LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE
Peter Borwein and Tamas´ Erdelyi´
Abstract. The results of this paper show that many types of polynomials cannot be small on subarcs of the unit circle in the complex plane. A typical result of the paper is the following. Let Fn denote the set of polynomials of degree at most n with coefficients from {−1, 0, 1}. There are absolute constants c1 > 0, c2 > 0, and c3 > 0 such that
exp (−c1/a) ≤ inf kpkL1(A) , inf kpkA ≤ exp (−c2/a) 06=p∈Fn 06=p∈Fn
for every subarc A of the unit circle ∂D := {z ∈ C : |z| = 1} with length 0 The lower bound results extend to the class of f of the form n j f(z)= X aj z , aj ∈ C , |aj |≤ M, |am| = 1 j=m with varying nonnegative integers m ≤ n. It is shown that functions f of the above form cannot be arbitrarily small uniformly on subarcs of the circle. However, this does not extend to sets of positive measure. It shown that it is possible to find a polynomial of the above form that is arbitrarily small on as much of the boundary (in the sense of linear Lebesgue measure) as one likes. An easy to formulate corollary of the results of this paper is the following. Corollary. Let A be a subarc of the unit circle with length ℓ(A) = a. If (pk) is a sequence of monic polynomials that tends to 0 in L1(A), then the sequence H(pk) of heights tends to ∞. The results of this paper are dealing with (extensions of) classes much studied by Littlewood and many others in regards to the various conjectures of Littlewood concerning growth and flatness of unimodular polynomials on the unit circle ∂D. Hence the title of the paper. 1991 Mathematics Subject Classification. 11J54, 11B83. Key words and phrases. Transfinite diameter; integers; diophantine approximation; Cheby- shev; polynomial; restricted coefficients; −1, 0, 1 coefficients; 0, 1 coefficients. Research is supported, in part, by the National Science Foundation of the USA under Grant No. DMS–9623156 and conducted while an International Postdoctoral Fellow of the Danish Research Council at University of Copenhagen (T. Erd´elyi). Research is supported, in part, by NSERC of Canada (P. Borwein). Typeset by AMS-TEX 1 2 PETERBORWEINANDTAMAS´ ERDELYI´ 1. Introduction Littlewood’s well-known and now resolved conjecture of around 1948 concerns polynomials of the form n kj p(z) := aj z , j=1 X where the coefficients aj are complex numbers of modulus at least 1 and the ex- ponents kj are distinct non-negative integers. It states that such polynomials have L1 norms on the unit circle ∂D := z C : z =1 { ∈ | | } that grow at least like c log n with an absolute constant c> 0. This was proved by Konjagin [Ko-81] and independently by McGehee, Pigno, and Smith [MPS-81]. Pichorides, who contributed essentially to the proof of the Littlewood conjecture, observed in [Pi-83] that the original Littlewood conjecture (when all the coefficients are from 0, 1 would follow from a result on the L1 norm of such polynomials on sets E {∂D of} measure π. Namely if ⊂ n zkj dz c E | |≥ Z j=0 X for any subset E ∂D of measure π with an absolute constant c > 0, then the original Littlewood⊂ conjecture holds. Throughout the paper the measure of a set E ∂D is the linear Lebesgue measure of the