A nonlinear extremal problem on the Hardy space

Transcription

1 University of Richmond UR Scholarship Repository Math and Computer Science Faculty Publications Math and Computer Science 9 A nonlinear extremal problem on the Hardy space William T Ross University of Richmond, wross@richmondedu Stephan Ramon Garcia Follow this and additional works at: Part of the Algebra Commons This is a pre-publication author manuscript of the final, published article Recommended Citation Ross, William T and Garcia, Stephan Ramon, "A nonlinear extremal problem on the Hardy space" (9) Math and Computer Science Faculty Publications This Post-print Article is brought to you for free and open access by the Math and Computer Science at UR Scholarship Repository t has been accepted for inclusion in Math and Computer Science Faculty Publications by an authorized administrator of UR Scholarship Repository For more information, please contact scholarshiprepository@richmondedu

2 Computational Methods and Function Theory Volume (), No, XXYYYZZ A nonlinear extremal problem on the Hardy space Stephan Ramon Garcia and William T Ross Keywords Extremal problem, truncated Toeplitz operator, Toeplitz operator, Clark operator, Aleksandrov-Clark measure, reproducing kernel, complex symmetric operator, conjugation MSC 47A5, 47B35, 47B99 ntroduction n this paper, we discuss the nonlinear extremal problem ( ):= sup f dz () fh i kfk= and its relationship to the classical linear extremal problem ( ):= sup F dz () F H i kf k = and the compression of Toeplitz operators to the model spaces H H n the above, D denotes the open unit disk, denotes the unit circle, is a rational function whose poles do not lie on, H p is the classical Hardy space [, 9, 4], and is a finite Blaschke product We refer to as the kernel of the extremal problems () and () Since the focus of this paper is primarily on H,weletk k (with no subscript) denote the norm on H while k k denotes the norm on H Despite the nonlinear nature of (), our investigation will involve the theory of linear operators, specifically the relatively new field of truncated Toeplitz operators and complex symmetric operators [7, 8, 39] The linear extremal problem () and its natural generalization to H p ( apple p apple) have a long and storied history dating back to the early twentieth century [, 9,, 3, 7, 34] For the moment, we single out several notable classical results which motivate this paper We first mention a theorem of Macintyre and Rogosinski [7, 34] (see also [3, p 33]) which asserts that for any given rational kernel having no poles on there exists an F e in the unit ball of H for which ( ) = F e dz i This function F e is called an extremal function for () n general, F e is not necessarily unique ndeed, one need only consider the extremal problem z = sup F () = F H kf k = First author partially supported by National Science Foundation Grant DMS SSN /$ 5 c XX Heldermann Verlag

3 Stephan Ramon Garcia and William T Ross CMFT to see that the functions (z )( z) +, C, (3) all serve as extremal functions (On the other hand, the analogous extremal problem for H p with p> always has a unique solution [, Thm 8]) Nevertheless, for general rational we can always choose an extremal function F e which is the square of an H function [3, p 33], from which it follows that ( ) = ( ) Using the theory of truncated Toeplitz operators and the structure of the underlying Jordan model spaces, we give a new proof of this fact in Proposition A second result worth mentioning here is due to Fejér [3], who showed that for any complex numbers c,c,,c n one has c z + + c n z n+ = khk, where H is the Hankel matrix (blank entries to be treated as zeros) c c c c n c c c n H = c c n C A c n and khk denotes the operator norm of H (ie, the largest singular value of H) n the special case when c = c = = c n =, Egerváry [] provided the explicit formula z + z + + z n+ = (n + ) sec (4) n +3 and showed that the polynomial 4 n +3 apple (n + ) n sin + z sin n +3 n zn sin n +3 is an extremal function (n particular, note that this is the square of an H function) Golusin in [] finds an extremal function for the original (general) Fejér extremal problem as well as some others [] We refer the reader to [, 9, 3] for further references Fejér s result can also be phrased in terms of Toeplitz matrices, which turn out to represent the simplest type of truncated Toeplitz operator a class of operators which figures prominently in our approach to () (see Subsection 4) ndeed, if T is the following lower-triangular Toeplitz matrix c ṇ T = Bc c n, c c c n A c c c c n then UT = H, whereu is the permutation matrix and thus kt k = khk U = C A, (5)

4 (), No A nonlinear extremal problem on the Hardy space 3 n Theorem and Corollary we extend Fejér s result and show that ( ) = ( ) is equal to the norm of a certain truncated Toeplitz operator on a canonically associated model space K = H H, where is a certain finite Blaschke product associated with To relate ( ) to the norm of this associated truncated Toeplitz operator, we develop a new variational characterization of the norm of a complex symmetric operator, something interesting in its own right For a given rational, the discussion above tells us that there is an extremal function f e ball(h ) n other words, the function f e satisfies ( ) = fe dz i Using the theory of complex symmetric operators, we will show in Corollary that f e belongs to a certain model space K associated with ( ) More importantly, we establish a procedure to compute f e along with necessary and su cient conditions, in terms of truncated Toeplitz operators, which determine when the extremal function f e is unique (up to a sign) Next, we consider the obvious estimate ( ) = ( ) apple max ( ) Using some results from [37], we will show in Corollary 5 that equality holds if and only if = B B,where C and B,B are finite Blaschke products with no common zeros and such that deg B < deg B n particular, this demonstrates the utility of relating ( ) tothe norm of a truncated Toeplitz operator Let us now suggest another application Since, for n =,,, and D, n! c n, = (z ) n+ is the best constant in the inequality F (n) ( ) applec n, kf k, F H, we can use operator theoretic techniques to explicitly determine the constants c n, as well as the associated extremal functions Although this problem has been well-studied [,, 6, 7], we obtain the same results (see Theorem 3) using the new language of truncated Toeplitz operators n Section 9 we briefly explore how our results can be extended to handle certain kernels L () which are not necessarily rational t should be remarked that () is not the first nonlinear extremal problem on the Hardy space to be explored The reader is invited to consult [, 3, 5, 4, 8, 4] for other examples of nonlinear extremal problems on H p as well as other spaces of analytic functions We thank D Khavinson for pointing out these papers to us and for several enlightening conversations n addition, we thank David Sherman for pointing out the identity in (6) Preliminaries As mentioned in the introduction, we will relate the extremal problems () and () to the emerging study of truncated Toeplitz operators and complex symmetric operators This section lays out all of the appropriate definitions and basic results

