ON GENERALIZED SCHWARZ-PICK ESTIMATES

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ON GENERALIZED SCHWARZ-PICK ESTIMATES J. M. ANDERSON AND J. ROVNYAK. Introduction By Pick s invariant form of Schwarz s lemma, an analytic function B(z) which is bounded by one in the unit disk D = {z : z < } satisfies the inequality () B (α) B(α) at each point α of D. Recently several authors [, 0, ] have obtained more general estimates for higher order derivatives. Best possible estimates are due to Ruscheweyh []. Below in we use a Hilbert space method to derive Ruscheweyh s results. The operator method applies equally well to operator-valued functions, and this generalization is outlined in 3. Theorem. If B(z) is analytic and bounded by one in the unit disk, then () ( α ) n B (n) (α) B(α) for each α in D and n =,,.... For fixed α and n, (3) sup ( α ) n B (n) (α) B B(α) =, where the supremum is over all nonconstant analytic functions B(z) which are bounded by one on D. The case α = 0 in () asserts that if B(z) = B 0 + B z + B z +, then B n B 0 for every n. This result is classical and due to F. Wiener; see e.g. [, 4, 9]. Results for functions of several variables have also been obtained in [] and []. In particular, in [] rather intricate estimates in several variables are obtained from the above inequality of Wiener. Theorem. (i) When n =, equality holds in () for some α in D if and only if it holds for all α in D, and this occurs if and only if (4) B(z) = γ z β βz for some numbers γ and β such that γ = and β <. (ii) If α = 0, then for any n equality holds in () if and only if (5) B(z) = γ zn β βz n, J. M. Anderson acknowledges the support of the Leverhulme Trust.

where γ = and β <. (iii) If α D \ {0} and n, equality holds in () only for a constant of absolute value one. Theorem (ii) can also be proved from the fact that the extremal function f(z) = n=0 a nz n of modulus at most one which maximizes Re a n subject to the constraint that f(0) = a 0 is unique. A proof of this may be deduced as a special case of [7, Theorem.5, p. 67]. See also [3] for related estimates of Taylor coefficients of functions in the unit ball of H p, p. We thank Catherine Bénéteau, Dmitry Khavinson, and Barbara MacCluer for helpful discussions of this work.. Hilbert space method Given a function B(z) which is analytic and bounded by one in D, let be the Hilbert space of functions on D with reproducing kernel (see [6, 3]) K B (w, z) = B(z)B(w), w, z D. z w Every such space is a linear subspace of the Hardy class H, and the inclusion mapping from into H is contractive. A space is contained isometrically in H if and only if B(z) is an inner function, and in this case = H BH. For any space, the transformation R(0): f(z) [f(z) f(0)]/z is a contraction from into itself and satisfies (6) f(z) f(0) z f(z) f(0) for every f(z) in (see [6, Problem 47]). More generally, for each α in D, the transformation R(α): f(z) [f(z) f(α)]/(z α) is everywhere defined on into itself, and (7) R(α) α. This follows from [6, Problem 8, p. 45]. In fact, (8) R(α) = [I αr(0)] R(0) = α k R(0) k+, and therefore R(α) k=0 α k = /( α ). We remark that equality holds in (6) for every f(z) in if and only if one of the following equivalent conditions holds (see [6, Theorem 6] and [3, Chapters IV and V]): (i) B(z) ; (ii) B(z) is an extreme point of the unit ball of H ; (iii) π 0 log [ B(e iθ ) ] dθ =. The extremal functions in Theorem both satisfy (iii) above, and we raise the question if there is a better bound in () when B(z) is subject to the constraint π 0 k=0 log [ B(e iθ ) ] dθ = K >.

Lemma 3. If α <, then [B(z) B(α)]/(z α) belongs to and B(z) B(α) z α B(α). Proof. Write B(z) = B( z) for all z in D. It is known that [B(z) B(w)]/(z w) belongs to for every w in D, and that there is a contractive transformation W from into H( B) such that W f = g is given by g(w) = f(z), B(z) B( w) z w, w D. This follows from the scalar case of Theorem 5, p. 350, in the Appendix of [5]; for a different proof, see [, p. 93]. If f(z) = K B (α, z), then g(z) = [ B(z) B(ᾱ)]/(z ᾱ). Since W is a contraction, B(z) B(ᾱ) K B (α, z) z ᾱ = B(α). H( e B) The lemma follows from this inequality on replacing B(z) by B(z) and α by ᾱ. Proof of Theorem. Fix α D. We first prove (). Set f (z) = B(z) B(α) z α and f n = R(α) n f, n =, 3,.... For each n, f n (z) = k=n pointwise in a neighborhood of α, and so B (k) (α) k! f n (α) = B(n) (α) (z α) k n Hence by the Schwarz inequality, Lemma 3, and the inequality (7), B (n) (α) = fn (9) (z), K B (α, z) = R(α) n f (z), K B (α, z). This is equivalent to (). R(α) n f (z) K B (α, z) B(α) B(α) ( α ) n = B(α) ( α ) n. 3

