 2 1 +  w − a      Thus in this above theorem, the expression m0(r, a) replaces m(r, a) of the Theorem 2, where, 2π 2 iθ 2 1 1 log[(1 + |a| )(1 + | f (re )| )] 2 m0(r, a) = dθ 2π Z | f (reiθ) − a| 0

Unlike theorem 2 theorem 2’ is exact. Notice that m0(r, a) is always non- negative because the chordal distance between any two points is always is less or equal to one. 22 Basic Theory

Corollary. If f (z) is meromorphic in the plane then r r dt dt n(t, a) ≤ A(t) + C. Z t Z t 1 1 for all a where C is independent of a and r(> 1) but may depend on f . Proof. r dt A(t) = N(r, a) + m0(r, a) − m0(0, a) Z t 0 1 dt A(t) = N(1, a) + m0(1, a) − m0(0, a) Z t 0 r dt subtracting, since N(r, a) = n(t, a) , t R0 r r dt dt n(t, a) = A(t) + m0(1, a) − m0(r, a) Z t Z t 1 1 r dt ≤ A(t) + max m0(1, a) Z t a 1

23 since m0(r, a) is greater or equal to zero. The maximum on the right is finite. In fact, for variable a and fixed rm0(r, a) is continuous in a and is bounded as a moves over the Riemann sphere. This is evident if the function f (z) , a on |z| = r. If f (z) = a, at a finite number of points the argument is similar to that in the Poisson- Jensen formula. Putting C = Maxa m0(1, a) the result is obtained. Remark. By an inequality for real positive functions due to W.K. Hay- man and F.M. Stewart [1] and using corollary 1 we deduce that if f (z) is meromorphic and not constant in the plane and n(r) = sup ·n(r, a) and ǫ > 0, a Basic Theory 23 then

(1.8) n(r) < (e + ǫ)A(r) ...... on a set E of r having the following property. If E(r) is the part of E in dt [1, r] then ≥ c(ǫ) log r for all large r, where c(ǫ) depends only on t ER(r) ǫ.

The following are some open problems:

(a) Can e be replaced by a smaller constant and in particular by one in (1.8)?

(b) Does (1.8) hold for all large r; or for all large r except a small set? 24

r dt n(t) < (constant). A(r) Z t 1 for some arbitrarily large r?

1.7 Functions in the plane 25 Let S (r) be a real function ≥ 0, and increasing for 0 ≤ r ≤∞. The order k and the lower order λ of the function S (r) are deﬁned as

k = lim [log S (r)]/ log r. λ) r→∞ The order and the lower orders of the function always satisfy the relation, 0 ≤ λ ≤ k ≤∞. If 0 < k < ∞, we distinguish the following possibilities,

(a) S (r) is maximal type if

C = limS (R)/Rk = inﬁnity. 24 Basic Theory

(b) S (r) is of mean type of order k if 0 < C < ∞.

(d) S (r) is of convergence class if,

∞ S (t)dt/tk+1 converges. Z 1

Note that if S (r) is of order k and ǫ > 0,

S (r) < rk+ǫ for all large r.

and S (r) > rk−ǫ for some large r. (This follows from the deﬁnition of order of S (r).) It can be seen that if S (r) is of order k and of convergence type i.e., (d) then it is of minimal type, (c). In fact in this case

∞ S (r) dr < ǫ if r0 > t(ǫ) Z rk+1 r0

26 Then 2r0 S (r) dr < ǫ Z rk+1 r0 and since S (r) increases with r,

2r0 S (r) S (r ) dr ≥ 0 r k+1 k+1 0 Z r (2r0) r0 and −(k+1) S (r0)2 /r0k < ǫ for r0 > t(ǫ) that is, lim S (r)/rk = 0. r→∞ Basic Theory 25

So if (d) holds S (r) is of minimal type. We also note that if k′ is greater than the order of S (r) then,

∞ ′ S (r)dr/rk +1 converges. Z 1

∞ S (r) and if k′ is less than k dr diverges. rk′+1 R1

We have next Theorem 4. If f (z) is a regular function for |z|≤ R and

M(r, f ) = Max|z|=r f (z) then, r + R T(r, f ) ≤ log+ M(r, f ) ≤ T(R, f ) for 0 < r < R. R − r Proof. Since f (z) is regular in |z|≤ R and r < R we have,

2π 1 T(r, f ) = T(r) = m(r, f ) = log+ | f (reiθ)|dθ 2π Z 0 1 + iθ ≤ Max = log | f (re )|2π 2π |z| r = log+ M(r, f )

To prove the other side of the above inequality of the theorem we dis- 27 tinguish between the two cases (i) M(r) < 1 or (ii) M(r) ≥ 1. In the case (i), log+ M(r) = 0 and T(r, f ) × (R + r)/(R − r) being non-negative the inequality holds good. Hence we can now suppose that M(r) ≥ 1, and in which case log+ M(r) = log M(r). The Poisson-Jensen formula gives, for z = reiθ

2π 1 (R2 − r2) log | f (z)| = log | f (Reiφ)| dφ. 2π Z R2 + 2Rr cos(φ − θ) + r2 0 26 Basic Theory

2 (R − aµz) − log+ R(z − aµ) X

in the above equality aµ runs through all the zeros of f (z). For, as regards the zeros with modulus less or equal to R

2 2 2 (R − aµz) (R − aµz) (R − aµz) log+ = log since ≥ 1. R(z − aµ) R(z − aµ) (z − aµ)R

and for the zeros aµ with modulus greater than R,

2 (R − aµz) log+ = 0 R(z − aµ)

and Poisson’s formula is unaﬀ ected.

Clearly,

(R2 − r2) (R2 − r2) log | f | ≤ log+ | f | R2 − 2Rr cos(φ − θ) + r2 R2 − 2Rr cos(φ − θ) + r2 R + r R + r ≤ log+ f − log+ f R − r R − r 28

This being true for all z, |z| = r, holds good for the z at which f (z) takes the maximum M(r, f ). Hence, log M(r, f ) ≤ (R + r/R − r)T(R, f ). Basic Theory 27

Corollary. If f (z) is an integral function, the order and type (a, b, c or d deﬁned in the last article) of T(r) and log+ M(r) are the same. From the left hand side of the inequality of Theorem 4, it follows that the order and type of T(r, f ) are not bigger than those of log+ M(r). In order to prove the reverse, let us substitute R = 2r in the right side of the theorem. We get,

log+ M(r, f ) ≤ 3T(2r, f )

Suppose k is the order of T(r, f ), then given ǫ > 0, T(r, f ) < ǫrk+ǫ for all r > r0 by the deﬁnition of order. Hence combining the two inequalities,

+ k+ǫ log M(r, f ) ≤ 3T(2r, f ) < 3ǫ(2r) for r > r0. This implies that the order of log+ M(r, f ) is less or equal to k. If T(r) has mean type, we may take ǫ = 0. If T(r) has minimal type, we may 29 in addition take c arbitrarily small. Hence T(r, f ) and log+ M(r, f ) both have same order and belong to the same order type a, b or c. In order to complete the corollary, we have to consider the case when T(r, f ) is of (d) i.e., convergence type. Consider,

∞ log+ M(r) dr Z rk+1 r0 ∞ ∞ log+ M(r) 3T(2r) dr ≤ dr Z rk+1 Z rk+1 r0 r0 k being the order of T(r, f ).

∞ T(r)dr = 3.2k Z rk+1 2r0 by change of variable. Since T(r) is of convergence type, it follows from the above inequality that log+ M(r) also belongs to the same class. It is clear that this relation holds good in the reverse direction. 28 Basic Theory

We now proceed to deﬁne the order and the order type of a function f (z) meromorphic in the plane on the above analogy.

Deﬁnition . The order and type of a function f (z) meromorphic in the plane are deﬁned as the order and type of T(r, f ).

We note that this coincides with that of order and type of log+ M(r, f ) if f (z) is an entire function.

30 Example. (1) If the function f (z) = ez, log M(r) = r and T(r) = r/π. The function has order one, mean type. In this case log+ M(r) and T(r) are of the same type, viz., mean type, though the values of the type are diﬀerent. More precisely, for T(r), lim T(r)/r = 1/π r→∞ (the order being 1) and for log+ M(r), lim (1/r) log+ M(r) = 1. r→∞ In the above example the ratio of the two super. limites that is, T(r) lim r→∞ k r = 1 log+ M(r) lim r→∞ rk In general case for a function of order k and mean type this ratio is bounded by 1. But it is still an open problem to ﬁnd the best possible lower bound. It can be shown easily from theorem 4, taking R = r[1 + (1/k)], that a lower bound is 1/e(2k + 1). In this connection P.B. Kennedy [1] has given a counter example of a function of order k mean type in | arg z| < π/2k, and bounded for π/2k ≤ arg z ≤ π, provided 1 k > 2 , such that

log+ | f (reiθ)|∼ rk cos kθ for |θ| < π/2k π log+ | f (reiθ)| = 0(1) ≤ |θ| <π 2k Hence T(r, f ) ∼ rk/k, log M(r) ∼ rk. Thus e(2k + 1) cannot be replaced 1 by any constant less than kπ if k > 2 . Basic Theory 29

Remark . For functions of inﬁnite order T(r) and log+ M(r) no longer have necessarily the same order of magnitude. Actually it has been shown by W.K. Hayman and F.M. Stewart [1] that if ǫ > 0, log M(r) < z (2e+ǫ)[T(r)+ A(r)] for some large r. As an example consider f (z) = ee 31

log M(r) = er and r 3 1 1 T(r) ∼ e (2π ) 2 /r 2 .

1.8 Representation of a function in terms of its zeros and poles 32 Deﬁnition. For any complex number z, and any integer q > 0,

z+ 1 z2+···+ 1 zq E(z, q) = (1 − z)e 2 q .

Theorem 5 (Nevanlinna). If f (z) is meromorphic in the plane with zeros aµ and poles bν and f (z) being of order at most q of the minimal type then, z E , q − 1 1aµ|

Note that the theorem only asserts that the limit on the right exists, but it does not indicate whether the products are convergent or divergent. In fact if f (z) only satisfies the weaker condition that lower limit as r tends to infinity of T(r)/rq is equal to zero (instead of the condition assumed in the theorem viz., upper limit of the same is zero.), the result still holds provided that R is allowed to tend to infinity through a suitable sequence of values, instead of all values. To start with let us assume that the function f (z) has no zero or pole 33 30 Basic Theory

at zero. The Poisson-Jensen formula gives,

R2 − b z z = reiθ + log ν R(z − bν) |bXν|

Both the sides are equal harmonic functions v(z) (say) of z near the point ∂v ∂v z = reiθ where f (reiθ) , 0, ∞. Let us operate −i on both the sides. ∂x ∂y We assume that R is such that f (z) , 0, ∞ on |z| = R. Diﬀerentiating under the integral sign and observing that

Reiφ + z (R2 − r2) Real = Reiφ − z! R2 − 2Rr cos(φ − θ) + r2 We deduce,

2π f ′(z) 1 2Reiφ = log | f (Reiφ)| dφ f (z) 2π Z (Reiφ − z) 0 1 a − − µ 2 "(aµ − z) (R − aµz)# |aXµ|

2π q−1 ′ | iφ | iφ d f (z) = 1 log f (Re ) Re dz! f (z) ! π Z (Reiφ − z)q+1 0 q 1 b + (q − 1) − ν (b − z)q 2 q  |bX|

q 1 aµ + (q − 1) − q 2 q (aµ − z) (R − zaµ)  |aXµ|

2Rk dt T(2Rk, f ) ≥ N(2Rk, f ) ≥ n(t, f ) ≥ n(Rk, f ) log 2. Z t Rk q q Thus n(Rk, f )Rk tends to zero as k tends to ∞. Similarly, n(Rk, 1/ f )/Rk = + q → 0, as k →∞) since, T(Rk, 1/ f ) T(Rk, f ) 0(1) and T(Rk, f )/Rk → 0. Our aim now is to show that some of the terms on the right of the equation 11.9 including the integral tend to zero uniformly for z on any bounded set as k tends to infinity. Then letting k tend to infinity, we get a modified equation integrating which q times we will get the result. 1 1 Now suppose that |z| < R . Then, |b z| < R · R for |b | < R . 35 2 k ν 2 k k ν k and 1 |R2 − b z|≥ R2 − |b z|≥ R2 k ν k ν 2 k Hence, q q b R q ν k = 2 < q 2 − 1 R (Rk bνz) ( R2)q k 2 k this inequality being true for all poles bν with |bν| < Rk. Therefore, summing up for all poles bν, |bν| < Rk

q q q bν bν n(Rk, f )2 ≤ < q 2 − 2 − q R X (Rk bνz) X (Rk bνz) k and hence the right hand side tends to zero uniformly as k tends to in- ﬁnity for z in any bounded set. A similar result holds good in the case of the zeros of the function. 32 Basic Theory

Now coming to the integral, 1 1 |R eiφ − z)|≥ R for |z| < R . k 2 k 2 k Hence the modulus of the integral on the right of (1.9) is at most,

2π q+1 Rk 2 iφ (q!) + log | f (Rke )| dφ π Rq 1 Z k 0

2π 2π q+1 (q!) 2 + iφ + + 1 q  log | f (Re )|dφ log dφ π R Z Z | f (Reiφ)| k    0 0  q+2  2 + < (q!) q [m(Rk, f ) m(Rk, 1/ f )] Rk T(Rk, f ) T(2Rk, f ) < (Const.) q < (Const.) q → 0 Rk Rk as k →∞.

36 Now the equation (1.9) takes the form,

d q−1 f ′(z) = lim S k(z) dz! f (z) k→∞ Where

= 1 1 S k(z) (q − 1)!  q − q  (bν − z) (aµ − z)  |bXν|

f ′(z) 1 1 z zq−2 = lim − − −···−   2 q−1  f (z) k→∞ (bν − z) bν bν b |bXν|

1 1 zq−2 − − −···− + p (z)  q−1  q−2 (aµ − z) aµ a  |aXµ|

2 q−1 z + z +···+ z z aµ 2a2 q−1 1 − e µ (q−1)aµ |aµ|

1.9 Convergence of Weierstrass products 38 Let a1, a2,... an ... be a sequence of complex numbers (none 0) with moduli r1, r2,..., rn,... in the increasing order of magnitude. Let n(r) be deﬁned as n(r) = Sup. k. rk

R 1 1/rk = dn(t) n Z tk rXn

R dt = k n(t) + n(R)/Rk because n(t) = 0 Z tk+1 0 ∞ dR near 0. Suppose now I = n(R) is less than inﬁnity. n(R) has at 1 Rk+1 R0 most convergence type of order k. (Hence it is also of minimal type). k = k Therefore upper limit of n(r)/r as r tends to inﬁnity zero, and 1/rn converges to k(I1). The convergence of the series implies the conver-P gence of the integral, because n(R)/Rk ≥ 0.

39 Now consider the integral, R R dr N(r) = N(R)/(−k)Rk + dN(r)/krk Z rk+1 Z 0 0 R n(r) = −N(R)/kRk + dr/rk+1 Z k 0 ∞ From this inequality we get that the integrals N(r)dr/rk+1 and 0 ∞ R n(r)dr/rk+1 will converge or diverge together, due to the following 0 R ∞ dr k = reason. The convergence of N(r) + implies that lim N(R)/R 0, rk 1 R→∞ R0 ∞ dr hence the convergence of n(r) follows. On the other hand if rk+1 R0 ∞ R dr n(r)dr/rk+1 converges N(r) at once by comparison. rk+1 R0 R0 Therefore it remains to be proved that n(r) and N(r) have the same order and type. Suppose n(r) has order k, then given δ> 0, n(r) < crk+δ for r greater than r0 r dt N(r) = n(t) Z t 0 Basic Theory 35

r0 r dt dt = n(t) + n(t) for r > r0 Z t Z t 0 r0 r0 r dt k−1+ N(r) ≤ n(t) + cr dr for r > r0 Z t Z 0 r0 k+δ ≤ constant + cr /(k + δ) for r > r0. This implies that order of N(r) is not greater than that of n(r), and if n(r) has got mean or minimal type so has N(R). The result for convergence or divergence class follows from earlier inequalities. The estimate 40 for n(r) in terms of N(r) follows from 2r dt n(r) ≤ (1/ log 2) n(t) ≤ N(2r)/ log 2. Z t r From this inequality it can be derived in the same way as before, that the order of n(r) is not greater than that of N(r) etc. Deﬁnition . The order and type of n(r) or N(r) (being the same) are called the order and type of the sequence (an). The order of n(r) is also sometimes called exponent of convergence of the sequence. k We note that 1/rn converges if and only if n(r) has order less than k, or order k of convergenceP type. (This is a consequence of lemma 3)

Next let us state the theorem,

Theorem 6. Suppose that an is a sequence having at most order q + 1 ∞ (a positive integer) convergence class. Then the product π(z) = n=1 E(z/an, q) converges absolutely and uniformly in any bounded region,Q and for |z| = r, r ∞ dt dt log |π(z)| < A(q) rq n(t) + rq+1 n(t)  q+1 q+2   Z t Z t   0 0      36 Basic Theory

Proof.

q u+ 1 u2+···+ u E(u, q) = (1 − u)e 2 q 1 uq log E(u, q) = log(1 − u) + u + u2 + ··· + 2 q = − uk/k Xq+1

k |u| + 1 41 so that | log E(u, q)|− ≤ 2|u|q 1 if |u|≤ since log |E(u, q)| is the q+1 k 2 P 1 real part of log E(u, q) we get log |E(u, q)| ≤ q + 1 if |u| ≤ . Suppose 2 1 ≤ |u|≤ 1. Then 2 1 |u|q log |E(u, q)|≤ log |(1 − u)| + |u| + |u|2 + ··· + 2 q 1 |u|q ≤ |u| + |u| + |u|2 + ··· + . 2 q 1 Also |u| ≥ , |u|q−1 ≥ 1/2q−1, 2q−1|u|q−1 ≥ |u|. So that since |u| ≤ 1, 2 |u|q ≤ |u|≤ 2q−1|u|q and log |E(u, q)|≤ 2q−1|u|q + q2q−1|u|q = A(q)|u|q < 2A(q). A(q) being a constant depending only on q. Thus for |u|≤ 1, log |E(u, q)|≤ A(q)|u|q+1.

