Factorizations Of Functions In H (T n ) By Takahiko Nakazi * This research was artially suorted by Grant-in-Aid for Scientific Research, Ministry of Education of Jaan 2000 Mathematics Subject Classification : 32 A 35, 46 J 15 Key words and hrases : Hardy sace, olydisc, factorizatio, extreme oint 1
Abstract. We are interested in extremal functions in a Hardy sace H (T n ) (1 ). For examle, we study extreme oints of the unit ball of H 1 (T n ) and give a factorization theorem. In articular, we show that any rational function can be factorized. 2
1. Introduction Let D n be the oen unit olydisc in C n and T n be its distiguished boundary. The normalized Lebesgue measure on T n is denoted by dm. For 0 <, H (D n ) is the Hardy sace and L (T n ) is the Lebesgue sace on T n. Let N(D n ) denote the Nevanlinna class. Each f in N(D n ) has radial limits f defined on T n a.e.dm. Moreover, there is a singular measure dσ f on T n determined by f such that the least harmonic majorant u(log f ) of log f is given by u(log f )(z) = P z (log f + dσ f ) where P z denotes Poisson integration and z = (z 1, z 2,, z n ) D n. Put N (D n ) = {f N(D n ) ; dσ f 0}, then H (D n ) N (D n ) N(D n ) and H (D n ) = N (D n ) L (T n ) N(D n ) L (T n ). These facts are shown in [5, Theorem 3.3.5]. Let L be a subset of L (T n ). For a function f in H, ut L f = {φ L ; φf H }. When L f H, f is called an L-extremal function for H. When L = L (T n ), L = L R (T n ) or L = L U (T n ) is the set of all unimodular functions, such L-extremal functions have been considered in [3]. In [3], the author studied functions which have harmonic roerties (A),(B),(C). For examle, the roerty (A) is the following : If f H and f g a.e. on T n, then f g on D n. It is easy to see that f is an L-extremal function for H and L = L (T n ) if and only if f has the roerty (A). The roertis (B) and (C) are related to L = L R (T n ) and L = L U (T n ), resectively. In this aer, as L we consider only the above three sets. Definition. When f is not L-extremal for H, if there exists a function φ in L such that φf = h is an L-extremal function for H, we say that f is factorized as f = φ 1 h. In this aer, we are interested in when f is factorized for L = L (T n ) or L = L R (T n ). The function h in N(D n ) is called outer function if log h dm = log hdm >. T n T n The function q in N (D n ) is called inner function if q = 1 a.e.dm on T n. When L L, a L -extremal function is always L-extremal. If f is an outer function, then f is L-extremal for L = L (T n ). In fact, if φf is in H then φ belongs to f 1 H and f 1 H N. Hence if φ is bounded then φ belongs to H because N L (T n ) = H. When n = 1, f is L-extremal if and only if f is an outer function. This is known because f has an inner outer factorization. In this aer, for a subset S in L we say that S is of finite dimension if the linear san of S is of finite dimension. We use the follwing notations. z = (z j, z j), z j = (z 1,, z j 1, z j+1,, z n ). 3
D n = D j D j, D j = l j D l where D n = n l=1 D l and D l = D. T n = T j T j, T j = l j T l where T n = n l=1 T l and T l = T. m = m j m j, m j = l j m l where m = n l=1 m l and m l is the normarized Lebesgue measure on T l. 2. L = L R In this section, we assume that L = L R. When n = 1, any nonzero function in H has a L R -factorization in H by Proosition 1. Even if n > 1, we have a lot of L R -extremal functions for H. Proosition 1. If f = qh where q is inner and h is L R -extremal for H, then f has a L R -factorization in