5 4 Stephan Ramon Garcia and William T Ross CMFT Model spaces Given a rational function whose poles do not lie on, consider the associated finite Blaschke product ny z j (z) = (6) jz j= whose zeros,,, n, repeated according to multiplicity, are precisely the poles of which lie in D The degree, deg, of the Blaschke product in (6) is defined to be n, the total number of zeros counted according to multiplicity Note that might have other poles which lie outside of the closed unit disk D However, these poles are not counted amongst the j s We also assume that has at least one pole in D, since otherwise ( ) = ( ) =, which is of no interest With the above, we form the corresponding model space K := H H More precisely, K is the closed subspace of H defined by H H := H \ ( H )? (7) where the implicit inner product is the standard L inner product: Z hu, vi := u( )v( ) d, u,v L := L (, d ) (8) The term model space stems from the important role that K plays in the model theory for Hilbert space contractions see [3, Part C] Let us briefly recall several important facts about K First note that Beurling s theorem [, p 4] asserts that H is a typical proper invariant subspace of finite codimension (since is a finite Blaschke product) for the unilateral shift operator [Sf](z) =zf(z), f H t follows from a standard duality argument that K is a typical nontrivial, finite-dimensional invariant subspace for the backward shift operator [S f(z) f() f](z) =, f H z For further information on various function-theoretic aspects of the backward shift operator, we refer the reader to the seminal paper [] and the recent texts [7, 35] When the zeros,,, n of are distinct, it is easy to verify that K is spanned by the Cauchy kernels, jz j =,,,n (9) Thus each function f in K can be written uniquely as f(z) = p(z) Q n j= ( jz), () where p(z) is an analytic polynomial of degree at most n Conversely, by partial fractions, every such function belongs to K n particular, each function in K is analytic in a neighborhood of D t is also not hard to see that the preceding discussion is still valid even if the zeros of are not all distinct, so long as the terms in the denominator of () are repeated according to multiplicity and the Cauchy kernels in (9) are replaced by ( jz), apple j apple n, apple m apple m j, () m where m j denotes the multiplicity of j as a zero of

6 (), No A nonlinear extremal problem on the Hardy space 5 From the representation (7) we find that K carries a natural isometric, conjugate-linear involution C : K! K defined in terms of boundary functions by Cf := fz () Although, at first glance, the expression fz in () does not appear to correspond to the boundary values of an H function, let alone one in K, a short computation using (7) reveals that hcf, hi and Cf,zh both vanish for all h H and f K,whenceCf indeed belongs to K An important aspect of the spaces K involves the reproducing kernels k w (z) = which enjoy the so-called reproducing property (w) (z), z,w D, (3) wz hf,k w i = f(w), w D,f K (4) n light of the fact that is a finite Blaschke product, note that k w belongs to K for all w in D and that the analyticity of on ensures that (3) is well-defined when both z and w belong to That (4) holds is a straightforward consequence of the Cauchy ntegral Formula and the fact that hf,z n i = for all n Letting P denote the orthogonal projection of L onto K, we also observe that Finally, another short computation reveals that [P f](w) =hf,k w i, f L,w D (5) (Ck w )(z) = (z) z (w), z,w D (6) w Takenaka-Malmquist-Walsh bases Suppose that the finite Blaschke product (6) has n zeros,,, n repeated according to their multiplicity Although the functions from () form a basis for K, they are not orthonormal However, there is a well-known orthonormal basis that can be constructed from these functions ndeed, for w D we let B w denote the disk automorphism B w (z) = z w wz and consider the unit vectors 8p >< if k =, v k (z) := z Qk p (7) >: i= B k if apple k apple n i kz By (), v k K for all apple k apple n A short computation based on the reproducing property and the fact that B w B w = on shows that for j<kwe have * Qk p p + hv k,v j i = i=j B k i kz, j = jz (since B j ( j ) = ) while for j = k we have hv j,v j i = Thus {v,v,,v n } constitutes an orthonormal basis for the model space K A standard text refers to this observation as the Malmquist-Walsh Lemma [9, V] but these functions appeared as early as 95 in a paper of Takenaka [4] n light of this, we shall call {v,v,,v n } the Takenaka-Malmquist-Walsh basis for K

7 6 Stephan Ramon Garcia and William T Ross CMFT We do wish to mention a particular important case that will be used in Section 8 Suppose that = = = n = so that = B n n this case, the functions p v k (z) = ( z) (z k )k, apple k apple n (8) form an orthonormal basis for K Note that these vectors are simply an orthonormalization of the functions in () 3 Aleksandrov-Clark bases t turns out that each model space K comes equipped with another natural family of orthonormal bases Let us briefly describe their construction Note that = on Also note that > on : write ny (z) = B j (z) j= and use formal logarithmic di erentiation and the fact that to get the identity When z =, we get (z) = (z) B j (z) = j ( jz) nx j= ( ) = ( ) j ( jz)(z j) nx j= j j Since ( ) = on, the result follows Fix some and note that since = and > on, the equation ( ) = has precisely n distinct solutions,,, n on Here, as before, n denotes the degree of the Blaschke product (6) A short computation now shows that k j = q ( j ), apple j apple n Next we define unimodular constants!,!,,! n by! j := e i (arg arg j), apple j apple n (9) However the arguments of and,,, n are selected, they are to remain consistent throughout the following Now consider the functions v j (z) :=! j k j (z) kk j k t is readily verified that hv j,v k i = j,k and, moreover, that! j = p ( j ) (z) j z () Cv j = v j, apple j apple n () We refer to the basis in () as a modified Aleksandrov-Clark basis for K The terminology stems from the fact that the v j are the eigenvectors of a certain Aleksandrov-Clark operator on K [6, 8, 3, 38] Such bases are discussed in further detail in [5, Sect 8] One notable advantage of using Aleksandrov-Clark bases over the Takenaka-Malmquist-Walsh basis is the fact that all Aleksandrov-Clark bases are C-real in the sense that () holds