Let 1 uq |u|≥ 1, then log |E(u, q)| ≤ |u| + |u| + |u|2 + ··· + 2 q ≤ (q + 1)|u|q

Now u = z/zn. Let |z| = r, |zn| = rn and N the least integer for which rn ≥ r. Then |u|≥ 1. Thus N−1 N−1 q −q log |E(z/zn, q)|≤ A(q)r rn Xn=1 X1 Basic Theory 37 and ∞ ∞ q+1 −q−1 log E(z/zn, q) ≤ B(q)r rn Xn+1 Xn+1 Thus

∞ N−1 ∞ | |≤ q −q + q+1 −q−1 log E(z/zn, q) C r rn r rn  X1  X1 XN     r ∞  = q 1 + q+1 1 C r dn(t) r + dn(t)  Z tq Z tq 1   r   0   r ∞  = q 1 + q+1 1 K r + n(t)dt r + n(t)dt  Z tq 1 Z tq 2   r   0    K depending only on q. This proves the theorem. 42 We also see that for large n,

+ z r q 1 log E , q < A(q) , zn ! rn !

−(q+1) and, the product converges since rn converges. Our Theorem 6 shows that if in Theorem 5 f (z) hasP at most order q convergence class, then the two products converge separately, uniformly and absolutely on every bounded set.

As a consequence of the theorem we have

Theorem 7. If a sequence an deﬁned as in the last theorem, has order n=∞ ρ, q − 1 ≤ ρ< q, q an integer, then πa(z) = E(z/an, q − 1) has order n=1 ρ and further if ρ is not an integer, (z) hasQ the same type class as (an). a Q Hence if f (z) is meromorphic of ﬁnite non-integral order, then the roots of f (z) = a, have the same order and type class as f (z) except for at most one value of a on the Riemann sphere. 38 Basic Theory

ρ+ǫ Proof. If n(t) < ct for t > t0 where ǫ > 0

r t0 r dt rq−1 n(t) ≤ rq−1 n(t)dt/tq + rq−1 ctρ+ǫdt/tq Z tq Z Z 0 0 t0 rρ+1−q+ǫ Crρ+ǫ − 0(rq−1) + C rq−1 = 0 rq−1 + ρ + 1 + −q + ǫ ρ + ǫ + 1 − q!

43 and the result regarding the order of the ﬁrst integral follows. The second integral is treated similarly and then the ﬁrst part follows from Theorem 6. If ρ> q − 1 and f (z) is of mean type or minimal type of order ρ we can take ǫ = 0, and in the case of minimal type c small and the results for the type at once follow. Suppose for instance that n(t) has order less ρ+ǫ than q so that n(t) < cr for t > t0. Then for large r,

∞ ∞ n(t)dt tρ+ǫ crρ+ǫ r < crq dt = if ǫ < q − ρ. Z tq+1 Z tq+1 (q − ρ − ǫ) r r

and the integral is 0(rρ+ǫ ) and is 0(rρ) if n(r) = 0(rρ) as required.

Suppose now that f (z) is a meromorphic function of ﬁnite non- integral order ρ and q such that q − 1 <ρ< q. Then,

Pq−1(z) f (z) = e Π1(z)/Π2(z)

where Π1(z) and Π2(z) are the products over the zeros and poles respectively. Now if both have a smaller order and type than f (z), so does their 44 ratio. since

T(r, Π1/Π2) ≤ T(r, Π1) + T(r, 1/Π2)

= T(r, Π1) + T(r, Π2) + 0(1).

and ePq−1(z) has order q − 1 < ρ, so we should get a contradiction. Hence at least one of the Π1(z) or Π2(z) and so either zeros or the poles have the same order and type as the function f (z). Basic Theory 39

If for instance the poles do not have the order and type of f (z) then since f (z) − a has the same poles as f (z) (for every finite a), the roots of f (z) = a that is the zeros of f (z) − a have the order and type of f (z). This kind of argument was used to show that Riemann-Zeta function has an infinity of zeros. We now illustrate the above by means of an example. If f (z) is meromorphic in the plane and lim T(r, f )/ log r<+∞ then µ→∞ f (z) is rational. In this case f (z) has lower order zero, and lower limit of N(r, f )/ log r as r tends to ∞ is less than ∞. [T(r, f ) ≥ N(r, f )]. Similarly, N(r, 1/ f ) lim < ∞ ν→∞ log r further, R dt R N(R, f ) ≥ n(t, f ) ≥ n(r, f ) log Z t r r N(r2, f ) so that n(r, f ) ≤ = 0(1) for a sequence of n → ∞. So f (z) 1 2 log r has only a finite number of zeros and poles. Hence from theorem 5 45 p f (z) = z (Π(1 − z/aµ)/(1 − z/bν) where both the products are finite. Thus for all transcendental meromorphic functions T(r, f )/ log r tends to infinity as r tends to infinity.

Part II Nevanlinna’s Second Fundamental Theorem

2.1 46 As the name indicates this theorem is most fundamental in the study of meromorphic functions. It is an extension of Picard’s Theorem, but goes much farther. We develop the theorem in theorems 8 and 9 of this chapter, and then proceed systematically to explore some of its consequences.

Theorem 8. Suppose that f (z) is a non-constant meromorphic function in |z| ≤ r. Let a1, a2,..., aq be distinct ﬁnite complex numbers, δ > 0 such that |aµ − aν|≥ δ for 1 ≤ µ ≤ ν ≤ q. Then,

ν=q m(r, f ) + m(r, aν) ≤ 2T(r, f ) − N1(r) + S (r) Xν=1 where N1(r) is positive and is given by

′ ′ N1(r) = N(r, 1/ f ) + 2N(r, f ) − N(r, f ) and

νq = ′ + ′ − S (r) m(r, f / f ) m r, f /( f aν)  Xν=1      41 42 Nevanlinna’s Second Fundamental Theorem

3q + q log+ + log 2 + log+ 1/| f ′(0)| δ ! with modiﬁcations if f (0) = 0 or ∞, f ′(0) = 0.

Proof. Let a new function F be deﬁned as follows

ν=q F(z) = 1/[ f (z) − aν] Xν=1

and suppose that for some ν, | f (z) − aν| < δ/3q. Then for µ , ν,

Therefore,

1/| f (z) − aµ| < 3/2δ for µ , ν

< (1/2q)[1/| f (z) − aν|]

47 Again,

|F(z)|≥ 1/| f (z) − aν|− 1/ f (z) − aµ| Xµ,ν

≥ [1/| f (z) − aν|][1 − (q − 1)/2q]

≥ 1/2| f (z) − aν|

Hence + + log |F(z)|≥ log 1/| f (z) − aν|− log 2. In this case,

Taking integrals we deduce, q 3q (2.1) m(r, F) ≥ m(r, 1/ f − a ) − q log+ − log 2 ν δ Xν=1 Again to get an inequality in the other direction, 1 f m(r, F) = m r, f ′F f f ′ ! By equation (1.2′) of 1.2 i.e. ,

T(r, f ) = T(r, 1/ f ) + log | f (0)| m(r, f / f ′) = m(r, f ′/ f ) + N(r, f ′/ f ) − N(r, f / f ′) + log[| f (0)|/| f ′(0)|] m(r, 1/ f ) = T(r, f ) − N(r, 1/ f ) + log 1/| f (0)|

Hence we get ﬁnally, 48

m(r, F) ≤ T(r, f ) − N(r, 1/ f ) + log 1/| f (0)| + m(r, f ′/ f ) + N(r, f ′/ f ) − N(r, f / f ′) + log | f (0)|/| f ′(0)| + m(r, f ′F)

q + The above inequality, combined with (2.1) gives m(r, aν) − q log ν=1 3q/δ − log 2 ≤ Right hand side of the above inequality.P Add to both the sides m(r, f ), we get the inequality, q m(r, f ) + m(r, aν) ≤ T(r, f ) + [m(r, f ) + N(r, f )] − N(r, f ) Xν=1 44 Nevanlinna’s Second Fundamental Theorem

+ m(r, f ′/ f ) + log 1/| f ′(0)| + log 2 q + ′ − + + − m r, f /( f aν) q log 3q/δ N(r, 1/ f )  Xν=1    + N(r, f ′/ f ) − N(r, f /f ′). 2 = 2T(r, f ) − N (r) + m(r, f ′/ f ) + log 1 | f ′(0)| q 3q f ′ + q log+ + m r, δ  ( f, a )  Xν=1 ν     ′  ′ where, N1(r) = N(r, f ) + N(r, 1/ f ) + N(r, f / f ) − N(r, f / f ). Now for any two functions f (z) and g(z) Jensen’s formula gives

49 Thus

′ N1(r) = N(r, f ) + N(r, 1/ f ) + N(r, 1/ f ) + N(r, f ) − N(r, f ′) − N(r, 1/ f ) = 2N(r, f ) + N(r, 1/ f ′) − N(r, f ′)

as required.

2.2 Estimation of the error term We shall ﬁrstly prove some lemmas. Nevanlinna’s Second Fundamental Theorem 45

Lemma 1. Suppose that f (z) is meromorphic in the |z| ≤ R and that = 1 + 0 < r < R. Let ρ 2 (r R) and δ(z) the distance of z from the nearest pole or zero of f (z) in |z| < ρ. Then, m(r, f ′/ f ) < log+ T(R, f ) + 2 log+[1/(R − r)] + 2 log+ R 2π 1 1 + log+ dθ + 0(1) 2π Z δ(reiθ) 0 0(1) depending only on the behaviour of f (z) at z = 0. Proof. We have by diﬀerentiation of Poisson formula as in theorem 5,

Hence 50 a 1 1 b µ − + − ν 2 2 "(ρ − aµz) (aµ − z)# (bν − z) ρ − bνz X X   1 1  ≤ [n(ρ, f ) + n(ρ, 1/ f )] + " δ(z) (ρ − r)# We now estimate n(ρ, f ) and n(ρ, 1/ f ). For this we have,

R n(t, f )dt/t ≤ N(R, f ) ≤ T(R, f ) Z ρ 46 Nevanlinna’s Second Fundamental Theorem

R and since n(t, f )/t ≥ n(ρ, f )/R for t greater than ρ, n(t, f )dt/t ≥ Rρ n(ρ, f )(R − f )/R. Therefore [n(ρ, f )](R − ρ)/R ≤ T(R, f ) or n(ρ, f ) ≤ RT(R, f )/(R − ρ) = 2RT(R, f )/(R − r) since 2ρ = (R + r). Similarly n(ρ, 1/ f ) ≤ 2RT(R, 1/ f )/(R−r) = 2R[T(R, f )+0(1)]/(R− r). 0(1) depends only on the behaviour of f (z) at z = 0. Thus

4R n(ρ, f ) + n(ρ, 1/ f ) ≤ [T(R, f ) + 0(1)] (R − r) and so

a 1 1 b µ − + − ν 2 2 "ρ − aµz (aµ − z)# (bν − z) (ρ − bνz) X X   4R  1 2  ≤ [T(R, f ) + 0(1)] + (R − r) " δ(z) (R − r)#

Further,

2π 1 2ρeiφ log | f (ρeiφ)| dφ 2π Z (ρeiφ − z)2 0

2π 1 2ρ ≤ log | f (ρeiφ)|dφ 2π (ρ − r)2 Z 0

51 for,

|ρeiφ − z| ≥ ||ρeiφ| − |z|| = (ρ − r) 2π 2π 1 4ρ + iφ + 1 =  log | f (ρe )dφ + log dφ π (R − r)2 Z Z | f (eiφρ)|    0 0  8ρ  8ρ  = [m(ρ, f ) + m(ρ, 1/ f )] ≤ [2T(ρ, f ) + 0(1)] (R − r)2 (R − r)2 16ρ 16R = [T(ρ, f ) + 0(1)] ≤ [T(R, f ) + 0(1)] (R − r)2 (R − r)2 Nevanlinna’s Second Fundamental Theorem 47

0(1) depending only on the behaviour of f (2) at z = 0. Thus from the above inequalities and the equation for f ′(z)/ f (z) we get ﬁnally, 4R 2 1 | f ′(z)/ f (z)|≤ [T(R, f ) + 0(1)] + (R − r) " (R − r) δ(z)# 16R + [T(R, f ) + 0(1)] (R − r)2 4R (R − r) R − r = 6T(R, f ) + T(R, f ) + 0(1) + 0(1) (R − r)2 " δ(z) δ(z) # 4R R − r ≤ 6 + [T(R, f ) + 0(1)] (R − r)2 " δ(z) # Hence, 4R R − r log+ | f ′(z)/ f (z)|≤ log+ + log+ 6 + + log+ T(R, f ) (R − r)2 δ(z) ! 1 (R − r) + 0(1) ≤ log+ R 2 log+ + log+ T(R, f ) log+ + 0(1) (R − r) δ(z) 1 1 ≤ 2 log+ R + log+ + 2 log+ + log+ T(R, f ) + 0(1) δ(z) (R − r)

0(1) depending only on the behaviour of f (z) at z = 0. Integrating the 52 above inequality on the circle |z| = r,

2π 2π 1 1 1 log | f ′(reiθ)/ f (reiθ)|dθ ≤ 2 log+ R + log+ dθ 2π Z 2π Z δ(reiθ) 0 0 1 +2 log+ + log+ T(R, f ) + 0(1) (R − r) which gives the lemma.

Lemma 2. Let z be any complex number and 0 < r < ∞. Let Ek be the set of all θ such that |z − reiθ| < kr where 0 < k < 1. Then 1 log[r/|(z − reiθ)|]dθ<π log + 1 Z " k ! # Ek 48 Nevanlinna’s Second Fundamental Theorem

Proof. We may after rotation assume z real and positive. For if z = 0 Ek is obviously void and there is nothing to prove. So let z > 0. Then for θ iθ iθ in Ek |z − re | ≥ | Im(z − re )| = r sin θ and Ek is contained in an interval of the form |θ| <θ0 where r sin θ0 ≤ kr, that is, sin θ0 ≤ k. So,

θ0 sin θ 2 iθ π ≥ |θ| <θ0 we get log[r/|re − z|]dθ ≤ 2 log dθ θ π Z Z 2θ Ek 0

θ0 π = 2 log dθ Z 2θ 0 π π because θ < or > 1. 0 2 2θ

θ0 θ0 π log[r/|(reiθ − z)|]dθ ≤ 2 log dθ − 2 log θdθ Z Z 2 Z Ek 0 0 π = 2θ log − 2θ log θ + 2θ 0 2 0 0 0

Lemma 3. With the hypothesis of lemma 1, m(r, f ′/ f ) ≤ 3 log+ T(R, f )+4 log+ R+4 log+[1/(R−r)]+log+(1/r)+0(1) Proof. Notations being the same as in the proof of lemma 1, write, 1 1 r [1/δ(z)] = [r/δ(z)] · (1/r). Then log ≤ log − log . So δ(z) r δ(z)

2π 2π 1 1 1 1 r log+ dθ ≤ log+ + log+ dθ 2π Z δ(reiθ) r 2π Z δ(reiθ) 0 0 Nevanlinna’s Second Fundamental Theorem 49

r Let E denote the set of θ in (0, 2π) for which δ(reiθ) < where n = n2 n(ρ, f ) + n(ρ, 1/ f ). (If n = 0 so that there are no zeros and poles we can put δ(z) =+∞ and there is nothing to prove. So assume n ≥ 1). For each iθ iθ 2 point θ in E there is a zero or pole zν such that δ(re ) = |zν −re | < r/n 54 and then + + iθ log [r/δ(z)] = log [r/|re − zν|] = 2 = Now write log0 x log x if x ≥ n and log0 x 0 otherwise. Then since n ≥ 1, log0 x ≥ 0 always. + + iθ Also for θ in E log [r/δ(z)] = log [r/|(re − zν)|] for some ν Since, 2 iθ 1 < n < [r/|re − zν|], r r r log+ = log ≤ log |reiθ − z | 0 |reiθ − z | 0 |reiθ − z| ν ν Xµ where the sum is taken over all zeros and poles zµ in |z| < ρ. Thus r r log+ dθ ≤ log dθ iθ 0 | iθ − | Z δ(re ) µ Z re zµ E X E π π ≤ [log n2 + 1] = [2 log n + 1] ≤ A. n2 n Xµ by lemma 2 with k = 1/n2, and where A is an absolute constant. Also on the complement of E, δ(reiθ) ≥ r/n2 and,

2π r log+ dθ ≤ log n2dθ = 2π log n2 Z δ(reiθ) Z compl. of E 0

Adding we obtain

2π 1 r log+ dθ ≤ [2 log n + A] 2π Z δ(reiθ) 0 2π 2π 1 1 1 1 r log+ dθ = log+ dθ 2π Z |δ(reiθ)| 2π Z r |δ(reiθ)| 0 0 50 Nevanlinna’s Second Fundamental Theorem

2π 1 r 1 ≤ log dθ + log+ 2π Z |δ(reiθ)| r 0 ≤ 2 log+ n + log+ 1/r + A.