set ⊂ t [ π, π) : eit E . { ∈ − ∈ } Konjagin [Ko-96] gives a lovely probabilistic proof that this hypothesis fails. He does however conjecture the following: for any fixed set E ∂D of positive measure there exists a constant c = c(E) > 0 depending only on E⊂such that n zkj dz c(E) . E | |≥ Z j=0 X In other words the sets E ∂D of measure π in his example where ǫ ⊂ n zkj dz <ǫ E | | Z ǫ j=0 X must vary with ǫ> 0. We show, among other things, that Konjagin’s conjecture holds on subarcs of the unit circle ∂D. Additional material on Littlewood’s conjecture and related problems concerning the growth of polynomials with unimodular coefficients in various norms on the LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 3 unit disk is to be found, for example, in [Bou-86], [Be-95], [Ka-85], [Li-86], [Ma-63], [Ne-90], [Od-93], and [So-95]. All the results of this paper concern how small polynomials of the above and related forms can be in the Lp norms on subarcs of the unit disk. For 1 p the results are sharp, at least up to a constant in the exponent. ≤ ≤ ∞ An interesting related result is due to Nazarov [Na-93]. One of its simpler ver- sions states that there is an absolute constant c> 0 such that cm(I) n max p(z) max p(z) z I | |≤ m(A) z A | | ∈ ∈ n for every polynomial p of the form p(z)= a zkj with k N and a C and j=0 j j ∈ j ∈ for every A I, where I is a subarc of ∂D with length m(I) and A is measurable P with Lebesgue⊂ measure m(A). This extends a result of Tur´an [Tu-84] called Tur´an’s Lemma, where I = ∂D and A is a subarc. 2. Notation µ For M > 0 and µ 0, let M denote the collection of all analytic functions f on the open unit disk≥D := zS C : z < 1 that satisfy { ∈ | | } M f(z) , z D . | |≤ (1 z )µ ∈ − | | We define the following subsets of 1. Let S1 n := f : f(x)= a xj , a 1, 0, 1 Fn j j ∈ {− } j=0 X and denote the set of all polynomials with coefficients from the set 1, 0, 1 by {− } ∞ := . F Fn n=0 [ More generally we define the following classes of polynomials. For M > 0 and µ 0 let ≥ n µ := f : f(x)= a xj , a C , a Mjµ , a =1 , n N . KM j j ∈ | j |≤ | 0| ∈ j=0 X On occasion we let := 1, := 1 , and := 0 . S S1 SM SM KM KM We also employ the following standard notations. We denote by n the set of P c all polynomials of degree at most n with real coefficients. We denote by n the set of all polynomials of degree at most n with complex coefficients. The heightP of a polynomial n p (z) := a zj , a C , a =0 , n j j ∈ n 6 j=0 X 4 PETERBORWEINANDTAMAS´ ERDELYI´ is defined by a H(p ) := max | j | : j =0, 1,... ,n . n a | n| Also, p A := sup p(z) k k z A | | ∈ and 1/q p := p(z) q dz k kLq(A) | | | | ZA are used throughout this paper for measurable functions (in this paper usually polynomials) p defined on a measurable subset of the unit circle or the real line, and for q (0, ). ∈ ∞ 3. New Results The first two results concern lower bounds on subarcs in the supremum norm. Theorem 3.1. Let 0 c (1 + log M) f exp − 1 k kA ≥ a for every f (:= 1 ) that is continuous on the closed unit disk and satisfies ∈ SM SM f(z ) 1 for every z C with z = 1 . | 0 |≥ 2 0 ∈ | 0| 4M Corollary 3.2. Let 0 c (1 + log M) f exp − 1 k kA ≥ a for every f (:= 1 ). ∈ KM KM The next two results show that the