8 (), No A nonlinear extremal problem on the Hardy space 7 4 Truncated Toeplitz operators As mentioned in the introduction, we will show in Corollary that the extremum ( ) in the nonlinear problem () coincides with the operator norm of a certain truncated Toeplitz operator on an associated model space Let us now consider these operators Definition For ' L = L (, d ) and an inner function, the operator A ' : K! K defined by A ' f = P ('f) () is called the truncated Toeplitz operator on K with symbol ' Truncated Toeplitz operators are sometimes referred to as compressed Toeplitz operators since A ' is simply the compression of the standard Toeplitz operator T ' on H to the model space K Recall that if P : L! H denotes the orthogonal projection from L onto H (called the Riesz or Szegö projection), then T ' f = P ('f) The operator norm of a truncated Toeplitz operator A ' on K will be denoted by ka ' k K!K or, when the context is clear, simply by ka ' k t is easy to see from the definition () that A ' is a bounded operator on K satisfying ka ' kapplek'k For further information, we direct the reader to the recent article [39], which appears destined to become the standard reference for truncated Toeplitz operators Maintaining the same notation as in (6), we observe that if the symbol ' of a truncated Toeplitz operator A ' : K! K belongs to H, then its eigenvalues are precisely the numbers '( ),'( ),,'( n ) t is perhaps easier to prove the corresponding statement for the adjoint A ' = A ' To see this, note that since ( i ) =, then k i (z) =( iz) and so, by the Cauchy ntegral Formula, Thus h'f, k i i = '( i )f( i ) 8f K ha ' k i,fi = hp ('k i ),fi = hk i,'fi = '( i )f( i )= holds for all f in K whence D E '( i )k i,f A ' k i = '( i )k i, i =,,,n (3) We approach our nonlinear extremal problem ( ) by considering a related system of approximate anti-linear eigenvalue problems (see Lemma 3) corresponding to certain truncated Toeplitz operators This technique, introduced in [6, Thm ], can be thought of as a complex symmetric adaptation of Weyl s criterion [33, Thm V] To make this more precise, we require a brief discussion of complex symmetric operators Throughout the following, H will denote a separable complex Hilbert space and (T ) will denote the spectrum of a bounded linear operator T : H!H Definition A conjugation on H is a conjugate-linear operator C : H! H that is both involutive (ie, C = ) and isometric (ie, hcx,cyi = hy, xi for all x, y H) The canonical example of a conjugation is simply entry-by-entry complex conjugation on a l -space n fact, each conjugation is unitarily equivalent to the canonical conjugation on a l -space of the appropriate dimension [7, Lem ] From our perspective, the most pertinent example is the conjugation () on the model space K, which in light of (), is easily seen to be entry-by-entry complex conjugation with respect to any of the modified Aleksandrov-Clark bases {v,v,,v n } defined by () Definition A bounded linear operator T : H!His C-symmetric if T = CT C and complex symmetric if there exists a conjugation C with respect to which T is C-symmetric

9 8 Stephan Ramon Garcia and William T Ross CMFT t turns out that T is a complex symmetric operator if and only if T is unitarily equivalent to a symmetric matrix with complex entries, regarded as an operator acting on an l -space of the appropriate dimension (see [5, Sect 4] or [7, Prop ]) n fact, if C denotes the conjugation () on K, then any C-symmetric operator on K has a symmetric matrix representation with respect to any modified Aleksandrov-Clark basis For further details, we refer the reader to [5, 7, 8] The connection between complex symmetric operators and our nonlinear extremal problem ( ) is furnished by Theorem and Corollary below The proofs will depend on the following three lemmas Lemma For ' L and inner, the truncated Toeplitz operator A ' on K is C- symmetric with respect to the conjugation Cf = fz on K The proof, which is a straightforward computation, can be found in [7, Prop 3] or [5, Thm 5] n particular, the preceding remarks ensure that a truncated Toeplitz operator A ' has a complex symmetric (ie, self-transpose) matrix representation with respect to any modified Aleksandrov-Clark basis () This property of truncated Toeplitz operators will be apparent when we consider several numerical examples later on We also mention that the identity CA ' C = A ' along with (3) will show that if ' H,then where i are the zeros of A ' Ck i = '( i )Ck i, 5 The norm of a complex symmetric operator n order to compute the quantity ka ' k K!K, we require a few words concerning the polar decomposition of a complex symmetric operator Recall that the polar decomposition T = U T of a bounded linear operator T : H!Hexpresses T = U T uniquely as the product of a positive operator T = p T T and a partial isometry U which satisfies ker U =ker T and maps ran T (the closure of the range of T ) onto ran T [9, p 48] f T is a C-symmetric operator, then we can decompose the partial isometry U as the product of C with a partial conjugation We say that an conjugate-linear operator J is a partial conjugation if J restricts to a conjugation on (ker J)? (with values in the same space) n particular, the linear operator J is the orthogonal projection onto the closed subspace ran J =(kerj)? The following lemma is [8, Thm ] Lemma f T : H!His a bounded C-symmetric operator, then T = CJ T where J is a partial conjugation, supported on ran T, which commutes with T = p T T Now recall that Weyl s criterion [33, Thm V] from the spectral theory of selfadjoint operators states that if A is a bounded selfadjoint operator, then belongs to (A) if and only if there exists a sequence u n of unit vectors so that lim k(a )u nk = n! The following result [6, Thm ] characterizes approximate anti-linear eigenvalue problem Lemma 3 f T is a bounded C-symmetric operator and ( T ) in terms of what one might call an 6=, then (i) ( T ) if and only if there exists a sequence of unit vectors u n which satisfy the approximate anti-linear eigenvalue problem lim n! k(t C)u nk = (4) Moreover, the u n can be chosen so that Ju n = u n for all n