To prove (3) it is enough to suppose that n (when n =, equality in () is attained for any automorphism of D). Consider B(z) = (z c)/( cz) for a fixed c in D. A short calculation yields Hence ( α ) n B (n) (α) B (n) (z) = c n ( c )( cz) n. B(α) = ( α )n ( ) c n ( c )( cα) n (α c)/( cα) = ( α )n ( ) c n ( c ) cα n+ cα α c = ( α )n ( ) c n ( c ) cα n+ ( )( c ) ( ) n α = c n. cα The last expression tends to as c e iθ where θ is chosen such that e iθ α = α, and (3) follows. Proof of Theorem. (i) This is well known. (ii) The case n = is included in (i), so we may suppose n. If B(z) has the form (5), then B(z) = γ [ β + ( β )z n + ], and so B (n) (0) = ( β ) = [ B(0) ]. This is () with equality. Conversely, assume that equality holds in () with α = 0. If B(z) = B 0 + B z + B z +, then in the notation of the proof of Theorem, B(z) B(0) f (z) = = B + B z +. z By (9) with α = 0, B (n) (0) = R(0) n f (z), K B (0, z) (0) R(0) n f (z) K B (0, z) f (z) K B (0, z) B(0). Thus equality holds throughout. In particular, R(0) n f (z) = f (z). By iterating the difference-quotient inequality (6), we get f (z) = R(0)n f (z) f (z) B B n, 4

and therefore B = = B n = 0. Hence f (z) = B n z n + B n+ z n +, so R(0) n B(z) B(0) f (z) = z n. For equality to hold in (), equality must also hold in the Schwarz inequality in (0), which implies that for some constant γ. Hence and so R(0) n f (z) = γk B (0, z) B(z) B(0) z n = γ[ B(z)B(0)], B(z) = γzn + B(0) + γb(0)z n = B(0) + γ[ B(0) ]z n +. In particular, B (n) (0)/ = γ[ B(0) ]. Therefore since equality holds in (), γ =. Thus B(z) has the form (5) with β = γb(0). (iii) Fix α D \ {0} and n. Assume that B(z) is analytic and bounded by one in D and () B (n) (α) = B(α) ( α ) n. By (8), R(α) = k=0 αk R(0) k+. Thus since n, we can write the chain of relations (9) in the form B (n) (α) = R(α) n R(α)f (z), K B (α, z) ( α ) n R(α)f B(α) ( α ) n ( α ) n k=0 k=0 B(α) ( α ) n. α k R(0) k+ f B(α) α k f B(α) By (), equality holds at every stage. Since α 0 and R(0) is a contraction, we deduce that () R(0) k+ f = f, k = 0,,,.... Let f (z) = k=0 a kz k. Iteration of (6) yields R(0) k+ f f (z) a 0 a k, k = 0,,,.... Hence by (), f (z) = [B(z) B(α]/(z α) is identically zero. But then B(z) is constant, and this constant has absolute value one by (). 5