55 where n = n(ρ, f ) + n(ρ, 1/ f ). We have from (2.2)

n = n(ρ, f ) + n(ρ, 1/ f ) ≤ 4R[T(R, f ) + 0(1)]/(R − r).

Hence,

log+ n ≤ log+ R + log+ 1/(R − r) + log+ T(R, f ) + 0(1),

giving

2π 1 1 1 1 log dθ ≤ log+ + 2 log+ + 2 log+ R 2π Z |δ(reiθ)| r R − r 0 + 2 log+ T(R, f ) + 0(1)

From this and lemma 1, lemma (3) follows. That is,

m(r, f ′/ f ) < 3 log+ T(R, f ) + 4 log+ 1/(R − r) + 4 log+ R + log+ 1/r 0(1).

Lemma 4 (Borel). (i) Suppose T(r) is continuous, increasing and T(r) ≥ 1 for r0 ≤ r < ∞. Then we have

(2.3) T[r + 1/T(r)] < 2T(r)

outside a set E of r which has length (that is) linear measure at most 2.

56 (ii) If T(r) is continuous and increasing for r0 ≤ r < 1 and T(r) ≥ 1 then we have

(2.4) T[r + (1 − r)/(eT(r))] < 2T(r) Nevanlinna’s Second Fundamental Theorem 51

outside a set E of r such that dr/(1 − r) ≤ 2. In particular ER T[r + (1 − r)/eT(r)] < 2T(r) for some r such that ρ < r < ρ′ if ′ 2 r0 <ρ< 1 and 1 − ρ < (1 − ρ)/e .

Proof. We prove first (i) [that is in the plane]. Let r1 be the lower bound of all r for which (2.3) is false. If there are no such r there is nothing to prove. We now define by induction a sequence of numbers rn. Suppose ′ = + that rn has been defined and write rn rn 1/T(rn). Define then rn+1 as ′ the lower bound of all r ≥ rn for which (2.3) is false. We have already defined r1 and so we obtain the sequence (rn). Note that by continuity of T(r) (2.3) is false for r = rn, for n = 1, 2, 3,... that is rn belongs to E0, E0 being the exceptional set. From the definition of rn+1 is follows that ′ there are no points of E in (rn, rn+1) and that the set of closed intervals ′ [rn, rn] contains E0. If there are an infinity of rn, (rn) cannot tend to a ′ ′ finite limit r. For then since rn < rn ≤ rn+1, rn tends to r also. But ′ = rn − rn 1/T(rn) which is greater or equal to 1/T(r) > 0, since T(r) is increasing, for all m which is a contradiction. ′ ′ = + It remains to be shown that (rn − rn) ≤ 2. Now T(rn) T[rn ′ 1/T(rn)] ≥ 2T(rn) since rn belongsP to E. And so T(rn+1) ≥ T(rn) ≥ n n 2T(rn). Therefore, T(rn+1) ≥ 2T(rn) ≥ ... ≥ 2 T(r1) ≥ 2 since T(r) ≥ 57 n−1 ′ = 1−n 1. Thus T(rn) ≥ 2 . Now (rn − rn) 1/T(rn) ≤ 2 ≤ 2. To prove part (ii) of the theorem,P set RPlog[1/(1 − rP)] getting r = 1 − −R e and put T(r) = ϕ(R)·ϕ(R) then is continuous and increasing for R0 = log[1/(1 − r0)] ≤ R < ∞ and ϕ(R) ≥ 1. Apply then the first part to ϕ(R). Then we have ϕ[R+1/ϕ(R)] < 2ϕ(R) for R > log[1/(1−r0)] outside a set ′ = = E of R such that 2 ≥ (Rn −Rn) dR dr/(1−r). Translating back ER R to r, R′ = R + 1/ϕ(R)P becomes log[1/(1 − r′)] = log[1/(1 − r)] + 1/T(r). − 1 That is (1 − r′) = (1 − r)e T(r) and T(r′) < 2T(r). By the first mean value theorem f (b) = f (a) + (b − a) f ′(x), where a ≤ x ≤ b. Since T(r) increases with r, T[r + (1 − r)/eT(r)] < 2T(r) outside the exceptional set E of r for which dr/(1 − r) ≤ 2. If E contains the whole of the E R ′ ′ ρ − ≤ − ≤ interval ρ < r < ρ then ρ dr/(1 r) dr/(1 r) 2, that is R ER log(1 − ρ)/(1 − ρ′) ≤ 2, and so (1 − ρ)/(1 − ρ′) ≤ e2 as required. 52 Nevanlinna’s Second Fundamental Theorem

2.3 Theorem 9. If f (z) is meromorphic and non-constant in the plane and S (r, f ) denotes the error term in Theorem 8 then we have

(2.5) S (r, f ) = O[log T(r, f )] + O(log r)

as r tends to infinity through all values if f (z) has finite order, and through all values outside a set of finite linear measure otherwise. (ii) If f (z) is meromorphic (non-constant) in |z| < 1 and the lim{T(r, f )/ log[1/(1 − r)]} = ∞, then we have S (r, f ) = O[T(r, f )] r→1 as r tends to one on a set E such that dr/(1 − r) = ∞. ER q Proof. If ϕ(z) = [ f (z) − aν] then ν=1 Q S (r, f ) = m(r, f ′/ f ) + m[r,ϕ′/ϕ] + O(1)

q q ′ ′ ′ 58 because S (r, f ) = m(r, f / f )+m(r, f /( f −aν))+O(1) and f /( f − ν=1 ν=1 ϕ′ P P a ) = by logarithmic diﬀerentiation. Therefore from the lemma 3, ν ϕ for any R > r we have,

S (r, f ) ≤ 3 log+ T(R, f ) + 4 log+ R + 4 log+[1/(R − r)] + log+(1/r) + 3 log+ T(R,ϕ) 4 log+ R + 4 log+ 1/(R − r) + log+(1/r) + 0(1)

Also q (2.6) T(r,ϕ) ≤ T(r, f − aν) ≤ T(r, f ) + O(1),... Xν=1 Thus

S (r, f ) ≤ 3(1 + q) log+ T(R, f ) + 8 log+ R + 8 log+[1/(R − r)] +2 log+(1/r) + O(1) Nevanlinna’s Second Fundamental Theorem 53

If r is greater than 1 (we are considering S (r, f ) for large r) log+(1/r) = 0. Now suppose that f (z) is meromorphic of ﬁnite order so that T(r, f ) < K 2 r for r greater than r0. Also choose R = r and r ≥ 2 so that R − r > 1. 1 Then log+ = 0. We get R − r !

S (r, f ) ≤ 3(1 + q) log+ T(R, f ) + 8 log+ R + O(1), log+ T(R, f ) < k log+ R < 2k log+ r = 2k log r since R = r2 and r > 1. We thus have ﬁnally, S (r, f ) ≤ 6(1 + q)K log r + 16 log r + O(1) showing that S (r, f ) = 0(log r) which gives (2.5) since log+ T(r, f ) = O(log r).

Note that by our examples T(r, f )/ log r tends to infinity unless f (z) 59 is a rational function in which case S (r, f ) is bounded because f ′/ f → 0 as z →∞ for any polynomial and hence for any rational function. Thus if f (z) has finite order S (r, f )/T(r, f ) → 0 as r →∞. If f (z) has infinite order take R = [r + 1/T(r)], then log+ R ∼ log r[since T(r) → ∞] · log+ 1/(R − r) = log+ T(r) and log 1/r = 0 finally. Outside the exceptional set of lemma 4, T(R, f ) < 2T(r, f ) and so, log+ T(R, f ) ≤ log+ T(r, f ) + log 2. Hence again we have (2.5) outside the exceptional set. This completely proves i) In order to prove 1 (ii) denote by rn a sequence such that T(rn, f )/ log → ∞ as 1 − rn ! n → ∞ and by taking a sub sequence if necessary we can assume that 1 − rn ′ 1 − r + < . Then let r be defined by 1 − r = (1 − r )/10 so that n 1 10 n n n ′ rn ′ dr r < r < r + < 1. Then since = log(1−r )/(1−r ) = log 10 > n n n 1 1 − r n n rRn dr 2 the union E of all the intervals (r , r′ ) is such that =+∞. 1 n n 1 − r ER1 Further each such interval contains a point not in the exceptional set E, for T(r, f ) because by lemma 4 dr/(1 − r) ≤ 2, pro- ER vided only that T(r1) > 1. For a not exceptional point r of E1 take 54 Nevanlinna’s Second Fundamental Theorem

R = r + (1 − r)/eT(r), then 1 eT(r) 1 log+ = log+ < log+ T(r) + log+ + 1 R − r 1 − r 1 − r + + 60 and log T(R) < log T(r) + log 2, Thus by (2.6) 1 S (r, f ) < 3(1 + q) log+ T(R, f ) + 8 log+ R + 8 log+ + O(1) (R − r) 1 < (3 + 3q + 8) log+ T(r) + 8 log+ + O(1) (1 − r) Also, log+[1/(1 − r)] < log+ 1/(1 − r′ ) + log+ 10 < log+ 1 + 0(1) = n (1−rn) 1−rn O[T(rn)] + O(1) = O[T(r)]. So since log T(r, f ) = OT(r) we get S (r, f ) = OT(r). This proves dr (ii) for a set E of r such that =+∞ and containing at least one 1 1 − r E1 R ′ point in each interval rn < r < rn. In face E comprises all the r in the ′ sequence of intervals [rn, rn], n > 1 except possibly a set E0 such that dr ≤ 2. 1 − r ER0

2.4 Applications 61 Definition . Let n(t, a) denote the number of roots of f (z) = a in |z| ≤ t, the multiple roots being counted according to their multiplicity and n(t, a) the number of roots of f (z) = a in |z| < t with the multiple roots r counted simply. Further let N(t, a) = n(t, a) − n(0, a)dt/t. [ n(0, a) is R0 equal to one if f (0) = a and zero otherwise]; and N(r, a) as before with n(t, a) instead of n(t, a). Let now the function f (z) be meromorphic, and non-constant in |z| < R, 0 < R ≤ ∞ and suppose that f (z) satisfies the hypotheses of theorem 9, so that lim[S (r, f )/T(r)] = 0 and T(r) tends r→R to infinity as r tends to R. Now write δ(a) = lim[m(r, a)/T(r)] = 1− lim[N(r, a)/T(r)] because r→R r→R [m(r, a) + N(r, a)]/T(r) = [T(r) + 0(1)]/T(r) → 1; Nevanlinna’s Second Fundamental Theorem 55 thus −N(r, a) lim[m(r, a)/T(r)] = lim 1 + = 1 + lim −N(r, a)/T(r) r→R r→R T(r) r→R = 1 − lim N(r, a)/T(r) r→R Again write, θ(a) = lim[N(r, a) − N(r, a)]/T(r) r→R and Θ(a) = 1 − lim N(r, a)/T(r) = lim[1 − N(r, a)/T(r)]. r→R r→R Clearly, δ(a), θ(a) and Θ(a) lie in the closed interval [0, 1]. Also N N N − N 1 − = 1 − + T T T N N N − N lim 1 − ≥ lim 1 − + lim r→R T r→R T r→R  T    i.e. Θ(a) ≥ δ(a) + θ(a).   62 The quantity δ(a) is called the defect of a and θ(a) the Branching index (Verzweigungsindex) of a. Now we have the defect relation of Nevanlinna. This is the second fundamental theorem in its most effec- tive form and is very important in the theory. Theorem 10. If f (z) satisfies the hypotheses of the Theorem 9, then Θ(a) = 0 except possibly for a finite or countable sequence aν of values of a and for these Θ(aν) ≤ 2.

Proof. From theoremP 8, for any q distinct values aν, of a including a1 = ∞ q m(r, aν) < 2T(r, f ) − N1(r) + S (r) Xν=1 q and adding N(r, aν) to both sides and using ﬁrst fundamental theorem ν=1 T(r, a) = T(Pr, f ) + O(1), we get q qT(r, f ) < 2T(r, f ) − N1(r) + N(r, aν) + S (r, f ) + 0(1) Xν=1 56 Nevanlinna’s Second Fundamental Theorem

q (q − 2)T(r, f ) < N(r, aν) − N1(r) + S (r) + 0(1). Xν=1

′ ′ Now, N1(r) = 2N(r, f ) − N(r, f ) + N(r, 1/ f ), and since by deﬁnition ′ N(r, f ) = N(r, ∞), N(r, ∞) − N1(r) = N(r, a1) − N1(r) = N(r, f ) − N(r, f ) − N(r, 1/ f ′). 63 So,

q ′ ′ (q−2)T(r, f ) < N(r, aν)+ N(r, f )− N(r, f )− N(r, 1/ f )+S (r)+0(1) Xν=1

′ If f (z) has a pole of order p at z0, f (z) has a pole of order p + 1 at z0 so that n(t, f ′) − n(t, f ) = n(t, f ) and so N(r, f ′) − N(r, f ) = N(r, f ),. Similarly if a is anyone of a2, a3,..., aq (ﬁnite) and f (z) = a has a root of multiplicity p, f ′(z) has there a zero of order p − 1. Thus

q q ′ ′ N(r, aν) − N(r, 1/ f ) = N(r, aν) − N0(r, 1/ f ) Xν=2 Xν=2

′ ′ where N0(r, 1/ f ) refers to zeros of f (z) at points other than the roots of f (z) = a. Hence we get,

q ′ (q − 2)T(r, f ) < N(r, aν) − N0(r, 1/ f ) + S (r) + N(r, ∞) + 0(1) Xν=2 q ′ i.e. (q − 2)T(r, f ) < N(r, aν) − N0(r, 1/ f ) + S (r, f ) + 0(1) Xν=1 q N(r, a ) S (r, f ) + 0(1) i.e. ν ≥ (q − 2) − since N (r, 1/ f ′) ≥ 0 T(r) T(r) 0 Xν=1 since lim S (r, f )/T(r) = 0 and T(r, f ) →∞ as r → R r→R

q O(1) + S (r) N(r, a ) lim − = 0. Thus lim ν ≥ (q − 2) r→R " T(r) # r→R T(r) Xν=1 Nevanlinna’s Second Fundamental Theorem 57 and afortiori q N(r, a ) lim ν ≥ q − 2 r→R T(r) Xν=1 that is, 64 q [1 − Θ(aν)] ≥ (q − 2) as required. Xν=1 This shows that Θ(a) > (1/n) at most for 2n values of a and so Θ(a) > 0 for at most a countable number of a. If these a’s are arranged in a q sequence Θ(ar) ≤ 2 for any ﬁnite q, and so to inﬁnity. r=1 P Consequences

(1) Since δ(a) ≤ Θ(a) we have δ(aν) ≤ 2 and thus there exists at most two values of a for whichP δ(a) = 1, or more generally 2 = δ(a) > 3 . Thus if the equation f (z) a has only a finite number of roots in the plane, N(r, a) = 0(log r) as r tends to infinity, and we should have T(r) lim log[r/T(r)] > 0, i.e. lim < ∞. r→∞ r→∞ log r i.e., f (z) is rational. Thus if f (z) is transcendental in the plane δ(a) < 1 the equation f (z) = a has infinitely many roots. The same is true in all cases if R is finite, since otherwise N(r, a) = 0(1) and so lim T(r) < +∞ that is T(r) = 0(1) as r tends to R, r→R since T(r) is increasing. This would contradict

1 lim T(r)/ log =+∞. r→R 1 − r ! This result thus contains Picard’s theorem as a special case.

(2) Θ in relation to N and N. Suppose that the function f (z) = a has 65 only multiple roots of multiplicity k ≥ 2. Then

N(r, a) ≤ (1/k)N(r, a) ≤ (1/k)[T(r, f ) + 0(1)] 58 Nevanlinna’s Second Fundamental Theorem

Hence in this case, lim N(r, a)/T(r) ≤ (1/k) r→R 1 Θ(a) = 1 − lim N(r, a)/T(r) ≥ 1 − (1/k) ≥ r→R 2

If a1, a2,..., aq are q such values with k = kν for aν, we have since Θ(aν) ≤ 2, [1 − (1/kν)] ≤ 2. ν ν P P In particular there can be at the most four such values aν of a for 1 [1 − (1/kν)] ≥ 2 . If f (z) is regular m(r, f ) = T(r, f ) and δ(∞) = limm(r, f )/T(r, f ) = 1. and so Θ(∞) = 1 because Θ(∞) ≤ δ(∞) = 1. So, Θ(aν) ≤ 1 for ν any finite number of finite aν’s. So that there can be onlyP two such values aν which are taken multiply. Such values are called fully branched (Vollständig Verzweight). These results are best possible, for sin z and cos z, have the values ±1 “fully branched”. Again for the Wierstrassian elliptic function P(z) which satisfies the differential equation ′ 2 [P (z)] = (P(z) − e1)(P(z) − e2)(P(z) − e3)

where e1, e2, e3 are distinct finite numbers the values e1 e2 and e3 are evidently fully branched and so is infinity. If P(z) has a pole of order k, 66 P(z) ∼ A(z−ζ)−k and [P′(z)]2 ∼ P(z)3 ∼ A3(z−ζ)−3k from the differential equation. But [P′(z)]2 ∼ A′z−2k−2 i.e., − 2k − 2 = −3k Hence we have k = 2 so all the poles are double and infinity is fully branched. 2 = q We also note that the equation w ν=1(z − aν) can have no para- metric solution z = ρ(t), w = ψ(t) whichQ are integral functions of t if q ≥ 3 or which are meromorphic if q ≥ 5. Because if ϕ(t0) = a1 for 2 2 2 instance then w = ψ (t) has a zero at t0 and so ψ(t) has a zero and ψ (t) −a has at least a double zero at t0. Hence also ϕ(t) 1 has at least a double zero at t0 and all roots of ϕ(t) = aν therefore will be multiple and q Θ(aν) ≥ 2 for ϕ(t), a contradiction. ν=1 P Nevanlinna’s Second Fundamental Theorem 59

We just remark that this result extends to a general equation g(z, w) = 0 of genus greater than 1.

