ON THE GROWTH OF ENTIRE FUNCTIONS WITH ZERO SETS HAVING INFINITE EXPONENT OF CONVERGENCE

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1 Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 7, 00, ON THE GROWTH OF ENTIRE FUNCTIONS WITH ZERO SETS HAVING INFINITE EXPONENT OF CONVERGENCE Joseph Miles University of Illinois, Department of Mathematics 409 West Green Street, Urbana, IL 680, U.S.A.; Abstract. Let Z be a sequence of complex numbers tending to infinity having infinite exponent of convergence. For r > 0, let nr) denote the number of members of Z of modulus at most r. If log log nr) µ = lim r log log r, it is shown that inf f lim r log Mr, f) nr) min π + log, log µ + ), µ where the infimum is over all entire f vanishing precisely on Z and Mr, f) denotes the maximum of fz) on z = r. This bound strengthens earlier results of Bergweiler.. Introduction Let Z = {z j : j =,, 3,...} be a sequence of complex numbers, not necessarily distinct, tending to infinity and ordered so that z z. We consider entire functions with zero set precisely Z, i.e., entire f with a zero of multiplicity m at a complex number β provided that z j = β for exactly m different values of j. We compare the rate of growth of the maximum modulus function Mr, f) = max z =r fz) of such f with nr), the number of distinct values of j such that z j r. The exponent of convergence σ of the sequence Z is defined to be the infimum of all α > 0 for which z j α <, or equivalently by z j 0 log nr).) σ = lim r log r. 000 Mathematics Subject Classification: Primary 30D5; Secondary 30D0, 30D35.

2 70 Joseph Miles It is an easy consequence of Jensen s theorem that for any entire f vanishing precisely on Z, the order ρ of f, defined by ρ = lim r log log Mr, f), log r satisfies ρ σ. We shall be concerned exclusively with sequences Z with infinite exponent of convergence, and thus all entire f we consider will be of infinite order. One measure of the growth of nr) for sequences with infinite exponent of convergence is log log nr).) µ = lim r log log r. It is elementary to verify that µ if σ > 0 and that σ = if µ >. Bergweiler has studied the growth of entire functions vanishing precisely on a sequence Z with infinite exponent of convergence. Let Lµ) := sup inf Z f lim r log Mr, f), nr) where the supremum is over all sequences Z satisfying.) and the infimum is over all entire f vanishing precisely on Z. In the case µ =, we additionally require of Z that σ =.) Bergweiler [] proved the following two theorems. Theorem A. For every Z satisfying.), there exists an entire f vanishing precisely on Z such that 3µ log Mr, f) 4µ ), < µ <, lim αµ) := r nr) 3 4, µ =. Theorem B. For < µ < there exists Z satisfying.) such that for every entire f vanishing precisely on Z, log Mr, f) lim r nr) βµ) := µ ) log µ µ + µ + ) log µ + µ. We note that Theorem A implies that Lµ) αµ) and Theorem B implies Lµ) βµ). We also note that αµ) tends to infinity as µ + and that Theorem A gives no information about L). It is elementary that βµ) is a decreasing function of µ on, ) with βµ) log as µ + and βµ) 0 as µ. Bergweiler [] has asked i) whether Lµ) is a bounded function for µ and in particular if L) is finite and ii) whether Lµ) 0 as µ. We answer these questions by proving the following two theorems. We remark that our results give information about the minimum modulus of f as well as the maximum modulus.

