journal of functional analysis 160, 2841 (1998) article no. FU983338
The Failure of Fatou's Theorem on Poisson Integrals of Pettis Integrable Functions* Francisco J. Freniche, Juan Carlos Garcia-Vazquez, and Luis Rodriguez-Piazza
Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Sevilla, Apdo. 1160, Sevilla 41080, Spain E-mail: frenicheÄcica.es, garciaÄcica.es, piazzaÄcica.es Received May 22, 1997; accepted May 21, 1998
In this paper we prove that for every infinite-dimensional Banach space X and every 1 p<+ there exists a strongly measurable X-valued p-Pettis integrable function on the unit circle T such that the X-valued harmonic function defined as its Poisson integral does not converge radially at any point of T, not even in the weak topology. We also show that this function does not admit a conjugate func- tion. An application to spaces of vector valued harmonic functions is given. In the case that X does not have finite cotype we can construct the function with the addi- tional property of being analytic, in the sense that its Fourier coefficients of negative frequency are null. In the general case we are able to give a countably additive vector measure, analytic in the same sense. 1998 Academic Press
INTRODUCTION
Let + be a finite Borel measure on the unit circle T. The classical Fatou's theorem states that the Poisson integral of + converges radially on every point t at which + is differentiable with respect to the normalized Lebesgue measure *. In particular, because of Lebesgue's differentiation theorem, if F is integrable on T, then its Poisson integral converges almost everywhere. The Poisson integral of + is the harmonic function on the unit disc D, 2 defined as the convolution with the Poisson kernel Pr(t)=(1&r )Â (1&2r cos t+r2), that is,
Pr V +(t)= Pr(t&s) d+(s). |T On the other hand, as for any summability kernel, it is also a well known p p fact that Pr V F converges to F in the norm of L (T), whenever F # L (T),
* Research supported by DGICYT grant PB96-1327. Part of this work is from the second author's Doctoral Thesis written at the Universidad de Sevilla, under the supervision of the first author. 28 0022-1236Â98 25.00 Copyright 1998 by Academic Press All rights of reproduction in any form reserved. PETTIS INTEGRABLE FUNCTIONS 29 for 1 p<+ . It is also well known that every integrable function F has a conjugate function or Hilbert transform F .If1
Pr V F(t)= Pr(t&s) F(s) d*(s), |T where the integral is now considered in the Bochner sense, converges to F p in the norm of the Bochner space L (T, X )asr Ä 1 and Pr V F(t) con- verges also pointwise to F(t) for almost all t # T [3], [7]. We also notice that it is well known that the boundedness of the Riesz projection from L p(T, X ) onto H p(T, X )(1
X, and for every measurable set A, there exists A F(t) d*(t)#X, called the
Pettis integral of F on A, satisfying x* A F(t) d*(t)= A x*F(t) d*(t). If 1 p< , we shall say that F is p-Pettis integrable when it is Pettis integrable and x*F # L p(T) for every x*#X*, and the operator x*#X* Ä p x*F # L (T) is compact. We shall denote by Pp(T, X ) the normed space of (equivalence classes of) Pettis p-integrable functions with the norm
1Âp &F& =sup |x*(F(t))| p d*(t) : &x*& 1 . Pp {\|T + =
In Theorem 2 we give for every infinite-dimensional Banach space X and for every 1 p< a p-Pettis integrable function F such that for every t # T, Pr V F(t) does not converge (not even in the weak topology). Our construction is inspired in that of [5] but we need more precise estimates, 30 FRENICHE, GARCIA-VAZQUEZ, AND RODRIGUEZ-PIAZZA which leads us to prove Lemma 1, in which we built a suitable sequence of Cantor-like sets. The function F that we construct in Theorem 2 does not have a con- jugate function, that is, there is no G: T Ä X** such that x*G is the conjugate function of x*F, for every x*#X*. As a consequence we also obtain that the spaces of harmonic vector valued functions hp(D, X ) and w hp (D, X ) considered in [2] do not coincide when X is infinite dimensional (see Section 1 below for the definitions of these spaces). Regarding the convergence in the norm, we also prove in Section 1 that if F # Pp(T, X ), then Pr V F converges to F in the norm of Pp(T, X ) when r Ä 1. These results are closely related to that of [2] if we think of p F # Pp(T, X ) as an operator from X* into L (T). a In Section 2 we study the normed space Pp(T, X ) of analytic p-Pettis integrable functions consisting of those functions F # Pp(T, X ) which are &int analytic, that is, F (n)= T F(t) e d*(t)=0 for every n<0 (the integral is defined in the Pettis sense). As a consequence of Theorem 3, we obtain that, in contrast with the M. Riesz theorem in the scalar case, if X is a infinite-dimensional then the space Pp(T, X ) is not complemented in Pp(T, X ) by the natural Riesz projection, even if 1
Pr V +(t) has no limit (not even weakly). We do not know if there exists an analytic, strongly measurable Pettis integrable function with that property. We point out that although the results stated in Theorem 2, Proposi- tion 5, and Theorem 10 also hold for the Fejer kernel, the techniques used in Theorem 9 do not work in this case. Along this work we shall use standard notation as can be found in [10], [4], [9].
