13-dual-ban

13-dual-ban - Lecture 13 4. The weak dual topology In this...

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Unformatted text preview: Lecture 13 4. The weak dual topology In this section we examine the topological duals of normed vector spaces. Be- sides the norm topology, there is another natural topology which is constructed as follows. Definition. Let X be a normed vector space over K (= R , C ). For every x X , let x : X * K be the linear map defined by x ( ) = ( x ) , X * . We equipp the vector space X * with the weak topology defined by the family = ( x ) x X . This topology is called the weak dual topology , which is denoted by w * . Recall (see Section 3) that this topology is characterized by the following property ( w * ) Given a topological space T , a map f : T X * is continuous with respect to the w * topology, if and only if x f : T K is continuous, for each x X . Remark that all the maps x : X * K , x X are already continuous with respect to the norm topology. This gives the fact that the w * topology on X * is weaker than the norm topology. Remark 4.1. The w * topology is Hausdorff. Indeed, if , X * are such that 6 = , then there exists some x X such that x ( ) = ( x ) 6 = ( x ) = x ( ) . Proposition 4.1. Let X be a normed vector space over K . For every > , X * , and x X , define the set W ( ; x, ) = X * : | ( x )- ( x ) | < . Then the collection W = W ( ; x, ) : > , X * , x X is a subbase for the w * topology. More precisely, given X * , a set N X * is a neighborhood of with respect to the w * topology, if and only if, there exist > and x 1 ,...,x n X , such that N W ( ; ,x 1 ) W ( ; ,x n ) . Proof. It is clearly sufficient to prove the second assertion, because it would imply the fact that any w * open set is a union of finite intersections of sets in W . If we define the collection S =- 1 x ( D ) : x X , D K open , then we know that S is a subbase for the w * topology. 89 90 LECTURE 13 Fix X * . Start with some w * neighborhood N of , so there exists some w * open set E with E N . Using the fact that S is a subbase for the w * topology, there exist open sets D 1 ,...,D n K , and points x 1 ,...,x n , such that n k =1- 1 x k ( D k ) E. Fix for the moment k { 1 ,...,n } . The fact that - 1 x k ( D k ) means that ( x k ) D k . Since D k is open in K , there exists some k > 0, such that D k B k ( ( x k ) ) . Then if we have an arbitrary W ( ; k ,x k ), we will have | ( x k )- ( x k ) | < k , which gives - 1 x k ( D k ). This proves that W ( ; k ,x k ) - 1 x k ( D k ) . Notice that, if one takes = min { 1 ,..., n } , then we clearly have the inclusions W ( ; ,x k ) W ( ; k ,x k ) - 1 x k ( D k ) ....
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13-dual-ban - Lecture 13 4. The weak dual topology In this...

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