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CONTINUITY in the ALEXIEWICZ NORM Dedicated to Prof. J 131 (2006) MATHEMATICA BOHEMICA No. 2, 189{196 CONTINUITY IN THE ALEXIEWICZ NORM Erik Talvila, Abbotsford (Received October 19, 2005) Dedicated to Prof. J. Kurzweil on the occasion of his 80th birthday Abstract. If f is a Henstock-Kurzweil integrable function on the real line, the Alexiewicz norm of f is kfk = sup j I fj where the supremum is taken over all intervals I ⊂ . Define I the translation τx by τxfR(y) = f(y − x). Then kτxf − fk tends to 0 as x tends to 0, i.e., f is continuous in the Alexiewicz norm. For particular functions, kτxf − fk can tend to 0 arbitrarily slowly. In general, kτxf − fk > osc fjxj as x ! 0, where osc f is the oscillation of f. It is shown that if F is a primitive of f then kτxF − F k kfkjxj. An example 1 6 1 shows that the function y 7! τxF (y) − F (y) need not be in L . However, if f 2 L then kτxF − F k1 6 kfk1jxj. For a positive weight function w on the real line, necessary and sufficient conditions on w are given so that k(τxf − f)wk ! 0 as x ! 0 whenever fw is Henstock-Kurzweil integrable. Applications are made to the Poisson integral on the disc and half-plane. All of the results also hold with the distributional Denjoy integral, which arises from the completion of the space of Henstock-Kurzweil integrable functions as a subspace of Schwartz distributions. Keywords: Henstock-Kurzweil integral, Alexiewicz norm, distributional Denjoy integral, Poisson integral MSC 2000 : 26A39, 46Bxx 1. Introduction ¡ ¡ For f : ¡ define the translation by τxf(y) = f(y x) for x; y . If ! − 2 f Lp (1 6 p < ) then it is a well known result of Lebesgue integration that f is 2 1 continuous in the p-norm, i.e., lim τxf f p = 0. For example, see [4, Lemma 6.3.5]. x!0 k − k In this paper we consider continuity of Henstock-Kurzweil integrable functions in Supported by the Natural Sciences and Engineering Research Council of Canada. 189 Alexiewicz and weighted Alexiewicz norms on the real line. Let be the set of HK ¡ functions f : ¡ that are Henstock-Kurzweil integrable. The Alexiewicz norm of ! f is defined f = sup I f where the supremum is over all intervals I ¡ . 2 HK k k I j j ⊂ Identifying functions almost evRerywhere, becomes a normed linear space under HK that is barrelled but not complete. See [1] and [5] for a discussion of the Henstock- k·k Kurzweil integral and the Alexiewicz norm. It is shown below that translations are continuous in norm and that for f we have τxf f > osc f x where osc f 2 HK k − k j j is the oscillation of f. For particular f the quantity τxf f can tend to 2 HK k − k 0 arbitrarily slowly. If F is a primitive of f then τxF F 6 f x . An example k − k k kj j shows that if f then the function defined by y τxF (y) F (y) need not be in 1 12 HK 7! − L but if f L then τxF F 6 f x . For a positive weight function w on the 2 k − k1 k k1j j real line, necessary and sufficient conditions on w are given so that (τxf f)w 0 k − k ! as x 0 whenever fw is Henstock-Kurzweil integrable. The necessary and sufficient ! conditions involve properties of the function gx(y) = w(y + x)=w(y). Sufficient conditions are given on w for (τxf f)w 0. Applications to the Dirichlet k − k ! problem in the disc and half-plane are given. All of the results also hold when we use the distributional Denjoy integral. Define to be the completion of with respect to . Then is a subspace of the space A HK k·k A of Schwartz distributions. Distribution f is in if there is function F continuous on A the extended real line such that F 0 = f as a distributional derivative. For details on this integral see [6]. First we prove continuity in the Alexiewicz norm. Theorem 1. Let f . For x; y ¡ define τxf(y) = f(y x). Then 2 HK 2 − τxf f 0 as x 0. k − k ! ! ¢¤£¦¥§¥©¨ β β−x β . Let x; α, β ¡ . Then (τxf f) = f f. Write F (x) = 2 α − α−x − α x R R R −∞ f. Taking the supremum over α and β, R τxf f 6 sup F (β x) F (β) + sup F (α x) F (α) k − k β2 j − − j α2 j − − j 0 as x 0 since F is uniformly continuous on ¡ : ! ! Notice that for each x ¡ , the translation τx is an isometry on , i.e., it is a 2 HK homeomorphism such that τxf = f . It is also clear that we have continuity at k k k k each point: for each x ¡ , τxf τx0 f 0 as x x . 