haarmeasure

haarmeasure - 1 Construction of Haar Measure Definition...

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Unformatted text preview: 1 Construction of Haar Measure Definition 1.1. A family G of linear transformations on a linear topological space X is said to be equicontinuous on a subset K of X if for every neighborhood V of the origin in X there is a neighborhood U of the origin such that the following condition holds if k 1 ,k 2 ∈ K and k 1- k 2 ∈ U, then G ( k 1- k 2 ) ⊆ V that is T ( k 1- k 2 ) ∈ V for all T ∈ G . 1 Theorem 1.2 (Kakutani). Let K be a compact, convex subset of a locally convex linear topological space X , and let G be a group of linear mappings which is equicontinuous on K and such that G ( K ) ⊆ K . Then there exists a point p ∈ K such that T ( p ) = p ∀ T ∈ G Proof. By Zorn’s lemma, K contains a minimal non-void compact convex subset K 1 such that G ( K 1 ) ⊆ K 1 . If K 1 contains just one point then the proof is complete. If this is not the case, the compact set K 1- K 1 contains some point other than the origin. 2 Thus, there exists a neighborhood V of the origin such that ¯ V 6⊇ K 1- K 1 . There is a convex neighborhood V 1 of the origin such that αV 1 ⊆ V for | α | ≤ 1 . By the equicontinuity of G on the set K 1 , there is a neighborhood U 1 of the origin such that if k 1 ,k 2 ∈ K 1 and k 1- k 2 ∈ U 1 then G ( k 1- k 2 ) ⊆ V 1 . 3 Because each T ∈ G is invertible, T maps open sets to open sets (open mapping theorem) and T ( A ∩ B ) = TA ∩ TB for any sets A,B . Since T is linear, T convex-hull ( A ) = convex-hull T ( A ) for any set A . Because G is a group, G ( G A ) = G A for any set A . 4 Thus U 2 := convex-hull ( G U 1 ∩ ( K 1- K 1 )) = convex-hull ( G ( U 1 ∩ ( K 1- K 1 ))) ⊆ V 1 is relatively open in K 1- K 1 and satisfies G U 2 = U 2 6⊇ K 1- K 1 . By continuity, G ¯ U 2 = ¯ U 2 . Define ∞ > δ := inf { a : a > ,aU 2 ⊇ K 1- K 1 } ≥ 1 and U := δU 2 . For each < < 1 , (1 + ) U ⊇ K 1- K 1 6⊆ (1- ) ¯ U. 5 The family of relatively open sets { 2- 1 U + k } ,k ∈ K 1 , is a covering of K 1 . Let { 2- 1 U + k 1 ,..., 2- 1 U + k n } be a finite sub-covering and let p = ( k 1 + ...k n ) /n . If k is any point in K 1 , then k i- k ∈ 2- 1 U for some 1 ≤ i ≤ n . Since k i- k ∈ (1 + ) U for all i and all > , we have p ∈ 1 n ( 2- 1 U + ( n- 1) · (1 + ) U ) + k. For = 1 4( n- 1) , we have p ∈ (1- 1 4 n ) U + k for each k ∈ K 1 . Let K 2 = K 1 ∩ k ∈ K 1 (1- 1 4 n ) ¯ U + k 6 = ∅ . 6 Because (1- 1 4 n ) ¯ U 6⊇ K 1- K 1 , we have K 2 6 = K 1 . The closed set K 2 is clearly convex. Further since T ( a ¯ U ) ⊆ a ¯ U for T ∈ G , we have T ( a ¯ U + k ) ⊆ a ¯ U + Tk for all T ∈ G ,k ∈ K 1 . Recalling TK 1 = K 1 for T ∈ G , we find that G K 2 ⊆ K 2 , which contradicts the minimality of K 1 ....
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This note was uploaded on 02/28/2012 for the course MATH 251C taught by Professor N.r.wallach during the Winter '11 term at Colorado.

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haarmeasure - 1 Construction of Haar Measure Definition...

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