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The Failure of Fatou's Theorem on Poisson Integrals of Pettis Integrable Functions* Francisco J

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The Failure of Fatou's Theorem on Poisson Integrals of Pettis Integrable Functions* Francisco J 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 1p<+ 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 1p<+. It is also well known that every integrable function F has a conjugate function or Hilbert transform F .If1<p<, the M. Riesz theorem states that the Hilbert transform is bounded from L p(T) into itself, which is equivalent to the fact that the Riesz projection is bounded from L p(T) onto H p(T). The problem of extending classical theorems in Harmonic Analysis to the setting of Banach space valued functions or measures has been con- sidered by several authors [1], [2], [3], [7], [8], .... It is known that for every F in the Bochner space L p(T, X ), 1p<, the convolution 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<p<) is equivalent to the fact that X has the UMD property. If X has this property, then F exists for every F # L1(T, X ) as a consequence of the weak inequality [11]. In connection with our work, in [7] some examples of Bochner integrable functions without conjugate function are given. Recently, in [5] some examples are given of Pettis integrable functions which do not satisfy Lebesgue's differentiation theorem which is on the basis of Fatou's theorem for both the scalar case and the vector valued case. In this paper we study the convergence of the Poisson integral of Pettis integrable functions: the integral in the definition of the convolution must be now taken in the Pettis sense. It is said that a strongly measurable function F: T Ä X is Pettis integrable when x*F # L1(T) for every x*#X*, the dual of 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 1p<, 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 1p< 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<p<+. Nevertheless, we cannot conclude from this fact that it is not complemented by any other projection, because it is not complete (as we prove in Proposition 7) thus Rudin averaging method is not available in this context. The Poisson integral of the function F constructed in Theorem 2 is harmonic but it is not analytic. We consider the problem of giving an analytic function F failing Fatou's theorem. If X is a Banach space without finite cotype we succeed in giving such a function in Theorem 9. In the general case, we construct in Theorem 10 an analytic countably additive, absolutely continuous vector measure (without any Pettis density) + defined on the Lebesgue measurable sets of T such that for every t # T, 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 =<if(k, n){(j, m), and n &nb (2) *(Ak)C2 . 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+1mn+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 .
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