Lecture06 - Lecture 6 Miscellany [ (1) If S : X × Y → Z...

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Unformatted text preview: Lecture 6 Miscellany [ (1) If S : X × Y → Z is (separately) continuous, then S is bounded when X is complete. Proof. This uses the Principle of Uniform Boundedness (PUB) . PUB : If the collection { A λ : λ ∈ Λ } ⊂ L ( X,Z ) is pointwise bounded in the sense that if x ∈ X then the set { A λ ( x ) : λ ∈ Λ } ⊂ Z is bounded (i.e. for all x ∈ X there is an M x such that λ ∈ Λ ⇒ k A λ x k ≤ M x ) , then it is uniformly bounded in the sense that there exists M such that λ ∈ Λ ⇒ k A λ k ≤ M provided X is complete. Recall from last time that we had T x ( y ) = S ( x,y ) and T x ∈ L ( Y,Z ). So we have T : X → L ( Y,Z ). Consider the set { S ( · ,y ) : y ∈ Y, k y k ≤ 1 } ⊂ L ( X,Z ) . This is pointwise bounded: if x ∈ X is fixed, then y ∈ Y, k y k ≤ 1 ⇒ k S ( · ,y )( x ) k = k S ( x,y ) k = k T x ( y ) k ≤ k T x k . So (PUB) there is M such that y ∈ Y, k y k ≤ 1 ⇒ k S ( · ,y ) k ≤ M. 1 Thus y ∈ Y ⇒ k S ( · ,y ) k ≤ M k y k . and so x ∈ X,y ∈ Y ⇒ k S ( x,y ) k = k S ( · ,y )( x ) k ≤ k S ( · ,y ) kk x k ≤ M k y kk x k . In the proof above we assumed X is complete to be able to use PUB. However notice that we could just as well have assumed Y was complete and changed the roles of X and Y ....
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.

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Lecture06 - Lecture 6 Miscellany [ (1) If S : X × Y → Z...

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