THE p-bohr RADIUS OF A BANACH SPACE

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1 THE p-bohr RADIUS OF A BANACH SPACE O. BLASCO Abstract. Following the scalar-valued case considered by Djakow and Ramanujan in [0] we introduce, for each complex Banach space X and each 1 p <, the p-bohr radius of X as the value r p(x) = sup{r 0 : x n p r np sup f(z) p } z <1 where x n X for each n N {0} and f(z) = xnzn H (D, X). We show that for a complex (possibly infinite dimensional) Banach space X the condition r p(x) > 0 for some p and is equivalent to X being p- uniformly C-convex. We analyze the p-bohr radius in the cases X = L q (µ) for different values of p and q showing that for p < and dim(l q (µ)) one has r p(l q (µ)) = 0 while for p one has r p(l q (µ)) = 1 whenever p q p. We also provide some lower estimates for r (L q (µ)) for 1 q <. 1. Introduction and preliminaries Let us start by recalling the remarkable discovery of H. Bohr of a universal constant r 1 = 1 3 (denoted the Bohr radius) satisfying (1.1) a n ( 1 3 )n f, for any f(z) = a nz n H (D, C). The reader is referred to the paper by H. Bohr [10] which includes Wiener s proof showing that r 1 = 1 3 is sharp. A bit later some other proofs of such inequality were obtained (see [, 6]). Throughout the decades several variations of Bohr s inequality (1.1) have appeared. Djakov and Ramanujan in [0] (see also [4] for further considerations replacing the H -norm by the H p -norm) studied, for each 1 p <, the best constant r p such that (1.) ( a n p (r p ) np) 1/p f H, where f(z) = a nz n. Notice that although Bohr s result establishes that r 1 = 1/3 and clearly r p = 1 for p due to Haussdorf-Young s inequality, however computing the precise value of r p for 1 < p < seems to be rather complicated. As far as we know the best 1991 Mathematics Subject Classification. Primary 46E40; Secondary 46B0, 30B10, 30D55. Key words and phrases. Bohr radius, PL-uniform convexity. Partially supported by Project MTM P(MECC). Spain. 1

2 O. BLASCO known estimates were obtained in [0, Theorem 3] and are given by (1 + ( ) p p ) 1 p p (1.3) r p inf 0 a<1 (1 a p ) 1/p ((1 a ) p + a p (1 a p )) 1/p. A bit later V. Paulsen, G. Popescu, D. Singh [, Corollary.7] gave the following modification of (1.1) (1.4) a 0 + a n ( 1 )n 1 whenever f(z) = a nz n and f H 1. Also the value 1/ is sharp. More recently the author (see [6, Proposition.4]) extended such a result to 1 p showing that if f(z) = a nz n and f H 1 one has (1.5) a 0 p p + a n ( + p )n 1. p +p Also the value is sharp. Several authors (see [4, 11, 1, 19, 0, 8]) have found some other extensions of Bohr s estimate in different directions. For instance, after the paper by Dineen and Timoney [18] some multi-dimensional analogues of Bohr s inequality where the disc D is replaced by a domain Ω C m were considered in [9]. Since then several applications and connections with local Banach space theory and other topics were shown by different authors (see for instance [1,, 3, 14, 15, 16]). In this paper we are interested in the vector-valued analogue of (1.) and to show its possible connection with Banach space theory. Let us fix our notation and give some definitions first. Throughout this paper H (D, X) stands for the space of bounded holomorphic functions from the unit disc D into a complex Banach space X and we write f H (D,X) = sup z <1 f(z). As usual for 1 p <, H p (D, X) stands for the space of holomorphic functions from D into X such that f Hp (D,X) = sup ( 0<r<1 π 0 f(re it ) p dt π )1/p <. In [6] the author defined the Bohr s radius of a Banach space X as the value R(X) = sup{r 0 : x n r n f H (D,X)}. Embedding C into X we obtain the trivial upper bound R(X) 1 3 for any Banach space X. However it was shown that R(X) = 0 for X = C m p whenever m, where C m p, for 1 p, stands for the space C m endowed with the norm w p = ( m i=1 w i p ) 1/p for 1 p < and w = sup m i=1 w i. This fact led the author to consider the vector-valued analogue of (1.4) and to introduce for a given Banach space X and parameters 0 < p, q <, the quantities (see [6, Definition 1.3]) (1.6) R p,q (f, X) = sup{r 0 : x 0 p + ( x n r n ) q 1} where f(z) = x nz n with f H (D,X) 1, and (1.7) R p,q (X) = inf{r p,q (f, X) : f H (D,X) 1}.

