A TECHNIQUE OF BEURLING FOR LOWER ORDER

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1 Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 34, 29, A TECHNIQUE OF BEURLING FOR LOWER ORDER P. C. Fenton and John Rossi University of Otago, Department of Mathematics and Statistics P.O. Box 56, Dunedin, New Zealand; pfenton@maths.otago.ac.nz Virginia Tech, Department of Mathematics Blacksburg, VA , U.S.A.; rossij@vt.edu Abstract. Suppose that φ = u v, where u and v are subharmonic in the plane, with u nonconstant. Suppose also that lim r (log (B(r, u) + B(r, v)) / log r λ, for some λ satisfying < λ < 1/2, and that the deficiency δ of φ satisfies 1 δ < cos πλ. Given σ satisfying λ < σ < 1/2 and 1 δ < cos πσ, we have A(r, φ)/b(r, φ) > κ = κ(σ, δ) := cos πσ (1 δ) 1 (1 δ) cos πσ for all r in a set of upper logarithmic density at least 1 λ/σ. Here A(r, φ) = inf z =r φ(z) and B(r, φ) = sup z =r φ(z). 1. Introduction This note concerns δ-subharmonic functions φ and the relationship between A(r, φ) and B(r, φ), where A(r, φ) = inf φ(z), z =r B(r, φ) = sup φ(z). z =r If φ = u v, where u and v are subharmonic in the plane, the deficiency of φ is defined to be δ := 1 lim r N(r, v)/t (r, φ), where T (r, φ) = N(r, φ + ) + N(r, v). Here N(r, v) = 1 2π v(re iθ ) dθ v(), 2π and N(r, u) and N(r, φ + ) are defined similarly. Let µ(e, u) be the Riesz measure of u for any Borel set E, and define µ (t, u) := µ({z : z t}, u), with a similar definition for v. We will prove: Theorem 1. Suppose that u and v are subharmonic in the plane, with u nonconstant, and that, for some λ satisfying < λ < 1/2, log (B(r, u) + B(r, v)) (1) lim r log r λ. Let φ = u v and suppose that the deficiency δ of φ satisfies 1 δ < cos πλ. Given σ satisfying λ < σ < 1/2 and 1 δ < cos πσ, we have (2) A(r, φ) B(r, φ) > κ = κ(σ, δ) := cos πσ (1 δ) 1 (1 δ) cos πσ for all r in a set of upper logarithmic density at least 1 λ/σ. 2 Mathematics Subject Classification: Primary 3D15. Key words: δ-subharmonic, lower order, deficiency.

2 38 P. C. Fenton and John Rossi The upper (or lower) logarithmic density of a set S (, ) is the upper (or lower) limit as r of (log r) 1 S (1,r) t 1 dt. It is known that (2) holds in a set of lower logarithmic density at least 1 λ/σ if the lower limit in (1) is replaced by the upper limit [4, Theorem 1.1]. We remark that if v in Theorem 1 (implying δ = 1), then the result was obtained by Barry [1] as an extension of the classical cos πλ theorem for subharmonic functions. (In fact his result holds for < λ < 1.) Original estimates in Barry s paper involve A(t, u) cos πσb(t, u) dt. t σ+1 Progressing from these estimates to estimates of the integrand naturally introduces exceptional sets. Hayman [6] subsequently constructed a subharmonic function u of order λ, < λ < 1, such that, for given σ with λ < σ < 1, (3) A(r, u ) B(r, u ) cos πσ for all r in a set with upper (and lower) logarithmic density equal to λ/σ. This established that the size of the exceptional set in Barry s result is sharp. Similarly, by setting φ(z) = u (z) (1 δ)u ( z), we see that (2) also fails on E and so the size of the exceptional set is sharp in Theorem 1 as well. These examples raise the question as to whether we can find classes of functions with no exceptional sets. In the proof of Theorem 2, we will show that the structure of our exceptional set in Theorem 1 is directly related to the behaviour of the Riesz measures of u and v. It will follow that if, for all large values of t, µ (t, u) and µ (t, v) are continuous functions of t of the form (4) µ (t, u) = ε 1 (t)t λ, µ (t, v) = ε 2 (t)t λ, where ε i (t) as t for i = 1, 2, then there exists a positive number R such that (2) holds for all r > R. Effectively then, there is no exceptional set. As far as we are aware, this is the first time such an observation has been made about a general class of functions. We will suppose that u and v are modified so as to be harmonic in z < 1, which does not affect the generality of our results. If u has order less than one, then [5, p. 311] u(z) = u() + log 1 z/ζ dµ(ζ, u) and, with we have u (z) := u() + ζ < log 1 + z/t dµ (t, u), (5) u ( r) A(r, u) B(r, u) u (r), r <. Theorem 1 is a consequence of the following result, which adapts a technique of Beurling [2, p. 762] to situations in which something is known about lower rather than upper growth.

