The Krzyż conjecture: an extremal problem for non-vanishing bounded analytic functions

The Krzyż conjecture: an extremal problem for non-vanishing bounded analytic functions Dragan Vukotić Universidad Autónoma de Madrid Joint work with María J. Martín, Eric T. Sawyer, and Ignacio Uriarte-Tuero CIRM, Luminy, Function Spaces and Harmonic Analysis 30th October, 2014

Notation D = {z C : z < 1}: unit disk. T = {z C : z = 1}: unit circle. H 2 : the space of all functions such that f (z) = n=0 a nz n in D and ( ) 1/2 f 2 = a n 2 <. n=0 H : f = sup z D f (z) <.

Linear Extremal Problems Let 1 p. Find sup Φ(a 0,..., a n ) : f H p 1, f (z) = a j z j, j=0 where Φ is a continuous functional. Landau, Carathéodory, Szegö: concrete problems of this type. Unified approach (dual extremal problems): Rogosinski- H. Shapiro and S.Ya. Khavinson.

A non-linear extremal problem Let B denote the class of all f analytic in D such that 0 < f (z) 1 for all z en D. Consider the extremal problem of determining the supremum: M n = sup { a n : f B }, n 1. Obviously, M n 1 for all n 1. Uniform bounds on M n smaller than one exist.

Some basic observations Use normal families to show that the supremum M n is attained for some function f (an extremal function). Clearly, if f (z) = j=0 a jz j is extremal and λ = µ = 1, then so is λf (µz) = λ j=0 µj a j z j (rotations of f ). For every n 1, the function f n (z) = e (zn 1)/(z n +1) = 1 e + 2 e zn 2 3e z3n +... shows that M n 2/e = 0.73575888....

The normalized problem By using rotations again, we can study instead the equivalent normalized problem of finding for n 1. M n = sup{re a n : f B, a 0 > 0}, The n-th coefficient of an extremal function for the above problem must actually satisfy the condition Re a n = a n. That is, a n > 0 actually must hold for such functions. (Otherwise use rotations to get a function in B for which Re a n > M n.)

The Krzyż conjecture Conjecture. (Krzyż, 1968) M n = 2/e for all n 1. For a fixed n, equality is attained only for the function f n as above: f n (z) = e (zn 1)/(z n +1) = 1 e + 2 e zn 2 3e z3n +... and its rotations. (With our normalization, this should be the unique extremal function.)

Some known uniform estimates Horowitz (1978): M n 1 1 3π + 4 π sin 1 12 = 0.99987..., n N. This was improved to 0.9991... in Ermers thesis (1990). Asymptotic bounds: Prokhorov, Romanova (2006). Being non-linear, the problem is harder than most linear ones.

Further observations No more than one coefficient of f can be 2/e since if a n 2/e and a m 2/e, m n, then 1 f 2 f 2 2 = ( ) 2 2 a n 2 2 = 8 e e 2 > 1. n=0 For the purported extremal function, we have a 0 = 1 e, a 1 = a 2 =... = a n 1 = 0, a n = 2a 0.

Partial results Case n = 1: relatively easy (known since 1934). n = 2 is already non-trivial: Krzyż, Reade, MacGregor (unpublished). n = 3: Hummel, Scheinberg, Zalcman (1977). n = 4: Tan (1983?), Brown (1987). n = 5: Samaris (2003). Lewandowski, Szynal, Peretz: various partial contributions.

The structure of extremal functions Hummel, Scheinberg, and Zalcman (1977) established the structure of extremal functions, indicating various proofs of this fact. This structure can be deduced from an earlier work of S.Ya. Khavinson. Other proofs: Kortram (1993), based on the study of extreme points of the unit ball in H. Marshall (2011, unpublished), based on the Carathéodory-Toeplitz theorem (from 1911).

All extremal functions have the form N j=1 f (z) = e r α j z 1 j α j z+1, 1 N n, r j > 0, α j = 1, j = 1, 2,..., N. There are many parameters to handle and it is not at all obvious that we must have a complete symmetry: N = n, r j = 1 n, α j = e 2πji/n. Note that the above f is actually f = e B 1 B+1, where B is a finite Blaschke product of degree N n. NEED TO SHOW: B(z) = z n.