3 On the growth of entire functions 7 Theorem. If Z satisfies.) for < µ, then.3) inf f lim r log fre iθ ) nr) log µ + µ, < µ <, 0, µ =, where the infimum is over all entire f vanishing precisely on Z. Theorem. then If Z is any sequence with infinite exponent of convergence, inf f lim r log fre iθ ) nr) π + log, where the infimum is over all entire f vanishing precisely on Z. We note that the combination of Theorems and yields Lµ) min π + log, log µ + ), µ <, µ 0, µ =. We further note that Theorem, in conjunction with Theorem B, shows that.4) o) µ Lµ) + o), µ. µ By a refinement of his proof of Theorem B, Bergweiler was able to show [, p. 03] that lim inf µ Lµ) >.6. This result together with Theorem yields log Mr, f).5).6 < sup inf lim Z f r nr) π + log, where the supremum is over all Z with infinite exponent of convergence and the infimum is over all entire f vanishing precisely on Z. We have been unable to narrow the gap between the upper and lower bounds in either.4) or.5). Our results involve upper bounds for log Mr, f) in terms of nr) on a sequence tending to infinity. It is observed in [] that in general it is not possible to obtain such bounds on a large set of r -values, for example on a set of positive logarithmic density. Suppose, for example, that Z has infinite exponent of convergence, all members of Z are positive real numbers, and lim r r E log nr) log r <

4 7 Joseph Miles for some set E [, ) of upper logarithmic density. It is well known [5] that any entire f vanishing precisely on Z has infinite lower order. Thus lim r r E log Mr, f) nr) = for any such f. It is perhaps worth remarking that no results such as Theorems and are possible for the ratio log Mr, f)/nr), where Nr) = r 0 nt) t dt is the integrated counting function of value distribution theory. For in Bergweiler s examples in Theorem B, nr)/nr) as r and thus log Mr, f)/nr) for every entire f vanishing precisely on Z. There is a vast literature concerning comparisons of the growth of log Mr, f) for an entire function f to the distribution of its zeros. An excellent collection of references appears in [].. Preliminaries The following lemma, used in the proof of Theorem, is due in its essential form to Newman [8]. See also [3].) We are indebted to J. Fournier for bringing the lemma to our attention and for several helpful communications. For completeness we include Fournier s proof of the lemma in the precise form required for our purposes. Lemma. Suppose 0 < ε < and that P θ) = is a real trigonometric polynomial with.) c k and.) c 0 M k= M c k e ikθ ), k M, M + ε k M + ε.

5 On the growth of entire functions 73 Then there exists a real trigonometric polynomial Qθ) = L k= L d k e ikθ where L > M such that d k = c k for k M and Q < π ε + ε. Proof. Let Rθ) = k= M c k e ikθ + c 0 and let b j = c j M for j M and b M+ = c 0. For 0 j M, we have by.).3) b j+ = c j M ) < M + ε + j M From.) we conclude ε)j + )..4) b M+ = c o M + ε) < ε)m + ). We define a linear functional T : H C by By a theorem of Hardy [4, p. 48], T f) ) T f) = T a j z j = j=0 j=0 j=0 M a j b j+. j=0 M a j M bj+ a j ε)j + ) π f ε), where we have used.3) and.4). The Hahn Banach theorem implies that T extends to a bounded linear functional on L [ π, π] with norm at most π/ ε), and thus there exists a function h on [ π, π] bounded by π/ ε) such that T f) = π fθ)hθ) dθ π π

6 74 Joseph Miles for all f H, where fθ) = lim r fre iθ ). Letting fθ) = e in )θ H for n, we get T f) = π { e in )θ bn, n M +, hθ) dθ = π π 0, n > M +. Taking conjugates we obtain π { e inθ bn, n M +, Hθ) dθ = π π 0, n > M +, where Hθ) = e iθ hθ). Setting H θ) = e im+)θ Hθ), we have.5) π { e in M )θ H bn, n M +, θ) dθ = π π 0, n > M +. Using c n M = b n for n M and c o = b M+, we rewrite.5) as.6) π e ikθ H θ) dθ = π π c k, M k, c o, k = 0, 0, k > 0. Clearly H = h π/ ε) and we note by.6) that H has the same Fourier coefficients as does R for k M. We set Gθ) = H θ) + H θ). Thus G π/ ε) and G has the same Fourier coefficients as does P for k M. For all integers k, let B k = π e ikθ Gθ) dθ. π π Consider the L th Cesàro mean σ L of G, i.e., let σ L θ, G) = L k= L k ) B k e ikθ. L +

