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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 finite-dimensional. 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 component-wise vector space operations and (say) the norm k ( x,y ) k = k x k k y k ....
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- Summer '10