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GENERALIZATIONS OF GONÇALVES’ INEQUALITY 5 0 PETERBORWEIN,MICHAELJ.MOSSINGHOFF,ANDJEFFREYD.VAALER 0 2 n Abstract. LetF(z)= Nn=0anznbeapolynomialwithcomplexcoefficients a and roots α1, ..., αN, lPetkFkp denote its Lp norm over the unit circle, and J letkFk0 denoteMahler’smeasureofF. Gonçalves’ inequalityassertsthat 1 N N 1/2 1 kFk2≥|aN| max{1,|αn|2}+ min{1,|αn|2} n=1 n=1 ! Y Y ] 2 1/2 CA =kFk0 1+ |ak0FaNk40| ! . Weprovethat . h N N 1/p at kFkp≥Bp|aN| max{1,|αn|p}+ min{1,|αn|p} m n=1 n=1 ! Y Y [ for1≤p≤2,whereBp isanexplicitconstant, andthat v1 kFkp≥kFk0 1+ p24|ak0FakN40|2!1/p 3 for p ≥ 1. We also establish additional lower bounds on the Lp norms of a 6 polynomialintermsofitscoefficients. 1 1 0 5 0 1. Introduction / h Let∆ Cdenotetheopenunitdisc,∆itsclosure,andlet (∆)denotethealge- at braofcon⊂tinuous functionsf :∆ C thatare analyticon∆A. Then (∆), m is a Banach algebra,where → {A k·k∞} : v f =sup f(z) :z ∆ =sup f(e(t)) :t R/Z , k k∞ | | ∈ {| | ∈ } i X and e(t) denotes the func(cid:8)tion e2πit. If f(cid:9) (∆) and 0<p< , we also define ∈A ∞ r a 1 1/p f = f(e(t))pdt , k kp | | (cid:18)Z0 (cid:19) and we define 1 f =exp log f(e(t)) dt . k k0 | | (cid:18)Z0 (cid:19) It is known (see [2, Chapter 6]) that for each f in (∆) the function p f is A → k kp continuous on [0, ], and if 0<p<q < , then these quantities satisfy the basic ∞ ∞ 2000 MathematicsSubjectClassification. Primary: 30A10,30C10;Secondary: 26D05,42A05. Key words and phrases. Lp norm, polynomial, Gonçalves’ inequality, Hausdorff-Young inequality. ResearchofP.BorweinsupportedinpartbyNSERCofCanadaandMITACS. 1 2 PETERBORWEIN,MICHAELJ.MOSSINGHOFF,ANDJEFFREYD.VAALER inequality (1.1) f f f f . k k0 ≤k kp ≤k kq ≤k k∞ Clearly, equality can occur throughout (1.1) if f is constant. On the other hand, if f is not constant in (∆), then the function p f is strictly increasing on A → k kp [0, ]. N∞ow suppose that F is a polynomial in C[z] of degree N 1, and write ≥ N N (1.2) F(z)= a zn =a (z α ). n N n − n=0 n=1 X Y In this case, the quantity F is Mahler’s measure of F, and by Jensen’s formula k k0 one obtains the well-known identity N (1.3) F = a max 1, α . k k0 | N| { | n|} n=1 Y Thus a special case of (1.1) is the inequality (often called Landau’s inequality) N F a max 1, α . k k2 ≥| N| { | n|} n=1 Y For polynomials of positive degree, the sharper inequality 1/2 N N (1.4) F a max 1, α 2 + min 1, α 2 k k2 ≥| N| { | n| } { | n| }! n=1 n=1 Y Y was obtained by Gonçalves [1]. Note that equality occurs in (1.4) for constant multiples of zN 1. Alternatively, the inequality (1.4) may be written in the less − symmetrical form 1/2 a a 2 0 N (1.5) F F 1+ | | . k k2 ≥k k0 F 4 ! k k0 For a positive real number p, define the real number B by p 1 1 1/p Γ(p+1) 1/p (1.6) B = 1 e(t)p dt = , p 2 | − | 2Γ(p/2+1)2 (cid:18) Z0 (cid:19) (cid:18) (cid:19) and note that B = 2/π and B = 1. In this article we establish the following 1 2 generalizations of Gonçalves’ inequality. Theorem 1. Let F(z) C[z] be given by (1.2). If 1 p 2, then ∈ ≤ ≤ 1/p N N (1.7) F B a max 1, α p + min 1, α p k kp ≥ p| N| { | n| } { | n| }! n=1 n=1 Y Y and if p 1, then ≥ 1/p p2 a a 2 0 N (1.8) F F 1+ | | . k kp ≥k k0 4 F 4 ! k k0 GENERALIZATIONS OF GONÇALVES’ INEQUALITY 3 Equality occurs in (1.7) for constant multiples of zN 1. The inequality (1.8) − is never sharp for p =2, but since B <1 for 1 p <2 it is clearly stronger than p 6 ≤ (1.7) in this range when F 2/ a a is large. For example, one may verify that k k0 | 0 N| (1.8) produces a better bound in the case p=1 whenever F 2 π2 k k0 > =1.1576382... a0aN 2 2 √4+2π π2 | | − − Also,forfixedF therightside(cid:0)of(1.8)achievesa(cid:1)maximumatp=2c F 2/ a a , k k0 | 0 N| where c = 1.9802913... is the unique positive number satisfying 2c2 = (1 + c2)log(1+c2). In view of (1.1), inequality (1.8) is therefore only of interest when 1 p 2c F 2/ a a . ≤ ≤ k k0 | 0 N| To prove Theorem 1, we first establish some lower bounds on the L norms of p a polynomial in terms of two of its coefficients a and a , provided M L is L M | − | sufficientlylarge. Theseinequalitieshavesomeindependentinterest,andwerecord the results in the following theorem. Theorem 2. Let F(z) C[z] be given by (1.2), and let L and M be integers ∈ satisfying 0 L<M N and M L>max L,N M . Then ≤ ≤ − { − } (1.9) F a + a . k k∞ ≥| L| | M| Further, if 1 p 2 then ≤ ≤ (1.10) F B (a p+ a p)1/p, k kp ≥ p | L| | M| and if p 1 and a and a are not both 0, then L M ≥ 2 1/p pmin a , a L M (1.11) F max a , a 1+ {| | | |} . k kp ≥ {| L| | M|} (cid:18)2max{|aL|,|aM|}(cid:19) ! At this point, it is instructive to recall the Hausdorff-Young inequality. If p=2 and F(z) is given by (1.2), then by Parseval’s identity we have 1/2 (1.12) F = a 2+ a 2+ + a 2 . k k2 | 0| | 1| ··· | N| (cid:16) (cid:17) If p=1, then the inequality (1.13) F max a , a ,..., a k k1 ≥ {| 0| | 1| | N|} follows immediately from the identity 1 a = F(e(t))e( nt)dt. n − Z0 Now suppose that 1 < p < 2 and let q be the conjugate exponent for p, so p−1+ q−1 =1. Then the Hausdorff-Young inequality [3, p. 123] asserts that (1.14) F (a q+ a q + + a q)1/q, k kp ≥ | 0| | 1| ··· | N| and so interpolates between (1.12) and (1.13). If p = 2, then (1.10) and (1.11) are equivalent and clearly follow from the identity (1.12). But for 1 < p < 2, the inequalities (1.10) and (1.11) are not immediate consequences of (1.14). In fact, it iseasytoseethatthelowerboundsin(1.10),(1.11),and(1.14)arenotcomparable. If p=1, the same remarks apply to (1.10), (1.11), and (1.13). 