1. NORM AND POINTWISE CONVERGENCE OF POISSON INTEGRALS OF PETTIS INTEGRABLE FUNCTIONS
n n Let I k be the sequence of the dyadic intervals, n=1, 2, ..., k=1,2,...,2 , n n n that is I k=[2?(k&1)Â2 ,2?kÂ2 ). PETTIS INTEGRABLE FUNCTIONS 31
n n Lemma 1. Let b>1. There exist measurable sets Ak/I k , n=1, 2, ..., k=1, 2, ..., 2n, and a constant C depending on b such that:
n m (1) Ak & A j = n Proof.LetN be a nonnegative integer to be determined. Let Bk/ n+N n n &N &nb n n m I (k&1) 2N+1/I k such that *(Bk)=2 2 .LetAk=Bk " [Bj : m>n, j=1, ..., 2m]. Condition (1) in the statement of the lemma is clearly satisfied by construction. m Let us notice that, if n+1 m n+N, there is at most one of the Bj n cutting Bk . Thus we have N j n n 1 2 *(Ak) *(Bk)& : N (n+ j) b& : N (n+N+ j) b j=1 2 2 j=1 2 2 1 N 1 2 j N nb 1& : jb& : (N+ j) b 2 2 \ j=1 2 j=1 2 + 1 1 1 1 N nb 1& b & Nb : (b&1) j . 2 2 \ 2 &1 2 j=1 2 + Taking N big enough, we obtain condition (2). K We recall that Y= X is the injective tensor product, the completion of the space of weak* to weak continuous finite rank operators from X* into Y, with the uniform norm of operators. Theorem 2. Let 1 p< and let X be infinite dimensional. There exists F: T Ä X Pettis p-integrable such that lim &Pr V F(t)&=+ r Ä 1 uniformly in t # T. Proof. Let (xj)j=1 be a basic sequence in X. Let (Mn)n=1 be a sequence of finite intervals of N with max Mn 2n 1Â2 2n 2n 1Â2 1 2 n 2 2 : |ak | : ak ek 2 : |ak | \k=1 + " k=1 " \k=1 + for every ak # C. 32 FRENICHE, GARCIA-VAZQUEZ, AND RODRIGUEZ-PIAZZA Let 0 2n 1 1 n F(t)= : : /An e , na n k k n=1 2 k=1 *(Ak) n n where the sets Ak are given by Lemma 1 for b. As the Ak are pairwise disjoint, it is plain that F is a well-defined strongly measurable function. Let us see that F is p-Pettis integrable. Let q be the conjugate exponent of p. For every g # Lq(T) such that &g&q 1, we have 2n 2n 2 1Â2 1 n 2 nb n : g/An d* ek 2 : g/A d* n | k | k " k=1 *(Ak) \ T + " C \k=1 \ T + + 2n 2 nb 2 sup : ak /An &(a )& 1 k C k 2 " k=1 "p 2 nb 1 2 nbÂp sup &(ak)&p . C 2 &(ak)&2 1 If 1 p<2 then we can bound this by (2ÂC)2nbÂq(2n)(1Âp&1Â2). If we take b>1 and a<1 such that (1Âp)&(1Â2)+(bÂq) 2n 1 1 n n : : g/A d* ek na n | k n=1 2 " k=1 *(Ak) \ T + " converges uniformly in &g&q 1, so that there exists an operator p u # L (T)= X defined by 2n 1 1 n n u(g)= : : g/A d* ek na n | k n=1 2 k=1 *(Ak) \ T + satisfying x*F=u*(x*) and it follows that F # Pp(T, X ). If 2 p< then we obtain the bound (2ÂC)2nbÂq. Then it suffices to take b>1 and a<1 such that (bÂq) 2? 1&r 4? , 2n 4 2n n and we take k such that t # I k . PETTIS INTEGRABLE FUNCTIONS 33 2 It is easy to check that Pr(0)=(1+r)Â(1&r) and |Pr $(s)| 2rÂ(1&r) , n for every |s| 2?Â2 . It follows that Pr(s) 1Â(1&r) for those s. We have 2m 1 1 m &Pr V F(t)&= : : Pr V /Am (t) e ma m j j " m=1 2 j=1 *(A j ) " 1 1 1 Pr(t&s) d*(s) na n | n 2c 2 *(Ak) } Ak } 1 1 1&r a 1 , 2c \4? 