3 A technique of Beurling for lower order 381 Theorem 2. Suppose that < ν < 1, that u is a nonconstant subharmonic function in the plane and that (6) lim B(r, u)/r ν =. r (i) There are sequences ε n and R n such that (7) B(8R n, u) < ε n R ν n, N(4R n, u) < ε n R ν n and µ (4R n, u) < ε n R ν n for all n. (ii) Given C 1 and n N, define (8) U n (z) = U n (z, ν) := u()+ For any α satisfying 4Rn (9) α α n := 5Cν 1 ε n, let log 1+ z/t dµ (t, u)+8cε n R ν n log 1+ z/(4r n ). (1) a α,n := max{n(r, u) αr ν : r [, R n ]}, and let r α,n be any point in [, R n ] at which (11) a α,n = N(r α,n, u) αr ν α,n. Then a α,n, (12) A(r α,n, u) > U n ( r α,n ) > πν(cot πν)n(r α,n, u) + (1 πν cot πν)a α,n and (13) B(r α,n, u) < U n (r α,n ) < πν(csc πν)n(r α,n, u) + (1 πν csc πν)a α,n. Further, for any α >, the part of the set (14) T = T (α ) := {r α,n : α α 5ε n /ν} n=1 contained in [1, ) has infinite logarithmic measure. If u has lower order λ, where λ < ν, then T has upper logarithmic density at least 1 λ/ν. It is always possible to choose α in such a way that a α,n as r α,n in T. The following theorem is a lower order analogue of Theorem 3 in [3]. The constant (cos πλ k)/(1 k cos πλ) is sharp; see the remark following Theorem 3 in [3]. Theorem 3. Suppose that u and v are subharmonic in the plane, with u nonconstant, and that, for some λ satisfying < λ < 1/2, B(r, u) + B(r, v) lim = r r λ and N(r, v) kn(r, u) + O(1) as r, for some k such that k cos πλ. Let φ(z) = u(z) v(z). Then, with r α,n as defined in (11), (15) A(r α,n, φ) where b rα,n as r α,n. cos πλ k 1 k cos πλ B(r α,n, φ) + b rα,n,

4 382 P. C. Fenton and John Rossi Acknowledgment. The authors are most grateful for the referee s careful reading of the original manuscript, which brought to light a dismaying number of errors and confusions. 2. Proof of Theorem 2 Concerning the sequences ε n and R n, choose ε n arbitrarily, with ε n, and R n such that B(8R 1, u) u(), N(R 1, u) >, R n, and B(8R n, u) < 1 2 ε nr ν n log 2 for all n, which is possible from (6) and the fact that u is conconstant (so that B(r, u) and N(r, u) as r ). The first part of (7) is satisfied and, from Jensen s theorem, N(4R n, u) N(8R n, u) u() + B(8R n, u) 2B(8R n, u), so the second part is satisfied also. For the third part, µ (4R n, u) log 2 8Rn 4R n µ (t, u) dt/t N(8R n, u). Notice that, since u is harmonic in z < 1, we have N(r, u) = for r 1, and thus R 1 > 1. According to a result of Kjellberg [7, formulas (6), (8) and (18)], (see also [1, pp ]), if u 1 (z) = u 1 (z, R n ) := u() + log 1 z/ζ dµ(ζ, u), ζ <4R n then (16) u(z) u 1 (z) z R 1 n B(8R n, u) for z R n. Write so that for all r, u 2 (z) = u 2 (z, R n ) := u() + 4Rn log 1 + z/t dµ (t, u), (17) u 2 ( r) A(r, u 1 ) B(r, u 1 ) u 2 (r), from (5) (or by direct computation). Since B(8R n, u) < ε n R ν n, we deduce from (16) and (17) that, for z R n, and therefore u(z) > u 1 (z) z Rn 1 B(8R n, u) > u 2 ( z ) ε n z Rn ν 1 (18) A(r, u) > u 2 ( r) ε n rr ν 1 n, r R n. Similarly (19) B(r, u) < u 2 (r) + ε n rr ν 1 n, r R n. With (8) in view, we have, for r R n, (2) U n ( r) = u 2 ( r) + 8Cε n R ν n log(1 r/(4r n )) u 2 ( r) 2Cε n rr ν 1 n < A(r, u),