7 Because Fejér s kernel is positive, we have On the growth of entire functions 75 σ L G < π ε. Finally, set Qθ) = σ L θ, G) + M k= M k L + c ke ikθ. Since c k < / ε) for k M, we have Q σ L + MM + ) L + ) ε) < π ε + ε provided that L + > MM + )/ε ε). Furthermore, for k M, since B k = c k, we see that π e ikθ Qθ) dθ = c k, π π proving Lemma. The following elementary growth lemma is used in the proof of Theorem. Lemma. Suppose g: [0, ) [0, ) is a nondecreasing function continuous from the right and suppose that 0 < α < lim x gx)/x. Then there exists a nondecreasing sequence x j such that gx) gx j ) αx x j ), 0 x x j. Proof. Let b j be an increasing sequence and let x j = inf{x 0 : gx) αx + b j }. Clearly x j is a well-defined sequence of real numbers such that x j+ x j. The continuity of g from the right ensures that gx j ) αx j + b j. The definition of x j implies that gx) < αx + b j, 0 x < x j, and hence gx) gx j ) < αx x j ), 0 x < x j, completing the proof of Lemma.

8 76 Joseph Miles The proofs of both Theorems and are based on an analysis of the Fourier series of log Hre iθ ) where H is entire. Suppose that H is entire with H0) =. Suppose that H has zero sequence {z ν } with due regard to multiplicity and that z ν r for ν =,, 3,.... Let nt) be the number of zeros of H counting multiplicity of modulus no more than t. We write log Hre iθ ) = m= c m r, H)e imθ, where the Fourier coefficients, first studied by F. Nevanlinna [7] see also [6]), are given by the following formulas:.7).8) where r nt) c 0 r, H) = Nr) = dt, 0 t c m r, H) = β m rm + ) m r m z ν <r.9) log Hz) = z ν β m z m m= near 0 for some branch of the logarithm, and zν r c m r, H) = c m r, H), m. ) m, m, We collect certain estimates see.4) and.5)) involving the Fourier series of log Hre iθ ) which are common to the proofs of Theorems and. Suppose z ν 0, z ν, and H is the convergent product ) z.0) Hz) = E, ν, z ν ν= where we recall that the logarithm vanishing at z = 0 of the usual Weierstrass factor Ez, p) is given by.) log Ez, p) = zp+ p + zp+, z <. p + Suppose that H has no zero of modulus r. From.8),.9), and.) we conclude that if < r, then.) c m r, H) = ) m r ) m zν, m m r ν>m z ν <r z ν z ν <r

9 and if > r then.3) c m r, H) = m On the growth of entire functions 77 ν m r< z ν r z ν ) m m z ν <r Since the modulus of each term on the right side of.3) is at most /m and there are at most m such terms, we see that if > r, then.4) c m r, H). We now suppose that < r. Taking into account the possibility of H having more than one zero of modulus, we see from.) that c m r, H) r ) m ) m ) r t dnt) m t r + ) m ) m ) r m n z zm m ) m) + ) m zν.5) r m r ν m r ) m ) m ) r t nt) + dt + t r t, where we have used integration by parts and the fact that the summation in the middle expression contains at most m terms, each of modulus no more than /m. A critical role in the proof of both of our theorems is played by.5). 3. Proof of Theorem We first suppose < µ <. With no loss of generality, we may presume that z > e, for multiplication of f by a polynomial leaves.3) unaffected. Let ε in 0, ) be such that ε) µ > + ε. With the convention log + 0 = 0, we apply Lemma for t e with x = log log t, gx) = log + log + ne ex ), and α = ε)µ to conclude that there exists an increasing sequence rj such that zν r ) m. log + log + nt) log log nr j ) ε)µlog log t log log r j ), e t r j. By continuity it is immediate that there exists r j > r j such that nr j) = nr j ) and 3.) log + log + nt) log log + ε)nr j ) We rearrange 3.) to obtain ε)µlog log t log log r j ), e t r j. 3.) ) log + ε)µ nt) log t log + ε)nr j ), e t r j. log r j