3 THE p-bohr RADIUS OF A BANACH SPACE 3 Several results concerning the values R p,q (X) for X = C, X = C m p and X = L p were analyzed. Let us consider now the vector-valued analogue of the approach due to Djakov and Ramanujan. Definition 1.1. Let 1 p < and let X be a complex Banach space. We write (1.8) r p (f, X) = sup{r 0 : x n p r np 1} where f(z) = x nz n with f H (D,X) 1 and define the p-bohr radius of X r p (X) = inf{r p (f, X) : f H (D,X) 1} ( = sup{r > 0 : x n p r np) 1/p f H (D,X)}. Of course the quantities r p (X) and R p,q (X) are related. Actually for 1 p < and 1/p + 1/p = 1 we have that (1.9) R p,p (X) r p (X) 1/p R p,p (X) Indeed, clearly R p,p (X) r p (X) using the estimate x n p r np ( x n r n ) p. On the other hand, if f H (D, X) and we denote r p (f, X) = r then for each 0 < s < 1 one has x 0 p + ( x n r n s n ) p x 0 p + ( x n p r np s p )( ) p 1 1 s p s p x 0 p + (1 x 0 p )( ) p 1. 1 s p Choosing s = 1/p one gets r p (X) 1/p R p,p (X). In particular using (1.9) and (1.5) we obtain a lower estimate for r p, namely (1.10) r p R p,p R p,1 = p + p. The reader should notice that (1.10) is sharp for p = 1 while for p = one only gets r 1. Remark 1.. For any Banach space X the function p r p p(x) is increasing, that is (1.11) r p1 p 1 (X) r p p (X), p 1 p. Indeed, first recall that x n f H (D,X) for all n (this can be seen composing with functionals x X and using the scalar-valued case). Hence if f(z) = x nz n H (D, X) has norm 1 and p 1 p then x n p (r p1 (f, X) p1/p ) np x n p1 r p1 (f, X) np1 1. This gives r p1 p 1 (f, X) r p p (f, X) and we obtain (1.11). The estimate (1.11) can be easily improved using interpolation as the following lemma shows.

4 4 O. BLASCO Proposition 1.3. Let X be a complex Banach space, 1 p 1 < p < p < and + θ p. Then 1 p = 1 θ p 1 (1.1) r p (X) r p1 (X) 1 θ r p (X) θ. Proof. Let us show that for each f H (D, X) with norm 1 we have that r p (f, X) r p1 (f, X) 1 θ r p (f, X) θ. Denote r 1 = r p1 (f, X) and r = r p1 (f, X). Hence ( x n p1 r np1 1 ) 1/p1 1 and ( x n p r np ) 1/p p 1. Now setting 1 (1 θ)p = q 1, p θp = q and r = r (1 θ) 1 r θ we can use Hölder s inequality to obtain ( x n p r np ) 1/p = ( x n p(1 θ) r n(1 θ)p 1 x n pθ r npθ ) 1/p ( x n p1 r np1 1 ) (1 θ)/p1 ( x n p r np ) θ/p 1. This shows that r p (f, X) r (1 θ) 1 r θ. As a consequence of (1.1) one gets the lower estimate (1.13) r p ( 1 3 )/p 1, 1 < p <. Note that since 3 y + y for 0 y 1, choosing y = /p 1, we obtain p +p ( 1 3 )/p 1 and then (1.13) improves (1.10). One may wonder whether r p = ( 1 3 )/p 1 for 1 < p <. However the already known lower estimate given in (1.3) is better than the one obtained in (1.13). Indeed, the inequality x < x + 1 for 0 < x < 1 implies, choosing x = p 1, that ( p ) 1 p < and then (1 + ( p ) 1 p p ) p > ( 1 3 ) p p, 1 < p <. Of course 0 r p (X) 1 for any Banach space (just take f(z) = xz with x = 1) and r p (X) r p for any complex Banach space X. Due to (1.9) and [6, Theorem.] one can not expect r p (X) > 0 for dim(x). However it is not difficult to find examples with r p (X) > 0 or even r p (X) = 1 for values p. We would like to mention two well-known properties which appear naturally when considering the p-bohr radius and which allow us to see that if 1 < q < then r p (L q (µ)) = 1 for certain values of p and r p (L (µ)) = 0 for any p 1. We first recall the notion of Fourier type p first introduced by J. Peetre Definition 1.4. (see [5]) Let 1 < p. A complex Banach space X is said to have Fourier type p if there exists F p (X) > 0 such that for any f L p (T, X) ( ˆf(n) p ) 1/p F p (X) f Lp (T,X) where 1/p + 1/p = 1. n= Proposition 1.5. Let 1 < p and 1/p + 1/p = 1. If X has Fourier type p > 1 with F p (X) = 1 then r p (X) = 1. In particular (1.14) r p (L q (µ)) = 1, 1 < q <, p max{q, q }.