Recurrence relations Lemma. If f and g are analytic in D and f = e g, f (z) = a j z j, g(z) = j=0 b j z j, j=0 then and k=0 a 0 = e b 0 n 1 ( a n = 1 k ) a k b n k, n 1. n

More precisely, (b1 2 a 1 = a 0 b 1, a 2 = a 0 2 + b 2 ) a 3 = a 0 ( b 3 1 6 + b 1b 2 + b 3 a 4 = a 0 ( b 4 1 24 + b2 1 b 2 2 + b 1b 3 + b2 2 2 + b 4, ), ),...

In general, where, for each n 1, P n (b 1, b 2,..., b n ) = a n = a 0 P n (b 1, b 2,..., b n ), 1 m n, m j=1 i j (n)=n c i1 (n),i 2 (n),...,i m(n)b i1 (n)b i2 (n)... b im(n), where all i j (n) N and c i1 (n),i 2 (n),...,i m(n) > 0. Whenever 1 m n and m j=1 i j(n) = n, the term containing the product b i1 (n)b i2 (n)... b im(n) actually appears in the expression for P n with a non-zero coefficient in front of it.

Some relevant observations If f (z) = j=0 a jz j is extremal for the Krzyż problem (with a 0 > 0) then we know that a n > 0. Also, one can use a variational method to show that P(λ) = a n + 2a n 1 λ +... + 2a 1 λ n 1 + 2a 0 λ n has positive real part on the unit circle (equivalently, on the closed unit disk). Consequence: a n 2a 0.

Can also show: if f is extremal, f = e g, and the coefficients of g are denoted by b j, we have Writing Re {a n b 0 + a n 1 b 1 +... + a 0 b n } = 0. P(z) = 2a 0 n (z λ k ), k=1 and computing P(0), it follows that the zeros λ k of the polynomial P(λ), 1 k n, satisfy the condition ( 1) n n λ k 1. k=1

A trick of F. Wiener If f 1 in D, f (z) = k=0 a kz k, and ω = e 2πi/n is the primitive n-th root of unity, define the function W n f (z) = 1 n 1 f (ω k z) = n H(z) = k=0 a nk z nk = H(z n ). k=0 a nk z k = a 0 + a n z + a 2n z 2 +..., k=0 also bounded by one and with H (0) = a n.

How can Wiener s trick help? Let f be extremal. If the function H does not vanish in D, the conjecture follows by applying Wiener s trick to reduce the problem on M n to that of M 1 (known). If it vanishes, there exist a Blaschke product B (possibly a constant of modulus one) and a function G, analytic and non-vanishing in D, such that H = BG, G = H 1.

Since for any α with α = 1 we have H = (αb)(αg) and H(0) = a 0 > 0, we can replace B by αb and so without loss of generality we may assume that B(0) is real and positive. Thus, W n f (z) = a 0 + a n z n +... = B(z n )G(z n ) = (B 0 + B n z n +...)(C 0 + C n z n +...). When W n f does not vanish in the disk, can think of B B 0 = 1 so as to include both cases. Can show: if f is extremal and 0 < B 0 1, we have a n a 0 1 + 1 B 0.

Some special results One can prove the Krzyż conjecture under some additional hypotheses on an extremal function such as: a i = 0 for all i belonging to some I {1, 2,..., n 1}. Typically, about a half of these initial coefficients are assumed to vanish. J.E. Brown s (1985): if a i = 0 whenever 1 i < (n + 1)/2 then the conjecture follows. Peretz (1991) proved the following: (a) if n is odd and a 1 = a 3 =... = a n 2 = 0 then a 0 1/e, (b) if, in addition to this, a 0 = 1/e then a n = 2/e.

Note: Brown s assumptions a i = 0 whenever 1 i < (n + 1)/2 imply that also b i = 0 whenever 1 i < (n + 1)/2. Also simple: for n odd, Peretz s assumptions a 1 = a 3 =... = a n 2 = 0 imply that a k = a 0 b k for each k {1, 2,..., n}, hence also b 1 = b 3 =... = b n 2 = 0.

Inductive sets In what follows, the sets I of indices as in the papers by Brown and Peretz will be called inductive sets. This will lead to further examples and a unified proof of the conjecture under other similar assumptions. Fix n N, n 2. Given K {1, 2, 3,..., n 1}, define C n K = {c = (c 1, c 2,..., c n ) C n : c i = 0 for all i K }. By an additive semigroup or simply semigroup we will mean a subset of N closed under addition. For K {1, 2, 3,..., n 1}, denote by G(K ) the additive semigroup generated by (K {n}) c = N \ (K {n}).