10 78 Joseph Miles Hence or log + nt) log t ) ε)µ ) log t log + ε)nrj ), e t r j, log r j log r j 3.3) log + nt) log t log + ε)nr j) log r j, e t r j. To simplify notation, we define 3.4) u j := log + ε)nr j) log r j and rewrite 3.3) as 3.5) nt) t u j, 0 t r j. Since Z has infinite exponent of convergence we observe from.) and 3.3) that u j as j. For a fixed α >, consider the function gy) = y α, 0 y. Clearly g0) =, g) = 0, and g is increasing and convex on 0, ). If y 0 in 0, ) is determined by y α 0 = ε, then gy) ) y 0 gy 0) ) y 0 0 = ε, 0 < y < y 0, and gy) 0 y gy 0) 0 y 0 = αỹ α 0 > α ε), y 0 < y <, for some ỹ 0 in y 0, ). Recalling that α = ε)µ > and setting y = log t)/ log r j, we conclude that there exists r j in [, r j ) with log r j / log r j = y 0 such that 3.6) and 3.7) log t log r j log t log r j ) ε)µ < ε) log t log r j, < t < r j, ) ε)µ ) log t < ε) µ, r j < t < r j. log r j

11 From 3.) and 3.6) we have or On the growth of entire functions 79 log + nt) log + ε)nr j ) < ε) log t log r j, e t < r j, 3.8) nt) < t ε)u j, 0 < t < r j. From 3.) we have 3.9) log + nt) log + ε)nr j ) < log + ε)nr j ) ) ) ε)µ log t ), log r j for e t r j. The combination of 3.7) and 3.9) yields or log nt) log + ε)nr j ) < ε) µu j log t log r j ), r j < t < r j, ) ε) t µu j 3.0) nt) < + ε)nr j ), r j < t < r j. r j We define the entire function H by.0) and proceed to estimate c m r j, H) from above. First we note from.7), 3.4), and 3.5) that 3.) c 0 r j, H) = rj 0 nt) t dt rj 0 t u j dt = ru j j = + ε)nr j) = o nr j ) ). u j u j For > r j, i.e., for m > nr j ), we have.4) with r = r j. Thus we consider m such that < r j. First suppose m > + ε)u j. From.5) and 3.5) we have 3.) c m r j, H) r m j < r m j rj rj rj { ) m nt) dt + t t t u j m dt + rm j εu j + m /+ε)) rj m { zm u j m + m u j rj m }. }

12 80 Joseph Miles Writing 3.3) c m r j, H) = rj m β m j), + ε)u j < m nr j ), we see from 3.) that there exists δ j 0 such that 3.4) β m j) /m < δ j, + ε)u j < m nr j ). We now suppose that m [ + ε)u j ]. We have from.5) that 3.5) c m r j, H) + rj rj r j ) m ) m ) t nt) + dt t r j t ) m ) m ) rj t nt) + dt + t t rj r j = c a mr j, H) + c b mr j, H) +, where of course the first term is omitted if r j. From 3.0) we have c b mr j, H) + ε)nr j) + ε)nr j) rj r j ) m ) m ) ) ε) rj t t µu j dt + t r j r j t ε) µu j m + ε) µu j + m Estimating the sums by the corresponding integrals we obtain for large j ). 3.6) [+ε)u j ] m= c b mr j, H) + ε)nr j) + ε)nr j) log ε) µu j + + log ) ε) µu j + + ε)u j ε) µu j + ε)u j ) ε) µ + + ε ε) + o nr j ) ). µ + ε) By 3.5) we may write 3.7) c m r j, H) = c α mr j, H) + c β mr j, H) where c α mr j, H) c a mr j, H) and c β mr j, H) c b mr j, H) +. Thus by 3.4) and 3.6) 3.8) [+ε)u j ] m= c β mr j, H) + ε)nr j) ε) ) µ + + ε log ε) + o nr j ) ). µ + ε)