5 THE p-bohr RADIUS OF A BANACH SPACE 5 Proof. The inequality ( x n p ) 1/p f H p (D,X) f H (D,X) for any f = x nz n H (D, X), gives that r p (f, X) 1. (1.14) follows using that L q (µ) has Fourier type p = min{q, q } and F p (L q (µ)) = 1 (see [5]). Besides the notion of Fourier type bigger than one, there is another geometrical property of the Banach space that plays some important role to have r p (X) > 0. Definition 1.6. (see [7]) A complex Banach space X is said to satisfy the strong maximum modulus theorem if f(z) has no maximum in D for any non-constant bounded analytic function f : D X. Proposition 1.7. If r p (X) > 0 for some 1 p < then X does satisfy the strong maximum modulus theorem. In particular X = C m for m, X = c 0 and X = C([0, 1]) satisfy that r p (X) = 0 for any 1 p <. Proof. Assume that there exists a non constant f H (D, X) of norm 1 and z 0 D such that f(z 0 ) = 1. Using a Moebius transformation we may assume that z 0 = 0. Hence r p (f, X) = 0 and therefore r p (X) = 0 for any p 1. To finish the proof it suffices to recall that C m for m, c 0 and C([0, 1]) do not satisfy the strong maximum modulus theorem (see [7]). The strong maximum modulus theorem is related with the strict c-convexity (see [1, 7]). Let us mention certain notions on C-convexity that are particularly interesting in our situation. Definition 1.8. (see [1, 13]) Let p <. A complex Banach space X is called p-uniformly C-convex if there exists a constant λ > 0 such that (1.15) ( x p + λ y p ) 1/p max x + e iθ y θ for all x, y X. Denote A p (X) the supremum of the constants λ satisfying (1.15). There are some equivalent formulations of such a concept. One is the so-called p- uniformly P L-convexity (we refer to [17, 1, 3, 4] for information on that) where the max θ x + e iθ y is replaced by ( π x + e iθ y q dθ 0 π )1/q for some 1 q < and another one is given in terms of Littlewood-Paley inequalities (see [7]). Our main theorem gives another interesting characterization of such a convexity property. Theorem 1.9. Let X be a complex Banach space and p. X is p-uniformly C-convex if and only if the p-bohr radius r p (X) > 0. This result is closely related to another description of p-uniformly C-convexity achieved in [7, Proposition.1] which states the existence of a constant λ > 0 such that (1.16) ( f(0) p + λ f (0) p ) 1/p f H (D,X) for any f H (D, X).