previous results are, up to constants, sharp. Theorem 3.3. Let 0 c1 inf f A exp − 0=f k k ≤ a 6 ∈F whenever ℓ(A)= a c . ≤ 2 LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 5 Theorem 3.4. Let 0 The next two results extend the first two results to the L1 norm (and hence to all L norms with p 1). p ≥ Theorem 3.5. Let 0 Corollary 3.7. Let A be a subarc of the unit circle with length ℓ(A)= a. If (pk) is a sequence of monic polynomials that tends to 0 in L1(A), then the sequence H(pk) of heights tends to . ∞ The final result shows that the theory does not extend to arbitrary sets of positive measure. Theorem 3.8. For every ǫ > 0 there is a polynomial p 1 such that p(z) < ǫ everywhere on the unit circle except possibly in a set of line∈ Kar measure at| most| ǫ. The above results should be compared with earlier result of the authors [Bor- 96] on approximation on the interval [0, 1]. These state that there are absolute constants c1 > 0 and c2 > 0 such that exp c1√n inf p [0,1] exp c2√n − ≤ 0=p n k k ≤ − 6 ∈F for every n 2. ≥ 6 PETERBORWEINANDTAMAS´ ERDELYI´ 4. Lemmas Lemma 4.1. Let 0 Lemma 4.2. Let a> 0 and E := z C : 0 Im(z) a . a { ∈ ≤ ≤ } Suppose g is an analytic function in the interior of Ea, and suppose g is continuous on E . Then a ∪ ∞ 1/2 1/2 max g(z) max g(z) max g(z) . z: Im(z)=a/2 | |≤ z: Im(z)=0 | | z: Im(z)=a | | { } { } { } The next lemma, that plays a crucial role in the proof of Theorem 3.1, can be easily derived from Lemma 4.2. Lemma 4.3. Let 0 To prove Theorem 3.3, we need some corollaries of the Hadamard Three Circles Theorem. Suppose f is regular inside and on z C : r z r . { ∈ 1 ≤ | |≤ 2} For r [r1, r2], let ∈ M(r) := max f(z) . z =r | | | | Then M(r)log(r2/r1) M(r )log(r2/r)M(r )log(r/r1) . ≤ 1 2 LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 7 Corollary 4.4. Let a (0, 1/8]. Suppose f is regular inside and on the ellipse Ea with foci at 1 8a and∈1 and with major axis − [(1 4a) 17a , (1 4a)+17a] . − − − Let E be the ellipse with foci at 1 a and 1 and with major axis a − [(1 4a) 10a , (1 4a)+10a] . e − − − Then 1/2 1/2 max f(z) max f(z) max f(z) . e z E | |≤ z [1 8a,1] | | z Ea | | ∈ a ∈ − ∈ Corollary 4.5. We have 1/2 max f(z) ((n + 1)exp(13na))1/2 max f(z) e z E | |≤ z [1 8a,1] | | ∈ a ∈ − for every f and a (0, 1/8]. ∈Fn ∈ In one of our proofs of Theorem 3.3 we will need the upper bound of the result below proved in [Bor-96]. An application of this lemma makes the proof of Theorem 3.3 shorter in the special case when the subarc A of the unit circle is symmetric with respect to the real line and contains 1. Our second proof of Theorem 3.3 is longer. This self-contained second proof does not use the lemma below and eliminates the extra assumption on the subarc A. Lemma 4.6. There are absolute constants c4 > 0 and c5 > 0 such that exp c4√n inf p [0,1] exp c5√n − ≤ 0=p n k k ≤ − 6 ∈F for every n 2. ≥ To prove Theorem 3.4 we need two lemmas. Lemma 4.7. Suppose n p(x)= a xj , a 9 , a C , j | j |≤ j ∈ j=0 X n k − p(x) = (x 1)kq(x) , q(x)= b xj , b C . − j j ∈ j=0 X Then n k − en k 1 en k q b 9(n + 1)e − 9(n + 1) k k∂D ≤ | j|≤ k ≤ k j=0 X (where ∂D denotes the unit circle). As a consequence, if A denotes the subarc of the unit circle that is symmetric to the real line, contains 1, and has length 2k/(9n), then e k p 9(n + 1) . k kA ≤ 9 To prove Theorem 3.4 our main tool is the next lemma due to Hal´asz [Tu-84]. 