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Lecture06 - Lecture 6 Miscellany(1 If S X Y Z is(separately...

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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
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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 . Remark : PUB does not hold without some condition on X . Example: let Z = R and X any infinite-dimensional vector space equipped with the norm associated to a basis.
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