4 + 1&r where c is twice the basis constant. Consequently, &Pr V F(t)& tends to as r goes to 1 indepently on t. K Remark. The analogous result for the Fejer kernel n | j| ijt Kn(t)=: 1& e &n \ n+1+ is true. We take the same function as in Theorem 2 and we take into account the following facts on the Fejer kernel: (a) Kn(0)=n and (b) 2 |K$n(s)| n for every s # T. The proof then follows the same lines as for the Poisson kernel. Let 1 p< .Letu: X* Ä L p(T) be a bounded operator and F: T Ä X** be a function. We say that u is representable by F if u(x*)= x*F for every x*#X*. Let H be the Hilbert transform taking f into H( f )=f , the conjugate function. Given an operator u: X* Ä L p(T), we shall say that u~: X* Ä L p(T) is the conjugate operator of u if u~(x*)=H(u(x*)) for every p x*#X*. Let F # Pp(T, X ), and uF : X* Ä L (T) the operator induced by F, p that is, uF (x*)=x*F; by definition, uF # L (T)= X.IfuF has a conjugate operator and it is representable by a function, we will say that F admits conjugate function. Theorem 3. Let 1 p< and F the function constructed in the proof of Theorem 2. Then F has a conjugate operator which is not representable by a function, hence F does not admit conjugate function. Proof.Let1 Let p=1 and 0 Let 1 p< . Assume that G: T Ä X** represents u~F in the sense that u~F (x*)=x*G for every x*#X*. Then as in the proof of Lemma 2.6 in [12] there exists a measurable h: T Ä R such that |u~F (x*)(t)| &x*& h(t) almost everywhere. Let Y be the closed linear subspace generated by the basic sequence (xj). p The operator u~F can be seen as a member of L (T)= Y satisfying |u~F ( y*)(t)| &y*&h(t) almost everywhere. Let Bn=[t # T: n&1 h(t) 2n 1 1 n ?n**G= : /~An e na n k k 2 k=1 *(Ak) almost everywhere, since for every y*#Y* 2n 1 1 n y*(?n**G)=(?n*y*) G=u~F (?n*y*)= : /~An ?n*y*(e ). na n k k 2 k=1*(Ak) na 2n n n Let t # T such that ?**n G(t)=1Â2 (1Â*(A )) /~An(t) e for every n. k=1 k k k Let n>1 such that 2?Â2n t&s t&s /~An(t)= lim cot /An(s) d*(s)= cot /An(s) d*(s) k = Ä 0 ||t&s|>= 2 k |T 2 k Then we have &?n**&&G(t)& &?**(n G(t))& 2n 1 1 n = : /~An (t)e & na n j j 2 " j=1 *(Aj ) 1 1 t&s cot /An(s) d*(s) na n k 2 *(Ak) |T 2 1 1 1 n n na n 2 *(Ak) 2 *(Ak) 2 PETTIS INTEGRABLE FUNCTIONS 35 which tends to infinity since a<1. We arrive to &G(t)&= for almost every t # T which is a contradiction. K Remarks. (1) Let us observe that the conjugate operator, although it is not representable by a function, has an X-valued measure + which represents u~F in the sense that the measure x*+ has the density u~F (x*) with respect to Lebesgue measure for every x*#X*. The measure + is given by 2n 1 1 n n +(A)= : : /~A (t) d*(t) ek . na n | k n=1 2 k=1 *(Ak) A p n ijt (2) Let 1 Observe that the operator Sn=sn Id, where Id is the identity on X, n ijt acting on Pp(T, X )bySn(F)= &n F ( j)e , also satisfies that Sn(F) con- verges to F in Pp(T, X ). For, Sn(F)(x*)=Sn b uF (x*) which tends to uF (x*) uniformly in &x*& 1 because uF is a compact operator. It is also clear that the sequence (Sn) is uniformly bounded. Contrary to the scalar case, Theorem 3 shows that the Hilbert transform is not an p operator from Pp(T, X ) into itself. It takes values into L (T)= X. Now we give one application to spaces of vector valued harmonic func- tions. Recall that hp(D, X ) is the