10 (), No A nonlinear extremal problem on the Hardy space 9 (ii) is an eigenvalue of T (ie, a singular value of T )ifandonlyiftheanti-linear eigenvalue problem Tu = Cu has a nonzero solution u To compute the norm of a truncated Toeplitz operator, and ultimately following general formula for the norm of a complex symmetric operator Theorem f T : H!His a bounded C-symmetric operator, then ( ), we require the (i) kt k =sup kxk= htx,cxi (ii) f kxk =, then kt k = htx,cxi if and only if Tx =!kt kcx for some unimodular constant! (iii) f T is compact, then the equation Tx = kt kcx has a unit vector solution Furthermore, this unit vector solution is unique, up to a sign, if and only if the kernel of the operator T kt k is one-dimensional Proof To prove (i) observe that since htx,cxi apple ktxkkcxkapplektk whenever kxk apple, it su ces to prove that kt kapplesup kxk= htx,cxi Let := kt k (which belongs to ( T )) and note that by Lemma 3, there exists a sequence u n of unit vectors such that Ju n = u n for all n and such that lim n! ktu n Cu n k = Thus we have kt k = hu n,u n i (ku n k = ) = h T u n,u n i h T u n u n,u n i = h T Ju n,u n i h T Ju n u n,u n i (Ju n = u n ) = hj T u n,u n i hj T u n u n,u n i (J T = T J, Lemma ) apple hj T u n,u n i + hj T u n u n,u n i = hcj T u n,cu n i + hcj T u n Cu n,cu n i (C is isometric, ) = htu n,cu n i + htu n Cu n,cu n i (CJ T = T, Lemma ) apple sup htx,cxi + ktu n Cu n k (kcu n k = ku n k = ) kxk= Since the second term in the previous line tends to zero as n!, the desired inequality follows This proves (i) Let us now consider (ii) First observe that the (() implication is obvious For the ()) implication, suppose that kxk = and kt k = htx,cxi By the CSB inequality we have kt k = htx,cxi applektxkkcxkapplekt kkcxkapplekt kkxk = kt k whence htx,cxi = ktxkkcxk By the condition for equality in the CSB inequality, we find that Tx = htx,cxi Cx n other words, Tx =!kt kcx for some! =, concluding the proof of (ii) For (iii), first note that T is compact and thus kt k is an eigenvalue of T whence, in light of Lemma 3, the equation Tx = kt kcx has a unit vector solution We now show that this solution is unique, up to sign, if and only if the kernel of the operator T kt k is one-dimensional For one direction, we employ the following representation of compact C-symmetric operators from [8, Thm 3]: d n X T = X n Cu n,k u n,k, n k= where n are the distinct eigenvalues of the selfadjoint operator T (which, since T is compact, form a countable set whose only possible limit point is zero) and {u n,,,u n,dn } is a certain orthonormal basis for the eigenspace of T corresponding to n which also satisfies the auxiliary

11 Stephan Ramon Garcia and William T Ross CMFT condition Tu n,k = n Cu n,k Thus if the unit vector solution to Tx = kt kcx is unique up to sign, then the kernel of T kt k is one-dimensional On the other hand, if Tx = kt k Cx,then Tx = kt k Cx ) CTx = kt k x (kt kr, C = ) ) CTCTx = kt k x ) T Tx = kt k x (CTC = T ) ) T x = kt k x since T = p T T Suppose now that x, y are two unit vectors such that Tx = kt k Cx, Ty = kt k Cy, and x 6= ±y n this case, the conjugate-linearity of C, along with the facts that kxk = kyk = and x 6= ±y, ensures that x and y are not scalar multiples of each other n particular, x and y are linearly independent and satisfy T x = kt k x and T y = kt k y This implies that the kernel of the operator T kt k has dimension at least two, as claimed 3 A reduction As before, let denote the finite Blaschke product (6) whose zeros, repeated according to multiplicity, are precisely the poles of the rational function which lie in D Writingf as f = P f + h where h H, one can show that our original nonlinear extremal problem ( ) in () can be reduced to the new problem sup fk kfk= i f dz (5) posed on the model space K ndeed, for f H we have f dz = (P f) + (P f) h + h dz i i = (P f) dz (6) i We obtain (6) by noting that the function belongs to H whence the second and third terms in the preceding line vanish by Cauchy s Theorem We gather from this reduction two important observations The first is that an extremal function for the problem ( ) exists since the nonlinear functional f 7! f dz i is a continuous map and the supremum in (5) is over the (compact) unit ball of the finite dimensional space K The second is that an extremal function for ( ) liesink and thus from the representation () has the form f(z) = p(z) Q n j= ( jz), where p(z) is an analytic polynomial of degree at most n Putting together the material we have developed so far, we obtain the following corollary to Theorem, which, in terms of computing the norms of truncated Toeplitz operators, is interesting in its own right Corollary Let be a nonconstant inner function

12 (), No A nonlinear extremal problem on the Hardy space (i) f ' L, then (ii) f ' H, then ka ' k K!K ka ' k K!K = sup fk kfk= = sup fh kfk= i i 'f dz (7) 'f dz (8) Proof Lemma asserts that the operator A ' : K! K is C-symmetric with respect to the conjugation Cf = fz on K Next we note that ha ' f,cfi = hp ('f),cfi = h'f, Cfi = 'f, fz = 'f z, holds for any f in K Using the identity dz = iz dz to write the expression 'f z, as a contour integral on and noting that =/ ae on, yields (upon an application of Theorem ) the desired formula (7) This establishes (i) Let us now prove (ii) t follows from assertion (i) and the reduction discussed in (6) that 'f i dz = '(P f) dz, f H i n other words, if the function ' belongs to H, then the supremum in (7) can be taken over ball(h ) This yields (ii) The following corollary to Theorem and Corollary provides the fundamental connection between truncated Toeplitz operators and the nonlinear extremal problem ( ) Corollary Suppose that is a rational function having no poles on and poles,,, n lying in D, counted according to multiplicity Let denote the associated Blaschke product (6) whose zeros are precisely,,, n and note that ' = belongs to H We then have the following: (i) The nonlinear extremal problems equal to ka ' k K!K (ii) There is a unit vector f K satisfying ( ) from () and (5) are equal Moreover, both are A ' f = ka ' kcf (9) and any such f is an extremal function for ( ) n other words, f dz = ka ' k i K!K (iii) Every extremal function f for ( ) belongs to K and satisfies A ' f = ka ' kcf (iv) An extremal function for ( ) is unique, up to a sign, if and only if the kernel of the operator A ' ka ' k is one-dimensional Proof Use Theorem, Corollary, the reduction in (6), along with the fact that K is finite dimensional so that the C-symmetric operator T = A is compact