67 Theorem 11. Suppose f (z) is meromorphic of ﬁnite order in the plane and Θ(a1) = Θ(a2) = 1, a1 , a2. Then if a is not equal to a1, a2, N(r, a) ∼ T(r) as r tends to inﬁnity.

Proof. In fact we have if f (z) is not rational

T(r, f ) < N(r, a1) + N(r, a2) + N(r, a) + S (r, f ) + O(1) and in this case S (r, f ) = O[T(r)], (even S (r) = 0(log r)). By hypothesis since Θ(ai) = 1 − lim[N(r, a)/T(r)], N(r, ai) = 0[T(r)] i = 1, 2. Therefore [1 + 0(1)]T(r, f ) < N(r, a) as r →∞. N(r, a) That is lim [N(r, a)/T(r)] ≥ 1, and evidently lim ≤ 1 that r→∞ r→∞ T(r) is, N(r, a)/T(r) → 1, as r tends to inﬁnity. Similarly since N(r, a) ≥ N(r, a), lim N(r, a)/T(r) ≥ 1 and again N(r, a) ∼ T(r). If f (z) is rational r→∞ these results follow by elementary methods; in fact in this case there is at most one a namely a = f (∞) for which N(r, a) = OT(r) unless f is a constant.

Theorem 12. If aν(z) for ν = 1, 2, 3 are three functions satisfying T(r, aν) = O[T(r, f )] as r → R, and f (z) is as in theorem 10, then we have 3 1 [1 + O(1)]T(r, f ) < N r, f − a (z)! Xi=1 i as r tends to R through a suitable sequence of values.

Proof. Set 68 f (z) − a (z) a (z) − a (z) ϕ(z) = 1 2 3 f (z) − a3(z) a2(z) − a1(z)

It is easy to see using T(r, ai) = O[T(r, f )] that T(r, ) = T(r, f ) = [1 + O(1)]T(r, f ), also ϕ = 0, 1, ∞, only if f − a1(z), f − a2(z), f − a3(z) 60 Nevanlinna’s Second Fundamental Theorem

are zero or if two suitable a’s are equal. 1 N r, ≤ T[r, 1/(a2 − a3)] = T[r, a2 − a3] + O(1) " a2 − a3 #

≤ T(r, a2) + T(r, a3) + O(1) = O[T(r)] 1 So, N(r, 1/ϕ) ≤ N r, + O[T(r, f )] etc. by Jensen’s formula and f − a1 ! hypothesis.

In the course of the proof of theorem 10 we obtained, q ′ (q − 2)T(r, f ) < N(r, ai) − N0(r, 1/ f ) + S (r, f ) + O(1) Xi=1 q < N(r, ai) + S (r, f ) + O(1). Xi=1

Take f = ϕ, and a1 = ∞, a2 = 0, a3 = 1 to get 1 1 T(r,ϕ) < N(r,ϕ) + N r, + N r, + O[T(r,ϕ)] ϕ! ϕ − 1! That is, 1 1 [1 + O(1)]T(r,ϕ) < N(r,ϕ) + N r, + N r, ϕ! ϕ − 1!

for a suitable sequence of r tending to R. ϕ = 0, only if either f = a1, 1 1 1 69 or a2 = a3. So N r, ≤ N r, + N r, which is equal ϕ! f − a1 ! a2 − a3 ! 1 1 to N r, + O[T(r)]. Similar reasoning gives N r, + N(r,ϕ) + f − a1 ! ϕ! 1 3 1 N r, < N r, + O[T(r)]. Thus since T(r,ϕ) = [1 + ϕ − 1! i=1 f − ai ! O(1)]T(r) the resultP follows. Remark. Note that the same reasoning cannot be applied to more than three functions. In fact it is not known whether the analogous result is still true if we take more than three functions. Nevanlinna’s Second Fundamental Theorem 61

2.5 Picard values of meromorphic functions and their derivatives: 70 A function f (z) will be called admissible if it satisﬁes the hypothesis of theorem 9 in a circle |z| < R and also f (z) is transcendental if R = ∞. Note that if f (l)(z) lth derivative of f (z), all the poles of f (1)(z) have multiplicity at least l + 1. Therefore, N[r, f (l)] ≤ N(r, f (l))/(l + 1) and

N(r, f (l)) 1 N(r, f (l)) l Θ(∞) = lim 1 − ≥ lim 1 − ≥ T(r, f (l)) ! (l + 1) T(r, f (l)) (l + 1)

q (l) We obtain for f (z), since Θ(aν) ≤ 2, if a1,..., aq, are distinct and 0 q 1 P finite, Θ(aν) ≤ 1 + . Thus there can be at most one finite value 1 l + 1 whichP is taken only a finite number of times or more generally for which 3 Θ(a) > . 4 l Write now ψ(z) = a0(z) + f (z) + ··· + al(z) f (z)ai(z) being functions satisfying T[r, ai(z)] = OT(r, f ) and we assume ψ(z) is not identically constant. Then we have the following sharpened form of a theorem of Milloux.

Theorem 13. If f (z) is admissible in |z| < R then,

1 ′ T(r, f ) < N(r, f ) + N(r, 1/ f ) + N r, − N0(r, 1/ψ ) + S 1(r, f ) " ψ − 1#

′ ′ where N0(r, 1/ψ ) indicates that zeros of ψ corresponding to the re- peated roots of ψ = 1 are to be omitted, and lim S 1(r, f )/T(r, f ) = 0. r→R Note that this reduces to the second fundamental theorem for q = 3 when ψ = f.

71 Firstly let us prove some lemmas. 62 Nevanlinna’s Second Fundamental Theorem

Lemma 5. If l is a positive integer and f (z) admissible in |z| < R, such 0 < r <ρ< R then,

f (l) 1 m r, < A(l) log+ T(ρ, f ) + log+ ρ + log+ + log+(1/r) + O(1) " f # " ρ − r # A(l) being a constant depending only on l. Proof. The proof is by induction. The result is true for l = 1, by Lemma = 1 + 3 of section 2.1. Take ρ1 2 (ρ r) and assume that result for l. Now, 1 l l T[ρl, f (z)] = m[ρ1, f (z)] + N[ρ1, f (z)] (l) ≤ m[ρ1, f ] + m(ρ1, f / f ) + (l + 1)N(ρ1, f ) since at a pole of f of order k, f (l) has a pole of order k + l ≤ k(l + 1). By our induction hypothesis the right hand side is less than

+ + (l + 1)T(ρ1, f ) + A(l) log T(ρ, f ) + log ρ " 1 1 + log+ + log+ + O(1) ρ − ρ1 ρ1 # 1 1 < [l + 1 + A(l)]T(ρ, f ) + A(l) log+ ρ + log+ + log+ + O(1) " ρ − ρ1 ρ1 # So,

+ (l) + + + + + 1 log T[ρ1, f ] < log T(ρ, f ) + log (log ρ) + log log ρ − ρ1 ! 1 + log+ log+ + O(1) ρ1 ! 1 1 < log+ T(ρ, f ) + log+ ρ + log+ + log+ + O(1) ρ − ρ1 ρ1

72 Also by Lemma 3 applied to f (l),

f l+1 1 m r, < 3 log+ T ρ , f (l) + 4 log+ ρ + log+ f l 1 1 r ! h i 1 + 4 log+ + O(1) ρ1 − r Nevanlinna’s Second Fundamental Theorem 63

1 1 < 3 log+ T(ρ, f ) + 7 log+ ρ + 7 log+ + 2 log+ + O(1) ρ − r r 1 1 < A log+ T(ρ, f ) + log+ ρ + log+ + log+ + O(1) " ρ − r r # because, 1 ρ − r = ρ − ρ = (ρ − r)(ρ > ρ > r) 1 1 2 1 Therefore, since f l+1 f l+1 m r, < m r, + m r, f (l)/ f . f f (l) " # " # h i f l+1 m r, is less than f l ! 1 [A + A(l)] log+ T(ρ, f ) + log+ + log+(1/r) + O(1) + log+ ρ " ρ − r # completing the inductive proof.

Lemma 6. If ψ(z) is deﬁned as in theorem 13, and 0 < r <ρ< R, then 1 m(r, ψ/ f ) < O[T(r, f )] + A(l)[log+ T(ρ, f ) + log+ + log+ ρ] + O(1) ρ − r as r → R in any manner. Proof.

l (z) f (i) m(r, ψ/ f ) ≤ m r, a + log(l + 1).  i f  Xi=0   l   l (i) ≤ m[r, ai(z)] + m[r, f / f ] + log(l + 1) + (l + 1) log 2. Xi=0 Xi=0 l = O[T(r)] + m[r, f (i)/ f ] + O(1) Xi=0 1 < O[T(r)] + A(l) log+ T(ρ, f ) + log+ + log+ ρ + O(1) " ρ − r # from the lemma 5 and because m[r, ai(z)] ≤ T[r, ai(z)] = O[T(r)], and 73 log+(1/r) remains bounded as r tends to R. 64 Nevanlinna’s Second Fundamental Theorem

Lemma 7. If ψ(z) is deﬁned as above then as r → R in any manner while 0 < r <ρ< R, T(r, ψ) < [l + 1 + O(1)]T(r, f ) + A(l) 1 log+ T(ρ, f ) + log+ + log+ ρ + O(1) . " ρ − r ! # Proof. m(r, ψ) ≤ m(r, ψ/ f )+m(r, f ). If f has a pole of order K at a point (ν) and aν(z) a pole of order Kν then aν(z) f has a pole of order ν + K + Kν, and so ψ(z) has a pole or of order Max . (ν + K + Kν) ≤ (l + 1)K + Kν. This gives P

N(r, ψ) ≤ (l + 1)N(r, f ) + N(r, aν) Xν ≤ (l + 1)N(r, f ) + O[T(r, f )] Adding the above two inequalities, T(r, ψ) ≤ [l + 1 + O(1)]T(r, f ) + m(r, ψ/ f ), and now the lemma follows from the previous lemma. Lemma 8. If S (r, ψ) is deﬁned as in Theorem 8 with ψ instead of f then if 0 < r <ρ< R and r tends to R, 1 S (r, ψ) < A log+ T(ρ, f ) + log+ + log+ ρ + O(1) . " ρ − r ! # = 1 + Proof. Let ρ1 2 (ρ r). Lemma 3 gives then

+ + 1 + S (r, ψ) < A log T(ρ1, ψ) + log + log ρ1 + O(1) . " ρ1 − r ! # By lemma 7 since log+ x ≤ x, 1 T(ρ1, ψ) < A(l) T(ρ, f ) + + ρ + O(1) " ρ − ρ1 ! #

+ + + 1 + 74 that is, log T(ρ1, ψ) < log T(ρ, f ) + log + log ρ + O(1). ρ − ρ1 Substituting this and remembering that r < ρ1 < ρ, ρ − ρ1 = ρ1 − r = 1 2 (ρ − r) we get the result. Now we are ready to prove theorem 13. Nevanlinna’s Second Fundamental Theorem 65

Proof of theorem 2.13. We have from theorem 8,

1 1 m(r, ψ) + m r, + m r, ≤ 2T(r, ψ) − N1(r, ψ) + S (r, ψ) ψ! ψ − 1! i.e. 1 1 T r, + T r, ≤ T(r, ψ) − N1(r, ψ) + S (r, ψ) ψ! ψ − 1! 1 1 + N(r, ψ) + N r, + N r, ψ! ψ − 1! i.e. 1 1 T(r, ψ) ≤ N(r, ψ) − N1(r, ψ) + N r, + N r, + S (r, ψ) + O(1) ψ! ψ − 1! also 1 N(r, ψ) − N1(r, ψ) = N(r, ψ) − N r, and ψ′ ! 1 1 1 1 N r, − N r, = N r, − N0 r, . Hence ψ − 1! ψ′ ! ψ − 1! ψ′ ! Thus,

1 T(r, ψ)N(r, ) + N(r, 1/ ) + N r, − N0(r, 1/1) + S (r, ) + O(1) −1!

′ ′ where N0(r, 1/ψ ) denotes the fact that zeros of ψ at multiple roots of ψ − 1 are omitted. Thus, since T(r, ψ) = m(r, 1/ψ) + N(r, 1/ψ) + 1 ′ O(1)m(r, 1/ψ) ≤ N(r, ψ) + N r, − N0(r, 1/ψ ) + S (r, ψ) + O(1). ψ − 1! Note again that poles of ψ occur only at poles of f or of aν(z), and in N 75 each pole is counted only once. Then

N(r, ψ) ≤ N(r, f ) + N(r, aν) Xν 66 Nevanlinna’s Second Fundamental Theorem

≤ N(r, f ) + OT(r, f ).

Again,

T(r, f ) = m(r, 1/ f ) + N(r, 1/ f ) + O(1) ≤ m(r, ψ/ f ) + m(r, 1/ψ) + N(r, 1/ f ) + O(1).

Substituting we obtain,

1 ′ [O(1) + 1]T(r, f ) ≤ N(r, f ) + N r, + N(r, 1/ f ) − N0(r, 1/ψ ) ψ − 1! +S (r, ψ) + m(r, ψ/ f ),

since f (z) being admissible O(1) = O[T(r)]. Now if 0 < r <ρ< R lemmas 6 and 8 give

1 m(r, ψ/ f ) + S (r, ψ) < A(l) log+ T(ρ, f ) + log+ + log+ ρ + O(1) . " ρ − r # and now the result is completed just as in theorem 9 by means of lemma 4. Hence theorem 13.

Consequences. Theorem 14 (Milloux). If f (z) is admissible in |z| < R and is regular there then either f (z) assumes every finite value infinitely often or every derivative of f (z) assumes every finite value except possibly zero, infinitely often.

Proof. Suppose f (z) = a, f (l)(z) = b have only a ﬁnite number of roots g(l)(z) f (l)(z) 76 where b , 0. Choose g(z) = f (z) − a, ψ(z) = = in theorem b b 13. Then since N(r, g) = 0, g being regular

1 [1 + 0(1)]T(r, g) < N r, + N(r, 1/g) + S 1(r) " g(l)(z) − b#

where lim S 1(r)/T(r, g) = 0. r→R Nevanlinna’s Second Fundamental Theorem 67

If R = ∞ this gives lim T(r, g)/ log r < +∞ by assumptions that r→∞ 1 n t, and n(t, 1/g) are finite as r tends to infinity so that g(z) " f (l)(z) − b# is rational and so is f (z). That is f (z) is a polynomial. If R is less than infinity we obtain, lim T(r, g) < ∞, and so since r→R T(r, g) increases with r lim T(r, g) < ∞ and the same result applies to f (z) giving a contradiction to admissibility. Theorem 15 (Saxer). If f (z) is meromorphic in the plane and f , f ′, f ′′ have only a finite number of zeros and poles then f (z) = P1(z)/P2(z) P (z) ′ ′′ e 3 P1,P2,P3 being polynomials. If f , f , f have no zeros and poles then f (z) = ea+bz where a, b are constants. Proof. Set g(z) = f (z)/ f ′(z). Then g′(z) = 1 − [ f (z) f ′′(z)/ f ′2(z)]. Sup- pose that g(z) is transcendental and so admissible in the plane. Now g = 0, ∞ only when f ′ = 0, on f = 0, ∞ that is a finite number of times and so, N(r, g) + N(r, 1/g) = O(log r). Next g′(z) = 1 only for f = 0 1 or f ′′ = 0 that is N r, = O(log r) by hypothesis. Now theorem g′ − 1! 13 applied to g(z) gives for a sequence of r tending to infinity, taking ψ = g′, [1 + O(1)]T(r, g) = O(log r). This implies g(z) is rational, i.e. a contradiction. Hence g(z) is rational 77 so that f ′/ f = g is rational. Now f ′/ f has simple poles with integer residues at the poles and zeros of f (z). Since f ′/ f is rational by expand- ing it in partial fractions we get, ′ f / f = kr/(z − zr) + P(z) Xr with kν integers and P(z) a polynomial. Integrating the above,

k f (z) = (z − zr) P(z)dz Z Yr e This proves the ﬁrst part and if f (z) has no zeros or poles the product term disappears and f = eP(z), f ′(z) = P′(z)eP(z) and f ′(z) , 0 implies P′(z) , 0, which gives P(z) = a + bz. 68 Nevanlinna’s Second Fundamental Theorem

Remark . Note that we cannot do the same thing with f (z) and f ′(z). For if f (z) = eg(z), f ′(z) = g′(z)eg(z) and if we put g′(z) = eh(z) where h(z) is an arbitrary integral function, so that g(z) = eh(z)dz and f (z) = exp · eh(z)dz . Then f is an integral function withR f , 0, and f ′ , 0. So inhR theoremi 15 we cannot leave out the restriction on f ′′. Further F(z) = f (z)dz is a function for which F′ , 0, F′′ , 0 and so we cannot leaveR the restriction on f .