Banach space of harmonic X-valued functions F on the unit disc such that 1Âp &F& =sup &F(reit)& p d*(t) < . hp 0 r<1 \|T + w The space hp (D, X ) is the Banach space of harmonic X-valued functions F on the unit disc such that 1Âp it p &F&hw=sup sup |x*F(re )| d*(t) < . p | 0 r<1 &x*& 1 \ T + These spaces where studied for instance in [2] where it was shown that w for every 1 p< there exists a Banach space X such that hp (D, X ){ hp(D, X ). In the next proposition we improve this result: Proposition 4. If X is infinite dimensional and 1 p< then w hp (D, X ){hp(D, X ). Proof. Let us consider the harmonic function Pr V F(t) where F is the function constructed in Theorem 2. As &Pr V F(t)& Ä uniformly in t,it 36 FRENICHE, GARCIA-VAZQUEZ, AND RODRIGUEZ-PIAZZA follows that Pr V F(t) is not in hp(D, X ). On the other hand, Pr V F(t)isin w hp (D, X ) since &x*(Pr *F)&p=&Pr V (x*F)&p &x*&&F&. K To finish this section we shall study the convergence in the norm of the Poisson integral. p Given a operator v: L (T) Ä X, we can consider the convolution Pr *v: p L (T) Ä X as the operator Pr V v( f )=v(Pr V f ). It is clear that &Pr *v& &v&. Proposition 5. Let v: L p(T) Ä X a compact operator, w* to w con- tinuous in the case p= . Then limr Ä 1 Pr V v=v in the uniform norm of operators. Proof.Forp>1 this was proved in [2]. For p=1 the adjoint operator v* takes values in L1(T) being v a w*to w-continuous operator. Let x*#X*; then (Pr V v)* (x*)=Pr V v*(x*). For, consider f # L (T) and observe that (Pr V v)* (x*)( f )=x*(v(Pr V f ))=v*(x*)(Pr V f )=(Pr V v(x*))( f ). m Given =>0, by compactness of v*, there exist (x*)j j=1 so that m (v*(x*))j j=1 is an =-net in v*(BX*). If &x*& 1 we can find xj*andr0 such that if r r0 then &v*(x*)&(Pr *v)* (x*)&=&v*(x*)&Pr V v*(x*)& &v*&&x*&xj*&+&v*(xj*)&Pr *v*(xj*)& +&Pr V v*(x*&xj*)& (&v*&+1+&v*&) =. Therefore we have &v&Pr *v&=&v*&(Pr V v)*& (&v*&+1+&v*&) =, for r r0 . K p Let 1 p< and F # Pp(T, X ). The adjoint operator vF=(uF)*: L (T)* Ä X given by vF ( f )= T f(t) F(t) d*(t), satisfies the hypothesis of the previous Proposition. Hence Pr V vF Ä vF . Let us observe that we can define Pr V F(t)= T Pr(t&s) F(s) d*(s) as a Pettis integral. Indeed, the function Pr V F is strongly measurable because the function t # T Ä p Pr(t&s)#L (T) is continuous and Pr V F is the composition of this PETTIS INTEGRABLE FUNCTIONS 37 continuous function with the bounded linear operator vF . Moreover the p function Pr V F is in Pp(T, X ). For, take f # L (T)* and observe that f(t) Pr V F(t) d*(t)= F(t) Pr V f(t) d*(t), |T |T which is an element of X. Corollary 6. If F # Pp(T, X ) and 1 p< then Pr V F converges to F in the norm of Pp(T, X ). Proof. It is enough to see that P V v =v , which is easy to r F Pr V F check. K Remarks.(1)If+ is a countably additive X-valued measure, absolutely continuous with respect to the Lebesgue measure, then the con- volution Pr V +(t)= T Pr(t&s) d+(s) can be defined using the Bartle integral [4]. This function is Pettis integrable and with the same ideas as above it can be seen that if + has relatively compact range then the vector measures with density Pr V + converge to + in the semivariation norm. (2) The Poisson kernel can be substituted both in Proposition 5 and Corollary 6 by any abstract summability kernel in the sense of [9, p. 9]. 