13 Stephan Ramon Garcia and William T Ross CMFT 4 Equality of the linear and nonlinear extrema For a rational function having no poles on, it turns out that the classical linear extremal problem ( ) given by () and our nonlinear extremal problem ( ) given by () yield the same answer We begin with the observation that for any f H with kfk =, we have f H and f = Therefore which implies that {f : f H, kfk apple} {F : F H, kf k apple }, ( ) apple ( ) (3) Recall from the introduction that Egerváry, in the special case where (z) = z + z + + z n+, showed (see (5)) that an extremal function for the linear extremal problem ( ) is the square of an H function whence ( ) = ( ) As a somewhat easier example, we remind the reader of the family of functions (3) which all serve as extremal functions for the linear extremal problem (/z ) n particular, the extremal function z + p is the square of an H function and we again have ( ) = ( ) As it turns out, the preceding examples are not peculiar in this regard ndeed, S Ya Khavinson [3] showed, in greater generality beyond rational, that one of the extremal functions for ( ) can be taken to be a function F with no zeros in D Thus f := p F will be (up to a unimodular constant) an extremal function for ( ) Here is a new proof of the identity ( ) = ( ) in the language of truncated Toeplitz operators Proposition f is a rational function with no poles on, then ( ) = words, f dz i i sup F H kf k = sup f,gh kfk=kgk= F dz = sup fh kfk= ( ) n other Proof Since every F H can be written as F = fg,wheref,g H and kf k = kfkkgk (see [9, Ex, Ch ] or [, Thm 35]), it follows that fg dz = F dz (3) i i sup F H kf k = As usual, let be the corresponding Blaschke product from (6) Writing f,g in the preceding as f = f + h and g = g + h,wheref,g K and h,h H, we obtain fg dz = (f + h )(g + h )dz = f g dz (3) i n particular, observe that all but one of the terms in the product integrate to zero since the function ' = belongs to H Putting this all together we find that sup F dz = F H i kf k = sup f,gh kfk=kgk= i fg dz (by (3))

14 (), No A nonlinear extremal problem on the Hardy space 3 = sup f,gk kfk=kgk= i = sup h f, zg i f,gk kfk=kgk= fg dz (by (3)) = sup h'f, Cgi (' = ) f,gk kfk=kgk= = sup h'f, gi (C = ) f,gk kfk=kgk= = sup hp ('f),gi f,gk kfk=kgk= = sup ha ' f,gi f,gk kfk=kgk= Thus ( ) = ( ) as claimed = ka ' k K!K = sup fh i kfk= f dz (Corollary ) 5 Computing the supremum Recall that Corollary ensures that ( ) =ka ' k K!K where ' denotes the H function We propose two simple and practical procedures for computing ka ' k Both methods involve computing matrix representations for the operator A ' n practice, these matrices are explicitly computable and hence their norms can be determined algebraically (for n =dimk small) or numerically via Mathematica or any other comparable piece of software 5 Using the Takenaka-Malmquist-Walsh basis Our first approach involves representing the truncated Toeplitz operator A ' : K! K as an n n matrix with respect to the Takenaka-Malmquist-Walsh basis (7) for K (i) Let ' := and note that ' H since the poles of lying in D cancel with the zeros of (ii) The jk-th entry [M A' ] jk of the matrix representation M A' of A ' with respect to the Takenaka-Malmquist-Walsh basis {v,v,,v n } is ha ' v k,v j i The resulting matrix is lower triangular (see [9, Lect V] and below) M A' (iii) The largest singular value of the matrix M A' is the desired quantity ka ' k = ( ) A straightforward computation confirms that the matrix representation M A' of A ' with respect to the Takenaka-Malmquist-Walsh basis is lower triangular ndeed, for j<kwe have [M A' ] jk = ha ' v k,v j i = hp ('v k ),v j i *! k p Y k = ' B i kz, i=! jy p + j B i jz i=

15 4 Stephan Ramon Garcia and William T Ross CMFT q ky = j i=j B i p + A k, = (33) kz jz {z } H since B j ( j ) = and thus the matrix is lower triangular The diagonal entries of our matrix are also relatively easy to compute: [M A' ] kk = '( k ), apple k apple n Since M A' is a lower triangular matrix this is to be expected since the eigenvalues '( ),'( ),,'( n ) of the operator A ' must appear along the main diagonal For j>kthecomputations are more involved: *! k p Y k [M A' ] jk = ' B i kz,! jy p + j B i i= i= jz * p! k = ' j p + kz, Y j B i i=k jz =( j ) ( k ) '(z) dz i Qj i=k B i ( kz)(z j) (34) Although the preceding can be evaluated explicitly using the residue calculus, the resulting expression is somewhat unwieldy and we choose not to write it here We will see the matrix representation of a truncated Toeplitz operator with respect to the Takenaka-Malmquist-Walsh basis again in Section 8 5 Using modified Aleksandrov-Clark bases A similar approach using the Aleksandrov- Clark bases can also be formulated (i) Let ' := and note that ' H since the poles of lying in D cancel with the zeros of (ii) Fix some on and let v,v,,v n be the corresponding modified Aleksandrov- Clark basis from () n particular, compute the unimodular constants,,, n and!,!,,! n from (9) (iii) The jk-th entry [M A' ] jk of the matrix representation M A' of A ' with respect to the Aleksandrov-Clark basis {v,v,,v n } is ha ' v k,v j i The resulting matrix M A' is complex symmetric (ie, self-transpose) (iv) The largest singular value of the matrix M A' is the desired quantity ka ' k = ( ) Let us derive a general formula for M A' in the simple case where the zeros,,, n are distinct (One can make adjustments for the general case where the zeros,,, n are not necessarily distinct) Using the fact that the functions ', v j, and v k are analytic in a neighborhood of the closed unit disk D, it follows from the residue calculus that ha ' v k,v j i = h'v k,v j i! k! = p p j ( k ) ( j )! k! = p p j ( k ) ( j ) z = z = '(z) (z) k z (z) j z dz iz '(z) (z) (z) dz k z (z)(z j ) i