But we will show that we can leave out the restriction on f ′. This is precisely Theorem 19. 78 In connection with theorem 15 we also quote the following extension by Csillag [1].

Theorem 16. If l and m are diﬀerent positive integers and f (z) an integral function such that f (z) , 0, f (l)(z) , 0 and f (m)(z) , 0, then f (z) is equal to eaz+b.

2.6 Elimination of N(r, f ) We shall prove the following theorem

Theorem 17. If f (z) is admissible in |z| < R, and l ≥ 1 then,

1 T(r, f ) < [2 + (1/l)]N(r, 1/ f ) + 2[1 + (1/l)]N r, + S 2(r, f ), f (l) − 1!

where lim S 2(r, f )/T(r) = 0. r→R

Let us set ψ(z) = f (l)(z) in theorem 13 (Th. of Milloux), to get (2.7) 1 1 T(r, f ) < N(r, f ) + N(r, 1/ f ) + N r, − N0 r, + S 1(r, f ) f (l) − 1! f (l+1) !

1 (l+1) (l) where in N0 r, zeros of f at multiple roots of f (z) = 1 are f (l+1) ! to be omitted. Further we need, Nevanlinna’s Second Fundamental Theorem 69

l+1 Lemma 9. If g(z) = f (l+1)(z) /[1 − f (l)(z)]l+2 then h i 1 1 ′ lN1(r, f ) ≤ N2(r, f ) + N r, + N0 r, + m(r, g /g) f (l) − 1! f (l+1) ! (2.8) + log |g(0)/g′(0)| where N1(r, f ) stands for the N-sum over the simple poles of f (z); 79 N2(r, f ) for the N sum over multiple poles of f (z), with each pole counted only once. Thus N(r, f ) = N1(r, f ) + N2(r, f ).

Now at a simple pole z0 of f (z), f (z) = O(1) + [a/(z − z0)] where a is not equal to zero. Diﬀerentiating l times,

al!(−1)l+1 1 − f (l)(z) = + O(1) l+1 (z − z0) This can be written as al!(−1)l+1 1 − f (l)(z) = [1 + O{(z − z )l+1}] l+1 0 (z − z0) The diﬀerentiation of both the sides again gives,

a(l + 1)!(−1)l+1 f (l+1)(z) = 1 + O{(z − z )l+2} 0 (z − z )l + 2 h i 0 Hence, (−1)l+1(l + 1)l+1 g(z) = [1 + O{(z − z )l+1}] a(l)! 0 ′ So, g(z0) , 0, ∞ but g (z) has a zero of order at least l at z = z0. Now we have

N(r, g/g′) − N(r, g′/g) = m(r, g′/g) − m(r, g/g′) + log |g(0)/g′(0)| from Jensen’s formula. As we saw at the end of section 2 the left hand 80 side is

N(r, g) + N(r, 1/g′) − N(r, g′) − N(r, 1/g) 70 Nevanlinna’s Second Fundamental Theorem

= N(r, 1/g′) − N(r, 1/g) − N(r, g) ′ = N0(r, 1/g ) − N(r, 1/g) − N(r, g)

′ ′ where in N0(r, 1/g ) only zeros of g which are not zeros of g are to be considered. By our above analysis,

′ lN1(r, f ) ≤ N0(r, 1/g ) N(r, 1/g) + N(r, g) + m(r, g′/g) + log |g(0)/g′(0)|

Then note that g = 0, ∞ only at poles of f (z) which must be multiple, at zeros of f (l)(z)−1, and zeros of f (l+1)(z) which are not zeros of f (l)(z)−1. This gives lemma 9. Now (2.7) gives on writing T(r, f ) = m(r, f ) + N(r, f ) (2.9) 1 1 N(r, f ) − N(r, f ) < N(r, 1/ f ) + N r, − N0 r, + S 1(r, f ) f (l+1) ! f (l+1) ! On the left the contribution of each multiple pole to the sum being counted once for N, but at least twice for N. So,

(2.10) N2(r, f ) ≤ N(r, f ) − N(r, f )

Also N(r, f ) = N2(r, f ) + N1(r, f ). Hence it follows from lemma 9 that,

N(r, f ) ≤ [1 + (1/l)]N2(r, f )

1 1 1 ′ 1 |g(0)| + N r, + N0 r, + m(r, g /g) + log l " f (l) − 1! f (l+1) ! # l |g′(0)|

81 By the inequalities (2.9) and (2.10) it is at the most,

1 (l) 1 1 + N(r, 1/ f ) + N(r, 1/( f − 1)) − N0 r, + S 1(r, f ) l !" f (l+1) ! #

1 1 ′ ′ + (1/l) N r, + N0 r, + m(r, g /g) + log |g(0)/g (0)| " f (l) − 1! f (l+1) ! # = (1 + 1/l)N(r, 1/ f ) + (1 + 2/l)N(r, 1/( f (l) − 1)) 1 − N0 r, + S 3(r) f (l+1) ! Nevanlinna’s Second Fundamental Theorem 71 where

′ ′ S 3(r) = [1 + (1/l)]S 1(r, f ) + (1/l)[m(r, g /g) + log |g(0)/g (0)|].

Substituting for N(r, f ) in (2.7) we obtain,

T(r, f ) < [2 + (1/l)]N(r, 1/ f ) + 2[1 + (1/l)] 1 1 N r, − 2N0 r, + S 1(r, f ) + S 3(r, f ) f (l) − 1! f (l+1) ! We observe that the above gives the result of theorem 17, provided we pose

S 2(r, f ) = S 1(r, f ) + S 3(r, f ) ′ ′ = [2 + (1/l)]S 1(r, f ) + (1/l)[m(r, g /g) + log |g(0)/g (0)|] and in order to complete the proof it is suﬃcient to prove that if r <ρ< R, and r tends to R,

m(r, g′/g) < A[log+ T(ρ, f ) + log+ 1/(ρ − r) + log+ ρ + 0(1)]

The above inequality is true for,

log g(z) = (l + 1) log f (l+1)(z) − (l + 2) log[1 − f (l)(z)] g′/g = (l + 1) f (l+2)/ f (l+1) + (l + 2) f (l+1)/(1 − f (l)) h+ + i + m(r, g′/g) ≤ m(r, f (l 2)/ f (l 1)) + m r, f (l 1)/( f (l) − 1) + O(1) + Now the result follows from Lemma 8, with ψ = f (l) −1 or f (l 1). Hence 82 the proof of theorem 17 is complete.

2.7 Consequences Theorem 18. The result of theorem 14 extends to meromorphic functions without any further hypothesis. To be precise, if f (z) is admissible in |z| < R and is meromorphic there, then either f (z) assumes every finite value infinitely often or every derivative of f (z) assumes every finite value except possibly zero infinitely often. 72 Nevanlinna’s Second Fundamental Theorem

The proof is as before by the consideration of g(z) = [ f (z) − a]/b instead of f (z); if the equations f (z) = a and f (l) = b have only a ﬁnite number of roots. For then g = 0 and g(l) = 1 have only ﬁnite number of roots and we can use Theorem 17.

Theorem 19. Theorem 15 (Saxer’s theorem) still holds if the hypothesis of f ′(z) is omitted. That is, if f (z) is meromorphic in the plane and f , and f ′′ have only a ﬁnite number of zeros and poles then,

P3(z) f (z) = P1(z)/P2(z)e

′′ P1,P2,P3 being polynomials. If f , f have no zeros and poles then f (z) = eaz+b, a and b being constants.

83 Proof. If f (z) and f ′′(z) have only a ﬁnite number of zeros and poles, put g(z) = f (z)/ f ′(z). Then g′(z) = 1 −{ f (z) f ′′/[ f ′(z)]2}g(z) has only ﬁnite number of zeros and so does g′(z) − 1 = f f ′′/ f ′2 namely at the poles of f and at the zeros of f and f ′′. Hence by Theorem 18, g(z) is rational and now the proof is completed as in Saxer’s theorem. If f (z) and f ′′(z) have no zeros or poles, f (z) = eP(z) where P(z) is a polynomial. Therefore, f ′′(z) equals [P′′(z) + P′(z)2]eP(z), and if P′(z) has degree n greater than or equal to 1, then P′(z)2, P′′(z) have degree 2n, n − 1, respectively, so that f ′′(z) has 2n zeros. Thus P′(z) = a = const. P(z) = az + b.

A slightly more delicate analysis shows that if f (z), f ′′(z) have no zeros but f (z) may have a ﬁnite number of poles, then f (z) = eaz+b or (az + b)−n, where n is a positive integer. Part III Univalent Functions

3.1 Schlicht functions 84 Definition. A function f (z) regular in a domain D is said to be univalent (Schlicht, Simple) if f (z) takes different values at different points of D. Then f (z) maps D, one-one conformally into a domain ∆.

Since f (z) takes no value more than once, if f (z) is Schlicht in the plane, the function A(r) (area on the Riemann sphere) is less or equal to one and T0(r) ≤ log r. So f (z) is rational and is in fact a polynomial which must be linear. We consider functions f (z) univalent in |z| < 1. ∞ n If f (z) = anz , ( f (z) − a0)/a1 is also univalent. In fact a1 cannot be n=0 zero, for otherwiseP f (z) would assume values at least twice near z = 0, a1 , 0. Hence we may assume f (z) to be of the form

∞ n (2) z + anz Xn=2

The class of functions, univalent in |z| < 1 with the expansion (2) is called S . The ﬁrst two results are,

Theorem 1. If f (z) ∈ S, |a2| ≤ 2, and equality is possible only for z f (z) = f (z) = , and θ [1 − zeiθ]2

73 74 Univalent Functions

1 Theorem 2. If f (z) ∈ S and w is a value not taken by f (z) then |w|≥ 4 , again equality is possible only for f (z) = fθ(z).

85 Both results in the above form are due to Biberbach, theorem 2 with 1 a smaller constant than 4 is due to Koebe though it seems we had the proof eventually form 1 Note that f (z) ∈ S , takes all values with modulus < 4 . We shall prove theorem 1 ﬁrst and then deduce theorem 2 from it. For that we need

Lemma 1. Suppose f (z) is regular and univalent in the annulus r1 < ∞ 2 2n 2n |z| < r2 then the area of the image of the annulus is π n|an| (r2 −r1 ), −∞ ∞ P n where f (z) has the expansion anz in the annulus. −∞ P

r2 2π Proof. The area of the image is clearly r dr | f ′(reiθ)|2dθ. The inte- rR1 R0 gral

2π 2π | f ′(reiθ)|2dθ = f ′(reiθ) f (reiθ)dθ Z Z 0 0 2π n−1 i(n−1)θ m−1 −i(m−1)θ = nanr e mamr e Z 0 hX i hX i

Since the multiplication of the two series and then term by term integra- 2π 0, m , n, tions are valid, and ei(m−n)θdθ = 2π, m = n. R0   2π ∞ ′ iθ 2 2 2 2n−2 | f (re )| dθ = 2π n |an| r Z −∞ 0 X

86 ∴ Area of image Univalent Functions 75

r2 2π = r dr | f (reiθ)|2dθ Z Z r1 0 r2 +∞ 2 2 2n−1 = 2π n |an| r dt. Z X−∞ r1 Again since integration term by term is valid,

+∞ = 2 2n 2n π n|an| (r2 − r1 ) X−∞ Hence the lemma.

We now proceed to prove the theorem. Consider the function f (z) ∈ S .

1 ∞ 1 2 2 n 2 2 f (z ) f (z) = z + anz , let F(z) = f (z ) = z " z2 # X2 h i 1 ∞ 2 = z 1 + a z2n−2  n  X2     f (z) f (z) Since is not zero has a one valued square root which is a z z 1 f (z2) 2 power series in z and hence is a power series in z2. " z2 # 1 F(z) = z + a z3 + ······ 2 2

F(z) is odd. Also F(z) is univalent. For if F(z1) = F(z2), then

1 1 2 2 = 2 2 f (z1) f (z2) h i h i i.e., 87 2 = 2 = 2 = 2 f (z1) f (z2) ⇒ z1 z2. 76 Univalent Functions

∴ z1 = ±z2. But F(−z1) = −F(z1) , F(z1) unless z1 = 0. So z1 = z2. Now set

1 − 1 1 g(z) = = f (z2) 2 = + b z + b z3 + ······ F(z) 2 1 3 h ∞i 1 = + b zn z n X1 = 1 where b1 − 2 a2. And g(z) is univalent in 0 < |z| < 1. Let J(r) be the curve which is the image of |z| = r by g(z), 0 < r < 1 and A(r) the area inside it. Then for 0 < r1 < r2 < 1 we have by the lemma ∞ 1 1 2 2n 2n A(r1) − A(r2) = ±π − + n|bn| (r − r ) r2 r2  2 1  1 2 X1     Since J(r1) and J(r2) do not cross, the left hand side and hence the right hand side is diﬀerent from zero. So A(r) is monotonic. Further for small π r, A(r) ∼ which tends to ∞ as r → 0. Hence A(r) decreases. The r2 left hand side is therefore positive and the quantity inside the brackets is positive and so we take the positive sign. 1 ∞ Set S (r) = − |b |2nr2n. 2 n r 1 88 Then A(r) = πS (Pr) + C, C being a constant. We want to prove that S (r) ≥ 0 for 0 < r < 1. iθ Suppose now that b1 i.e. a2 is real Otherwise if a2 = |a2|e we consider eiθ f (ze−iθ) in place of f (z) and then

iθ −iθ iθ −2iθ 2 e f (ze ) = z + e a2e z + ······ 2 = z + |a2|z + ······

Thus there is no loss of generality in assuming a2 real and positive. Now 1 g(z) = + b z + b z3 + ······ , so if, z = reiθ z 1 3 iθ 1 1 3 g(re ) = + b1r cos θ + i b1r − sin θ + O(r ). r ! r ! Univalent Functions 77

We have then

iθ 1 3 |Rl. g(re )|≤ + b1r + O(r ) r

iθ 1 3 1 3 | Im .g(re )|≤ − b1r| + O(r ) = − b1r + O(r ) r r !

(since a2 > 0, b1 < 0) so that the image of |z| = r by g(z) is contained 1 3 1 3 in the ellipse of semi-axes − b1r + O(r ), | + b1r| + O(r ). Hence r ! r area inside the image A(r) satisﬁes

1 π A(r) ≤ π − b2r2 + O(r2) = + O(r2). r2 1 ! r2

Thus 89 π πS (r) + C ≤ + O(r2). r2 But 1 S (r) = + O(r2) r2 so that 1 π + O(r)2 + C ≤ + O(r2) r2 r2 or C + O(r2) ≤ O(r2) letting r → 0 we see that C ≤ 0. This proves that −C + A(r) = πS (r) ≥ 0 since, −C ≥ 0, A(r) ≥ 0. Thus for 0 < r < 1 S (r) ≥ 0. [Actually a little more reﬁned argument shows that A(r) = πS (r)]. Therefore, ∞ 1 ≥ |b |2nr2n r2 n X1 ∞ 2 and letting r → 1, 1 ≥ n|bn| . Thus 1 P |b1|≤ 1 or |a2|≤ 2 78 Univalent Functions

1 Equality can hold only if b = 0 for n > 1 i.e. g(z) = − zeiθ. n z

z 2 1 i.e. F(z) = = [ f (z )] 2 1 − z2eiθ Thus z f (z) = = fθ(z). [1 − zeiθ]2

This proves the theorem provided we show fθ(z) ∈ S . To deduce theo- w f (z) rem 2, set g(z) = where f (z) , w for |z| < 1. w − f (z) Then w(z + a z2 + ······ ) z a z2 g(z) = 2 = (z + a z2 + ··· ) 1 + + 2 + ··· 2 2 w − (z + a2z + ··· .) w w !

90 by expansion

1 2 = z + a2 + z + higer powers of z. w!

Since the map is bi-linear, g(z) is also univalent. In fact g(z1) = g(z2) implies, f (z1) = f (z2) from which it follows that z1 = z2. Further g(z) ∈ S . Hence by the above theorem,

1 + a2 ≤ 2 w !

1 ≤ 2 + |a | w 2

≤ 4 1 ∴ |w|≥ and equality is possible only if |a | = 2, i.e. f (z) = f (z). 4 2 θ Note that z 1 (1 + z)2 f0(z) = = − 1 (1 − z)2 4 "(1 − z)2 # 1 + z if ζ = then by this linear transformation |z| < 1 corresponds to 1 − z real ζ > 0 i.e., Z = ζ2 gives the plane cut along the negative axis. So Univalent Functions 79

1 f (z) is univalent and in fact f (z) maps onto the plane cut from − 0 0 4 to ∞ along the real axis. Hence f0(z) is univalent possessing such an −iθ iθ expansion deﬁned for elements of S and so does fθ(z) = e f0(ze ). 1 1 Also f (z) , − e−i because f (z) , − . Note also that θ 4 0 4

∞ n i(n−1)θ fθ(z) = z + nz e X2 so that |an| = n for all n.

Theorem 3. Suppose f (z) ∈ S and |z| = r, 0 < r < 1 then the following 91 inequalities hold. r r ≤ | f (z)|≤ (1 + r)2 (1 − r)2 1 − r 1 + r ≤ | f ′(z)|≤ (1 + r)3 (1 − r)3 (1 − r) f ′(z) 1 + r ≤ ≤ r(1 − r) f (z) r(1 − r)

where equality is possible only for functions fθ(z) deﬁned already.

