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Unformatted text preview: Lecture 5 (Continuing from last time...) Conversely, suppose the bilinear map S : X × Y → Z is (separately) continuous. For x ∈ X we can define T x : Y → Z by y ∈ Y ⇒ T x ( y ) = S ( x,y ). Then T x is continuous and linear, and in fact T x ∈ L ( Y,Z ), and T : X → L ( Y,Z ); x → T x , is linear. If we knew that T were bounded (that is, that x ∈ X implies that k T x k ≤ M k x k ) then k S ( x,y ) k = k T x ( y ) k ≤ k T x kk y k ≤ M k x kk y k . So S is bounded as a bilinear map. As we know, T is bounded when X is finitedimensional. More generally, T is bounded when X is complete; this follows as an application of the ‘uniform boundedness principle’ (or Banach Steinhaus theorem, for those in the know). Theorem 1. If S : X × Y → Z is a bounded bilinear map, then S is differentiable at each ( a,b ) ∈ X × Y and S ( a,b ) ( h,k ) = S ( a,k ) + S ( h,b ) . Proof. We give X × Y the componentwise vector space operations and (say) the norm k ( x,y ) k = k x k ∨ k y k ....
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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.
 Summer '10
 Staff
 Algebra

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