8 PETERBORWEINANDTAMAS´ ERDELYI´ Lemma 4.8. For every k N, there exists a polynomial h such that ∈ ∈Pk 2 h(0) = 1 , h(1) = 0 , h(z) < exp for z 1 . | | k | |≤ Lemma 4.9. Let 0 Lemma 4.10. Let w = w C and let z := 1 (w + w ). Assume that J is an 1 6 2 ∈ 0 2 1 2 1 arc that connects w1 and w2. Let J2 be the arc that is the symmetric image of J1 with respect to the z0. Let J := J1 J2 be positively oriented. Suppose that g is an analytic function inside and on J.∪ Suppose that the region inside J contains the disk centered at z with radius δ > 0. Let g(z) K for z J . Then 0 | |≤ ∈ 2 2 1 g(z ) (πδ)− K g(z) dz . | 0 | ≤ | | | | ZJ1 5. Proofs of Theorems 3.1 – 3.6 Proof of Lemma 4.1. Suppose g 4M is continuous on the closed unit disk and g 1 1 . Let 2m 4 be the∈ smallest S even integer not less than 4π/a. Let 4M ≥ 4 ≥ 2πi ξ := exp 2m be the first (2m)th root of unity. We define 2m equally spaced points on ΓM by η := 1 + 1 1 ξk , k =0, 1,..., 2m 1 . k 4M − 4M − Then there is an absolute constant c5 > 0 such that 1 2 1 z c M − (ka) , k =1, 2,...m 1 , − | |≥ 5 − whenever z is on the smaller subarc of the circle ΓM with endpoints ηk and ηk+1 or with endpoints η2m k and η2m k 1, respectively. We define the function − − − 2m 1 − h(z) := g 1 + 1 1 ξj z 1 . 4M − 4M − 4M j=0 Y LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 9 Since g , we obtain ∈ S4M m 1 2 4m 4 4m 4 − 1 2 − M − 2 1 2 − 2 max h(z) g J 4M c4M − (ka) 4 g J z ΓM | |≤k k ≤ c5a ((m 1)!) k k ∈ k=1 − Y4m 4 4m 4 4m 4 m − Me − eM − g 2 e4 g 2 ≤ πc m 1 k kJ ≤ πc k kJ 5 − 5 c (1 + log M) exp 6 g 2 ≤ a k kJ with an absolute constant c6 > 0. Now the Maximum Principle yields that 1 2m 1 c6(1 + log M) 2 g 4M = h 4M max h(z) exp g J . ≤ z ΓM | |≤ a k k ∈ Since 2m 2+4π/a and g 1 1 , we obtain ≤ 4M ≥ 4 2m c (1 + log M) 2m c (1 + log M) 1 g 2 exp − 6 g 1 exp − 6 k kJ ≥ a 4M ≥ a 4 c (1 + log M) exp − 7 ≥ a with an absolute constant c7 > 0. This finishes the proof. Proof of Lemma 4.2. Let z0 be a complex number with Im(z)= a/2. We want to show that 2 g(z0) max g(z) max g(z) . | | ≤ z: Im(z)=0 | | z: Im(z)=a | | { } { } Without loss of generality we may assume that Re(z0) = 0; the general case follows by a linear transformation. So let z0 := ia/2. Applying the Maximum Principle to h(z) := g(z)g(ia z) on E , we obtain − a ∪ ∞ g(ia/2) 2 = g(ia/2)g(ia ia/2) = h(ia/2) max h(z) | | | − | | |≤ z ∂Ea | | ∈ = max g(z)g(ia z) max g(z) max g(z) . z ∂Ea | − |≤ z: Im(z)=0 | | z: Im(z)=a | | ∈ { } { } and the proof is finished. Proof of Lemma 4.3. This follows from Lemma 4.2 by the substitution α z w := log − . z β − Proof of Theorem 3.1. Without loss of generality we may assume that the arc A is the subarc of the unit circle with length ℓ(A)= a < π which is symmetric with respect to the real line and contains 1. Suppose f is continuous on the ∈ SM 10 PETERBORWEINANDTAMAS´ ERDELYI´ 1 1 open unit disk, and suppose f 4M 2 . Using the notation of Lemma 4.3, let g(z) := (z α)(z β)f(z). Then a straightforward≥ geometric argument yields that − − M (z α)(z β) 2M g(z) | − − | , z I | |≤ 1 z ≤ sin(a/2) ∈ 