16 (), No A nonlinear extremal problem on the Hardy space 5! k! j = p p ( k ) ( j )! k = p j! p j ( k ) ( j )! k! j = p p ( k ) ( j ) nx i= nx i= nx i= '( i ) ( i )( k i )( j i ) '( i ) ( i )( k i )( j i ) (35) '( i ) ( i )( k i )( j i ) (36) Observe that we employed the identity j! j =! j when passing from (35) to (36) Notice also from (36) that the n n matrix M A' =(ha ' v k,v j i) n j,k= is indeed complex symmetric as previously claimed t follows from the preceding calculations and the fact that v,v,,v n is an orthonormal basis for K that ka ' k K!K is equal to! k! j p p ( k ) ( j ) nx i= '( i ) ( i )( k i )( j i )! n j,k= (37) Example Let c,c,,c n be n arbitrary complex numbers and let,,, n be n distinct points in D Now form the rational function nx c i (z) = z i= Using the residue calculus and Corollary, we obtain i ( ) =ka ' k K!K where ' = and denotes the usual finite Blaschke product (6) Observe that nx (z) '(z) = (z) (z) = c i z i Employing the identity in (36) yields ( ) =! k '( i )=c i ( i ),! j p p ( k ) ( j ) nx s= i= apple i apple n c s ( k s )( j s ) Example To better demonstrate the procedure outlined above, let us work through a specific numerical example in detail Consider the nonlinear extremal problem where sup fh kfk= f () + f ( ) = ( ) =ka 'k K!K! n j,k= (z) =z z z, = z +, z and ' = as usual Letting = we find that the equation ( ) = has the solutions =, = From the preceding, we find that the desired unimodular constants! and! are given by! =,! = i

17 6 Stephan Ramon Garcia and William T Ross CMFT For a matrix A we have the identity kak = p tr(a A)+ det(a) + p tr(a A) det(a) (38) This formula follows from the polar decomposition A = U A where U is unitary and A = p A A is nonnegative f, denote the eigenvalues of A (ie, the singular values of A), note that tr(a A)= + and det A = since U is unitary Plugging this data into (36) and using (38) we find that! ka ' k K!K = n light of Proposition, we also have sup F H kf k = 5 4 7i 4 p 3 7i 4 p 3 3 F () + F ( ) = 6 (7 + p 37) Example 3 One could generalize the above example to sup fh kfk= = 6 (7 + p 37) 85 c f (a )+c f (a ) = ( ) =ka ' k K!K, where a,a D (a 6= a ), c,c C \{} n this case we have (z) = (z a )(z a ) ( a z)( a z), (z) = c z a + c z a, ' = We now represent our truncated Toeplitz operator with respect to the Takenaka-Malmquist- Walsh basis p a v (z) = a z, v (z) = z a p a a z a z We then use (33) and (34) to get ( ) c (a a ) ( a )( a a ) c p a p a ( a a ) + c p a p a ( a a ) c (a a ) ( a )( a a ) A (39) Example 4 Plugging a = p 4 + i p 4, a = 3, c =, c =, into (39) and using Mathematica reveals that sup F H kf k = = 48 F ( p 4 + i p 4 ) F ( 3 ) r p p p p Example 5 For another class of examples, suppose c,c,,c n are complex numbers and consider the rational function (z) = c z + c z + + c n z n+ The corresponding Blaschke product (6) is therefore (z) =z n+ and hence by Corollary we see that ( ) =ka z n+k K z n+!k z n+ = ka cn+c n z+ +c z nk K z n+!k z n+

18 (), No A nonlinear extremal problem on the Hardy space 7 The matrix representation of A z n+ with respect to the monomial basis {,z,,zn } for K z n+ is the lower triangular (n + ) (n + ) Toeplitz matrix c ṇ T = Bc c n (4) c c c n A c c c c n By Corollary we therefore obtain i sup fh kfk= c z + + c n z n+ f dz = kt k (4) Notice how this reproves the Fejér-Egerváry results mentioned in the introduction Example 6 By the above calculation, z + z = ka +z k Kz!K z = = +p 5 68 n particular, notice that this answer agrees with Egerváry s example (4) discussed in the introduction Example 7 This example generalizes the results of Example 6 and demonstrates the use of antilinear eigenvalue problems to calculate the norm of a complex symmetric operator Consider the nonlinear extremal problem a z + a z = sup a f()f () + a f() (4) fk kfk= where a and a are fixed To ensure that the problem (4) is nontrivial, we make the additional assumption that a 6= By the discussion in Example 5, it follows that a z + a = ka a+a z zk Kz!K = a z a a Although the norm of the above matrix can be computed explicitly using standard methods or the explicit formula (38), let us now illustrate our anti-linear eigenvalue technique The conjugation C on K z is given by Cf = fzz = fz n other words, C(c + c z)=c + c z, c,c C With respect to the monomial basis {,z} for K z we see that the anti-linear eigenvalue problem A a+a zu = Cu (which has a unit vector solution) is equivalent to the following R-linear problem: a u u = (43) a a u u The norm of the associated Toeplitz matrix coincides with the largest for which the preceding system is consistent

