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THICK SETS in BANACH SPACES and THEIR PROPERTIES 1. Definitions and General Results Kadets and Fonf [KF] Defined a Set in a Bana

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THICK SETS in BANACH SPACES and THEIR PROPERTIES 1. Definitions and General Results Kadets and Fonf [KF] Defined a Set in a Bana THICK SETS IN BANACH SPACES AND THEIR PROPERTIES OLAV NYGAARD Abstract. We make a survey over results involving the concepts thick and w¤-thick set and provide a lot of known such sets. We show that the long standing "separable quotient problem" and a problem from function theory is closely connected to thickness. 1. Definitions and general results Kadets and Fonf [KF] de¯ned a set in a Banach space to be thin if it can be represented as a countable non-decreasing union of non-norming sets. (A set in a Banach space is said to be non-norming if its closed absolutely convex hull does not contain any ball centered at the origin.) As in [N1] and [N2], let us say that a set is thick if it is not thin. One has the following "omnibus- theorem" describing thick sets (here recall that T : Y ! X Tauberian means ¤¤ ¡1 (T ) (X) ½ Y and note that for Tauberian operators TBY is closed in X): Theorem 1.1. Let A be a subset of a Banach space X. The following assertions are equivalent. (a) The set A is thick. (b) Whenever a family of continuous linear operators from the space X to some Banach space is pointwise bounded on A, then this family is norm bounded. (c) Whenever a sequence of functionals in the dual space X¤ is pointwise bounded on A, then this sequence is norm bounded. (d) Whenever Y is a Banach space and T : Y ! X is a continuous linear operator such that TY ⊃ A, then TY = X. (e) Whenever Y is a Banach space and T : Y ! X is a continuous linear Tauberian injection with T ¤¤ also a Tauberian injection, such that TY ⊃ A, then TY = X. (f) The span of A is dense and barrelled. (g) Whenever (­; §; ¹) is a measure space and a w¤-measurable function ¤ g : ­ ! X is such that, for all x 2 A, x ± g 2 L1(¹), then g is w¤-integrable. Date: August 25, 2005. 2000 Mathematics Subject Classi¯cation. Primary: 46B20. Key words and phrases. Thick set, w¤-thick set. 1 2 OLAV NYGAARD Proof. Let us give the main ingredients of a proof. We ¯rst concentrate on (a) ) (b) ) (c) ) (a). Then we look at (a) ) (d) ) (e) ) (a). After that we explain (f) , (b) and end the proof with (a) , (g). But at the very start, observe that a set W ½ X is non-norming if and only if its absolutely ¤ ¤ convex hull is non-norming if and only if infx 2SX¤ supx2W jx (x)j = 0. First the chain (a) ) (b) ) (c) ) (a): Let Y be some Banach space and denote as usual by L(X; Y ) the space of bounded linear operators from X to Y . Assume the family ¡ ½ L(X; Y ) is pointwise bounded on A and put An = fx 2 A : kT xk · n for all T 2 ¡g: By the pointwise boundedness A = [nAn, an increasing, countable union. Since A is thick, for some m there is a ± > 0 such that the absolute convex hull Am of Am contains ±BX . In other words (1=±)Am contains BX . Now 1 m sup kT xk · sup kT xk · ; x2BX ± x2Am ± so ¡ is bounded by m=±. Note that this is more or less the same argument as the standard proof of the Banach-Steinhaus theorem. That (b) ) (c) is obvious so we turn to (c) ) (a). For this assume A is thin. Then we can ¯nd an increasing, countable family of non-norming sets (An) such that ¤ ¤ A = [nAn. Since A1 is non-norming, we can ¯nd y1 with ky1k = 1 but ¤ ¡1 ¤ ¤ supx2A1 jy1(x)j < 2 . Since A2 is non-norming we pick y2 with ky2k = 1 ¤ ¡2 ¤ but supx2A2 jy2(x)j < 2 . We construct in this manner a sequence (yn) ¤ ¤ ¡n in the dual unit sphere SX with supx2An jyn(x)j < 2 . It remains to ¤ ¤ n ¤ explain why the unbounded family (xn) given by xn = 2 ¢ yn is pointwise bounded on A. For this, let x be some point of A. We want to show that ¤ supn jxn(x)j < 1. Find some k such that x 2 Ak and remember x 2 An whenever n ¸ k. There are two cases, n < k and n ¸ k. In the ¯rst case ¤ ¤ k ¤ n ¡n jxn(x)j · kxnkkxk < 2 kxk < 1. In the second case jxn(x)j < 2 ¢ 2 = 1. (We will improve this simple technique in the proof of (g) ) (a).) Now we