0 − | | (note that I is the line segment between α and β). Hence, with L := g (note 0 k kA that A = Ia), we conclude by Lemma 4.3 that 2ML 1/2 max g(z) . z Ia/2 | |≤ sin(a/2) ∈ Denote by Ga the open region bounded by Ia/2 and Ia. By the Maximum Principle 2L 1/2 max g(z) max L, . z Ga | |≤ sin(a/2) ∈ ( ) It is a simple elementary geometry to show that the arc K := Γ G has M ∩ a length at least a/2. Here, as in Lemma 4.1, ΓM denotes the circle with diame- ter 1+ 1 , 1 . Observe that f implies g . Also, since M 1, − 2M ∈ SM ∈ S4M ≥ 2 2 g 1 3 f 1 3 1 1 . 4M ≥ 4 4M ≥ 4 2 ≥ 4 Hence Lemma 4.1 can be applied with g . We conclude that ∈ S4M 2ML 1/2 c (1 + log M) max L , g exp − 4 , sin(a/2) ≥k kK ≥ 2a ( ) from which 1 L c (1 + log M) f g = exp − 1 k kA ≥ 4 k kA 4 ≥ a with an absolute constant c1 > 0. Proof of Corollary 3.2. Note that if f M , then f M and f is continuous on the closed unit disk. Also, if z = 1 ∈, then K ∈ S | 0| 4M z 2M 1 f(z 1 M | 0| 1 = . | 0|≥ − 1 z ≥ − 4M 2 − | 0| So the assumptions of Theorem 3.1 are satisfied and the corollary follows from Theorem 3.1. Proof of Corollary 4.4. This follows from the Hadamard Three Circles Theorem with the substitution z + z 1 w = (1 4a)+4a − . − 2 LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 11 The Hadamard Three Circles Theorem is applied with r1 := 1, r := 2, and r2 := 4. Proof of Corollary 4.5. This follows from Corollary 4.4 and the Maximum Princi- ple. We present two proofs of Theorem 3.3. The first one is under the additional assumption that the subarc A is symmetric with respect to the real line and contains 1. Proof of Theorem 3.3 in the above special case. By Lemma 4.6, for every integer n 2, there is a Q such that ≥ n ∈Fn Q exp c √n . k nk[0,1] ≤ − 5 Let n be chosen so that n := N , where N is defined by ⌊ ⌋ c a = 5 . 26√N Then, by Corollary 4.5, 1/2 1/2 1/2 max Qn(z) (n + 1) exp (c5/2)√n max f(z) e z E | |≤ z [1 8a,1] | | ∈ a ∈ − 1/2 1/2 (n + 1)1/2 exp (c /2)√n exp c √n ≤ 5 − 5 (n + 1)1/2 exp (c /4)√n ≤ − 5 c exp − 6 ≤ a whenever a c , where c > 0 and c > 0 are absolute constants. Now observe that ≤ 7 6 7 the unit circle intersects the ellipse Ea in an arc of length at least c8a, where c8 > 0 is an absolute constant. Therefore the Maximum Principle finishes the proof. e Our second proof of Theorem 3.3 is in the general case. In addition, it is self- contained. it Proof of Theorem 3.3. Let A := e : t [t1,t2] , where 0 t1 < t2 2π and { ∈ it0 } ≤ ≤ t1 t2 = a. Let t0 := (t1 + t2)/2, and w := e . We prove the following extension of− Lemma 4.6. For every sufficiently large n N, there is a Q such that ∈ n ∈Fn Q exp c √n . k nk[0,w] ≤ − 5 To see this we define k := 1 √n . Let ⌊ 2 ⌋ 1 k/(2n) =: y We use a counting argument to find a polynomial f n 1 with the property ∈F − 1 √n (5.1) f(y w) 2 − , j =0, 1,...,k, | j |≤ for sufficiently large n. Indeed, we can divide the 2(k + 1)–dimensional real cube Q := (z ,... ,z ) Ck : Re(z ), Im(z ) [ n 1,n + 1) { 0 k ∈ j j ∈ − − } into (2m(n + 1))2(k+1) subcubes by defining Qi0,i1,... ,i2k+1 i i +1 i i +1 := (z ,... ,z ) Ck : Re(z ) 2j , 2j , Im(z ) 2j+1 , 2j+1 , 0 k ∈ j ∈ m m j ∈ m m where (i0,i1,... ,i2k+1) are 2(k + 1)–tuples of integers with (n + 1)m ij (n + 1)m 1 for each j =0, 1,... , 2k +1. Let − ≤ ≤ − n 1 − j n 1 := f : f(x)= ajx , aj 0, 1 A − ∈{ } j=0 X denote the set of polynomials of degree at most n 1 with coefficients from 0, 1 . − { } Note that if P n 1, then ∈ A − M(P ) := (Re(P (y w)), Im(P (y w)),... , Re(P (y w)), Im(P (y w))) Q . 