19 8 Stephan Ramon Garcia and William T Ross CMFT n light of the assumption that a 6=, it follows that u 6=elseu = u = Solving the first equation in (43) for u and substituting the result in the second yields a a u = u (44) a Since the function 7! is increasing for >, it follows from (44) that the largest positive number satisfying (43) must satisfy a = a This in turn means that satisfies the quadratic equation a a = Solving this yields the explicit answer a z + a = a + p a +4 a, (45) z which agrees with (38) On the other hand, a direct attack on the problem would involve computing the eigenvalues of the matrix a a a a = a a a a a a a a + a and then explicitly computing their square roots This involves considerably more symbolic computation For instance, Mathematica yields the seemingly more complicated answer r a + a a p a +4 a, which is, of course, equal to (45) 6 Computing an extremal function By Corollary we have ( ) =ka k K!K = ha u, Cui, where u K is a unit vector solution to the anti-linear eigenvalue problem A u = ka kcu (such a vector exists) Here is a procedure which, at least in principle, can be used to compute an extremal function u for ( ) As before, let M A denote the matrix representation of the truncated Toeplitz operator A with respect to the modified Aleksandrov-Clark basis {v,v,,v n } in () n particular, recall that Cv j = v j for all apple j apple n by () f u K is a unit vector, then u = c v + c v + + c n v n, c + c + + c n = f, in addition, u is a solution to then the coe A u = ka kcu, cients c,c,,c n satisfy the equation M A B c n A = km A k c n C A (46)

20 (), No A nonlinear extremal problem on the Hardy space 9 where M A is a complex symmetric matrix f we write c j = x j + iy j where x j,y j R, the above equation can be written as a linear system in n real variables and then solved using linear algebra (then normalizing so that c + + c n = ) Remark f one only wishes to find the extremal function up to a unimodular constant, then one can avoid the conjugation C as follows: f {v,,v n } is any orthonormal basis for K and M is the matrix representation of A : K! K with respect to {v,,v n },it follows from the proof of part (iii) of Theorem that if (c,,c n ) C n is a unit vector and M M then, for some unimodular constant, is an extremal function for ( ) c n C A = kmk c n u = (c v + + c n v n ) Example 8 Let us return to the data of Example ( ) =ka k K!K 85, where C A, =z z / z/, = z + z / Moreover, we found that the desired supremum is equal to! 5 4 7i 4 p 3 7i 4 p 3 3 = 6 (7 + p 37) 85 n particular, recall that we have With = and =, =,! =,! = i, we can use () to compute the modified Aleksandrov-Clark basis v (z) = 3 z, v (z) = i p 3 i p 3 z for the model space K An extremal function u is any unit vector solution to A u = ka kcu To compute u = c v +c v, we need to find a unit vector solution to the anti-linear eigenvalue problem! 5 7i 4 4 p c 3 = 6 (7 + p c 37) 7i 4 p 3 A computation with Mathematica yields the coe cients s c = + p, c = s 37 i 3 c c p 37 Therefore, the desired unit vector u = c v + c v equals q q z p 37 + p 37 z z Another computation with Mathematica will show that A has two distinct singular values and hence the kernel of A ka k is one-dimensional By Corollary it follows that this extremal solution u for ( ) is unique up to a sign

21 Stephan Ramon Garcia and William T Ross CMFT When examining the extremal problem c z + c z + + c n z n+, we saw from our previous discussion that this supremum is equal to the norm of the lower triangular Toeplitz matrix T from (4), which is simply the matrix representation of A z n+ : K z n+! K z n+, where (z) = c z + + c n z n+, with respect to the orthonormal basis {,z,,z n } for K z n+ Observing that C(a + a z + + a n z n )=a n + a n z + + a z n, we see that in order to find a unit vector solution u to A z n+u = ka z n+kcu, we need to find a unit vector solution (a,a,,a n )to c ṇ a a Bc c n C B = ka z C n+k c c c n a n c c c c n a n a n a n a a (47) C A As before, writing a j = x j + iy j, x j,y j R, this can be solved for a,a,,a n using linear algebra (and then normalizing so that a + a + + a n = ) Example 9 We saw from Example 6 that z + z = ka +z k Kz!K z = = +p 5 68 To find an extremal function u = a + a z K z, we need to find a unit vector solution to a = +p 5 a a a A short computation yields the extremal function Recalling that u(z) = z + p 5+ q 5+ p 5 z + z = z + z, observe that u is the square root of the H extremal function from Egerváry s example (5) Another computation with Mathematica shows that A +z has two distinct singular values and so the kernel of A +z ka +z k is one dimensional By Corollary, we conclude that the extremal solution u for ( ) is unique up to a sign Example Consider the rational kernel =/z, which yields the nonlinear extremal problem f (z) i z dz = sup f()f () ( ) = sup fk kfk= fk kfk=

22 (), No A nonlinear extremal problem on the Hardy space According to our recipe, the corresponding Blaschke product is = z We therefore have ' = =whencea ' = and z = = n this case, f = p ( + z), f = p ( +z) (48) are linearly independent extremal functions for ( ) Despite the fact that the corresponding model space K z = W {,z} is two-dimensional, this does not imply that every function f(z) = a + a z satisfying a + a = is extremal ndeed, simply consider the functions or z This behavior is not unexpected, since the underlying extremal problem is nonlinear Example Recall the nonlinear extremal problem a z + a = a + p a +4 a z from Example 7 With denoting the above extremum, a computation with Mathematica shows that a solution to the corresponding antilinear eigenvalue problem (43) is given by u = i a a u = a a a + a, q a + a + + a a + a a q a + a + + a a This means that an extremal function is f = u + u z One can check that the singular values of a a are distinct if and only if a = Therefore f is the unique extremal function, up to sign, for (4) if and only if a = When a =, the supremum is equal to a and one can check by direct computation that (48) are two linearly independent solutions to (4) 7 Norm attaining symbols Since the truncated Toeplitz operator A ' : K! K is a multiplication operator followed by an orthogonal projection, it follows immediately that a ka ' k K!K applek'k, ' H (49) For a rational function having no poles on, we have shown that the nonlinear extremal problem ( ) given by () yields the supremum ( ) =ka ' k K!K where is the corresponding Blaschke product (6) and ' = n particular, it follows that ( ) = sup fh kfk= i f dz = ka ' k K!K applek'k (5) We now investigate conditions under which equality holds in (49) and (5) We say that a unit vector x is a maximal vector for a bounded linear operator T : H!H if ktxk = kt k t is clear that maximal vectors exist for any compact operator and thus for any operator on a finite-dimensional space From our perspective, the importance of maximal vectors stems from the following important lemma [37, Prop 5]:

