Factorizations Of Functions In H p (T n ) Takahiko Nakazi

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1 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

2 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

3 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

4 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

5 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 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

6 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 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 in [5]. For f z is in H (D), 1 and so f z is continuous on D. Now Theorem 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

7 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

8 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

9 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), T.Nakazi, Slice mas and multiliers of invariant subsaces, Can.Math.Bull. 39(1996), T.Nakazi, An outer function and several imortant functions in two variables, Arch.Math., 66(1996), T.Nakazi, On an invariant subsace whose common zero set is the zeros of some function, Nihonkai Math.J., 11(2000), 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), Takahiko Nakazi Deartment of Mathematics Hokkaido University Saoro , Jaan math.sci.hokudai.ac.j 9