13 On the growth of entire functions 8 We next consider c a mr j, H) for ε)u j < m [ + ε)u j ]. Writing 3.9) c a mr j, H) = r m j β m j), ε)u j < m [ + ε)u j ], we have from 3.8) and 3.5) that β m j) rj nt) dt tm+ < z m ε)u j m m ε)u j < Thus there exists ε j 0 such that rj t ε)u j m dt εu j m ε)/ ε/)). 3.0) β m j) /m < ε j, ε)u j < m [ + ε)u j ]. Finally, suppose m [ ε)u j]. Let m = [ 4 ε)u j]. Using 3.4) and 3.8) we have for large j Thus 3.) c a mr j, H) [ ε/)u j ] m= rj = r m j < r m j rj rj ) m nt) dt t t nt) t m + dt rm j rj rj ) m rj nt) dt t t t ε)u j m dt < r m j < + ε) ε/4) nr j ) ) ε/4. t ε)u j m dt c a mr j, H) < + ε) ε/4) nr j ) ) ε/4 ε)u j = o nr j ) ). From elementary considerations, because H has no zeros on z = r j, there exists M j > + ε)u j such that 3.) c m r j, H) < εnr j ). m >M j We define κ m j) for ε)u j < m M j by c α mr j, H) rj m, 3.3) κ m j) = ε) u j < m [ + ε)u j ], c m r j, H) rj m, + ε)u j < m M j.

14 8 Joseph Miles From.4), 3.3), 3.4), 3.9), and 3.0) we note that there exists γ j > 0 such that whether < r j or > r j, 3.4) κ m j) /m < γ j 0, ε)u j < m M j. We consider a subsequence r jk such that 3.5) u jk+ > M jk, k <, and ensuring by 3.4) that for k 3.6) κ m j p ) < We define 3.7) T j z) = γ jk+ < r jk, k <, r jk M j m=+[ ε/)u j ] ) m, ε)u j p < m M jp, p > k. κ m j)z m, T z) = T jk z), k= and note by 3.5) that the powers of z appearing in the various T jk We also note by 3.4) that T is entire. We set are distinct. and f k z) = Hz)e T j k z) fz) = Hz)e T z). From.7),.8), 3.7), 3.3), and 3.7) we see that 3.8) c o r jk, H), m = 0, c α mr jk, H) + c β mr jk, H), m [ ε)u j k ], c m r jk, f k ) = c β mr jk, H), ε)u j k < m [ + ε)u j ], 0, + ε)u jk < m M jk, c m r jk, H), m > M jk. From 3.), 3.8), 3.), 3.), and 3.8) we see that 3.9) ε) ) µ + + ε) c m r jk, f k ) < + ε)nr jk ) log ε) + ε + o) ) nr jk ). µ + ε) m=

15 On the growth of entire functions 83 Since κ m j k ) <, it follows from 3.5) that T jp r jk e iθ ) M jk r M j k j k < u j k ) + ε)nr jk ) ) / = o nrjk ) ). p<k Also by 3.6) implying that 3.30) T jp r jk e iθ ) p>k log e p k T jp r j k e iθ ) r jk )m r jk ) m <, m>m jk = o nr jk ) ). The combination of 3.9) and 3.30) yields log fr jk e iθ ) ε) ) µ + + ε < + ε)nr jk ) log ε) + ε + o) ) nr jk ). µ + ε) Letting ε 0 +, we obtain.3) for < µ <. If µ =, it is elementary from Lemma that for every µ < there exists a sequence satisfying 3.) with µ in place of µ. Thus the conclusion of Theorem holds with µ in place of µ for every µ in, ), proving the theorem in the case µ =. 4. Proof of Theorem In view of Theorem, we may restrict our attention to sequences Z for which µ < 3. With no loss of generality we may presume that z >. Suppose 0 < ε < 3. Let m j be an increasing sequence of positive integers. Let { rj = inf t z : log nt) log t } m j + ε. Because Z has infinite exponent of convergence, it is elementary that rj is a nondecreasing sequence of real numbers tending to infinity. It is immediate that 4.) nt) < t m j+ ε, t < r j. Because nt) is continuous from the right, for each j there exists ν such that z ν = r j and 4.) nr j ) r j ) m j+ ε.