0 0 k k ∈ n Also, there are exactly 2 elements of n 1. Therefore, if A − (2m(n + 1))2(k+1) < 2n holds, then there exist two different P1 n 1 and P2 n 1, and a subcube ∈ A − ∈ A − Qi0,i1,... ,i2k+1 such that both M(P1) = (Re(P1(y0w)), Im(P1(y0w)),... , Re(P1(ykw)), Im(P1(ykw))) and M(P2) = (Re(P2(y0w)), Im(P2(y0w)),... , Re(P2(ykw)), Im(P2(ykw))) are in Qi0,i1,... ,i2 +1 , and hence for 0 = f := P1 P2 n 1, we have k 6 − ∈F − 1 f(y w) √2 m− , j =0, 1,...,k. | j |≤ √n 1 Now choose m := 2− . This, together with k := √n , yields that the in- ⌊ ⌋ ⌊ 2 ⌋ equality (2mn)2(k+1) < 2n holds provided n is sufficiently large. This proves (5.1). In the rest of the proof let n be sufficiently large to satisfy (5.1). Associated with f satisfying (5.1), we define ∈Fn u(x) := Re(f(xw)) , x R , ∈ LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 13 and v(x) := Im(f(xw)) , x R . ∈ Obviously u(k+1) nk+2 and v(k+1) nk+2 . k k[0,1] ≤ k k[0,1] ≤ Let y [y0, 1] be an arbitrary point different from each yj . By a well-known formula for∈ divided differences, k u(y) u(y ) 1 + i = u(k+1)(ξ) k k (k + 1)! j=0 (y yj ) i=0 (yi y) j=0,j=i (yi yj) − X − 6 − for someQξ [y , 1]. Q ∈ 0 Combining (5.1) and the two observations above, we obtain 1 k u(y) u(k+1)(ξ) (y y ) | |≤ (k + 1)! | | − j j=0 Y k k (y yj) + u(y ) j=0 − | i | k i=0 (yi yQ) j=0,j=i (yi yj ) X − 6 − k 1 k+2 (k + 1)! 1 √n k! Q n +2 − ≤ (k + 1)! (2n)k+1 i!(k i)! i=0 X − (k+1) 1 √n k (1/2)√n 1 √n (1/2)√n 2− n +2 − 2 2− n +2 − 2 ≤ ≤ exp c √n ≤ − 4 with an absolute constant c4 > 0. Similarly v(y) exp c √n . | |≤ − 4 Hence f(yw) √2exp c √n | |≤ − 4 with an absolute constant c > 0. Since y [y , 1] is arbitrary, we have proved that 4 ∈ 0 f √2exp c √n , k k[y0w,w] ≤ − 4 1 1/2 where y =1 k/(2n) 1 n− for sufficiently large n. 0 − ≤ − 4 We conclude that the polynomial g(x) := xnf(x) satisfies g and ∈F2n g exp c √n k k[0,w] ≤ − 5 with an absolute constant c5 > 0. Now the proof can be finished by a trivial modification of the proof given in the special case when subarc A of the unit circle is symmetric with respect to the real line and contains 1. 14 PETERBORWEINANDTAMAS´ ERDELYI´ Proof Lemma 4.7. We have j 1 d k bj = (p(x)(x 1)− ) | | j! dxj − x=0 j 1 j k (k + m 1)! (j m) = ( 1) − p − (0) j! m − (k 1)! m=0 X − j j (k + m 1)! 1 (j m) (k + m 1)! = − p − (0) = − aj m (k 1)!m! (j m)! (k 1)!m! − m=0 m=0 X − − X − j k + m 1 k + j e(k + j) k = − aj m 9 9 m − ≤ k ≤ k m=0 X en k 9 ≤ k which proves the lemma. Proof of Theorem 3.4. Without loss of generality we may assume that A is the subarc of the unit circle with length ℓ(A) = a which is symmetric with respect to the real line and contains 1. Let k := 2 +1. Let h be the polynomial with 9a ∈Pk the properties of Lemma 4.8. Let u := k2. Let u k j Qu(x) := h (x) =: bj x . j=0 X Since 2 h(z) < exp , z 1 , | | k | |≤ u 4k b 2 = Q 2 exp = e4 . | j | k ukL2(∂D) ≤ k j=0 X So b =1 , b e2 9 , j =0, 1,...,u. | 0| | j |≤ ≤ Now let A denote the subarc