H : f = φ 1 k where φ = q+ q and k = (1+q 2 )h is L R -extremal for H. Proof. It is enough to show that (1 + q 2 )h is L R -extremal for H. If ψ L R (T n ) and ψ(1 + q 2 )h H then ψ(1 + q 2 ) belongs to H because h is L R -extremal for H. Since 1 + q 2 is outer and so 1 + q 2 is L R -extremal for H, ψ belongs to H. Proosition 2. Suose f is a nonzero function in H 1. f is L R -extremal for H 1 if and only if f/ f 1 is an extreme oint of the unit ball of H 1. Proof. It is well known. The degree of a monomial z α 1 1 zn αn (where α i Z + ) is α 1 + +α n. The degree of a olynomial P is the maximum of the degrees of the monomials which occur in P with non-zero coefficient. The degree of a rational function f = P/Q is the maximum of deg P, deg Q, rovided that all common factors of ositive degree have first been cancelled. Theorem 3. Let 0 < and L = L R. If f is a nonzero function in H and L f is of finite dimension then there exists a function φ in L such that f = φ 1 h and h is L-extremal for H. Proof. Suose that L f is of finite dimension. Then there exist s 1, s 2,, s n in L R such that {s j } n is a basis of L f, s 1 = 1 and s 1 n exists a real number λ such that (s n λ) 1 basis. that L snf When L snf / L. For if s 1 n L then there / L. Then {s 1, s 2,, (s n λ)} is also a = R, ut φ = s n and h = s n f, then the theorem is roved. Suose then l 1 s n is nonconstant because = R, ut φ = l 1 s n and R. Then there exists l 2 in R, we can roceed R. If l 1 is a nonconstant function in L snf s 1 n / L. We may assume that l 1 1 / L. When L l 1s nf h = l 1 s n f, then the theorem is roved. Suose that L l 1s nf L l 1s nf such that l 2 l 1 s n is nonconstant and l 1 2 / L. When L l 2l 1 s nf similarly. Put k j = l j l j 1 l 1 (j = 1, 2,, n) 4
where l 1 i / L (1 i j). Suose that L k js nf k j s n = α ij s i (j = 1, 2,, n) i=1 and so for j = 1, 2,, n α ij s i + (α nj k j )s n = 0. i=1 Hence α 11 α n 1 1 α n1 k 1 α 12 α n 1 2 α n2 k 2 = 0... α 1n α n 1 n α nn k n and so there exist γ 1,, γ n in C such that where Hence R for j = 1, 2,, n. Hence γ 1 (α n1 k 1 ) + γ 2 (α n2 k 2 ) + + γ n (α nn k n ) = 0 α 11 α n 1 1 γ j = α 1 j 1 α n 1 j 1 α 1 j+1 α < j n 1 j+1 α 1n α n 1 n γ j α nj = γ j k j. Here we need the following claim. t Claim For any t (1 t n), if (δ 1,, δ t ) (0,, 0) then δ = δ j k j can not be constant. Proof. Let s be the smallest integer such that δ s 0 and 1 s t. Then t δ = δ j k j. Hence j=s δ = δ s (l 1 l s ) + + δ t (l 1 l s )l s+1 l t. If δ = 0, then 0 = δ s + δ s+1 l s+1 + + δ t l s+1 l t and this contradicts that l 1 s+1 / L because δ s 0. If δ is a nonzero constant, then this contradicts that (l 1 l s ) 1 / L. Now we will rove that the equality : γ j α nj = γ j k j contradicts the definition of k j (1 j n). If γ n = 0, then there exist (δ 1,, δ n 1 ) (0,, 0) such that 5
δ 1 (α 11,, α n 1 1 ) + + δ n 1 (α 1 n 1,, α n 1 n 1 ) = (0,, 0). Hence because i=1 δ j α nj s n = δ j k j s n α ij s i + α nj s n = k j s n for j = 1, 2,, n. Hence δ j α nj = δ j k j because s n > 0. This contradicts the claim. Hence γ n 0. Thus (γ 1,, γ n ) (0,, 0) and γ j k j is constant. This also contradicts the claim. Thus L knsnf = R and so the theorem is roved. Lemma 1. Let 1 and f be in H. If f z (ζ) = f(ζz) is a rational function (of one variable) of degree k 0 <, for almost all z T n then f is a rational function (of n variables) of degree k and k k 0. Proof. There exist a nonnegative integer k k 0 and a closed set E k such that f z (ζ) is a rational function (of one