16 84 Joseph Miles We wish to focus on a sequence of circles containing no members of Z and thus choose r j > rj such that nr j) = nrj ) and 4.3) ) nrj ) rj rj < + ε. For j we define r j = inf { t z : log nt) log t } m j + ε), and note trivially that r j r j and 4.4) nt) < t m j+ ε)/, t < r j. Since log nt) is continuous from the right, we see from the definition of r j that log n r j ) log r j m j + ε. Since µ < 3, for large j we have log n r j) < log r j ) 3/ and thus 4.5) log r j m j > log r j m j + ε log r j) log n r j ) > log r j) / as j. We form the product.0) and now estimate c m r j, H). By.7), 4.), and 4.3) we have 4.6) c o r j, H) = r j 0 nt) t dt + rj r j nt) t dt < r j )m j+ ε m j + ε + log + ε). We next suppose m m j. We note from 4.) for large j that nr j ) > m j > m j m and thus < r j. From.5), 4.), and 4.3) we conclude c m r j, H) + r j rj r j rj < + ε)r j )m j+ ε ) m ) m ) t nt) + dt t r j t ) m ) m ) rj t nt) + dt + t r j t + + ε) log + ε) +. m j + ε m + m j + ε + m )

17 On the growth of entire functions 85 We conclude for m m j that there are expressions c α mr j ) and c β mr j ) satisfying 4.7) 4.8) c m r j, H) = c α mr j ) + c β mr j ), c α mr j ) + ε)r j )m j+ ε m j + ε m), and 4.9) c β mr j ) + ε)r j )m j+ ε m j + ε + m) +. We set c α mr j ) = c α mr j ) for m m j and also set c α 0 r j ) = c 0 r j, H). Now m j < log, m j + ε + m m= and thus by 4.9) for large j 4.0) m j m= c β mr j ) < + ε)log )r j )m j+ ε < + ε)log )r j )m j+ ε. + m j We now suppose m j < m nr j ) and note that < r j. From.5), 4.), 4.3), and 4.4) we have with the omission of an obvious term if > r j ) c m r j, H) rj + rj r j ) m nt) dt + t t r j ) m rj nt) dt + t t rj r J ) m rj nt) dt t t 4.) < r m j m m j+ ε)/ m m j + ε) ) + r m j r m m j +ε j m m j + ε) + + ε) log + ε) + = rj m β a m j) + βmj) b + βmj) c + βmj) d ).

18 86 Joseph Miles Trivially 4.) β c mj) + β d mj)) /m = + ε) log + ε) + ) /m/rj < r j 0, m j < m nr j ), j. Likewise for m j < m nr j ) we have for large j 4.3) β a m j) ) /m < m j+ ε)/m < z mj /4 0 as j. For m j < m nr j ) we have β b mj) ε r m m j +ε j and thus 4.4) log β b mj) ) /m log ε m m m j + ε m log r j. We consider two cases. If m j < m m j, then 4.5) log β b mj) ) /m < log ε m j ε m j log r j, j, by 4.5). For m > m j we have by 4.4) 4.6) log βmj) b ) /m log ε m + < log ε m j + + m j m + ε m + ε m j ) log r j ) log r j, j. For m > m j, including those for which > r j, the combination of.4), 4.), 4.), 4.3), 4.5), and 4.6) implies the existence of δ j 0 such that if 4.7) c m r j, H) = β m j)r m j, then 4.8) β m j) /m < δ j 0, j.

19 On the growth of entire functions 87 Using 4.6) and 4.8) we apply Lemma with M = m j to conclude for large j that there exists a trigonometric polynomial Q j θ) with where L j > m j, such that Q j θ) = L j m= L j d m r j )e imθ, 4.9) d m r j ) = c α mr j ), m m j, and 4.0) Q j < We write ) π ε + ε + ε)rj ) mj+ ε. 4.) d m r j ) = γ m r j )r m j, m j < m L j, yielding by 4.0) γ m r j ) ) π ε + ε + ε)r m j+ ε m j and hence log γ m r j ) /m log π/ ε) + ε ) + log + ε) m m + ε m j m log r j. For m j < m minl j, m j ), we have 4.) log γ m r j ) /m < log π/ ε) + ε ) + log + ε) m j ε m j log r j as j by 4.5). If L j > m j, for m j < m L j we have 4.3) log γ m r j ) /m < log π/ ε) + ε ) + log + ε) + m j + ε ) log r j m j as j. Combining 4.) and 4.3), we conclude there exists κ j > 0 such that 4.4) γ m r j ) /m < κ j 0, m j < m L j,