of the unit circle which is symmetric to the real line, 1 contains 1, and has length a 0, 9 . Since 2k/(9u)=2/(9k) a, Lemma 4.7 implies that ∈ ≥ c Q exp( c k) exp − 5 k ukA ≤ − 4 ≤ a with some absolute constants c4 > 0 and c5 > 0. Now let M 1. Without loss of generality we may assume that M = 9m with a nonnegative≥ integer m. When m = 0 the theorem follows from Theorem 3.3. So let m 1. Let n := um and let ≥ n m km j Pn(x) := Qu (x) := h (x) =: aj x . j=0 X Since 2 h(z) < exp , z 1 , | | k | |≤ LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 15 n 4km a 2 = P 2 exp = e4m . | j | k nkL2(∂D) ≤ k j=0 X So a =1 , a e2m M , j =0, 1,...,n. | 0| | j |≤ ≤ Also, using m 1, we obtain ≥ c m c (1 + log M P = Q m = exp − 5 exp − 2 k nkA k ukA a ≤ a with an absolute constant c2 > 0. This finishes the proof. Proof of Lemma 4.9. The proof is the same as that of Lemma 4.1 with trivial modifications. Proof of Lemma 4.10. Applying Cauchy’s integral formula with G(z) := g(z + (z z ))g(z (z z)) 0 − 0 0 − 0 − on J, we obtain 1 G(z) dz g(z ) 2 = G(z ) = | 0 | | 0 | 2πi z z ZJ − 0 2 G (z) dz 1 G(z) dz = | | | | 2π z z ≤ π z z ZJ1 − 0 ZJ1 | − 0| 1 g(z + (z z ))g(z (z z )) dz = | 0 − 0 0 − − 0 | | | π z z ZJ1 | − 0| 1 (πδ)− K g(z) dz . ≤ | | | | ZJ1 Proof of Theorem 3.5. Without loss of generality we may assume that the arc A is the subarc of the unit circle with length ℓ(A) = a<π/2 which is symmetric with respect to the real line and contains 1. Suppose f µ and ∈ SM 1 1 cos(a/4) 1 f − . 4M2µ − 2M2µ ≥ 2 Note that this is guaranteed by the assumption of the theorem since 1 1 cos(a/4) 1 0 − , ≤ 4M2µ − 2M2µ ≤ 4M2µ Let the region Ha,M,µ be defined by 1 cos(a/4) a a H := z = reiθ : cos(a/4) Let Γa,M,µ be the circle as in Lemma 4.9. It is a simple geometric argument to show that the arc I := Γa,M,µ Ha,M,µ has length greater than c5a with an absolute constant c > 0. Let z I ∩ H be fixed. Then we can choose w A and 5 0 ∈ ⊂ a,M,µ 1 ∈ w A such that z = 1 (w + w ). Let J be the arc connecting w and w on the 2 ∈ 0 2 1 2 1 1 2 unit circle. Note that J1 is a subarc of A. Let J2 be the arc which is the symmetric image of J1 with respect to the line segment connecting w1 and w2. Let g(z) := ((z w )(z w ))µf(z) . − 1 − 2 Then it is elementary geometry again to show that M (z w )(z w ) µ M2µ g(z) | − 1 − 2 | , z J . | |≤ (1 z )µ ≤ sinµ(a/2) ∈ 2 − | | By Lemma 4.10 we obtain 1 π(1 cos(a/4)) − M2µ (5.2) g(z ) 2 − g(z) dz . | 0 | ≤ M2µ sinµ(a/2) | | | | ZJ1 µ µ Observe that f implies g µ . Also, since M 1, ∈ SM ∈ SM4 ≥ 1 1 cos(a/4) µ 2 2µ 1 1 cos(a/4) g − 1 2− − f − 4M2µ − 2M2µ ≥ − 4M2µ − 2M2µ 2µ 1 1 1 cos(a/4) 1 f − ≥ − 8µ 4M2µ − 2M2µ 2 7 1 1 . ≥ 8 2 ≥ 4 µ Hence Lemma 4.9 can be applied with g M4µ . We conclude that there is a point z I H such that ∈ S 0 ∈ ⊂ a,M,µ c (µ + log M) g(z ) exp − 4 . | 0 |≥ a Combining this with (5.2) and J A gives 1 ⊂ 1 µ 1 µ f g g k kL1(A) ≥ 4 k kL1(A) ≥ 4 k kL1(J1) 1 µ π(1 cos(a/4)) sinµ(a/2) − g(z ) 2 ≥ 4 M2µ M2µ | 0 | c (µ + log M) exp − 1 ≥ a with an absolute constant c1 > 0. µ µ Proof of Corollary 3.6. Let f M . Then f M(µ!)2 and f is continuous on the ∈ K 1 ∈ S closed unit disk. Also, if z 2 , then | 0|≤ 4M(µ!) 2µ j µ ∞ 1 M(µ!)2 ∞ j f(z 1 M(µ!)2 jµ 1 | 0|≥ − 4M(µ!)22µ ≥ − 4M(µ!)2 