Sharp Inequalities for the Product of Polynomials

Let/j(z),... ,/m(z) be polynomials with complex coefficients, and let their product be of degree n. For any polynomial, let | |/ | | be the maximum of \J[z)\ on the unit circle. We determine constants Cm < 2 for which II/ill " ll/mll ^ C^H/j •••/„,|| for any n. The inequalities are asymptotically sharp as n ->oo. This improves earlier results of Gel'fond and Mahler, who gave the constants e and 2 respectively. \ifv...,fm have real coefficients, we show that | | / J ••• | | / J | < Q ||/x •••/J | for all m ^ 2 and that this is asymptotically sharp. That is, in the real case, the best constant does not depend upon m for m ^ 2. 0. Introduction Let/i(z),... ,fm(z) be polynomials with complex coefficients, let/(z) = /1(z) • • 'fm(z) denote their product, and suppose that the degree of / is n. Let || • || denote the maximum norm on the unit circle. It is obvious that ||/|| ^ H/J ••• ||/m||. Mahler [6], improving an earlier result of Gel'fond [2, p. 135], proved that, for all n, I I /J-II/JK2-II/H. (1) In a recent paper [1], if m = 2, we improved (1) by replacing 2 by the constant 5= exp (2G/n) = 1.79162..., where G denotes Catalan's constant The constant 6 is asymptotically best as «->oo. We conjectured there that the best constant in (1) would be given by Cm = exp((m/n)I(n/m)), where 1(0) = JJ log (2 cos (t/2))dt = C\2(n-6), Cl2 denoting Clausen's integral [5, Chapter 4]. Since G = I(n/2), we have S = C2. We shall prove this conjecture in the present paper. In the extreme case,/(z) can be taken to be z +1 and the zeros of each fk to be ~ n/m consecutive zeros of/. Then log | | /J ~ (n/m) log Cm for each k. In the case that the polynomials are assumed to have real coefficients, we show by an elementary argument that the correct constant in (1) is C2 for all m $s 2. One of the main ingredients of the proof in [1] is an integral representation of the function max (log \z—a\, log \z—b\) which follows from Jensen's formula: log \z\ = max (0, log \z\) = ~ f* log \z e»\ d. (2) Received 29 January 1992; revised 28 June 1993. 1991 Mathematics Subject Classification 30C10, 11C08. This research was supported in part by an operating grant from NSERC. Bull. London Math. Soc. 26 (1994) 449^54 450 DAVID W. BOYD We can regard (2) as expressing the subharmonic function log \z\ as a logarithmic potential, that is, as log \z\ = log \z — t\dv(t), (3) Jc where v is the uniform measure supported on the unit circle. By a change of variable, this gives a representation of log (max (\z — a\, \z-b\)) as a logarithmic potential of a Cauchy distribution supported on the perpendicular bisector of the segment ab. Our main new tool here is a generalization of this to deal with maxl!gA.^m (log \z—ck\) for any m > 1, that is, a representation of this subharmonic function as a logarithmic potential (Lemma 2). The remainder of the proof is then similar to the proof of Theorem 2 of [1]. 1. Preliminary results The following well-known result was also used in [1], and the proof is repeated here in order to make this paper self-contained. LEMMA 1. Let f be a polynomial of degree n. Then for any complex z, | / ( z ) | ^ Il/H max (l,|z|). Proof. If |z| ^ 1, this follows from the maximum modulus principle. If \z\ > 1, it follows from the maximum modulus principle applied to zf{\/z). LEMMA 2. Let S = {cv ...,cm), m > 1, be a finite set of complex numbers, and define u(z) = maxl!gfc^m log \z — ck\. Then there is a probability measure fi, whose support is contained in a finite rion of straight lines, such that u(z)= f log \z-t\dfx(t), (4) Jc forallzeC. Proof. Let Dk = {zeC: |z — ck\ > \z — ct\, i # k). Then Dk is an open set whose boundary is contained in the m— 1 perpendicular bisectors of the segments ckct, for / T* k. The function u is subharmonic, being the maximum of the subharmonic functions log \z — ck\. Furthermore, u is continuous everywhere and is harmonic in each of the sets Dk. Let z be fixed and define v(f) = log \z — t\, so that v is harmonic except at t = z. To begin with, let zeD := \JkDk. For e > 0 and R < oo, let De R denote the open disk with centre 0 of radius R from which is removed a closed disk of radius e with centre z. We apply Green's formula to De R, taking the distributional definition of the derivatives in question [3, Chapter 3]. Thus we have n denoting the outward normal, and a arc length. SHARP INEQUALITIES FOR THE PRODUCT OF POLYNOMIALS 451 As e -+ 0, a familiar calculation [3, pp. 111-112] shows that the integral over the small circle surrounding z tends to 27TM(Z). On the circle |/| = R, both u{t) and v(t) are log R + O(\/R). Also, du/dn and dv/dn are (l/R) + O(\/R), so that dv du ~/log u-—v— = O\ dn dn \ R hence the integral over the circle \t\ = R is O(log R/R), which vanishes as i? -•oo. Thus u(z) = — log \z — t\Audt. (6) 2nJc This proves (4) for zeD with d/iif) = {\/2n)Au{i)dt. To compute /a, we must calculate the Laplacian AM. Within D, AM = 0 since M is harmonic here. Suppose next that t is an interior point of a boundary segment of D separating D, from Dk. On this segment, choose local coordinates ( 0 without loss of generality. So M = \ log ((£ — a) + (|̂ | + b)). Hence du/d£, is continuous across this part of the boundary, but du/dn has a jump discontinuity of size 2b/((£ — a) + b) at (£,0). Thus //(£, rj) d£drj = — Au d^drj = ——^—^ 6{rj) d^ dn, (7) <5 denoting Dirac measure. So // is absolutely continuous with respect to linear Lebesgue measure along this boundary segment, with dn/d£ being essentially the Poisson kernel. We next observe that /n(C) oo shows that 1 = //(C), by dominated convergence. Finally, we can check that the right member of (4) is continuous in z, as in [4, pp. 275-276], for example. Since u(z) is also continuous and D is dense in C, this proves (4) for all zeC. REMARKS. 1. It is easy to give a proof of Lemma 2 independent of the theory of distributions. One replaces the set DsR by a union of Fk n De R where Fk <= Dk is a polygon with edges parallel to those of Dk at distance h from those of Dk (an inner parallel body). Then letting h-too, one obtains the density in (7).