0 2 k − k ! ! 0 The theorem also applies on any interval I ¡ . Restrict α and β to lie in I and ⊂ extend f to be 0 outside I. Or, one could use a periodic extension. The same results x also hold for the equivalent norm f = sup −∞ f . k k j j x2 R 190 Under the Alexiewicz norm, the space of Henstock-Kurzweil integrable functions x is not complete. Its completion with respect to the norm f = sup −∞ f is k k x2 j j the subspace of distributions that are the distributional derivative of a functionR in 0 ¡ ¡ ¡ C := F : ¡ ; F C ( ); lim F (x) = 0; lim F (x) , i.e., they are f ! 2 x→−∞ x!1 2 g distributions of order 1. See [6], where the completion is denoted . Thus, if f e A 2 A then f 0 (Schwartz distributions) and there is a function F C such that 0 2 D 0 1 0 2 1 F ; ' = F; ' = F ' = f; ' for all test functions ' = C ( ¡ ). −∞ e c h i −h i − h ib 2 D The distributional integralR of f is then f = F (b) F (a) for all 6 a 6 b 6 a − −∞ . We can compute the Alexiewicz normR of f via f = sup F (x) = F 1. If 1 k k x2 j j k k 0 0 0 f then τxf is defined by τxf; ' := f; τ−x' = F ; τ−x' = F; (τ−x') = 2 D 0 0 h 0 i h i h 1 i −h i F; τ x' = τxF; ' = (τxF ) ; ' . Of course we have L and each −h − i −h i h i ⊂ HK ⊂ A inclusion is strict. The theorem only depends on uniform continuity of the primitive and not on its pointwise differentiability properties so it also holds in . The same is true for the A other theorems in this paper. Corollary 2. Let f . Then τxf f 0 as x 0. 2 A k − k ! ! The following theorem gives us more precise information on the decay rate of τxf f . k − k Theorem 3. (a) Let : (0; 1] (0; ) such that lim (x) = 0. Then there is ! 1 x!0 1 f L such that τxf f > (x) for all sufficiently small x > 0. (b) If f 2 k − k 2 HK and f = 0 a.e. then the most rapid decay is τxf f = O(x) as x 0 and this is 6 k − k ! the best estimate in the sense that if τxf f =x 0 as x 0 then f = 0 a.e. The k − k ! ! implied constant in the order relation is the oscillation of f. ¢¤£¦¥§¥©¨ . (a) Given , define 1(x) = sup (t). Then 1 > and 1(x) 0<t6x decreases to 0 as x decreases to 0. Define (x) = (1=n) when x (1=(n + 1); 1=n] 2 1 2 for some n . Then > and is a step function that decreases to 0 as x 2 2 2 decreases to 0. Now let 1 3(x) = [ 2(1=(n 1)) 2(1=n)] n(n + 1) x + 2(1=n) − − − n + 1 when x [1=(n + 1); 1=n] for some n > 2. Define = on (1=2; 1]. Then > 2 3 2 3 and 3 is a piecewise linear continuous function that decreases to 0 as x decreases to 0. Define f(x) = 0 (x) for x (0; 1] and f(x) = 0, otherwise. For 0 < x < 1, 3 2 x x τxf f > f(y x) f(y) dy = f = 3(x) > (x): k − k Z − − Z 0 0 Since is absolutely con tinuous, f L1. 3 2 191 (b) Test functions are dense in , i.e., for each f and " > 0 there is HK 2 HK ' such that f ' < ". Let x ¡ . Then, since τx is a linear isometry, 2 D k − k 2 (τxf f) (τx' ') = τx(f ') (f ') < 2" and τx' ' 2" < k − − − k k − − − k k − k − τxf f < τx' ' + 2". It therefore suffices to prove the theorem in . Hence, k − k k − k y 2 D ¡ let f and let a; b ¡ . Write F (y) = f. Then, since F C ( ), 2 D 2 −∞ 2 R b (τxf f) = [F (b x) F (b)] [F (a x) F (a)] Za − − − − − − = F 0(b) x + F 00(ξ) x2=2 + F 0(a) x F 00(η) x2=2; − − for some ξ; η in the support of f. Now, 0 2 0 2 τxf f > sup f(a) f(b) x f 1x = osc f x f 1x : k − k a;b2 j − jj j − k k j j − k k The oscillation of f is positive unless f is constant, but there are no constant 2 D functions in except 0.
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