6 6 O. BLASCO The paper is divided into two sections. In the first one we prove Theorem 1.9 and in the second one we study r p (L q (µ) for different values of 1 p, q <. From (1.14) we know that for 1 < q < one has that r q (L q (µ)) = 1 and since L q (µ) is -uniformly C-convex (see [1, 13]) for 1 p Theorem 1.9 gives that actually r (L q (µ)) > 0. We shall get some lower estimates of r p (L q (µ) whenever 1 q < p p <.. Geometrical characterizations Let is introduce the following variation of the p-bohr radius motivated by (1.16). Definition.1. Let X be a complex Banach space and 1 p <. We denote (.1) r p (X) = sup{r > 0 : f(0) p + r p f (0) p f p H (D,X) } According to the result mentioned in the introduction p-uniformly C-convexity means r p (X) > 0 for p. Since r p (X) r p (X) one has that spaces with positive p-bohr radius are always p-uniformly C-convex. However we shall present an independent proof of Theorem 1.9 and get the characterization in (1.16) as a consequence. Let us first mention that contrary to the situation for r p (C) a precise value of r p (C) can be computed for all values of p. Proposition.. Let 1 p < and define (1 a p ) 1/p (.) γ p = inf 0<a<1 1 a. Then r p (C) = γ p. Proof. From Schwarz-Pick lemma we have that f (0) (1 f(0) ) for any f H (D) with norm 1. Hence for f(z) = a nz n and f H 1 we have a 0 p + a 1 p γ p p a 0 p + (1 a 0 ) p γ p p 1. Therefore we obtain r p (C) γ p. To see the other inequality consider φ a (z) = z a 1 az belongs to the unit ball of H. Then This shows that Hence r p (C) γ p. = a+ 1 a a a p + r p (C) p (1 a ) p φ a p H = 1, a < 1 r p (C) p 1 ap (1 a ) p, 0 < a < 1. The value of γ p is given in the following formula. Proposition.3. Let p 1. Then r 1 (C) = 1, r p(c) = 1 for p and r p (C) = x 1/p p + x 1/p p, 1 < p < where (1 x p ) /p 1 = 1 1+x p and 0 < x p < 1. an z n which

7 THE p-bohr RADIUS OF A BANACH SPACE 7 Proof. The case p = 1 and p are immediate from Proposition.. consider 1 < p < and denote (.3) h p (x) = (1 (1 x)/p ) p. x Clearly γp p = sup 0<x<1 h p (x). One observes that h p (1) = 1, lim x 0 + h p (x) = 0, h p(0 + h ) = lim p(x) x 0 ( p )p x lim p x 0 x = and h p(1 ) = 1. Since for 0 < x < 1 one has h p(x) = 1 x (1 (1 x)/p ) p 1( ) (1 x) /p 1 (1 + x) 1 Let us x = the function h p attains it maximum at 0 < x p < 1 such that h p(x p ) = 0 (that is to say (1 x p ) /p 1 = 1 1+x p ). Moreover (.4) γp p = h p (x p ) = p xp p 1 (1 + x p ) p and the result is achieved. Theorem.4. Let p 1. Then (.5) r p (X) ( r p p(x) + 1) 1/p r p(x) r p (X). A p (X) 1/p (.6) r p (X) A p (X) 1/p, p. Proof. Let f H (D, X) with norm 1 and f(z) = x nz n. Let n N and consider ξ = e πi n and define g(z) = 1 n n j=1 f(ξj z). Using that n j=1 ξj = 0 we obtain g(z) = x 0 + x n z n + x n z n +. Since g H (D, X) with norm g H (D,X) 1 we have that This gives the estimate Therefore x 0 p + g (0) p r p (X) p (1 g(0) p ). x n p r p (X) p (1 x 0 p ), n 1. x n p r pn x 0 p + r p (X) p (1 x 0 p )( r pn ) r p x 0 p + r p (X) p (1 x 0 p ) 1 r p max{1, r p (X) p r p 1 r p }. Now choosing r such that r p (X) p = rp 1 r we obtain (.5). p Let us now see (.6). Let f H (D, X) with f H (D,X) 1 and let ξ be in the unit ball of X. Since ξ, f H (D, C) with ξ, f H (D,C) 1 then Schwarz-Pick lemma gives ξ, f (0) 1 ξ, f(0) (1 ξ, f(0) ).