4 PETERBORWEIN,MICHAELJ.MOSSINGHOFF,ANDJEFFREYD.VAALER In section 2 we develop some preliminary results concerning lower bounds on L norms of binomials, and we use these facts to establish Theorems 1 and 2 in p section 3. 2. Norms of binomials For 0<r <1 and real t, recall that the Poisson kernel is defined by ∞ 1+re(t) 1 r2 P(r,t)= r|n|e(nt)= = − . ℜ 1 re(t) 1 re(t)2 n=X−∞ (cid:18) − (cid:19) | − | This is a positive summability kernel that satisfies 1 P(r,t)dt=1 Z0 and 1−ǫ lim P(r,t)dt=0 r→1− Zǫ for 0<ǫ<1/2. Lemma 3. If p>0 then 1 lim 1 re(t)pP(r,t)dt=0. r→1−Z0 | − | Proof. Let 0<ǫ<1/2 so that 1 ǫ 1−ǫ 1 re(t)pP(r,t)dt= 1 re(t)pP(r,t)dt+ 1 re(t)pP(r,t)dt | − | | − | | − | Z0 Z−ǫ Zǫ ǫ 1−ǫ 1 re(ǫ)p P(r,t)dt+2p P(r,t)dt ≤| − | Z−ǫ Zǫ 1−ǫ 1 re(ǫ)p+2p P(r,t)dt. ≤| − | Zǫ We conclude that 1 limsup 1 re(t)pP(r,t)dt 1 e(ǫ)p (2πǫ)p, | − | ≤| − | ≤ r→1− Z0 and the statement follows. (cid:3) For positive numbers p and r, we define 1 p (2.1) (r)= 1 r1/pe(t) dt. p I − Z0 (cid:12) (cid:12) It follows easily that r (r) is a co(cid:12)ntinuous, po(cid:12)sitive, real-valued function that →Ip (cid:12) (cid:12) satisfies the functional equation (2.2) (r)=r (1/r) p p I I forallpositiver. Thefollowinglemmarecordssomefurtherinformationaboutthis function. Lemma4. Foranypositivenumberp,thefunctionr (r) has acontinuousde- p →I rivativeateachpointof(0, )andsatisfiestheidentity ′(1)= (1)/2. Moreover, ∞ Ip Ip this function has infinitely many continuous derivatives on the open subintervals (0,1) and (1, ). ∞ GENERALIZATIONS OF GONÇALVES’ INEQUALITY 5 Proof. Suppose first that 0 < r < 1. Then 1 r1/pe(t) p/2 has the absolutely − convergentFourier expansion (cid:0) (cid:1) p/2 p/2 1 r1/pe(t) = ( 1)mrm/pe(mt). − m − (cid:16) (cid:17) mX≥0(cid:18) (cid:19) By Parseval’s identity, we have 2 p/2 (2.3) (r)= r2m/p. p I m m≥0(cid:18) (cid:19) X This shows that r (r) is represented on (0,1) by a convergent power series in p →I r1/p andthereforehasinfinitelymanycontinuousderivativesonthisinterval. Next, we observe that ∂ p 1 r1/pe(t) p 1 r1/pe(t) = − 1 P(r1/p,t) . ∂r − 2r − (cid:12) (cid:12) Itfollowsthatif0(cid:12)(cid:12)<ǫ 1/4an(cid:12)(cid:12)dǫ (cid:12)r 1 ǫ,th(cid:12)en(cid:16)thereexistsa(cid:17)positiveconstant (cid:12) ≤ (cid:12) ≤ ≤ − C(ǫ,p) such that ∂ p 1 r1/pe(t) C(ǫ,p). ∂r − ≤ tFhraotm the mean value theo(cid:12)(cid:12)(cid:12)(cid:12)rem(cid:12)(cid:12)(cid:12)and the do(cid:12)(cid:12)(cid:12)m(cid:12)(cid:12)(cid:12)(cid:12)inated convergence theorem, we find 1 1 p ′(r)= 1 r1/pe(t) 1 P(r1/p,t) dt, Ip 2r − − Z0 (cid:12) (cid:12) (cid:16) (cid:17) and therefore (cid:12) (cid:12) (cid:12) (cid:12) 1 p (r) 2r ′(r)= 1 r1/pe(t) P(r1/p,t)dt. Ip − Ip − Z0 (cid:12) (cid:12) Using the continuity of r (r) and(cid:12)Lemma 3, w(cid:12)e conclude that →Ip (cid:12) (cid:12) (2.4) lim ′(r)= (1)/2. r→1−Ip Ip Again using the mean value theorem, it follows that r (r) has a left-hand p → I derivative at 1 with the