variable) of degree k for all z E k and E k is a nonemty interior. We will use [5, Theorem 5.2.2]. In Theorem 5.2.2 in [5], we ut Ω = D n and E = E k. If f z (ζ) is a rational function (of one variable) of degree k for all z E, then f belongs to Y in Theorem 5.2.2 in [5]. For f z is in H (D), 1 and so f z is continuous on D. Now Theorem 5.2.2 in [5] imlies the lemma. Proosition 4. Suose 1. L f is of finite dimension if f is a rational function. Proof. Suose f = P/Q is a nonzero function in H where P and Q are olynomials. If s L f then sp/q H and so sp H. Hence s belongs to L P and so L f L P. It is enough to rove that L P is of finite dimension. Case n = 1. We have the inner outer factorization for n = 1, that is, P = qh where q is a finite Blaschke roduct and h is an outer function in H. Then it is easy to see that L P = L q. Since L q qh q H and L q L, L q qh 2 q H 2 = q(h 2 q 2 H2 ) = q(h 2 q 2 H 2 0). H 2 q 2 H 2 0 is of finite dimension because q is a finite Blaschke roduct. Case n 1. If s L P then sp H and by Case n = 1 (sp ) z (ζ) is a rational function (of one variable) of degree k 0 for almost all z T n. By Lemma 1, sp is a rational function (of n variables) of degree k and k k 0. This imlies that L P is of finite dimension. When f is a rational function in H, by Theorem 3 and Proosition 4 f has our factorization. The function h in N(D n ) is called z j -outer if T j T j log h(z j, z j) dm = T j (log h(z j, z j)dm j )dm j >. T j 6
Proosition 5. Fix 1 j n. If f(z 1,, z n ) is z i -outer in H for i j and 1 i n, then f has a factorization in H. Proof. We will generalize Theorem 2 in [3]. That is, when h is z i -outer in H for i j, L h = R if and only if the common inner divisor of {h α (z j )} α is constant, where h α (z j ) = h(z j, z j)z α j dm j T j, α = (α 1,, α j 1, α j+1,, α n ) and z α j = z α 1 1 z α j 1 j 1 z α j+1 j+1 z n αn. Note that h α (z j ) belongs to H (T j ). For the roof, we use the following notation : H (j) = {f L (T n ) ; ˆf(m 1,, m n ) = 0 if m i < 0 for all i j} and H (j) H (j) = L j = the Lebesgue sace on T j. If φ L h then g = φh and φ belongs to H (j) because h is z i-outer in H for i j. Since φ is real-valued, φ L j and so φ = φ(z j ). If the common inner divisor of {h α (z j )} α is constant, then for each α φ(z j )h α (z j ) = φ(z j )h(z j, z j)z α j dm j = g(z j, z j)z α j dm j T j belongs to H (T j ) and hence φ H (T j ). Therefor φ is constant. This imlies that L h = R. Conversely suose that L h = R. If {h α (z j )} α has a non-constant common inner divisor q(z j ), ut φ(z j, z j) = q(z j ) + q(z j ), then g = φh belongs to H. This contradiction shows the only if art. Now we will rove that f has a factorization in H. If {f α (z j )} α does not have common inner divisors, then by what was just rove L f = R and so we need not rove. If {f α (z j )} α have common inner divisors, let q(z j ) be the greatest common inner divisor. Put φ = q(z j ) + q(z j ) and h = φf, then h belongs to H and L h = R. This comletes the roof. T j 3 L = L In this section, we assume that L = L. If f is a L -extremal function for H then f is also a L R -extremal function for H. When n = 1, the converse is true. However this is not true for n 1. For examle, z 2w is a L R -extremal fnction but not a L -extremal. We can rove an analogy of Proosition 1 for L. Let M f be an invariant closed subsace generated by f in H and M(M f ) the set of multiliers of M f (see [1]). Then M(M f ) = L f for L = L. It is easy to see that (L ) f (L ) f = (L R ) f + i(l R ) f. It is easy to see that (L ) f is a weak closed invariant subsace which contains H. Then (L ) f /H is of infinite dimension (see [4, Theorem 1]). Thus we can not exect the analogy of Theorem 3. 7