8 8 O. BLASCO This shows that, for any ξ in the unit ball of X, Therefore, for any θ [0, π), Hence ξ, f(0) + 1 ξ, f (0) 1. f(0) + eiθ f (0) = sup ξ, f(0) + eiθ ξ =1 ξ, f (0) 1. f(0) p + A p(x) p f (0) p 1. This gives Ap(X)1/p r p (X). Proof of Theorem 1.9 Applying Theorem.4 one obtains (.7) A 1/p p (X) (A p (X) + p ) r p(x) A 1/p 1/p p (X). Taking into account that p-uniformly C-convexity means A p (X) > 0 the proof is complete. Combining (.5) in Theorem.4 and Proposition. one gets the following lower estimate for r p. Corollary.5. Let 1 < p <. Then (.8) r p (1 + γ p p ) 1/p. Combining (.6) and (.5) we also obtain the following lower estimate. Corollary.6. If X is -uniformly C-convex then r (X) A(X) A. (X)+4 3. p-bohr radius of L q -spaces In this section X = L q (µ) where µ is a measure space and 1 q. It was shown (see [6]) that r 1 (L q (mu)) = 0 for 1 q whenever dim(l q (µ)). We shall see that r p (L q (µ)) = 0 for 1 < p < while r p (L q (µ)) > 0 for p and 1 q p. Next result follows closely the ideas in [6] and it is included for sake of completeness. Theorem 3.1. Let (Ω, Σ, µ) a measure space such that there exists a couple of disjoint measurable sets A, B Σ with 0 < µ(a), µ(b) <. Then r p (L q (µ)) = 0, 1 p < q < or 1 q p <. In particular, if 1 p < then r p (C m q ) = 0 for m and r p (L q (T)) = 0. Proof. Assume first p < q. Since lim y y p/q (y 1) p/q = 0, one has that for each ε > 0 we can find 0 < γ < 1 such that Equivalently (γ 1 ) p/q (γ 1 1) p/q < ε p. (3.1) (1 γ) p/q + ε p γ p/q > 1.

9 THE p-bohr RADIUS OF A BANACH SPACE 9 Now define x 0 = (1 γ) 1/q and x µ(a) 1/q 1 = γ 1/q χ B χ A Clearly sup z <1 f(z) L q (µ) = 1 and f(z) = x 0 + x 1 z. x 0 p + x 1 p ε p > 1. µ(b) 1/q and set Hence r p (f, L q (µ)) ε and then r p (X) = 0. Assume now p < and q. We argue as above choosing now for each ε > 0 a value 0 < γ < 1 satisfying (3.) (1 γ) p/ + ε p γ p/ > 1. We now define and set x 0 = 1/q (1 γ) 1/( χ A µ(a) + χ ) B, 1/q µ(b) 1/q x 1 = 1/q γ 1/( χ A µ(a) χ ) B 1/q µ(b) 1/q f(z) = x 0 + x 1 z = 1/q ( 1 γ + γz) µ(a) + 1/q 1/q ( 1 γ γz) µ(b). 1/q Observe that f(z) = 1/q( 1 γ + γz q + 1 γ γz q) 1/q χ A 1/( 1 γ + γz + 1 γ γz ) 1/ 1. On the other hand, x 0 = 1 γ and x 1 = γ. The proof is finished using(3.) and arguing as in the previous case. Let us now analyze the case p and q p. Theorem 3.. Let p < and p q p. Then (3.3) r p (L q (µ)) = 1. Proof. We first consider p = (hence q = ). We can use Plancherel s theorem to obtain (3.4) ( x n L (µ) )1/ = f H (D,L (µ)) and, then r (L (µ)) 1. The other inequality is always true. Assume < p < and p q p. Select 0 < θ < 1 such that 1 p = 1 θ 1 β so that 1 q = 1 θ + θ β. Observe first that (3.5) max n x n Lβ (µ) f H1 (D,L β (µ)). Now we can use complex interpolation, and apply the result (see [8]) [H p1 (D, L α (µ)), H p (D, L β (µ))] θ = H p3 (D, L γ (µ)), χ B and for 1 p 3 = 1 θ p 1 + θ p and 1 γ = 1 θ α + θ β. We then conclude, due to (3.4) and (3.5) by considering p 1 =, p = 1 and α = (and hence p 3 = 1+θ and γ = q) ( x n p L q (µ) )1/p f H p 3 (D,L q (µ)).

