The next result is famous as the result that banach

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The next result is famous as ‘the result that Banach did not prove’. Theorem 45. Let ( U, k k ) be a complex normed vector space. If E is a subspace of U and there exists a continuous linear map T : E C then there exists a continuous linear map ˜ T : U C with k ˜ T k = k T k . We can now answer the question posed in the first sentence of this section. Lemma 46. If ( U, k k ) is normed space over the field F of real or complex numbers and a U with a 6 = 0 , then we can find a continuous linear map T : U F with Ta 6 = 0 Here are a couple of results proved by Banach using his theorem. 6 In fact the statement is marginally weaker than Zorn’s lemma but you need to be logician either to know or care about this. 15
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Theorem 47 (Generalised limits). Consider the vector space l of bounded real sequences. There exists a linear map L : l R such that (i) If x n 0 for all n then L x 0 . (ii) L (( x 1 , x 2 , x 3 , . . . )) = L (( x 0 , x 1 , x 2 , . . . )) . (iii) L ((1 , 1 , 1 , . . . )) = 1 . The theorem is illustrated by the following lemma. Lemma 48. Let L be as in Theorem 47. Then lim sup n →∞ x n L ( x ) lim inf n →∞ x n . In particular, if x n x then L ( x ) = x . Exercise 49. (i) Show that, even though the sequence x n = ( - 1) n has no limit, L ( x ) is uniquely defined. (ii) Find, with reasons, a sequence x l for which L ( x ) is not uniquely defined. Banach used the same idea to prove the following odd result. Lemma 50. Let T = R / Z be the unit circle and let B ( T ) be the vector space of real valued bounded functions. Then we can find a linear map I : B ( T ) R obeying the following conditions. (i) I (1) = 1 . (ii) If 0 if f is positive. (iii) If f B ( T ) , a T and we write f a ( x ) = f ( x - a ) then If a = If . Exercise 51. Show that if I is as in Lemma 50 and f is Riemann integrable then If = Z T f ( t ) dt. However, Lemma 50 is put in context by the following. Lemma 52. Let G be the group freely generated by two generators and B ( G ) be the vector space of real valued bounded functions on G . If f B ( G ) let us write f c ( x ) = f ( xc - 1 ) for all x, c G . There exists a function f B ( G ) and c 1 , c 2 , c 3 such that f ( x ) 0 for all x G and f ( x ) + f c 1 ( x ) - f c 2 ( x ) - f c 3 ( x ) ≤ - 1 for all x G . 16
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Exercise 53. If G is as in Lemma 52 then there is no linear map I : B ( G ) R obeying the following conditions. (i) I (1) = 1 . (ii) If 0 if f is positive. (iii) If c = If for all c G . It can be shown that there is a finitely additive, congruence respecting integral for R and R 2 but not R n for n 3. 7 Banach algebras Many of the objects studied in analysis turn out to be Banach algebras. Definition 54. An algebra ( B, + , ., × ) is a vector space ( B, + , ., C ) equipped with a multiplication × such that (i) x × ( y × z ) = ( x × y ) × z , (ii) ( x + y ) × z = x × z + y × z and z × ( x + y ) = z × x + z × y , (iii) ( λx ) × y = x × ( λy ) = λ ( x × y ) for all x, y, z B . [We shall write x × y = xy .] Note that there is no assumption that multiplication is commutative. In principle, we could talk about real Banach algebras (in which C is replaced by R ) but, though some elementary results carry over, our treatment will only cover complex Banach algebras.) Definition 55. A Banach algebra ( B, + , ., × , k k )
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