value (1)/2. p I From (2.2) and (2.3) we find that 2 p/2 (2.5) (r)=r r−2m/p p I  m  m≥0(cid:18) (cid:19) X   for r > 1. Thus r (r) is represented by r times a convergent power series p → I in r−1/p, and so has infinitely many continuous derivatives on the interval (1, ). ∞ Next, we differentiate both sides of (2.2) to obtain the identity ′(1/r) ′(r)= (1/r) Ip Ip Ip − r for r>1, and using the continuity of r (r) and (2.4), we conclude that p →I lim ′(r)= (1) lim ′(s)= (1)/2. r→1+Ip Ip −s→1−Ip Ip It follows that r (r) has a right-hand derivative at 1 with value (1)/2. p p →I I We conclude then that r (r) is continuously differentiable on (0, ) and p ′(1)= (1)/2. → I ∞ (cid:3) Ip Ip 6 PETERBORWEIN,MICHAELJ.MOSSINGHOFF,ANDJEFFREYD.VAALER From the proof of the lemma we obtain the following lower bound on the L p norm of a binomial. Corollary 5. Let 0 L<M be integers and let α and β be complex numbers, not ≤ both zero. If p>0 then pmin α , β 2 1/p αzL+βzM max α , β 1+ {| | | |} , p ≥ {| | | |} (cid:18)2max{|α|,|β|}(cid:19) ! (cid:13) (cid:13) with equal(cid:13)ity precisely(cid:13)when αβ =0 or p=2. Proof. The result is trivial if either α or β is zero, so we assume that this is not the case. We maythen assumeby homogeneitythatα=1, anditis clearfromthe definitionof f that we may assume that L=0,and thatβ is realand negative. k kp If β <1, then taking r = β p in (2.3) and keeping just the first two terms of the | | | | sum, we obtain 1+βzM p = 1+βz p 1+ p2|β|2. p k kp ≥ 4 If β >1, then (cid:13) (cid:13) | | (cid:13) (cid:13) 1+βzM p = β p β−1+z p = β p 1+z/β p, p | | p | | k kp so taking r= β(cid:13)−p, we ob(cid:13)tain in th(cid:13)e same w(cid:13)ay (cid:13) (cid:13) (cid:13) (cid:13) | | 1+βzM p β p 1+ p2 . p ≥| | 4 β 2! | | (cid:13) (cid:13) The case β = 1 follow(cid:13)s by conti(cid:13)nuity. For the case of equality, notice that the sum (2.3) has p−recisely two nonzero terms only when p=2. (cid:3) The next lower bound is obtained by establishing the convexity of the function r (r) for each fixed p in (0,2]. p →I Lemma 6. If 0<p 2, then the function r (r) satisfies the inequality p ≤ →I (1)(1+r) p (2.6) (r) I p I ≥ 2 for r >0. Proof. If p = 2 then (r) = 1+r and the result is trivial. Suppose then that 2 I 0 < p < 2. If r < 1, then we may differentiate the power series (2.3) termwise to obtain 2 p/2 2m ′(r)= r(2m/p)−1. Ip m p m≥0(cid:18) (cid:19) X As 0<p<2, it follows that r ′(r) is strictly increasing on (0,1), so r (r) →Ip →Ip is strictly convex on this interval. Thus, if r and s are in (0,1), then (2.7) (r) (s)+(r s) ′(s). Ip ≥Ip − Ip Letting s 1 and using Lemma 4, we obtain → − (1)(1+r) (2.8) ′(r) (1)+ ′(1)(r 1)= Ip . Ip ≥Ip Ip − 2 for 0<r <1. GENERALIZATIONS OF GONÇALVES’ INEQUALITY 7 In a similar manner, if r >1 we differentiate (2.5) termwise to obtain 2 p/2 2m ′(r)= 1 