Exercise 2.10

Prove that a bilinear mapping is continuous if it is continuous at the origin ( 0 , 0 ) .

Answers

Proof. Let ( X 1 , X 2 , Z ) be topological spaces and B a bilinear mapping

B : X 1 × X 2 Z . (1)

From now on, x = ( x 1 , x 2 ) denotes an arbitrary element of X 1 × X 2 . We henceforth assume that B is continuous at the origin ( 0 , 0 ) of X 1 × X 2 , given an arbitrary balanced open subset W of Z , there exists in X i ( i = 1 , 2 ) a balanced open subset U i such that

B ( U 1 × U 2 ) W . (2)

In such context, λ i ( x ) is chosen greater than μ i ( x i ) = inf { r > 0 : x i r U i } ; see [1.33] of Functional Analysis for further reading about the Minkowski functionals μ . In other words, x i lies in λ i ( x ) U i , since U i is balanced. Thus,

B ( x 1 , x 2 ) = λ 1 ( x ) λ 2 ( x ) B ( x 1 λ 1 ( x ) , x 2 λ 2 ( x ) ) (3) λ 1 ( x ) λ 2 ( x ) B ( U 1 × U 2 ) (4) λ 1 ( x ) λ 2 ( x ) W . (5)

Pick p = ( p 1 , p 2 ) in X 1 × X 2 , and let q = ( q 1 , q 2 ) range over X × Y , as a first step: It directly follows from (5 ) that

B ( p ) B ( q ) = B ( p 1 , p 2 q 2 ) + B ( p 1 , q 2 ) B ( q 1 , q 2 ) (6)  = B ( p 1 , p 2 q 2 ) + B ( p 1 q 1 , q 2 ) (7)  = B ( p 1 , p 2 q 2 ) + B ( p 1 q 1 , q 2 p 2 ) + B ( p 1 q 1 , p 2 ) (8)  λ 1 ( p ) λ 2 ( p q ) W + λ 1 ( p q ) λ 2 ( q p ) W + λ 1 ( p q ) λ 2 ( p ) W . (9) 

We now restrict q to a particular neighborhood of p . More specifically,

p i q i 1 μ 1 ( p 1 ) + μ 2 ( p 2 ) + 2 U i ; (10)

which implies

μ i ( q i p i ) = μ i ( p i q i ) 1 μ 1 ( p 1 ) + μ 2 ( p 2 ) + 2 (11)

(the equality at the left is valid, since U i = U i ). The special case

λ i ( p ) μ 1 ( p 1 ) + μ 2 ( p 2 ) + 1 , (12) λ i ( p q ) 1 μ 1 ( p 1 ) + μ 2 ( p 2 ) + 1 λ i ( q p ) (13)

implies that

B ( p ) B ( q ) W + W + W , (14)

since W is balanced. W being arbitrary, we have so established the continuousness of B at arbitrary p ; which achieves the proof. □

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2020-01-24 00:00
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