2. ANALYTIC PETTIS INTEGRABLE FUNCTIONS a Recall that Pp(T, X ) stands for the space of analytic p-Pettis integrable functions, consisting of those functions F # Pp(T, X ) which are analytic, &int that is, F (n)= T F(t) e d*(t)=0 for every n<0. a p Observe that Pp(T, X ) is a normed subspace of H (T)= X, the injec- tive tensor product of X and the classical Hardy space H p(T) by the map F Ä uF , where uF (x*)=x*F. Let 1 a Proposition 7. Let 1 p< . The space Pp(T, X ) is not complete if X a p is infinite dimensional. The completion of Pp(T, X ) is H (T)= X. 38 FRENICHE, GARCIA-VAZQUEZ, AND RODRIGUEZ-PIAZZA a Proof. The map F Ä uF allows us to identify Pp(T, X ) with a dense p subspace of H (T)= X.IfF is the function constructed in Theorem 2, p then we have that uF+iu~F is in H (T)= X but there is no function representing it, because u~F has no representing function. K In the following result we prove that there exists p-Pettis analytic functions which are not Bochner integrable. We denote by _n( f, t) the convolution with Fejer's kernel. Theorem 8. Let 1 p< . If X is infinite dimensional there exists a 1 F # Pp(T, X ) which is not in L (T, X ). Proof. As the identity on X is not a p-summing operator, we can take p p (xk)inX such that k=1 &xk& = , k=1 |x*(xk)| 1 for every &x*& 1, p and the operator x*#X* Ä (x*(xk)) # l is compact. q q Let us take (:k)#l such that k=1 &xk & :k= and k=1 :k=1. We choose a sequence of pairwise disjoint intervals Ak in T such that q *(Ak)=:k . Let g =*(A )&1Âp / . We can take n satisfying k k Ak k : &_ (g )&g & < . nk k k p k=1 inkt Define G(t)= k=1xk e gk(t). Since the choice of the xk and Ak ,itis easy to check that G is p-Pettis integrable but not Bochner integrable. We consider now F(t)= x einkt_ (g , t). Then F is a well defined k=1 k nk k strongly measurable function because |_ (g , t)|< almost k=1 nk k everywhere and (xk) is bounded. We have ( &F(t)&G(t)& p d*(t))p &x eink t(_ (g )&g )& < . T k=1 k nk k k p It follows that F is p-Pettis integrable but not Bochner integrable, the operator x* x*F being compact since (eink t_ (g )) is a basic sequence Ä nk k p equivalent to the canonical basis in l and the choice of (xk). K We notice at this point that, if 1 F ÂPp(T, X ). We could take F even with F (k)=0 for every k<0. For, it would be enough to take X and (xk) in the proof of Theorem 8 so that the operator x* Ä (x*(xk)) is not compact. It is easy to see that the Poisson integral Pr V F(t) of the function given in Theorem 2 is a harmonic non analytic function on the unit disk which does not have radial limits at any point. Now we give, for each X without finite cotype an analytic Pettis integrable function with values in X also failing the Fatou theorem. We notice that the proof relies on specific properties of the Poisson kernel and it is not valid for the Fejer kernel. PETTIS INTEGRABLE FUNCTIONS 39 Theorem 