5 from (18), and (21) from (19). Now, A technique of Beurling for lower order 383 U n (r) = u 2 (r) + 8Cε n R ν n log(1 + r/(4r n )) u 2 (r) + 2Cε n rrn ν 1 / ( 1 + r/(4r n ) ) u 2 (r) + Cε n rrn ν 1 > B(r, u), (22) N(r, U n ) = N(r, u 2 ) = N(r, u 1 ) = N(r, u), r 4R n, since µ (r, ) is the same for each function for r 4R n, and therefore (23) N(r, U n ) N(4R n, U n ) = N(4R n, u) < ε n R ν n < α n R ν n α n r ν for R n r 4R n, using (7) and (9). Also, since the third term on the right hand side of (8) arises from a point mass 8Cε n Rn ν at 4R n, we have, for r 4R n, (24) r µ (t, U n ) N(r, U n ) = N(4R n, U n ) + dt 4R n t = N(4R n, u) + ( ) ( µ (4R n, u) + 8Cε n Rn ν log r/(4rn ) ) < N(4R n, u) + 9Cε n Rn ν log ( r/(4r n ) ), using (7). The largest value of x ν log x is (eν) 1, at x = e 1/ν, and therefore R ν n log ( r/(4r n ) ) = 4 ν r ν (r/(4r n )) ν log ( r/(4r n ) ) < (eν) 1 r ν. From this, (7) and (24) we have, for r 4R n, (25) N(r, U n ) < (1 + 9C(eν) 1 )ε n r ν < (1 + 4Cν 1 )ε n r ν < 5Cν 1 ε n r ν α n r ν, using (9). Suppose now that α satisfies (9). From (1) and (11), N(r α,n, u) αrα ν n N(, u) =, while from (23) and (25), N(r, U n ) αr ν < for R n r <. We deduce that (26) a α,n = N(r α,n, u) αrα,n ν = N(r α,n, U n ) αrα,n ν = max{n(r, U n ) αr ν : r [, )}, using (22). From (26), it follows [2, p. 764] that (27) and U n ( r α,n ) > πν(cot πν)n(r α,n, U n ) + (1 πν cot πν)a α,n = πν(cot πν)n(r α,n, u) + (1 πν cot πν)a α,n (28) U n (r α,n ) < πν(csc πν)n(r α,n, u) + (1 πν csc πν)a α,n, which, combined with (2) and (21) at r = r α,n, give (12) and (13). It remains to prove the conclusions about T. Fix a positive integer n and consider, for that n, U n given by (8). Changing notation somewhat, for any α >, let t α be a point at which max{n(t, U n ) αt ν : t [, )} is attained. Following the argument of [4, pp ], the points t α increase as α decreases, and occupy all points of [, ) apart from exceptional intervals of the form (t α, t + α ), where (recalling that there may be several values of t α ) t α and t + α are the smallest and largest among the possible values of t α. Further, for all α, µ (t α, U n ) = αν(t α ) ν, and µ (t + α, U n ) = αν(t + α ) ν. As we have shown (cf. (26)), if

6 384 P. C. Fenton and John Rossi α α n = 5Cε n /ν then t α R n, so that µ (t α, U n ) = µ (t α, u). Returning to our earlier notation, we have t α = r α,n, so that (29) µ (r α,n, u) = αν(r α,n) ν and µ (r + α,n, u) = αν(r + α,n) ν. Thus the set {r α,n : α α α n } consists of all points in (rα,n, r α + n,n) outside an exceptional set of intervals of total logarithmic measure ν 1 log ( µ (r α,n, + u)/µ (rα,n, u) ) ν 1 log(µ (r α + n,n, u)/µ (rα,n, u)) (3) α α α n = log(r α + n,n/rα,n) + ν 1 log(α n /α ), using (29). Evidently, since T increases as α increases, we may assume without loss of generality that α is sufficiently small that α < N(R 1, u)r1 ν. In that case a α,n > for all n and all α satisfying α α α n, and, since N(t, u) α t ν = α t ν for t 1, we have r α,n > 1 for all n. Thus the logarithmic measure of the exceptional points in [1, r α + n,n] is at most log r α + n,n + ν 1 log(α n /α ), and consequently the logarithmic measure of the part of T