turn to (a) ) (d) ) (e) ) (a). First (a) ) (d): Let Y be a Banach space and assume A is thick and TY ⊃ A. Clearly TY also contains the absolute convex hull of A so we may just as well assume A is absolutely convex. We have A = [n(A \ n ¢ TBY ), a countable, increasing union of absolutely convex sets. Since A is thick there are an m and a ± > 0 with (A \ mT BY ) ⊃ ±T BX . In particular, TBY ⊃ (±=m)TBX , and the result follows from a classical result due to Banach. That (d) ) (e) is clear, we look at (e) ) (a). We do this proof the following way: We prove how to obtain a Banach space Y and an injection T : Y ! X such that TBY is closed, TY ⊃ A but T is not onto X whenever A is thin. After this we explain how to obtain the remaining properties in (e). Assume A is thin. If A is not even norming, then take Y as the span of A\ BX with A\ BX as unit ball (the Banach disc of A\ BX ) and let T be the embedding of Y into X. If A is norming the idea is to ¯nd an absolutely convex, bounded, closed set A~ such that A~ is non-norming but has at least the same span as A. THICK SETS IN BANACH SPACES AND THEIR PROPERTIES 3 Then we use the technique from the non-norming case. We now construct A~: Write A = [nAn, an increasing, countable union of non-norming sets. Note that, since [nAn = [n(An \ nBX ) and since (An \ nBX ) is just as non-norming as An is, we may assume each An to sit in nBX . Put B1 = A1, 2 Bn = An n An¡1, let B = [n(1=n )Bn and denote by B the closed, absolute convex hull of B. We take A~ as B. Closedness, absolute convexity and boundedness of A~ is then clear, it is also clear that the span of A~ contains the span of A so we need only prove that it is non-norming as well. We will prove that A~ contains no set of type "BX , " > 0. To see this, pick an m ¤ such that 1=m < " and use the non-normingness of Am to ¯nd an x 2 SX¤ ¤ ¤ with supx2Am jx (x)j < "=2. We want to show that supx2A~ jx (x)j < ". To show this, by linearity and continuity, it is naturally enough to show that ¤ supx2B jx (x)j < "=2. Take an arbitrary x 2 B. There are two possibilities, 2 either x is in some (1=n )Bn for n · m or it has to be that x sits in some 2 ¤ (1=n )Bn for n > m. In the ¯rst case, remember how x was chosen, ¤ 1 ¤ " " jx (x)j · 2 sup jx (y)j < 2 · : n y2Am 2n 2 ¤ In the second case, remember An ½ nBX and kx k = 1, ( ) ½ ¾ ¤ 1 ¤ 1 1 " jx (x)j · sup 2 sup jx (y)j · sup 2 ¢ n < < : n>m n y2Bn n>m n m 2 Now it is clear how to obtain a Banach space Y and an injection T with TBY closed which is onto A but not onto all of X. The additional properties in (e) are based on an application of the Davis-Figiel-Johnson-PeÃlczy¶nski procedure (see e.g [D, p. 227]) on A~ and the observation that this very procedure results in an isomorphic version of X if and only if the set on which it is performed is norming (see [N2, Prop. 2.5] for details). We next explain that (f) ) (b). For this note that, since the span of A is dense, its dual is X¤. Since it is barrelled (b) follows. We use (a) and (b) to get (f). Denseness follows from (a). Barrelledness follows from (b) and the already established denseness. In both directions we used the standard result that a normed space is barrelled if and only if the Uniform Boundedness Principle works on it. The proof why (a) ) (g) goes like this: Let A ½ X be thick, let (­; §; ¹) ¤ ¤ be a measure space, and let a w -measurable function fR: ­ ! X be such 1 that x ± f 2 L (¹) for all x 2 A. Denote Aj = fx 2 A: ­ jxfj d¹ · jg; j 2 1 N: Then A = [j=1Aj ", and the thickness of A implies the existence of some m 2 N and ± > 0 such that Am ⊃ ±BX . Thus it clearly su±ces to show that 1 x ± f 2 L (¹) for all x 2 Am. But this follows from a standard convexity and limit argument. (Note the importance that norm-closure goes via limits of sequences). 4 OLAV NYGAARD We now end the proof by using thinness to produce a function on the natural numbers N, with values in X¤, A-scalarly integrable but still not X- ¤ P ¤ scalarly integrable. Denoting f(j) = xj the point is to obtain j jxj (x)j < P ¤ 1 for all x 2 A but the existence of y 2 X such that j jxj (y)j diverges.
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