23 Stephan Ramon Garcia and William T Ross CMFT Lemma 4 (Sarason) Let be any inner function and T : K! K be a linear operator of unit norm that commutes with A z : K! K f T has a maximal vector, then there is a unique function ' H such that k'k =and A ' = T Moreover, ' is both an inner function and the quotient of two functions from K We say that a symbol ' L for a truncated Toeplitz operator A ' on K is norm-attaining if ka ' k = k'k t is important to note that this definition depends upon the particular inner function corresponding to the model space K upon which A ' acts For a finite Blaschke product, the next result says exactly when we have equality in (49): Theorem Let be a finite Blaschke product A symbol ' in H is norm-attaining with respect to K if and only if ' is a scalar multiple of the inner factor of a function from K Proof ()) Without loss of generality, we assume that ka ' k = k'k = Since is a finite Blaschke product, it follows that K is finite-dimensional whence the truncated Toeplitz operator A ' : K! K has a maximal vector Now recall the well-known fact [37, Thm ] that the commutant of A z : K! K is {A ' : K! K : ' H } whence A ' commutes with A z By Lemma 4, it follows immediately that ' is an inner function and ' = f/g, for some f,g K Thusf = 'g = ' g F g,where g is the inner factor of g and F g is the outer factor A short argument now reveals that A g f = P ( g ' g F g )=P ('F g )='F g (5) since the function 'F g belongs to H and is orthogonal to H ndeed, since f belongs to K we have, for any h H, h'f g, hi = h' g F g, ( g h)i = hf, ( g h)i = We conclude from (5) that ' is indeed the inner factor of a function from K (() Conversely, let ' be the inner factor of a function from K n other words, suppose that there exists an outer function F of unit norm such that 'F belongs to K This implies that ka ' F k = kp ('F )k = k'f k = kf k = whence ka ' k = Thus ' is a norm-attaining symbol with respect to Let us make a few important remarks concerning possible generalizations of the preceding theorem First observe that the (() implication of Theorem remains true for any inner function Furthermore, the ()) implication of the theorem remains true for general inner and ' H as long as the operator A ' : K! K has a maximal vector This occurs, for instance, if A ' is compact Corollary 3 Let be a finite Blaschke product f ' H is not a scalar multiple of a Blaschke product, then ka ' k < k'k Corollary 4 f is an inner function and ' is a finite Blaschke product of degree < dim K, then ' is a norm attaining symbol for K Proof Let,,, n denote the zeros of ' and note that n<dim K Consequently there exists a nonzero function f K such that f is orthogonal to the n kernel functions k,k,,k n n other words, there exists a nonzero function f K which vanishes at each,,, n Since inner factors of functions in K can be removed without leaving K, we may assume that the inner factor of f is precisely ' By the remarks preceding Corollary 3, it follows that ' is a norm-attaining symbol for K

24 (), No A nonlinear extremal problem on the Hardy space 3 Example f is a singular inner function, then there exists a Blaschke product ' having simple zeros such that ' is a norm-attaining symbol for K ndeed, an argument using Frostman s Theorem [3, Thm 3] (see also [5, Sec 33]) produces many such Blaschke products which occur as inner factors of functions in K Now employ the observations made prior to Corollary 3 The following corollary of Theorem (along with Corollary ) tells us when we have equality in (5) Corollary 5 For a rational function having no poles in, the following are equivalent: (i) ( ) = ( ) = max ( ), (ii) There exist a c C and finite Blaschke products B and B having no common zeros and satisfying deg B < deg B so that = c B B Proof f we assume (i), then, adopting the notation from Corollary, we see that the symbol ' = is norm-attaining with respect to K Thus by Theorem it follows that =cb where c C and B is the inner factor of some function f K Using the representation () we find that p(z) f(z) = Q n j= ( jz), (5) where p(z) is a polynomial with deg p<n= deg t follows that B is a finite Blaschke product with deg B < n The fact that and B have no common zeros follows from the definition of from (6) Conversely, if we assume (ii), then = c B,whereB and are finite Blaschke products with deg B<deg f,,, n are the zeros of (repeated according to multiplicity), then every f K takes the form (5) f we take p(z) to be a polynomial whose zeros are precisely those of B, which is possible since deg B<n,thenB will be the inner factor of a function from K By Theorem it then follows that the symbol is norm attaining with respect to K To conclude the proof, simply appeal to Corollary Example 3 n Example 6 we showed that z + z = +p 5 Observe that with (z) =/z +/z,wehave 68 max ( ) => ( ) This strict inequality is expected since the rational function (z) = z + z cannot be written as B /B where deg B < deg B ndeed, is the quotient of the nonconstant outer function z + and the finite Blaschke product z Example 4 n a very similar way, we know from Example that z + = z / 6 (7 + p 37) 85

25 4 Stephan Ramon Garcia and William T Ross CMFT With (z) =/z +/(z ), a brief calculation shows that max ( ) =3> ( ) As in the preceding example, the strict inequality is to be expected since (z) = z 4 z(z ) cannot be written in the form B /B where deg B < deg B Example 5 Let '(z) = z and Since ka z kapplekzk = and z (z) =z z (53) ka z ()k = kp (z)k = kzk =, it follows immediately that ka z k = That z is a norm-attaining symbol for K is to be expected, since z is indeed the inner factor of a function from K This can also be verified by a direct computation Using = in our recipe, we find that = and = whence! = and! = i Putting this all together yields the modified Aleksandrov-Clark basis v (z) = p (z) 3 (z), v (z) = i z +z with respect to which A z has the matrix representation! M Az = which, via (38), has norm, as expected 3 4 i p 3 4 i p Example 6 Let '(z) = z and let as in (53) Using the data from the previous example, we see that! M A' = 8 i p 3 8 i p which has norm equal to 4 ( + p 3) 6833 < =k'k This is to be expected since ' is outer 8 Best constants As an application of the identity ( ) = ( ) =ka k K!K, we can obtain the best constant in various pointwise estimates of the derivatives of H functions Such estimates have been studied before [,, 6, 7] (The papers [, 4, 5, 36] contain related results) using di erent methods and function theoretic language Our discussion uses operator theoretic language To begin, first observe that the Cauchy integral formula [, p 4] says that F (n) ( )= n! i, F (z) (z ) n+ dz, F H, D (54)