r−2m/p, Ip m − p m≥0(cid:18) (cid:19) (cid:18) (cid:19) X andagainr ′(r) is strictly increasingon(1, ), so r (r) is strictly convex →Ip ∞ →Ip on this interval. Thus (2.7) holds as well for r > 1 and s > 1, and letting s 1+ → we obtain (2.8) for r>1. We havethereforeverified(2.6)ateachpointr in(0,1) (1, ), anditistrivial at r =1. ∪ ∞ (cid:3) Usingthislemma,weobtainasecondlowerboundontheL normofabinomial. p Corollary 7. Let 0 L<M be integers and let α and β be complex numbers. If ≤ 0<p 2 then ≤ αzL+βzM B (αp+ β p)1/p. p ≥ p | | | | Proof. Theresultistri(cid:13)vialifeither(cid:13)αorβ iszero,soweassumethatthis isnotthe (cid:13) (cid:13) case. By homogeneity, we may assume then that α =1, and we may assume that | | L = 0 and that β is real and negative by the definition of f . Using Lemma 6, k kp we obtain 1+βzM p = 1+βz p p k kp (cid:13) (cid:13) = p(β p) (cid:13) (cid:13) I | | (1)(1+ β p) p I | | ≥ 2 =Bp(1+ β p). p | | (cid:3) 3. Proofs of the theorems The proof of Theorem 2 employs an averaging argument and makes use of the triangle inequality for L norms. We therefore require the restriction p 1 in the p ≥ statement of the theorem. Proof of Theorem 2. Suppose that F(z) = N a zn is a polynomial with com- n=0 n plex coefficients, L and M are as in the statement of the theorem, and 1 p 2. Set K =M L, and let ζ denote a primitPive Kth root of unity in C. Th≤en ≤ K − K N K 1 1 ζ−kLF ζkz = ζk(n−L) a zn K K K K K n ! k=1 n=0 k=1 X (cid:0) (cid:1) X X = a zn n 0≤n≤N X n≡L(modK) =a zL+a zM. L M Using the triangle inequality and the fact that the polynomials ζ−kLF(ζkz) all K K have the same L norm, we find that p K 1 (3.1) F ζ−kLF ζkz = a zL+a zM k kp ≥(cid:13)K K K (cid:13) L M p (cid:13)(cid:13)(cid:13) kX=1 (cid:0) (cid:1)(cid:13)(cid:13)(cid:13)p (cid:13)(cid:13) (cid:13)(cid:13) (cid:13) (cid:13) 8 PETERBORWEIN,MICHAELJ.MOSSINGHOFF,ANDJEFFREYD.VAALER for 1 p . The inequality (1.9) then follows by selecting a complex number ≤ ≤ ∞ z of unit modulus so that a zL and a zM have the same argument. Then in- L M equalities(1.10)and(1.11)areestablishedbycombining(3.1)withCorollary5and Corollary 7, respectively. (cid:3) TheproofofTheorem1proceedsbyapplyingTheorem2toapolynomialhaving the same values over the unit circle as the given polynomial F. Ostrowski [6] and Mignotte [5] (see also [4, p. 80]) employ a similar construction in their proofs of Gonçalves’ inequality (1.5) in the case p=2. Proof of Theorem 1. Suppose that F(z) = N a zn = a N (z α ) is a n=0 n N n=1 − n polynomial with complex coefficients. If F(z) has a root at z = 0, then (1.7) P Q and (1.8) follow immediately from (1.1), so we assume that a = 0. Let denote 0 6 E the collectionof all subsets of 1,2,...,N , and for each E in , let E′ denote the { } E complementofE in 1,2,...,N . ForeachsetE in ,wedefinethefiniteBlaschke { } E