Proof. Assume |z0| < 1 and set

z0 + z ϕ(z) = f = b0 + b1z + ······ "1 + z0z#

z0 + z Since = w is a bi-linear map of the unit circle onto itself. ϕ(z) is 1 + z0z ϕ(z) − b univalent in |z| < 1 and so 0 ∈ S . b1 Applying theorem 1 we deduce,

b 2 ≤ 2 b1

|b 2|≤ 2|b1| 80 Univalent Functions

we have z + z 1 (z0 + z)z ϕ′(z) = f ′ 0 − 0 2 1 + z0z!"1 + z0z (1 + z0z) # z + z 1 − (z )2 = f ′ 0 0 2 "1 + z0z# ((1 + zz0) )

92 and z + z 1 (z + z)z 2 ϕ′′(z) = f ′′ 0 − 0 0 2 1 + z0z!"1 + z0z (1 + z0z) # z + z (1 − |z |2)z − 2 f ′ 0 0 0 2 1 + z0z!" (1 + zz0) # Thus

′ 2 ′ b1 = ϕ (0) = (1 − |z0| ) f (z0) 1 1 b = ϕ′′(0) = [1 − |z |2]2 f ′′(z ) − f ′(z )z (1 − |z |2). 2 2 2 0 0 0 0 0

We have seen |b2|≤ 2|b1| i.e.

′′ 2 2 ′ 2 2 ′ (3.1) | f (z0)(1 − |O| ) − 2z0 f (z0)(1 − |z0| )|≤ 4(1 − |z0| )| f (z0)|

iθ If z0 = ρe this gives f ′′(z ) 2ρ2 4ρ 0 − ≤ ,ρ< ′ z0 2 2 1 f (z0) 1 − ρ 1 − ρ

∂w dw Now for any complex function = eiθ. If z = reiθ, and so ∂r dz ∂ f ′′ [log f ′(z)] = eiθ. Thus the above inequality is ∂r f ′

∂ 2ρ2 4ρ ρ log f ′(z) − ≤ , (z e−iθ = ρ), 2 2 0 ∂ρ 1 − ρ 1 − ρ

This gives bounds for f ′(z). 93 Again,

r r r 1 + ρ r | f (reiθ)| = f ′(ρeiθ)dρ ≤ | f ′(ρeiθ)|dρ ≤ dρ = Z Z Z (1 − ρ)3 (1 − r)2 0 0 0

= iθ 1 To get the lower bound for w f (ρe ), we suppose |w| < 4 , since for 1 |w| ≥ 1 the result is trivial ρ < . Thus by theorem 2 (Section 3) 4 (1+ρ)2 4! the line segment [0, w] lies entirely in the image of |z| < 1 by w = f (z). If λ is this line segment, γ the corresponding curve in the z-plane

ρ dw 1 − ρ |w| = |dw| = |dz|≥ dρ. Z Z dz Z (1 + ρ)3 λ γ 0

= ρ (θ + ρ)2 dw 1 − ρ because |dz|≥ dρ if dρ is positive since dρ< |dz|, if dρ< 0 dz (1 + ρ)3 the result is even more evident. ϕ( z) − b0 Since ∈ S , we obtain, 94 b1 ρ ϕ(ρeiθ) − b ρ ≤ 0 ≤ . + 2 2 (1 ρ) b1 (1 − ρ)

iθ Now z0 = ρe . 82 Univalent Functions

2 ′ Then ϕ(z0) = 0, b0 = f (z0), b1 = [1 − |z0| ] f (z0) |z | | f (z )| |z | 0 ≤ 0 | f ′(z )|≤ 0 2 2 0 2 1 + |z0| [1 − |z0| ] 1 − |z0| r(1 − r) f (z) r(1 + r) i.e. ≤ ≤ . 1 + r f ′(z) 1 − r

f ′ This gives the bounds for since equality holds in (3.1) only for f = ϕ(z) fθ(z) it can be shown that for all the inequalities of theorem 3 equality is possible only for this function. Theorem 4 (Littlewood, Paley, Spencer). Suppose f (z) ∈ S and for any λ → 0 set π 1 iθ λ (3.2) Iλ(r, f ) = | f (re )| dθ 2π Z −π d S (r) = r I (r). λ dr λ Then r π 2 λ iθ λ−2 ′ iθ 2 (3.3) S λ(r) = ρdρ | f (ρe )| | f (ρe )| dθ 2π Z Z 0 −π Thus λrλ (3.4) S (r) ≤ λM(r, f )λ ≤ , λ (1 − r)2λ r r S λ(ρ)dρ λ dρ Iλ(r) = ≤ λ M(ρ, f ) Z ρ Z ρ 0 0 1−2λ 1 A(λ)[1 − r] , λ> 2  1 1 (3.5) ≤ A log λ =  1 − r 2  1 A(λ) λ<  2 95   Univalent Functions 83

Proof. Suppose ∞ f (z) = z 1 + a zn−1  n   X2    Set   ∞ λ/2 1 λ n ϕ(z) = [ f (z)] = z 2 1 + b z  n   X1    f (z)   This is possible since is regular and non-zero in |z| < 1 and z λ so has a th power which is also regular. For deﬁniteness we take the 2 principal value of zλ/2, | arg z| <π. ∞ ′ λ n+ 1 λ−1 We have ϕ (z) = n + bnz 2 . 0 2 P π π ∞ 2 1 ′ iθ 2 1 ′ ′ λ 2 2n+λ−2 |ϕ (ρe )| dθ = ϕ ϕ dθ = n + |bn| ρ 2π Z 2π Z = 2 −π −π Xn 0 exactly as in lemma 1 (Section 3)

π π ∞ 1 iθ 2 1 2 2n+λ Iλ(r, f ) = |ϕ(re )| dθ = ϕϕdθ = |bn| ρ 2π Z 2π Z −π −π X0 d 2 2n+λ S λ(r, f ) = r Iλ(r, f ) = (2n + λ)|bn| r . dr X Further 96

r π ′ iθ π 2 2n+λ π ρdρ |ϕ (ρe )dθ = (2n + λ)|bn| r = S λ(r, f ). Z Z 2 2 0 −π X

′ λ ′ λ−1 Now ϕ = f f 2 2 λ2 |ϕ′|2 = | f ′|2| f λ−2| 4 84 Univalent Functions

Thus r π 2 λ ′ iθ 2 iθ λ−2 S λ(r, f ) = ρdρ | f (ρe )| | f (ρe )| dθ. 2π Z Z 0 −π giving (3.3). We now interpret the above formula. ρdρ dθ is an element of area in |z| < 1, | f ′|2ρdρdθ is the corresponding area in the image plane at w = Reiφ and | f |λ−2| f ′|2ρdρ dθ the mass of the area, if we imagine a mass density Rλ−2 on the circle |w| = R in the image plane. Then our integral is the total mass of the image. Since the image covers no point more than once and lies in |w| < M(r, f ) = M say, the mass of image ≤ total mass of the circle which is

M 2π 2πR · Rλ−2dR = Mλ and so Z λ 0 λ2 2π rλ S (r) ≤ Mλ = λMλ ≤ λ λ 2π λ (1 − r)2λ This gives (3.4), because from theorem 3, r M(r, f ) ≤ . (1 − r)2

97 Thus

r λ−1 ρ λ−1 Iλ(r) ≤ λ dρ. If λ ≥ 1, ρ ≤ 1 and so Z (1 − theta)2λ 0 r 1 λ 1−2λ Iλ(r) ≤ λ dρ ≤ (1 − r) . Z (1 − ρ)2λ 2λ − 1 0 Assume that λ< 1. Then

r r r λ−1 λ λ ρ = ρ ρ Iλ(r) ≤ λ dρ  − 2λ + dρ. Z (1 − ρ)2 (1 − ρ)2λ Z (1 − ρ)2λ 1  0 0 0  Univalent Functions 85

r rλ ρ ≤ − 2λ dρ since ρ ≤ ρλ if 0 ≤ λ ≤ 1 (1 − r)2λ Z (1 − ρ)2λ+1 0 r r rλ 1 1 = + 2λ dρ − 2λ dρ (1 − r)2λ Z (1 − ρ)2λ Z (1 − ρ)2λ+1 0 0 rλ 2λ 2λ 1 = + (1 − r)1−2λ − − + 1 if 2λ , 1. (1 − r)2λ 2λ − 1 2λ − 1 (1 − r)2λ 2λ 2λ 2λ ≤ (1 − r)1−2λ − + 1 ≤ (1 − r)1−2λ 2λ − 1 2λ − 1 2λ − 1

2λ Hence if 1 − 2λ> 0, (1 − r)1−2λ ≤ 1, I (r) ≤ . λ 2λ − 1 r r dρ If 2λ = 1, I (r) ≤ 1 1 dρ = 98 λ 2 1 2 ρ 2 (1−ρ) 1 − ρ R0 R0 1 1 + r 1 1 − r2 1 1 1 = log = log ≤ log = log 2 1 − r 2 (1 − r)2 2 (1 − r)2 (1 − r) Thus 2λ 1 (1 − r)1−2λ if λ> 2λ − 1 2  1 1 Iλ(r) ≤ log if λ =  1 − r 2  2λ 1  if λ< 2λ − 1 2  This proves the theorem completely. ∞ n Theorem 5 (Littlewood). If f (z) = z + anz ∈ S then |an| < e for 2 n ≥ 2. P

1 f (z)dz Proof. We have |a | = + n 2π zn 1 |z|R=ρ

π 1 I (ρ, f ) ≤ | f (ρeiθ)|dθ = 1 2πρn Z ρn −π 86 Univalent Functions

By inequality (3.5)

ρ dt ρ I1(ρ, f ) ≤ = . Z (1 − t)2 1 − ρ 0

1 1 Thus |a | < . We choose ρ so that is minimal n ρn−1(1 − ρ) ρn−1(1 − ρ) n − 1 n n−1 i.e. put ρ = 1− = to get |an| < n n n − 1 1 n−1 i.e., |an| < 1 + n < en n − 1!

99 This completes theorem 5. ρ Remark. Bazilevic [1] improved this to I (ρ, f ) < + 0.55 so that 1 1 − ρ2 we get 1 |a | < en + 1.51 n 2 Theorem 6. Suppose f (z) ∈ S and set

∞ = λ = λ n ϕ(z) [ f (z)] ϕ(z) z  an, λz  . X0      1 2λ−1 = + k+1 + Then if λ > 4 , |an,λ| < A(λ)n . In particular if f (z) z ak+1z kn+1 −1/3 ···+akn+1z +···∈ S , then |a2n+1| < A1 if k = 2 and |a3n+1| < A2n if k = 3,A1 and A2 are absolute constants.

ϕ′(z)dz Proof. Now (n + λ)|a | = 1 n,λ 2πi n+λ |z|=ρ z R

1 (3.6) ≤ I (ρ, ϕ′) ... ρn+λ−1 1

with the notation of theorem 4. Univalent Functions 87

Since ϕ′ = λ f λ−2 f ′

2π ′ λ ′ iθ iθ λ−1 I1(ρ, ϕ ) = | f (ρe )|| f (ρe )| dθ 2π Z 0 2π λ = | f ′(ρeiθ)|| f (ρeiθ)|t−1| f (ρeiθ)|λ−tdθ 2π Z 0 for any t

1 2π 2 ′ 1 ′ iθ 2 iθ 2t−2 I1(ρ, ϕ ) ≤  | f (ρe )| | f (ρe )| dθ × 2π Z    0   1   2π 2  1 iθ 2λ−2t (3.7) λ  | f (ρe )| dθ 2π Z    0    by Schwarz inequality. 100 1 Since λ > , we choose t suﬃciently small but positive such that 4 1 2λ − 2t > 1 ; for example t = 1 (λ − ) can be a choice of t. By theorem 2 2 4 4

1 1− 2n 2π ′ iθ 2 iθ 2t−2 π 1 ρdρ | f (ρe )| | f (ρe )| dθ ≤ S 2t 1 − , f Z Z 2t2 2n ! 1 0 1− n ≤ A(t)n4t

Suppose

π 2πµ = min. | f ′(ρeiθ)|2| f (ρeiθ)|2t−2dθ 1 1 1− n ≤ρ≤1− 2n Z 0 88 Univalent Functions

Hence A(t)n4t (3.8) µ ≤ ≤ A(t)n4t+1 1 1− 2n ρ dρ 1 1−R n Also by theorem 4,

2π 1 | f (ρeiθ)|2λ−2tdθ ≤ A(λ, t)(1 − ρ)−4λ+4t+1 since 2λ − 2t ≥ Z 2 0 (3.9) ≤ A(λ)n4λ−4t−1

101 since t depends upon λ. Hence for this value ρ from (3.7),

′ (4t+1+4λ−4t−1) 1 2λ I1(ρ, ϕ ) ≤ A(λ)n 2 = A(λ)n

Therefore from (3.6) it follows

2λ −n−λ A(λ)n 1 2λ (n + λ)|an,λ|≤ < 1 − A(λ)n ρn+λ−1 n! ≤ A(λ)n2λ

A(λ) just standing for a constant depending upon λ. Hence the result 2λ−1 |an,λ| < A(λ)n follows. ∞ kn+1 Suppose now that g(z) = z + akn+1z ∈ S . Then so does n=1 ∞ P k 1/k k n f (z) = [g(z )] = z 1 + akn+1z . For clearly f (z) is regular in n=1 ! ′ P |z| < 1. f (0) = 0, f (0) = 1. Suppose f (z1) = f (z2). Then

1 k 1 k 1 1 1 k = k = k = k = k g z1 g z2 ⇒ g z1 ωg z2 g z2 ω 1 1 k = k = where ω is a kth root of unity. But g(z) ∈ S , and we have z1 ωz2 ⇒ z1 = z2. Univalent Functions 89

Thus

1 1/k g z k = [ f (z)] ∞ 1/k n = z 1 + a + z  kn 1  Xn=1     Hence applying the ﬁrst part, with λ = 1/k (provided k = 1, 2, 3) 102

|akn+1| < A(k)n as required

i.e. |an+1| < A1n if k = 1

|a2n+1| < A2 if k = 2 −1/3 |a3n+1| < A3n if k = 3

This inequality is false for large k as was shown by Little wood [1] even if f (z) is continuous in |z| ≤ 1. We do not know whether it is ∞ n true for any k ≥ 4. If f (z) = z + anz is bounded and univalent 2 then the area of the image of |z| < 1P is at most πM2 where M is the 2 2 least upper bound of | f (z)|. Hence n|an| ≤ M and it follows that − 1 |an| = O(n 2 ) as n → ∞. NothingP stronger than this is known. For further discussion see Hayman Chapter 3. This is the correct order for mean-valent functions and probably for f (z) ∈ S also. The best example −13/14 due to Clunie (unpublished) gives |an| > n for some large n, so that 1 there is a gap between and 13/14. It can be shown that in theorem 6 2 the conclusion that |an| = O(1) obtains for all n if |an| = O(1) for some sequence of n with constant common diﬀerence. We do not know if this conclusion is till true i.e. if |an| = O(1) for a sequence n = nk such that nk+1 − nk < constant. In this connection Biernacki [1] has shown that for every f (z) ∈ S ,

3/2 ||an+1| − |an|| < A[log(n)] for n ≥ 2.

Hence if an = 0 for a sequence n = nk, with nk+1 − nk < const. |an| = 103 O(log n)3/2 for the intermediate coeﬃcients. It is still to be found out whether O(1) can replace O(log n)3/2. 90 Univalent Functions

Examples for theorem 6. z f (z) = (1 − z)2 z f (z) = [ f (zk)]1/k = = a zkn+1 k k 2/k n,k (1 − z ) X where 2 2 + 2 + k k 1 ...... k n − 1 a = n,k 1· 2 ...... n Γ + 2 2 −1 n n k = k ∼ Γ(n + 1)Γ(2/k) Γ(2/k) so that theorem 6 is best possible. ∞ n Theorem 7 (Rogosinski, Deudonne’, Szasz). If f (z) = z + anz be- 2 longs to S and has real coeﬃcients then |an|≤ n. P

Proof. Let f = u + iv. We have since an are real f (z) = f (z). z = x + iy, y , 0 implies v , 0 for otherwise if v = 0 for z = x + iy, y , 0 then f (z) = f (z) = u which contradicts uni valency. Further we assert that for z such that y > 0 v must have constant sign. For if instead we have v(z1) 104 positive and v(z2) negative z1 and z2 having the imaginary part positive, v(z) would be zero somewhere on the line joining z1 and z2, which is not possible. Clearly by considering f (z)/z for small z v > 0 for z with Im z > 0. The proof of the theorem essentially depends on this nature of the sign of v.

∞ iθ n Now v(re ) = anr sin nθ since an are real where a1 = 1. 1 P π π Hence we have on integration, v(reiθ) sin θn dθ = rna since the 2 n R0 other terms vanish. Also | sin nθ|≤ n sin θ for 0 ≤ θ ≤ π. π It is enough to show this inequality for 0 ≤ θ ≤ since | sin n(π − 2 θ)| = | sin nθ|, sin(π − θ) = sin θ. Univalent Functions 91

Sin θ π decreases in 0 <θ ≤ . θ 2 π π Hence if nθ ≤ i.e. θ ≤ 2 2n sin nθ sin θ ≤ nθ θ π which gives the inequality for θ ≤ . 2n In particular when sin π sin π = π 2 2n nθ π ≤ π 2 2 2n π or sin ≥ 1 . 2n n π π Hence if ≤ θ ≤ . 105 2n 2 π | sin nθ|≤ 1 ≤ n sin ≤ n sin θ 2n Hence follows the inequality | sin nθ|≤ n sin θ 0 ≤ θ ≤ π.