in L f. Proosition 6. Let 1 and f a nonzero function in H. Suose φ is (1) L f φl φf φh. (2) φ 1 is in L if and only if L f = φl φf (3) If L φf belongs to H. then L φf. L f then φ belongs to H. If L f L φf Proof. (1) If g L φf φ 1 L f L φf by (1). If L f = φl φf and φ 1 is in L then φ 1 then φgf = gφf H and so φg L f. (2) If φ 1 L then φ 1 L f = L φf and so φ 1 L φf. This L f. If k L f then k L φf imlies that φ 1 L. (3) Suose L φf and so kφf H. Hence φ 2 f H. Reeating this rocess, φ n L f and so φ n f H for all n 1. Thus φ belongs to H. If L f L φf was roved above for φ 1 assuming φ 1 (φf) = f. and φ 1 L, then φ 1 belongs to H. For aly what When f is a nonzero function in H, f is factorable in H if and only if there exists a nonzero function h in H such that f h a.e. on T 2 and L h = H. Proosition 7. Let 1 and f be a nonzero function in H. Suose φ is a nonzero function in L f. (1) If φ 1 is in L and L φf = H, then φ 1 belongs to H and L f = φh. (2) If L f = φh, then φ 1 belongs to H and L φf = H. (3) If L f is the weak closure of φh and φ = h a.e. for some function h in H, then L φ 0f = H for some inner function φ 0 and so f is factorable. (4) There exist f and φ such that φ is not the quotient of any two menbers of H (T n ). Proof. (1) By (2) of Proosition 6, φl φf = L f. Since L φf = H, φh = L f H and so φ 1 belongs to H. (2) Since L f φ L φf φh by (1) of Proosition 6 and L f = φh, L φf = H. It is clear that φ 1 H. (3) Since φ = h a.e., φ = φ 0 h and φ 0 = 1 a.e.. Then L f = [φh ] = φ 0 [hh ] H and so φ 0 is an inner function where [S] is the weak closure of S. (4) This is a result of [6]. Proosition 8. Let 1. Suose f and g are nonzero functions in H. (1) If L f = H and f g a.e., then there exists a function φ in H such that g = φf. (2) If L f = H and f = g a.e., then there exists an inner function φ such that g = φf. Proof. (1) Let φ = g/f, then φ L because f g a.e.. By (1) of Proosition 6, φ belongs to H because H = L f. (2) follows from (1). Proosition 9. Let 1. If f is homogeneous olynomial such that f(z 1,, z n ) = g(z, w) where z = z i, w = z j and i j then f is factorable in H. l l ( ) w j Proof. Since f(z 1,, z n ) = a j z l j w j, f(z 1,, z n ) = z l a j = j=0 j=0 z 8
l c (b j w c j z) where b j = 1 or c j = 1, and b j 1, c j 1. It is easy to see that j=0 L f = φh where φ = (αz βw) 1 and (α, β) ( D D) (D D) (cf. [2],[4]). By (2) of Proosition 7, f is factorable. Question (1) For any nonzero function f in H, does there exists a function φ such that L f L φf? (2) Describe φ in L such that L f L φf. (3) Describe φ in L such that [φh ] H. References 1. T.Nakazi, Certain invariant subsaces of H 2 and L 2 on a bidisc, Can.J.Math. XL(1988), 1272-1280. 2. T.Nakazi, Slice mas and multiliers of invariant subsaces, Can.Math.Bull. 39(1996), 219-226. 3. T.Nakazi, An outer function and several imortant functions in two variables, Arch.Math., 66(1996), 490-498. 4. T.Nakazi, On an invariant subsace whose common zero set is the zeros of some function, Nihonkai Math.J., 11(2000), 1-9. 5. W.Rudin, Function Theory in Polydisks, Benjamin, New York (1969) 6. W.Rudin, Invariant subsaces of H 2 on a torus, J.Funct.Anal. 61(1985), 378-384. Takahiko Nakazi Deartment of Mathematics Hokkaido University Saoro 060-0810, Jaan E-mail : nakazi @ math.sci.hokudai.ac.j 9