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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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 Three '10
 Smith
 Logic, Topology, Vector Space

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