π 2 iθ |an| = v(re ) sin nθ dθ πrn Z 0

Now making use of the fact that v > 0 for the range of θ

π 2 iθ |an|≤ v(re )| sin θn|dθ πrn Z 0 π 2 ≤ v(reiθ)n sin θ dθ πrn Z 0 n n = a = rn−1 1 rn−1 Let us make r → 1, |an|≤ n as required. 92 Univalent Functions

z The function ∈ S has real coefficients, and satisfies a = n. (1 − z)2 n Thus our inequality is sharp. We next proceed to prove another special case of Bieberbach’s con- jecture. For this we need the definition:

Deﬁnition. A domain D is said to be star-like w.r.t. the origin O if for any point P ∈ D, OP lies entirely in D.

∞ n 106 Theorem 8 (Nevanlinna). If w = f (z) = z+ anz ∈ S and maps |z| < 1 2 onto a domain D star-like with respect to wP= 0 then |an|≤ n. Equality is possible only when f (z) = fθ(z) deﬁned already.

Before proceeding with the proof of the theorem let us prove the following lemma due to Borel.

∞ n Lemma 2. If ϕ(z) = 1 + bnz satisﬁes Rlϕ(z) ≥ 0 for z ≥ 0 then n=1 P + ∞ 1 z n |bn|≤ 2, equality holds only for ϕ(z) = = 1 + 2z . 1 − z 1 P ϕ(z) − 1 Consider ψ(z) = . By hypothesis ϕ(z) has real part +ve ϕ(z) + 1 which clearly implies |ϕ(z) − 1| < |ϕ(z) + 1| and so |ϕ(z)| < 1 and further ψ(0) = 0. Hence now applying Schwarz’s lemma

|ψ′(0)|≤ 1. Equality holds only if ψ(z) = zeiθ.

Near ∞ n bnz = = 1 ! = b1 + z 0 ψ(z) P∞ z ······ n 2 2 + bnz n=1 P which gives that |b1| ≤ 2. In order to deduce the inequality for bn, consider

2 n ϕk(z) = 1 + bkz + b2kz + ······ + bnkz + ······ Univalent Functions 93

1 k = ϕ(w z1/k) k r Xr=1 1/k wr running through the kth roots of unity and for a ﬁxed value of z . r 2πi For wr = w where w = e k . If we set ∞ 1 k 1 ζ = z1/k, (wrz1/k) = 1 + b ζn[wn + ··· + wkn] k k n Xr=1 Xn=1 ∞ nk = 1 + bnkζ Xn=1 ∞ n = 1 + bnkz X1 as wn + ······ + wkn = 0 if k does not divide n = k if k divides n.

1/k Clearly Rl ϕk(z) ≥ 0 since Rlϕ(wrz ) ≥ 0 for every. Hence applying 107 previous part to ϕk(z), |bk|≤ 2.

Proof of theorem 3.8. Note ﬁrst that if Gr denotes the image of |z| < r by f (z) then if G1 is star-like so is Gr for 0 < r ≤ 1.

−1 Consider ζ = ϕ(z) = f [t f (z)] 0 < t < 1. Then since G1 is star-like t f (z) lies in the image of |z| < 1 by f (z) and so f −1[t f (z)] is a well deﬁned point in |ζ| < 1, and clearly ζ = 0 corresponds to z = 0. Thus |ϕ(z)| < 1 ϕ(0) = 0 and hence by Schwarz’s lemma

|ϕ(z)| ≤ |z| t f (z) = f (ζ) where |ζ| ≤ |z| and so if |z| < r, f (ζ) ∈ Gr i.e. t f (z) ∈ Gr. This being true for any t, 0 < t < 1, Gr is star-like. Since Gr is star-like its boundary Γr meets any ray from 0 to ∞ in 108 only one point. The limiting case when Γr contains a line segment in such a ray is excluded since Γr is a simple closed analytic curve. Thus arg . f (reiθ) increases with θ for any ﬁxed r, 0 < r < 1. Let log f (reiθ) = u + iv = log | f (reiθ)| + i arg . f (reiθ) 94 Univalent Functions

∂v and arg . f (reiθ) is increasing for any ﬁxed r implies ≥ 0. ∂θ ∂ i.e., Im . [log f (reiθ)] ≥ 0 ∂θ f ′(reiθ) Im . ireiθ ≥ 0 " f (reiθ) # reiθ f ′(reiθ) i.e., Rl. ≥ 0 f (reiθ) In other words, z f ′(z) Rl. ≥ 0, 0 ≤ |z| < 1 f (z) Now applying the lemma with ϕ(z) = z f ′(z)/ f (z) and observing, that this function satisﬁes the hypothesis of the lemma, (for ϕ(z) = 1 + O(z) near z = 0) Rl.ϕ(z) ≥ 0 and so if

∞ z f ′(z) ϕ(z) = = 1 + b zn f (z) n X1

then |bn|≤ 2. Again

∞ ∞ ∞ z f ′(z) = na zn = ϕ(z) f (z) = a zn 1 + b zn n  n   n  X1 X1 Xn=1         with a1 = 1. 109 Equating coeﬃcient of zn on either side,

nan = an + b1an−1 + b2an−2 + ······ + bn−1a1

i.e. (n − 1)an = b1an−1 + b2an−2 + ······ + bn−1a1

(n − 1)an ≤ 2[1 + |a2| + |a3| + ······ + |an−1|] by the lemma.

Suppose we now assume that |aν|≤ ν for ν ≤ n − 1 then we have,

(n − 1)|an|≤ 2[1 + 2 + ······ + n − 1] ≤ n(n − 1) Univalent Functions 95

|an|≤ n. and by theorem 1, |a2|≤ 2. Hence the proof is complete by induction on n. Further equality is possible for n ≥ 2 only if |a2| = 2 i.e. for functions = z fθ(z) (1−zeiθ)2 . For extensions of this result to a more general class of image do- mains see Kaplan, W. [1].

3.2 Asymptotic behaviour (1 − r)2 Theorem 9. If f (z) ∈ S and unless f (z) ≡ f (z) we have M(r, f ) θ r decreasing steadily with increasing r and so tends to α where 0 ≤ α< 1 as r → 1.

To prove this we require

Lemma 3. Suppose f (z) ∈ S and for ﬁxed θ f (reiθ) = R(r)eiλ(r). Suppose 110 further that 0 < r1 < r2 < 1 and r = r1, r2 correspond to R = R1,R2. Then

r2 R (1 − r )2 R (1 − r )2 r log 2 2 ≤ log 1 1 − 1 (1 − r)[λ′(r)]2dr. r2 r1 4 Z r1

Proof. We have

d d f ′(reiθ) log(Reiλ) = log f (reiθ) = eiθ . dr dr f (reiθ)

On the other hand d 1 dR log(Reiλ) = + iλ′(r) dr R dr By theorem 3

f ′(reiθ) 1 + r ≤ iθ so that f (re ) r(1 − r)

96 Univalent Functions

1 dR 2 (1 − r)2 + [λ′(r)]2 ≤ , " r dr # r2(1 − r)2

say a2 + b2 ≤ c2 i.e. |a|2 + |b|2 ≤ c2. This can be written as

(|c| − |a|)(|c| + |a|) ≤ b2 b2 b2 or |c| − |a|≥ ≥ |c| + |a| 2|c|

b2 since |c| ≥ |a|. So |a| ≤ |c|− . Since a ≤ |a| we get 2|c| 1 dR 1 − r r(1 − r) ≤ − [λ′(r)]2. R dr r(1 − r) 2(1 + r) 1 + r r ≤ − 1 (1 − r)[λ′(r)]2 r(1 − r) 4

111 since r1 ≤ r for the range r1 ≤ r ≤ r2 and 1 + r < 2. Integrating from r1 to r2 r r 2 + 2 1 r r1 ′ 2 log R2 − log R1 ≤ dr − (1 − r)[λ (r)] . Z r(1 − r) 4 Z r1 r1 r2 r2 r2 1 r = dr + dr − 1 (1 − r)[λ′(r)]2dr. Z r(1 − r) Z 4 Z r1 r1 r1 r2 r r r = log 2 − log 1 − 1 (1 − r)[λ′(r)]2dr. 2 2 (1 − r2) (1 − r1) 4 Z r1 Thus

r2 R (1 − r )2 R (1 − r )2 r log 2 2 ≤ log 1 1 − 1 (1 − r)[λ′(r)]2dr r2 r1 4 Z r1 giving the lemma Univalent Functions 97

(1 − r)2 Proof of theorem 3.9. Set ψ(r) = M(r, f ). Suppose that θ is so r iθ chosen that f (r2e ) = M(r2, f ) = R2. Then lemma 3 gives

r2 2 (1 − r1) R1 r1 ′ 2 log ψ(r2) ≤ log − (1 − r)[λ (r)] dr. r1 4 Z r1 r2 r1 ′ 2 ≤ log ψ(r1) − (1 − r)[λ (r)] dr. 4 Z r1 since R1 ≤ M(r1, f ). 112 Hence ψ(r2) ≤ (r1) showing that ψ(r) decreases (weakly) with r. If ψ(r1) = ψ(r2) then ψ(r) = a constant for r1 ≤ r ≤ r2 and

r2 (1 − r)[λ′(r)]2dr = 0. Z r1 ′ Since 1 − r > 0 therefore λ (r) = 0 or λ(r) = constant for r1 ≤ r ≤ r2. (1 − r)2 Also ψ(r) = constant gives | f (reiθ)| = constant = α (say) for r (1 − r)2 r ≤ r ≤ r . Because from the lemma | f (reiθ)| decreases with 1 2 r iθ increasing r for ﬁxed θ. Therefore if θ1 is such that f (r2e 1 ) = M(r2, f ) we see that (1 − r )2 (1 − r )2 2 2 iθ1 M(r2, f ) = | f (r2e )| r2 r2 (1 − r )2 (1 − r )2 1 iθ1 1 ≤ | f (r1e )|≤ M(r1, f ) r1 r1 2 (1 − r2) = M(r2, f ) because ψ(r) = constant. r2 (1 − r)2 Hence | f (reiθ)| is constant for r ≤ r ≤ r . r 1 2 Now αr α(reiθ)ei(λ−θ) f (reiθ) = Reiλ = eiλ = (1 − r)2 [1 − eiθre−iθ]2 98 Univalent Functions

αei(λ−θ)z 113 Or f (z) = for z = reiθ, r < r < r . (1 − ze−iθ)2 1 2 Hence by analytic continuation this equation holds in |z| < 1 and i(λ−θ) since f (z) ∈ S , αe = 1 and so f (z) ≡ f−θ(z) as required.

In all other cases ψ(r) decreases strictly with increasing r and

M(r, f ) lim ψ(r) = 1 = lim . r→0 r Thus ψ(r) < 1, 0 < r < 1 and since ψ(r) decreases α = lim ψ(r) exists r→1 1 and α ≤ ψ < 1. 2! Theorem 10. If f (z) ∈ S and α = 0 in theorem 9 then with the notation 1 of theorem 6 we have for λ > a = O(n2λ−1) as n → ∞ and in 4 n,λ a particular if a are the coeﬃcients of f , n → 0 as n tends to inﬁnity. n n Proof. We recall the proof of theorem 6 and the notations of that theorem. We proved

1 (n + λ)|a |≤ I (ρ, ϕ′) n,λ n + λ − 1 1 1 π 2 π ′ λ ′ iθ 2 iθ 2t−2 λ iθ 2λ−2t 1 I1(ρ, ϕ ) ≤  | f (ρe )| | f (ρe )|  f (ρe ) dθ ] 2 2π Z  2π Z  −π  −π     1 We can ﬁnd ρ such that 1 − ≤ ρ ≤ 1 − 1 and n 2n

π S 1 − 1 , f 1 ′ iθ 2 iθ 2t−2 2t 2n 1 | f (ρe )| | f (ρe )| dθ ≤ 2π Z 1 1 2 1 2 4t2 −π 2 1 − 2n − 1 − n 114 By theorem 4

2t −4t S 2t(r, f ) ≤ 2tM(r, f ) = O[(1 − r)] Univalent Functions 99 as r → 1. So in the above inequality the right hand side is n O[n4t] = O[n4t+1] instead of O(n4t+1). Just as before (Ch. (3.9) of section 3.1)

π | f (ρeiθ)|2λ−2tdθ = O n−1+4λ−4t Z −π h i and so we now get

+ 1 (1+4t+4t+4λ−1) n|an,λ|≤ (n + λ)|an,λ| = O n 2 = O(n2λ) as required.

The case α > 0. Suppose then from now on α> 0 in theorem 9 and we shall develop a series of asymptotic formulae by a form of the Hardy-Little wood method, culminating in a formula for the an,λ. We have ﬁrst

2 iθ Theorem 11. There exists θ0 in 0 ≤ θ0 ≤ 2π such that (1−r) | f (re 0 )|→ α as r → 1. 1 Proof. Set r = 1 − and choose θ so that f (r eiθn ) = M(r , f ). n n n n n Then from lemma 3 we have for 0 < r < rn (1 − r)2 (1 − r )2 iθn n iθn | f (re )|≥ | f (rne )| = Bn (say) r rn

Let now θ0 be a limit point of the θn and choose a ﬁxed r in 0 < r < 1. 115 Then (1 − r)2 (1 − r)2 | f (reiθ0 )| = lim | f (reiθn ) ≥ lim B r r n as n tends to inﬁnity through a suitable sequence for which θn → θ0, (1 − r)2 since after some n, r < r and | f (reiθn )|≥ B . Also lim B = α n r n n (1 − r)2 since M(r, f ) → α as r → 1. r 100 Univalent Functions

rα Hence | f (reiθ0 )|≥ and this is true for 0 < r < 1. Thus (1 − r)2

lim(1 − r)2| f (reiθ0 )|≥ α r→1

On the other hand lim(1 − r)2| f (reiθ0 )|≤ lim(1 − r)2 M(r, f ) = α. r→1 −−→ Thus lim(1 − r)2| f (reiθ0 )| = α as required. r→1

iθ iλ(r) Theorem 12. If θ0 is as in theorem 11 and f (re 0 ) = R(r)e then

1 (1 − r)[λ′(r)]2dr converges. Z 0 1 Hence if r → 1 and ρ → 1 while r <ρ< (1 + r) we have uniformly 2

(1 − r)2 f (reiθ0 ) → 1. (1 − ρ)2 f (ρeiθ0 )

Proof. We have by lemma 3 for 0 < r1 < r2 < 1

r2 r R (1 − r )2 R (1 − r )2 1 (1 − r)[λ′(r)]2dr ≤ log 1 1 − log 2 2 4 Z r1 r2 r1

116 Letting r2 → 1 and by theorem 11 we get

1 r R (1 − r )2 1 (1 − r)[λ′(r)]dr ≤ log 1 1 4 Z r1α r1

′ proving convergence, since λ (r) is bounded in 0 ≤ r ≤ r1. Now suppose that r is so near 1 that

1 1 (1 − t)[λ′(t)]2dt ≤ ǫ and r <ρ< (r + 1) Z 2 r Univalent Functions 101

Then

1 1 ρ ρ 2 ρ 2 ′ ′ 2 dt |λ(ρ) − λ(r)| = λ (t)dt ≤  (1 − t)[λ (t)] dt ×   Z Z  Z 1 − t  r  r   r        (by Schwarz inequality).  

1 1 − r 2 1 |λ(ρ) − λ(r)|≤ ǫ log ≤ (ǫ log 2) 2 " 1 − ρ#

So |λ(ρ) − λ(r)|→ 0 if r, ρ → 1 being related as in the theorem. Thus

(1 − r)2 f (reiθ0 ) arg = λ(r) − λ(ρ) → 0 (1 − ρ)2 f (ρeiθ0 ) and (1 − r)2 f (reiθ0 ) α → = 1. 2 iθ (1 − ρ) f (ρe 0 ) α by the previous theorem. The result follows. 117

Theorem 13. Suppose f (z) ∈ S . [By the hypothesis of theorem 9, α = (1 − r)2 lim M(r, f ) and]. Suppose α > 0 and θ0 as deﬁned in theorem r→1 r 2 iθ 11, [θ0 such that 0 ≤ θ0 ≤ 2π and (1 − r) | f (re 0 )|→ α as r → 1] and ∞ λ λ n 1 ϕ(z) = f (z) = z an,αz . Then if λ> n=0 4 P 1 ϕ 1 − eiθ0 + " n! # na ei(λ n)θ0 ∼ as n →∞. n,λ Γ(2λ)

Assuming the theorem let us ﬁrst deduce some corollaries.

αλn2λ−1 Corollary 1. We have |a |∼ as n →∞ where α ≤ 1. Equality n,λ Γ′(2λ) is possible only in the case when f (z) = fθ(z) deﬁned in theorem 1. 102 Univalent Functions

Proof. Taking moduli for the result of the theorem,

1 iθ0 |ϕ 1 − n e | n|a |∼ as n →∞ n,λ h Γ(2λ) i 1 iθ λ | f 1 − e 0 | (αn2)λ = n ∼ h Γ(2λ) i Γ(2λ)

after theorem 11.