3. Operator-valued functions The Hilbert space method applies with minor modifications to operator-valued functions. Fix a Hilbert space C, and write L(C) for the space of bounded linear operators on C. The norms on C and L(C) are both written, and the adjoint of an operator A L(C) is denoted A. We write v u = u, v C for the inner product of two vectors u and v in C. Let B(z) be an analytic function on D with values in L(C) such that B(z) on D. A particular case of Ginzburg s inequality (see e.g. [, p. 93]) or [8, p. 369]) states that B(β) B(α) v u β α u I B(α) B(α) u v I B(β)B(β) β v for any points α, β in D and vectors u, v in C. Letting β approach α, we deduce that B (α) I B(α) B(α) I B(α)B(α) for every α in D. This is an operator counterpart to the Schwarz-Pick inequality (). Theorem 4. Let B(z) be an analytic function with values in L(C) and satisfying B(z) on D. Then ( α ) n B (n) (α) I B(α) B(α) I B(α)B(α) for each α in D and n =,,.... The space is now a space of C-valued analytic functions on D having reproducing kernel (see [, 5]) K B (w, z) = I B(z)B(w). z w Thus for every w D and u C, K B (w, z)u belongs to as a function of z, and the identity f(z), K B (w, z)u = u f(w) holds for every f(z) in. As in the scalar case, [f(z) f(0)]/z belongs to whenever f(z) is in, and the inequality (6) holds. If α is any point of D, R(α): f(z) [f(z) f(α)]/(z α) maps into itself, and R(α) /( α ). Lemma 5. If α < and u C, then [B(z) B(α)]u/(z α) belongs to and B(z) B(α) u z α u B(α) B(α) u. Proof. This follows from [5, Theorem 5, p. 350] as in the proof of Lemma 3. Proof of Theorem 4. Fix α D. Let u and v be arbitrary vectors in C. Define f (z) = [B(z) B(α)]u/(z α) and f n = R(α) n f, n =, 3,.... For any n, 6

f n (α) = B (n) (α)u/. Thus v B(n) (α) u = fn (z), K B (α, z)v = R(α) n f (z), K B (α, z)v R(α) n f (z) K B (α, z) ( ( α ) n u ) B(α) B(α) ( u v ) B(α)B(α) v. The result follows on taking the supremum over all unit vectors u and v. References [] D. Alpay, A. Dijksma, J. Rovnyak, and H. de Snoo, Schur functions, operator colligations, and reproducing kernel Pontryagin spaces, Operator Theory: Advances and Applications, vol. 96, Birkhäuser Verlag, Basel, 997. [] C. Bénéteau, A. Dahlner, and D. Khavinson, Remarks on the Bohr phenomenon, Comput. Methods Funct. Theory 4 (004), no., 9. [3] C. Bénéteau and B. Korenblum, Some coefficient estimates for H p functions, Complex analysis and dynamical systems, Contemp. Math., vol. 364, Amer. Math. Soc., Providence, RI, 004, pp. 5 4. [4] H. Bohr, A theorem concerning power series, Proc. London Math. Soc. () 3 (94), 5, Collected Mathematical Works. Vol. III, paper #E 3, Dansk Matematisk Forening, København, 95. [5] L. de Branges and J. Rovnyak, Canonical models in quantum scattering theory, Perturbation Theory and its Applications in Quantum Mechanics (Proc. Adv. Sem. Math. Res. Center, U.S. Army, Theoret. Chem. Inst., Univ. of Wisconsin, Madison, Wis., 965), Wiley, New York, 966, pp. 95 39. [6], Square summable power series, Holt, Rinehart and Winston, New York, 966. [7] S. Ya. Khavinson, Two papers on extremal problems in complex analysis, Amer. Math. Soc. Transl. (), vol. 9, American Mathematical Society, Providence, RI, 986, Translated from the Russian manuscript by D. Khavinson. [8] M. G. Kreĭn and H. Langer, Über die verallgemeinerten Resolventen und die charakteristische Funktion eines isometrischen Operators im Raume Π κ, Hilbert space operators and operator algebras (Proc. Internat. Conf., Tihany, 970), North-Holland, Amsterdam, 97, pp. 353 399. Colloq. Math. Soc. János Bolyai, 5. [9] E. Landau and D. Gaier, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, third ed., Springer-Verlag, Berlin, 986. [0] B. D. MacCluer, K. Stroethoff, and R. Zhao, Generalized Schwarz-Pick estimates, Proc. Amer. Math. Soc. 3 (003), no., 593 599 (electronic). [], Schwarz-Pick type estimates, Complex Var. Theory Appl. 48 (003), no. 8, 7 730. [] St. Ruscheweyh, Two remarks on bounded analytic functions, Serdica (985), no., 00 0. [3] D. Sarason, Sub-Hardy Hilbert spaces in the unit disk, University of Arkansas Lecture Notes in the Mathematical Sciences, 0, John Wiley & Sons Inc., New York, 994, A Wiley-Interscience Publication. Department of Mathematics, University College London, Gower Street, London WCE 6BT, U. K. University of Virginia, Department of Mathematics, P.O.Box 40037,Charlottesville, VA 904-437, U. S. A. E-mail address: rovnyak@virginia.edu 7