2j j=1 j=1 X X 1 ∞ j 2 1 1 1 . ≥ − 4 2j ≥ − 4 ≥ 2 j=1 X LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 17 So the assumptions of Theorem 3.5 are satisfied with M replaced by M(µ!)2, and the corollary follows from Theorem 3.5. Proof of Corollary 3.7. Let nk p (z)= a zj, a C , a =0 , k j,k j,k ∈ nk ,k 6 j=0 X and let Mk := H(pk). Applying Corollary 3.6 with 1 nk 1 1 qk(z) := an ,k − z pk(z− ) M (= ) | k | ∈ K k KMk 1 and the arc B := z− : z A of length a, we obtain the corollary. { ∈ } 6. Proof of Theorem 3.8 Lemma 6.1. For every r (0, 1/2) there exists a trigonometric polynomial ∈ n j p(z)= cj z j= n X− such that c0 = 1, cj < r and p(z) < r everywhere on the unit circle except possibly in a set of linear| | measure| at most| r. Proof. The finite Riesz product N m m p(z)= (1 + rz j + rz− j ) j=1 Y j with mj := 4 and sufficiently large N is such an example. For r (0, 1/2) and j ∈ mj = 4 the Riesz products tend to 0 almost everywhere on the unit circle as N . See, for example, [Zy-59, p 208]. → ∞ The next lemma follows simply from the fact that the transfinite diameter of any closed proper subset of the unit circle is less than 1. (We remark that due to this fact the polynomial guaranteed by Lemma 6.2 can be chosen so that its coefficients are integers. We will not need this extra property.) Lemma 6.2. For every η > 0 there exists a polynomial L k g(z)= bkz kX=0 such that b0 = 1 and g(z) < η everywhere on the unit circle except possibly on a set of linear measure at| most| η. 18 PETERBORWEINANDTAMAS´ ERDELYI´ Proof of Theorem 3.8. For η := ǫ/2 we choose a polynomial L k g(z)= bkz kX=0 with the properties of Lemma 6.2, that is, b0 = 1 and ǫ (6.1) g(z) < everywhere on the unit circle except | | 2 ǫ possibly in a set of linear measure at most . 2 For every k with b > 1 we choose a trigonometric polynomial | k| nk j pk(z)= cj,kz j= n X− k so that ǫ c =1, c < 0,k | j,k| 2L b | k| and ǫ (6.2) pk(z) < everywhere on the unit circle except | | 2L bk | | ǫ possibly in a set of linear measure at most . 2L This can be done by Lemma 6.2. Now let L A Ak h(z) := g z = bkz with A :=1+2 max nk . k: b >1 k=0 | k| X Finally we define L Ak f(z) := h(z) bkz pk(z) . − k=1 |bXk|>1 It is straightforward from the construction that f 1. Also, (6.1) and the defini- tion of h imply that ∈ K ǫ (6.1) h(z) < everywhere on the unit circle except | | 2 ǫ possibly in a set of linear measure at most . 2 Finally (6.2) and the definition of f imply that L ǫ ǫ ǫ ǫ f(z) < + bk + L = ǫ | | 2 | |2L bk ≤ 2 2L k=1 | | |bXk |>1 everywhere on the unit circle except possibly in a set of linear measure at most ǫ ǫ + L = ǫ . 2 2L This finishes the proof. 7. Acknowledgment. We thank Fedor Nazarov for observing and proving Theorem 3.8. The content of Section 6 is due to him. We also thank him for several discussions about problems related to the paper. LITTLEWOOD-TYPE PROBLEMS ON SUBARCS OF THE UNIT CIRCLE 19 References Be-95. J. Beck, Flat polynomials on the unit circle – note on a problem of Littlewood, Bull. London Math. Soc. 23 (1991), 269–277. Bou-86. J. Bourgain, Sul le minimum d’une somme de cosinus, Acta Arith. 45 (1986), 381–389. Bor-95. P. Borwein and T. 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