2 iθ For, the choice of θ0 is such that (1 − r) f (re 0 ) → α as r → 1. 1 2 1 Precisely we get f 1 − eiθ0 → α as n →∞. n! " n! #

n2λ−1 Hence we get |a |∼ . This is corollary 1. n,α Γ(2λ)

∞ kn+1 Corollary 2. If fk(z) = z + akn+1z belongs to S , and if k = 1, 2 n=1 or 3 then, P Γ(n + 2/k) |a + |≤ for n ≥ n . kn 1 Γ(n + 1)Γ(2/k) 0 z 118 Equality holds for f (z) = and all n, otherwise strict in- (1 − zkeiθ)2/k equality holds for n > n0( f ).

To prove this note that

1/k k f (z) = [ fk(z )] ∈ S

and ∞ 1/k 1/k n 1/k fk(z ) = z ( akn+1z ) = [ f (z)] . X0 z Now either f (z) = and so (1 − zeiθ)2

z kn+1 inθ = = A + z e = f (z) k 1 2/k kn 1 k (1 − z e ) X Univalent Functions 103 where 2 2 + 2 + k k 1 ...... k n − 1 A + = kn 1 1· 2 ...... n Γ 2 + k n A + = kn 1 Γ(2/k)Γ(n + 1) 2 −1 n k ∼ Γ(2/k) Alternatively since in every other case α< 1 for f (z) so by corollary 1,

1 2 −1 α k n k |a + |∼ kn 1 Γ(2/k) and since α< 1, |akn+1| < Akn+1 for large n.

Note that if k = 1 we have |an+1| < n + 1 ﬁnally and if k = 2, |a2n+1| < 1. 119 Let us now go to the proof of theorem 13. We suppose without loss of generality that θ0 = (0) in theorem 13 for otherwise we can consider e−iθ0 f (zeiθ0 ) instead of f (z) and eiλθ0 [ f (zeiθ0 )]λ instead of ϕ(z). 1 Set rn = 1 − . Given ǫ > 0 we deﬁne a domain ∆n = ∆n(ǫ) = {z : n! ǫ 1 f (r ) < |1 − z| < | arg(1 − z)| < π − ǫ and α = n so that |α |→ α n nǫ 2 n n2 n as n →∞ by theorem 11. α Suppose now f (z) = n n (1 − z)2 104 Univalent Functions

Lemma 4. We have as n →∞ uniformly for z ∈ ∆n(ǫ), f (z) ∼ fn(z) and ′ ′ f (z) ∼ fn(z). 1 1 Proof. Put z = l (w) = r + w, so that (1 − z) = (1 − w). n n n n Then by this ∆n(ǫ) in the z-plane corresponds to ∆1(ǫ) in the w plane. Deﬁne a function gn(z) as

2 f [ln(w)] gn(w) = (1 − w) f (rn)

Note that for a fixed ǫ, ∆n(ǫ) lies in |z| < 1 for large n. Hence gn(w) is defined in ∆1(ǫ) for sufficiently large n.

Further |gn(w)| = O(1) for w ∈ ∆1(ǫ) and n > n0. In fact if ∆n(ǫ), ∆ 1 ∆ 1 ∆ ∆′ ∆′ 120 n( 2 ǫ), n( 4 ǫ) are denoted by n, n and d is the distance from 1 to ∆′′ ∆′ ∆′′ the outside of 1 , the distance between n and the exterior of n is exactly d/n. d So if ∆′′ lies in |z| < 1. ∆′ lies in |z| < 1 − and ∆′′ lies in |z| < 1 n n n n r d for all large n. Now | f (z)| ≤ ; |z| = r and so if |z| < 1 − < 1; (1 − r)2 n d 1 − |z|≥ . n 2 ′ n 2 Hence in ∆ | f (z)| = O = O(n ) as n → ∞ and f (rn) = n d2 ! 1 f 1 − ∼ αn2 by theorem 11 as n → ∞ and hence it follows that n! f [l (w)] O(n2) n = = O(1) as n → ∞ uniformly for w in ∆′ . i.e., g (w) 2 1 n f (rn) n ∆′ is bounded in 1 for large n. 1 1 Next choose −1 < w < 0 so that w is real and r + w = 1− (1−w) = n n n ln(w). After theorem 12. f [l (w)] (1 − r )2 1 n ∼ n = as n →∞ 1 2 f (rn) − 2 (1 − w) n2 (1 w) for the hypothesis of theorem 12 is satisﬁed because, 1 1 1 |z| = 1 − + w < rn < |(z + 1)|, rn → 1 as n →∞ n n 2

Univalent Functions 105 in this manner ensuring the above asymptotic equality. 2 Therefore it follows on bringing (1−w) to the left hand side gn(w) = f [l (w)] (1 − w)2 n tends to one as n → ∞, for real w satisfying 0 > w > f (rn) −1. = ∆′ Thus we have obtained gn(w) O(1) as w ranges in 1 and gn(w) → 121 ∆′ 1 as n → ∞ on a set of w having a limit in 1. Hence by Vitalis con- ′ vergence theorem (see Titchmarsh p.p. 168) gn(w) → 1, gn(w) → 0 ∆ ∆′ uniformly in 1 which is contained in 1 and satisfies the condition that ∆′ it is bounded by a contour that is interior to 1. f (z) Now translating back into z, n2(1 − z)2 tends to 1 as n tends to f (rn) infinity for z ∈ ∆n(ǫ). i.e. by the definition of fn(z) f (z) → 1 for z ∈ ∆n(ǫ) as n →∞. fn(z) Also d f [l (w)] (1 − w)2 n → 0 dw f (rn)

= 1 d 2 2 f (z) since n(1 − z) (1 − w); n n (1 − z) → 0 dz f (rn) d f (z) n (1 − z)2 → 0 dz f (rn) d (1 − z)2 f (z) = O(n) since f (r ) ∼ αn2 as n →∞. dz n for z in ∆n(ǫ). 1 Note that in ∆ (ǫ) and (1 − z) have the same order of magnitude. n n 2 ′ Thus, (1 − z) f (z) − 2(1 − z) f (z) = 0(n), z ∈ ∆n(ǫ) as n →∞ 122 2 f (z) f ′(z) = + O(n3), z ∈ ∆ (ǫ) as n →∞. (1 − z) n 2 f (z) = n + O(n3) (1 − z) 106 Univalent Functions

= ′ + 3 = ′ + fn(z) O(n ) fn(z)[1 O(1)] because 2αn f ′(z) = = O(n3). n (1 − z) ′ ′ ∆ Therefore, f (z) ∼ fn(z) uniformly for z ∈ n(ǫ) as n → ∞. This completes the proof of the lemma. Lemma 5. With the notation of theorem 4, for λ> 0,

λ −2 −2λ S λ(r, f ) = α S λ[r, (1 − z) ] + O(1 − r) as r → 1 1 and if λ> 2 λ −2 1−2λ Iλ(r, f ) = α Iλ[r, (1 − z) ] + O(1 − r) as r → 1 n2 Proof. Let R = |α | , ϕ (z) = f (z)λ = αλ(1 − z)−2λ and ∆ (ǫ) as n n ǫ2 n n n n 123 deﬁned in the previous lemma. Note that ϕn(z) maps ∆n(ǫ) onto the sector 4λ λ λ ǫ Rn < |w| < Rn, | arg(w) − λ arg αn| < (π − 2ǫ) = ′ 2 The area of the image |ϕn(z)| dx dy ∆!n(ǫ) = 2λ 8λ (3.2.1) λ(π − 2ǫ)Rn [1 − ǫ ] ∆ ′ ′ Also in n(ǫ), fn(z) ∼ f (z) and fn(z) ∼ f (z) and so ′ = ′ λ−1 ′ λ−1 = ′ ϕn(z) λ fn(z)[ fn(z)] ∼ λ f (z)[ f (z)] ϕ (z) So we have

(3.2.2) |φ′ (z)|2dx dy ∼ λ(π − 2ǫ)R2λ(1 − ǫ8λ) " n n ∆n(ǫ) λ ∆ Also since |ϕn(z)| < Rn in n(ǫ) we have by lemma 4, λ + |ϕ(z)| < Rn(1 ǫ) there for large n.

Now choose n so that Rn−1 < M(r, f ) < Rn. This is possible at least if r is suﬃciently near 1, since Rn →∞ with n. Univalent Functions 107

Let E0 be the set of points of |z| < r outside ∆n(ǫ). Then in E0, λ ∆ + λ |ϕ(z)| < Rn, and in n(ǫ), |ϕ(z)| < (1 ǫ)Rn. The image of |z| < 1 by f (z) contains no point more than once, and so the length of the image over any circle |w| = ρ is at the most 2πρ. If we cut the unit circle along the negative real axis, then the image by f (z)λ covers any circle, |w| = ρ, with length at the most λ2πρ. So the area of the image of the union of E0 and ∆n(ǫ) by ϕ(z) is at most 2λ + 2 πλRn (1 ǫ) . That gives, 124

|ϕ′(z)|2dx dy ≤ πλR2λ(1 + ǫ)2 " n E0∪∆n(ǫ) Also by (3.2.2), we have for large n

|ϕ′(z)|2dx dy > λπ(1 − ǫ)(1 − ǫ8λ)R2λ " n ∆n(ǫ) [Note the ǫ terms of r.h.s. of this inequality and (3.2.2)] So

|ϕ′(z)|2dx dy < R2λπλ (1 + ǫ)2 − (1 − ǫ)(1 − ǫ8λ) " n h i E0 1 2λ < ǫ Rn where ǫ1 is small if ǫ is small. Similarly, |ϕ′ (z)|2dx dy < ǫ′′R2λ " n n E0 By theorem 4 r π 2 λ iθ 2λ−2 ′ iθ 2 S 2λ(r, f ) = ρdρ | f (ρe )| | f (ρe )| dθ 2π Z Z 0 −π λ2 = | f (ρeiθ)|2λ−2| f ′(ρeiθ)|2dx dy 2π " |z|

1 = |ϕ′(z)|2dx dy. 2π " |z|

125 Hence 2 S (r, f ) − S (r, f ) = |ϕ′(z)|2 − |ϕ′ (z)|2 dx dy 2λ 2λ n π " n |z|

2 =  +  π Z Z  E E   1 0    ∆ ′ ′ where E1 is the part of |z| < r within n(ǫ). Since in E1ϕn(z) ∼ ϕ (z)

′ 2 ′ 2 = ′ 2 |ϕ (z)| − |ϕn(z)| dx dy O |ϕn(z)| dx dy Z Z E1 E1 = O |ϕ′ (z)|2dx dy Z n ∆n(ǫ) = 2 O(Rn) by (3.2.1)

Also

′ 2 ′ 2 ′ 2 + ′ 2 | |ϕ (z)| − |ϕn(z)| dx dy|≤ |ϕn(z)| dx dy |ϕ (z)| dx dy Z Z Z E0 E0 E0 ′′ + ′ 2 ≤ (ǫ ǫ )Rn.

Since ǫ′′ and ǫ′ can be made as small as we please by suitable choice of ǫ,

= + 2λ S 2λ(r, f ) S 2λ(r, fn) O(Rn ) = + 2λ S 2λ(r, fn) O(Rn−1) as Rn ∼ Rn−1 2λ = S 2λ(r, fn) + OM(r, f ) .

−4λ 126 i.e. S 2λ(r, f ) = S 2λ(r, fn) + O(1 − r) as r → 1 (by theorem 3). Univalent Functions 109

Hence we get

= 2λ −2 + −4λ S 2λ(r, f ) |αn |S 2λ[r, (1 − z) ] O(1 − r) as r → 1. which is the ﬁrst result. Integrating the above from 0 to r, w.r.t. r after dividing by r, we get 1 I (r, f ) = |α|2λI [r, (1 − z)−2] + O(1 − r)1−4λ as r → 1, if λ> . 2λ λ 4 Hence we get the lemma [replacing 2λ by λ]

λ −2 1−2λ Iλ(r, f ) = α Iλ[r, (1 − z) ] + O(1 − r) as r → 1 1 if λ> . 2 1 Lemma 6. If η > 0 and λ > we can choose k > 0 so that if 4 1 1 r < r < 1, 1 − ≤ r ≤ 1 − 0 n 2n

iθ 2 iθ 2 1−4λ | f (re )| + | fn(re )| dθ < η(1 − r) Z k(1−r)≤|θ|≤π

Let γ and γ ′ be the arcs |θ| ≤ k(1 − r) and k(1 − r) ≤ |θ| ≤ π on |z| = r. If k is any ﬁxed number, we can choose ǫ so small that the arc γ 127 1 1 lies in ∆ (ε) for 1 − ≤ r ≤ 1 − for large n (see Hayman [1]). n n 2n Hence 2λ 2λ | fn(z)| dθ ∼ | f (z)| dθ Z Z γ γ

2λ 2λ 2λ | fn| dθ − | f | dθ = O | f | dθ Z Z Z γ γ

= O{I2λ(r, f )} = O(1 − r)1−4λ 110 Univalent Functions

1 by the application of theorem 4, as 2λ> and also I (r, f )−I (r, f ) = 2 2λ 2λ n O[(1 − r)1−4λ] by the previous lemma. Hence subtraction gives,

2λ 2λ 1−4λ (3.2.3) | f | dθ − | fn| dθ = O[(1 − r) ] Z Z γ ′ γ ′

Now the integral

π 2λ 2λ |αn| dθ | fn| dθ Z Z (1 − 2r cos θ + r2)2λ γ ′ k(1−r) π dθ ≤ Z (r sin θ/2)4λ k(1−r) ∞ dθ A(λ) Z θ4λ k(1−r)

1 1 ≤ sin θ/2 ≥ Aθ in that range. (1 − 2r cos θ + r2) (r sin2 θ/2)2

2λ 1−4λ (4) | fn| dθ ≤ A(λ)[k(1 − r)] Z γ ′ 1 < η(1 − r)1−4λ 3

128 by properly choosing k suﬃciently large. Hence in virtue of the relation (2) we deduce same for f (z) and hence the lemma. We now proceed to prove theorem 13. We have as in lemma 5,

∞ = λ = λ −2λ = λ m ϕn(z) fn(z) αn(1 − z) αn bmz X0 2λ(2λ + 1) ...... (2λ + m − 1) b = m m! Univalent Functions 111

Γ(m + 1 + 2λ) m2λ−1 = ∼ Γ(2λ)Γ(m + 1) Γ(2λ) and as already deﬁned

∞ ϕ(z) = zλ 1 + a zm  m,λ   mX=1    ﬃ  ′  ′ Hence by considering the coe cients of ϕ (z) and ϕn(z), ′ ′ λ 1 ϕ (z) ϕn(z) (n + λ)an,λ − nbnα = − dz n 2Π Z zn+λ zn |z|=r 2π 1 1 (n + λ)a − nb αλ ≤ ϕ′(reiθ) − reλiλθϕ′ (reiθ) dθ n,λ n n 2π rn+λ−1 n Z 0

1 1 We shall choose r in the range 1 − ≤ r ≤ 1 − as usual. For a k to be n 2n determined, let γ and γ ′ be the arcs |θ|≤ k(1 − r) and k(1 − r) ≤ |θ|≤ π respectively on |z| = r. ′ ′ ′ λ iλθ = −1−2λ γ,ϕ ∼ ϕn ◦ (ϕn)(r e ) O(1 − r) as r → 1. and so 129 ′ ′ λ iλθ = −1−2λ = 2λ+1 ϕ − (ϕn)(r e ) O(1 − r) O(n ) as r → 1 or as n →∞ 1 The length of the path of integration of γ = O(1 − r) = O n!

′ iθ λ iλθ ′ iθ = 2λ ϕ (re ) − r e ϕn(re ) dθ O(n ) Z this being true for any ﬁxed k. Again,

| ϕ′(reiθ) − rλeiλθϕ′ (reiθ) dθ| Z n γ ′ 112 Univalent Functions

≤ |ϕ′(reiθ)|dθ + rλ|ϕ′ (reiθ)|dθ Z Z n γ ′ γ ′ 1 We now proceed as in theorem 6, choose t such that 2λ − 2t > and 2 using Schwarz’s Inequality.

|ϕ′(reiθ)|dθ = λ | f ′(reiθ)|| f (reiθ)|λ−1dθ Z Z γ ′ γ ′ 1 2 ≤ λ  | f ′(reiθ)|2| f (reiθ)|2t−2λdθ Z   γ   1   2   | f (reiθ)|2λ−2tdθ Z  γ ′      By the same method as of theorem 6, we can choose r such that the ﬁrst integral is O(n4t+1). By what we have just seen in lemma 6 the second integral can be made less that δη(1 − r)1−4λ+4t i.e. (const.) ηn4λ−4t−1 and η can be made as small as we please. So

|ϕ′(reiθ)|dθ < η′n2λ Z 130 using the inequality (8) of theorem 6. η′ can be made as small as we please by choosing k large enough ′ iθ ϕn(re )dθ can be dealt with similarly. γR′ So ﬁnally we see that

2π |[ϕ′(reiθ) − rλeiλθϕ ′ (reiθ)]dθ ≤ [2η ′ + O(1)]n2λ Z n 0 and since η′ can be made as small as we please the right hand side is O(n2λ). Univalent Functions 113

That gives

nb αλ a = n n + O(n2λ−1) n,λ n + λ n f (r )λ n2λ−1 = n + O(n2λ−1) n + λ n2λ Γ(2λ) Therefore, f (r )λ na ∼ n as required. n,λ Γ(2λ) This completes the proof of theorem 13. For an extension of the result to p-valent functions and further results see for example W.K. Hayman [1], [2].

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