Let n N, n 2, and a 0 > 0. A subset I of {1, 2, 3,..., n 1} is said to be n-inductive if a n = a 0 b n for all a C n I and b C n that satisfy the recursion formulas for a j and b j. A subset J of {1, 2, 3,..., n 1} is said to be exponentially n-inductive if a n = a 0 b n for all a C n and b C n J that satisfy the recursion formulas for a j and b j. We will sometimes suppress the integer n and simply say I is inductive or J is exponentially inductive when the value of n is understood.

The following lemma allows us to have enough examples. Lemma. Fix n N, n 2, a 0 > 0. (a) Let I = {i : 1 i n 1, a i = 0}. Then I is n-inductive if and only if n / G(I). (b) Let J = {j : 1 j n 1, b j = 0}. Then J is exponentially n-inductive if and only if n / G(J).

The situation considered by Peretz corresponds to the semigroup G(I) = 2N = {2, 4, 6, 8,...}. In Brown s result, G(I) is the semigroup generated by the set {i N : (n + 1)/2 i n 1}. In both cases, as observed before, I = J hence G(I) = G(J). There are other examples of inductive sets with density about 1/2 in 1, 2,..., n 1.

Summary of observations Let f = e g be extremal and let a j and b j be as before. Then a n > 0 (in fact, a n = M n 2/e). May assume: b 0 < 0. Know: (I) 2 an a 0 1 + 1 B 0, where B 0 is the constant term in the Blaschke factor of the earlier factorization normalized so that 0 < B 0 1. (II) The polynomial P(z) = a n + 2a n 1 z +... 2a 1 z n 1 + 2a 0 z n has non-negative real part on the closed unit disk D and strictly positive real part on D. (III) N n in the formula for the structure of extremal functions N j=1 f (z) = e r α j z 1 j α j z+1, and also Re b n 2 b 0 (by Carathéodory s lemma). (IV) The zeros λ j, 1 j n, of the polynomial P satisfy λ k 1 for 1 k n and ( 1) n n k=1 λ k 1.

The main result Theorem. Let n 2 and consider an arbitrary but fixed extremal function f for the normalized Krzyż problem. Write f (z) = j=0 a jz j, g(z) = j=0 b jz j, and f = e g as before, and consider the quantity B 0, the polynomial P and its zeros as described. (I) The following statements are equivalent: (1) a n = 2a 0 ; (2) a k = 0 whenever 1 k < n; (3) b k = 0 whenever 1 k < n;

(4) f (z) = e (zn 1)/(z n +1) (and, in particular, M n = 2/e); (5) the set I consisting of all indices i {1, 2,..., n 1} for which a i = 0 is n-inductive; (6) the set J consisting of all indices j {1, 2,..., n 1} for which b j = 0 is exponentially n-inductive; (7) there exists an analytic function H in D such that H(z) 1 for all z D and k N, k n/2, with g(z) = (z k H(z) 1)/(z k H(z) + 1) ; (8) the zeros of the polynomial P satisfy ( 1) n n k=1 λ k = 1; (9) the zeros of P all lie on the unit circle; (10) the zeros of P are actually the n-th roots of 1;

(11) a n b 0 + a 0 Re {b n } = 0; (12) Re {a 1 b n 1 + a 2 b n 2 +... + a n 1 b 1 } = 0; (13) a n = a 0 Re b n. (14) Re b n = 2 b 0, N = n, and r 1 = r 2 =... = r n ; (15) B 0 = 1; (16) W n f does not vanish in D; (17) W n f f ; (18) g = W n g. (II) In addition to the above, the following is true: there is a unique extremal function for the normalized problem if and only if every extremal function for this problem satisfies any one of the conditions (1) (18) from part (I), and therefore all of them.

According to the above findings, proving the conjecture essentially amounts to showing that N = n in N j=1 f (z) = e r α j z 1 j α j z+1, and the following three sets of numbers actually coincide: - {α 1,..., α N }, the rotation coefficients in the point masses in extremal functions, - {ω 1,..., ω n }, the n-th roots of 1, - {λ 1,..., λ n }, the roots of the polynomial P associated with the extremal function f, - the zeros of Re P = Q 2 on T. Alternatively, it suffices to show the uniqueness of the extremal function for the normalized problem. Although unpublished so far, this fact seems to be known to several experts.

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