10 10 O. BLASCO This shows that for each f(z) = x nz n belonging to the unit ball of H (D, L q (µ)) we have ( x n p L q (µ) )1/p 1. Hence r p (L q (µ)) 1. The only case that is left is 1 q < p p. Using Corollary.6 and Remark 1.11 one gets that r p (L q (µ)) r /q (L q (µ)) > 0. Let us see that in general also r p (L q (µ)) < 1. Proposition 3.3. Let X = L q (T) and 1 q <. Then 0 < r (X) < 1. Proof. As mentioned above the fact that r (X) > 0 follows from Corollary.6. We shall show that there exists f in the unit ball of H (D, L q (T)) with x n =. Therefore r (X) < 1. It suffices to select F H q (T)\H (T), say F (w) = a nw n, with F H q (T) = 1, that is sup 0<r<1 F r Lq (T) 1 where F r (e it ) = F (re it ) and consider the L q (T)- valued function f(z)(e it ) = F (ze it ) = a n e int z n. Hence x n (e it ) = a n e int and f(z) L q (T) = F z H q 1 for all 0 < z < 1 and x n = a n. Our aim is now to find lower estimates for r p (L q (µ) in the case 1 q < p p. We shall need the following lemma. Lemma 3.4. Let 1 q and F H q (D). Then (3.6) ( F (0) + ( 1 ) q q F (0) ) 1/ F H q (D) Proof. Let us first show that ( F (0) + 1 F (0) ) 1/ F H 1 (D). Assume that F H1 (D) = 1. Now, using factorization to write F = gh with g, h H (D) and h H (D) = g H (D) = 1. Hence F (0) = g(0)h(0), F (0) = h(0)g (0) + h (0)g(0) and g(0) + g (0) 1, h(0) + h (0) 1. Using that xy x + y we have F (0) + 1 F (0) g(0) h(0) + g(0) h (0) + g (0) h(0) Therefore which combined with = g(0) ( h(0) + h (0) ) + g (0) h(0) ( g(0) + g (0) )( h(0) + h (0) ) 1. ( F (0) + 1 F (0) ) 1/ F H1 (D) ( F (0) + F (0) ) 1/ F H(D)

11 THE p-bohr RADIUS OF A BANACH SPACE 11 gives also the case 1 < q < in (3.6) by invoking interpolation (see [5, Theorem 5.6.]). Theorem 3.5. Let (Ω, Σ, µ) a measure space and 1 q. Then r (L q (µ)) 1 1 q. Proof. Let f : D L q (µ) be a bounded holomorphic function, say f(z) = x nz n. We use Fubini s theorem and Minkowski s inequality, together with Lemma 3.4, to get the following estimates f q H (D,L q (µ)) f q H q (D,L q (µ)) = sup 0<s<1 = sup 0<s<1 sup 0<s<1 = sup 0<s<1 sup 0<s<1 = sup = 0<s<1 π 0 Ω Ω f(se iθ ) q dθ L q (µ) π ( π ) dµ(w) f(se iθ )(w) q dθ 0 π ( x 0 (w) + ( 1 ) q q x1 (w) s ) q/ dµ(w) ( x 0 (w) q, ( 1 Ω ) q x1 (w) q s q) C dµ(w) /q ( x 0 (w) q dµ(w), ( 1 ) q Ω ( x 0 L q (µ) + (1 ) q q x1 L q (µ) sq ) q/ ( x 0 L q (µ) + (1 ) q q x1 L q (µ)) q/. Ω x 1 (w) q s q dµ(w)) C /q Hence r (L q (µ)) ( 1 ) 1 q 1. Corollary 3.6. Let (Ω, Σ, µ) a measure space and 1 q < p p <. Then r p (L q (µ)) (1 + q 1 ) 1/p. In particular r p (L 1 (µ)) (1/3) 1/p. t Proof. Note that φ p (t) = is increasing. Combining now Theorem.4 and (t p +1) 1/p Theorem 3.5 one obtains r (L q (µ)) φ ( r (L q (µ))) φ ( 1 1 q ) = (1 + q 1 ) 1/. Finally use r p (X) r (X) /p to complete the result. References [1] L. Aizenberg Multidimensional analogues of Bohr s theorem on power series Proc. Amer. Math. Soc. 18 (1999), [] L. Aizenberg, A. Aytuna and P. Djakov Generalization of a theorem of Bohr for basis in spaces of holomorphic functions in several variables J. Math. Anal. Appl. 58 (001), [3] L. Aizenberg, I.B. Grossman, Yu.F. Korobeinik Some remarks on the Bohr radius for power series. (Russian)Izv. Vyssh. Uchebn. Zaved. Mat. 00, no. 10, 3 10; translation in Russian Math. (Iz. VUZ) 46 (00), no. 10, 18 (003) [4] C. Bénéteau, A. Dahlner, D. Khavinson, Remarks on the Bohr phenomenon. Comput. Methods Funct. Theory 4 (004), no. 1, 119.

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