20 88 Joseph Miles as j. By elementary considerations, there exists M j > L j such that 4.5) c m r j, H) < εnr j ). m >M j For m j < m M j, we define 4.6) b m j) = c m r j, H) + d m r j ) rj m, m j < m L j, c m r j, H) rj m, L j < m M j. From 4.7), 4.8), 4.), and 4.4), we conclude that there exists η j > 0 such that 4.7) b m j) /m < η j 0, m j < m M j, j. and Let T j z) = m j <m M j b m j)z m f j z) = Hz)e T jz). By.8), 4.7), and 4.6) c o r j, H) = c α o r j ), m = 0, c α mr j ) + c β mr j ), m m j, 4.8) c m r j, f j ) = d m r j ), m j < m L j, 0, L j < m M j, c m r j, H), m > M j. We also of course have c m r j, f j ) = c m r j, f j ) for m. From 4.), 4.0), 4.9), 4.0), 4.5), and 4.8), we have for all θ in [ π, π] and large j that log f j r j e iθ ) d m r j )e imθ m L j 4.9) + c β mr j ) + c m r j, H) m m j m >M j ) π < ε + ε + ε)nr j ) + log + 3ε)nr j ).

21 On the growth of entire functions 89 We consider a subsequence m jk satisfying 4.30) m jk+ > M jk, k, and 4.3) b m j p ) < where we have used 4.7). We set r jk ) m, k, p k +, m jp < m M jp, T z) = T jk z), k= and note by 4.30) that the powers of z in the various T jk are distinct. From 4.7) we see that T is entire and from 4.3) it follows that the Maclaurin coefficients of T have modulus less than. Thus by 4.), 4.3), and the fact that µ < 3, for π θ π and large k we have 4.3) k p= T jp r jk e iθ ) M jk r M j k j k < m jk r m j k j k < εnr jk ). 4.33) From 4.3) we have for π θ π that We set T jp r jk e iθ ) p>k p=k+ m j p <m M jp p=k+ m j p <m M jp fz) = Hz)e T z) b m j p ) r m j k ) m <. and note by 4.3) and 4.33) for large k and π θ π that log fr jk e iθ ) log f jk r jk e iθ ) < εnr jk ) +. We conclude by 4.9) that for large k log fr jk e iθ ) ) ) π < ε + ε + ε) + log + 5ε nr jk ). Since ε in 0, 3 ) is arbitrary, we obtain the conclusion of Theorem.

22 90 Joseph Miles References [] Beardon, A.F., and C.C. Yang: Open problems for the workshop on value distribution theory and its applications. - Hong Kong, July 996. J. Contemp. Math. Anal. 3, 997, [] Bergweiler, W.: Canonical products of infinite order. - J. Reine Angew. Math. 430, 99, [3] Boas, R.P.: Majorant problems for trigonometric series. - J. Anal. Math. 0, 96, [4] Duren, P.L.: Theory of H p Spaces. - Academic Press, New York, 970. [5] Gol dberg, A.A.: Meromorphic functions with separated zeros and poles. - Izv. Vysš. Učebn. Zaved. Mat. 7:4, 960, 67 7 Russian). [6] Miles, J., and D.F. Shea: An extremal problem in value distribution theory. - Quart. J. Math. Oxford ) 4, 973, [7] Nevanlinna, F.: Bemerkungen zur Theorie der ganzen Funktionen endlicher Ordnung. - Soc. Sci. Fenn. Comment. Phys.-Math. :4, 93, 7. [8] Newman, D.J.: Problem Amer. Math. Monthly 66, 959, 87; 67, 960, 8 8. Received 0 August 000