9. Let 1 p< . Let X be a Banach space without finite a cotype. There exists F # Pp(T, X ) such that lim supr Ä 1 &Pr V F(t)&= at any t # T. n Proof. For each n let J k be a partition of T into Mn subintervals of n n equal length, where Mn is to be fixed later. Let Ak be a subinterval of J k so that Mn n : : *(Ak)< . n=1 k=1 n Moreover, we take *(Ak) independent of k, so that there exists n $n=Mn *(Ak) for every k. n Mn There exists a closed subspace Y of X and a sequence (ek) k=1 of vectors in Y satisfying Mn 1 n max |ak | : ak ek 2 max |ak | k k 2 " k=1 " n Mn for every ak # C, and ?n : Y Ä [ek] k=1 be projections such that &?n & 2 m and ?n(ek )=0 for m{n. n n n 2 n Let .k(t)=:n for t # Ak and .k(t)=1Ân for t  Ak , where :n>0 is to be n fixed in connection with the Mn . We denote by f k the outer function satisfying n it n | f k(e )|=.k(t) almost everywhere on T, and n it n log | f k(re )|=Pr V log .k(t) for every r<1 and every t # T [6]. Mn n Mn n it n As n=1 k=1 *(Ak)< , the function F(t)= n=1 k=1 f k (e ) ek is defined almost everywhere. Let us see that the strongly measurable function 1Âp F is in Pp(T, X ) if we impose that n=1 :n($n ÂMn ) < . Indeed, if q is the conjugate exponent of p, then for every g # Lq(T) we have Mn n n n : g(t) f k(t) d*(t) ek 2 &g&q max & f k &p | k " k=1 \ T + " 1 =2 &g& : *(An)1Âp+ . q \ n k n2+ n Let us notice that F is analytic because the functions f k are analytic. Let rn>0 given by 1ÂMn=(1&rn)Â4. Let t # T. We choose k such that n t # J k . Then we have: 40 FRENICHE, GARCIA-VAZQUEZ, AND RODRIGUEZ-PIAZZA n 2 P V log . (t) (log : ) P (t&s) / n(s) d*(s)&log n rn k n rn A |T k n 1 2 (log :n) *(Ak) &log n 1&rn 1 = (log : ) $ &log n2. 4 n n Therefore 1 n 1 P V log .n(t) 1 $ Â4 &P V F(t)& |P V f (t)|= e rn k (: ) n . rn 2 rn k 2 2n2 n n3 In order that F satisfies all the requirements it is enough to take :n=2 , 2 pn3 2( p&1) $n=1Ân and Mn>2 n . K Finally, for general X, we give an analytic vector measure failing both Fatou's theorem and the corresponding result for the Fejer kernel. Theorem 10. If X is infinite dimensional there exists an X-valued analytic countably additive vector measure & such that limr Ä 1 &Pr V &(t)& = at any t # T. Proof.Let+ the X-valued measure with density F with respect to *, where F is the function constructed in Theorem 2. Let +~the measure associated to u~F as in Remark (1) following Theorem 2. We define &=++i+~. It is clear that & is analytic. The inequality &Pr V &(t)& |Pr V /An(t)+iP r V /An(t)| |Pr */An (t)|, k k k allows us to finish the proof as in Theorem 2. K REFERENCES 1. O. Blasco, Boundary values of functions in vector valued Hardy spaces and geometry on Banach spaces, J. Funct. Anal. 78 (1988), 346364. 2. O. Blasco, Boundary values of vector valued harmonic functions considered as operators, Studia Math. LXXVI (1987), 1933. 3. A. V. Bukhvalov, Hardy spaces of vector valued functions, J. Soviet Math. 16 (1981), 10511059. 4. J. Diestel and J. J. Uhl, ``Vector measures,'' Math. Surveys, Vol. 15, Amer. Math. Soc., Providence, 1977. 5. S. 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