Is an algebra b c such that b c k k is a banach space

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• 100000464160110_ch
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is an algebra ( B, + , ., × , C ) such that ( B, + , ., C , k k ) is a Banach space and such that the map ( x, y ) 7→ xy is continuous. Note that the two definitions above are not to be memorized; so far as this course is concerned the following definition is all that is required. Definition 56. A Banach algebra ( B, k k ) is a Banach space equipped with a continuous multiplication which makes it an algebra. As usual there is a little amount of playing about with the definition. Lemma 57. The following statements about ( B, k k ) a Banach space equipped with a multiplication are equivalent. (i) Multiplication is left and right continuous (that is, the map x 7→ xy is continuous for all y and the map y 7→ xy is continuous for all x ). (ii) There exists a K such that k xy k ≤ K k x kk y k for all x and y . (iii) ( B, k k ) is a Banach algebra. 17

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Lemma 58. If ( B, k k ) is a Banach algebra we can find a norm k k B on B which is equivalent to k k (that is, there exists a C > 0 such that C - 1 k x k ≤ k x k B C k x k ) such that k xy k B ≤ k x k B k y k B for all x, y B . Unless specifically indicated otherwise you may assume both in the rest of the notes and in the literature generally that the norm on a Banach algebra has been chosen to satisfy k xy k ≤ k x kk y k for all x and y . Definition 59. We say that a Banach algebra B has a unit e if xe = ex = x for all x B . The following remarks forms part of the course but are left as an exercise. Exercise 60. (i) If a Banach algebra has a unit that unit is unique. (ii) If ( B, k k ) is a Banach algebra with unit e we can find a norm k k B on B which is equivalent to k k such that k xy k B ≤ k x k B k y k B for all x, y B and k e k B = 1 . Unless specifically indicated otherwise you may assume both in the rest of the notes and in the literature generally that the norm on a Banach algebra with unit e has been chosen to satisfy k e k = 1 . for all x and y . Example 61. (i) A Banach space ( B, k k ) becomes a commutative Banach algebra if we define xy = 0 for all x, y B . If B is non-trivial the resulting algebra has no unit. (ii) Consider the Banach space l 1 of sequences a = ( a 0 , a 1 , . . . ) . If we define a * b = c with c r = X k + j = r, k 0 ,j 0 a j b k 18
then * is a well defined multiplication and l 1 is a Banach algebra with this multiplication. As a Banach algebra, l 1 is commutative with a unit. (ii) Consider the Banach space l 1 of sequences a = ( a 1 , a 2 , . . . ) . If we define a * b = c with c r = X k + j = r, k 1 ,j 1 a j b k then * is a well defined multiplication and l 1 is a Banach algebra with this multiplication. As a Banach algebra, l 1 is commutative but has no unit. (iii) Consider the Banach space l 1 of i two sided sequences a = ( . . . , a - 2 , a - 1 , a 0 , a 1 , . . . ) . If we define a * b = c with c r = X k + j = r a j b k then * is a well defined multiplication and l 1 is a Banach algebra with this multiplication. As a Banach algebra, l 1 is commutative and has a unit.

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• Fall '08
• Groah
• Math, Compact space, Banach space, Banach, Banach algebra, commutative Banach

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