Exercise 2.12

Let X be the normed space of all real polynomials in one variable, with

f = 0 1 | f ( t ) | dt .

Put B ( f , g ) = 0 1 f ( t ) g ( t ) dt , and show that B is a bilinear continuous functional on X × X which is separately but not continuous.

Answers

Proof. Let f denote the first variable, g the second one. Remark that

B ( f , g ) < f max [ 0 , 1 ] g ; (1)

which is sufficient (1.18) to assert that any f B ( f , g ) is continuous. The continuity of all g B ( f , g ) follows (Put C ( g , f ) = B ( f , g ) and proceed as above). Suppose, to reach a contradiction, that B is continuous. There so exists a positive M such that,

B ( f , g ) Mfg . (2)

Put

f n ( X ) 2 n X n R [ X ] ( n ) , (3)

so that

f n = 2 n n + 1 n∞ 0 . (4)

On the other hand,

B ( f n , f n ) = 4 n 2 n + 1 > 1 . (5)

Finally, we combine (4 ) and (5 ) with (2 ) and so obtain

1 < B ( f n , f n ) M f n 2 n∞ 0 . (6)

Our continuousness assumption is then contradicted. So ends the proof. □

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2020-01-24 00:00
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