in [1, r α + n,n] is at least ν 1 log(α /α n ) as n. Suppose that u has lower order λ, where λ < ν, and that ν 1 satisfies λ < ν 1 < ν. There is a sequence of positive numbers R n such that B(8R 1, u) u(), N(R 1, u) > and (31) B(8R n, u) < R ν 1 n, N(4R n, u) < R ν 1 n and µ (4R n, u) < R ν 1 n, for all n, as we have shown. Thus we have (7), with ε n = R ν 1 ν n, and (12) and (13) hold. From (29) and (3), again assuming as we may that α < N(R 1, u)r1 λ, the set {r α,n : α α α n } contains all points in [1, r α + n,n] outside an exceptional set of logarithmic measure at most ν 1 log µ (r α + n,n, u) ν 1 log(α ν). Now, from (29), (9) and the fact that r α + n,n R n, µ (r + α n,n, u) = α n ν(r + α n,n) ν = 5CR ν 1 ν n (r + α n,n) ν 5C(r + α n,n) ν 1, and therefore the logarithmic measure of the part of T in [1, r + α n,n] is at least (1 ν 1 /ν) log r + α n,n ν 1 log(5c/(α ν)). Since r + α n,n as n, the upper logarithmic density of T is thus at least 1 ν 1 /ν, for any ν 1 satisfying λ < ν 1 < ν, and so is at least 1 λ/ν. Concerning a choice of α that ensures that a α,n as r α,n in T, suppose that it is possible to find sequences α(j) and n j, with 1 α(j) 5ν 1 ε nj, such that r α(j),nj, while a α(j),nj = O(1) as j. Write γ j := a α(j),nj and γ := sup γ j, so that γ <. From the definition of a α(j),nj, we have (32) N(t, u) α(j)t ν + γ j, t R nj, with equality at t = r α(j),nj. Write α := lim j α(j), so that α 1, and suppose that α =. In that case, α(j) on a subsequence, and since R nj r α(j),nj as j, we conclude from (32) that N(t, u) γ, a contradiction, since u is nonconstant and of order less than 1. So α > and, from (32), N(t, u) α t ν +γ, for t. Thus, since N(r α(j),nj, u) = α(j)rα(j),n ν j + γ j α(j)rα(j),n ν j, we have (33) lim t N(t, u)/t ν = α.

7 A technique of Beurling for lower order 385 It follows that if < α < α then a α,n as r α,n in T. For otherwise, repeating the above argument, we would obtain lim t N(t, u)/t ν = α, for some α with < α α. This completes the proof of Theorem Proofs of Theorems 1 and 3 The argument conflates the proofs of Theorems 1 and 3 in [3]. To prove Theorem 1, given ν satisfying λ < ν < σ (so that 1 δ < cos πν), choose such that < < δ, 1 < cos πν and (34) K := κ(ν, ) > κ(σ, δ). This is possible, and in addition can be chosen so that δ as ν σ. Following [2, p. 394], we may assume that u(z) v(z) for all z and that lim r N(r, v)/n(r, u) 1 δ. Thus, for some positive constant c, (35) N(r, v) (1 )N(r, u) + c, r. (In what follows, we use c to denote a positive constant that depends only on u, v and the parameters of Theorem 1, not necessarily the same at each occurrence.) From (1), lim r (B(r, u) + B(r, v))/r ν =. We may thus find sequences ε n and R n such that the inequalities (7), with ε n /2 instead of ε n, hold for both u and v, and define U n by (8) with C = 9(1 ) 1 /8, and V n by (8) with C = 1 (and v replacing u). Let w(z) := u(z) + Kv(z), where K is given by (34). Then w is subharmonic, since K >, and the inequalities (7) hold for w also, since K < 1. Also, if W n is defined by (8), with C = 9(1 ) 1 /8 + K and w instead of u, then (36) W n (z) = U n (z) + KV n (z). Write ψ n (z) = U n (z) V n ( z). For r R n, we have A(r, φ) A(r, u) B(r, v) > U n ( r) V n (r) = ψ n ( r) and B(r, φ) B(r, u) A(r, v) < U n (r) V n ( r) = ψ n (r), from (2) and (21), so that, for r R n, (37) A(r, φ) KB(r, φ) > ψ n ( r) Kψ n (r) Also, integrating twice by parts, (38) V n (r) = v() + and further = W n ( r) KW n (r) (1 K 2 )V n (r). log 1 + r/t dµ (t, V n ) = v() + r N(t, V n ) (t + r) dt, 2 N(t, V n ) = N(t, v) (1 )N(t, u) + c = (1 )N(t, U n ) + c, t 4R n, using (35), while for t 4R n, using (35) again and also (7), N(t, V n ) = N(4R n, v) + (µ (4R n, v) + 8ε n R ν n) log(t/(4r n )) (1 )N(4R n, u) + 9ε n R ν n log(t/(4r n )) + c (1 )N(4R n, u) + (1 ) [ µ (4R n, u) +

8 386 P. C. Fenton and John Rossi + 8(9(1 ) 1 /8)ε n R ν n] log(t/(4r n )) + c = (1 )N(t, U n ) + c. Thus, from (38) and a similar expression for U n (r), V n (r) (1 )U n (r) + c. It follows from (36) that, for all r, W n (r) ((1 ) 1 + K)V n (r) c. From (37) then, A(r, φ) KB(r, φ) > W n ( r) K (1 )K W n(r) c (39) = W n ( r) cos πνw n (r) c, for r R n. We apply Theorem 2 (ii) to w, with C = 9(1 ) 1 /8 + K and α chosen so that a α,n as r α,n. From (12) and (13), (4) (41) W n ( r α,n ) > πν(cot πν)n(r α,n, w) + (1 πν cot πν)a α,n, W n (r α,n ) < πν(csc πν)n(r α,n, w) + (1 πν csc πν)a α,n. Recalling that r α,n R n, and substituting these inequalities into (39), we obtain (42) A(r α,n, φ) KB(r α,n, φ) > (1 cos πν)a α,n c > for all large r α,n. Thus A(r, φ)/b(r, φ) > K > κ(σ, δ), from (34), for all r in a set of upper logarithmic density at least 1 λ/ν. Since we may take ν as close to σ as we please, we conclude that (2) holds in a set of upper logarithmic density at least 1 λ/σ, which completes the proof of Theorem 1. To prove that (2) holds for all sufficiently large r under the assumption (4), note that the exceptional set E where (2) fails comes from the exceptional set for w = u + Kv. By (4), µ (t, w) = ε(t)t λ is continuous for all large t, and ε(t) as t. Thus, if α is sufficiently small and t (R, ) for some positive number R, then, provided R is sufficiently large, the equation µ (t, w) = ανt ν has precisely one solution. By the argument surrounding (29), we have r α,n = r + α,n for all such α and all large enough n. Thus, since the exceptional set is contained in the intervals (r α,n, r + α,n), E (R, ) = for sufficiently large R. The proof of Theorem 3 is almost identical to that of Theorem 1. One needs to set ν = λ in Theorem 2, make use of (14) and set k = 1 in (35). References [1] Barry, P. D.: On a theorem of Kjellberg. - Quart. J. Math. (Oxford) 15, 1964, [2] Fenton, P. C., and J. Rossi: Phragmén-Lindelöf theorems. - Proc. Amer. Math. Soc. 132, 23, [3] Fenton, P. C., and J. Rossi: cos πρ theorems for δ-subharmonic functions. - J. Anal. Math. 92, 24, [4] Fenton, P. C.: Density in cos πρ theorems for δ-subharmonic functions. - Analysis (Munich) 26, 26, [5] Hayman, W. K.: Subharmonic functions, Vol Academic Press, [6] Hayman, W. K.: Some examples related to the cos πρ theorem. - Mathematical essays dedicated to A. J. MacIntyre, Athens, Ohio, 197. [7] Kjellberg, B.: On the minimum modulus of entire functions of order less than one. - Math. Scand. 8, 196, Received 22 January 28