product B (z) by E 1 α z n B (z)= − E z α n n∈E − Y and the polynomial G (z) by E N G (z)=B (z)F(z)= b (E)zn. E E n n=0 X Clearly, b (E)=a ( α ) 0 N m − m∈E′ Y and b (E)=a ( α ). N N n − n∈E Y If z =1 then the Blaschke product satisfies B (z) =1, so G (z) = F for | | | E | k E kp k kp 0 p and every E in . Now select L = 0 and M = N for the polynomial ≤ ≤ ∞ E G (z) in Theorem 2. Then from (1.10) we obtain E 1/p (3.2) F B a α p+ α p k kp ≥ p| N| m∈E′| m| n∈E| n| ! Y Y for1 p 2. Also,assuming withoutloss ofgeneralitythat b (E) b (E), we 0 N ≤ ≤ | |≥| | find from (1.11) that 1/p p2 b (E)2 N F b (E) 1+ | | k kp ≥| 0 | 4 b0(E)2 ! (3.3) | | 1/p p2 a a 2 0 N = b (E) 1+ | | | 0 | 4 b0(E)4! | | for p 1. Inequalities (1.7) and (1.8) then follow from (3.2) and (3.3) by choosing E = ≥n: α 1 . (cid:3) n { | |≤ } GENERALIZATIONS OF GONÇALVES’ INEQUALITY 9 WeremarkthatthechoiceofE intheprecedingproofproducesthebestpossible inequality in (3.2). To establish this, suppose that E has b (E) = F /r ∈ E | 0 | k k0 for some real number r, so b (E) = r a a / F . Then certainly 1 r | N | | 0 N| k k0 ≤ ≤ F 2/ a a , and it is easy to check that k k0 | 0 N| F r a a a a k k0 + | 0 N| F + | 0 N| r F ≤k k0 F k k0 k k0 in this range, with equality occurring only at the endpoints. It is possible, however, that a different choice for E in (3.3) could produce a bound better than (1.8) for a particular polynomial. Specifically, if E has ∈ E b (E) = F /r, again with 1 r F 2/ a a , then we obtain an improved | 0 | k k0 ≤ ≤ k k0 | 0 N| bound whenever 4 F 4+p2 a a 2 rp <4 F 4+p2 a a 2r4, k k0 | 0 N| k k0 | 0 N| and this may oc(cid:16)cur when p is small.(cid:17)For example, the polynomial F(z) = 18z2 − 101z+90 has roots α = 9/2 and α = 10/9; choosing E = with p = 1 yields 1 2 {} F 90.9, but selecting E = 1 (so r = 9/2) produces a lower bound slightly k k1 ≥ { } larger than 102. References [1] J. V. Gonçalves, L’inégalité de W. Specht, Univ. Lisboa Revista Fac. Ci. A (2) 1 (1950), 167–171. MR12,605j [2] G.H.Hardy,J.E.Littlewood,andG.Po´lya,Inequalities,CambridgeUniv.Press,Cambridge, 1988. [3] Y.Katznelson,Anintroductiontoharmonicanalysis,3rded.,CambridgeUniv.Press,Cam- bridge,2004. [4] M.MignotteandD.S¸tefa˘nescu,Polynomials:analgorithmicapproach,Springer-Verlag,Sin- gapore,1999. [5] M.Mignotte,Aninequalityaboutfactorsofpolynomials,Math.Comp.28(1974),1153–1157. MR50#7102 [6] A. M. Ostrowski, On an inequality of J. Vicente Gon¸calves, Univ. Lisboa Revista Fac. Ci. A(2)8(1960), 115–119. MR26#2585 DepartmentofMathematicsandStatistics,SimonFraser University,Burnaby,B.C. V5A1S6Canada E-mail address: [email protected] DepartmentofMathematics,DavidsonCollege,Davidson,NorthCarolina28035USA E-mail address: [email protected] Departmentof Mathematics,University ofTexas, Austin, Texas78712USA E-mail address: [email protected]