Lemma 89 under the weak topology m is a compact

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Lemma 89. Under the weak topology M is a compact Hausdorff space. If x B and M ∈ M we now write ˆ x ( M ) = M ( x ). Lemma 90. Let B be a commutative Banach algebra with unit. The mapping x 7→ ˆ x is an algebraic homomorphism of B into C ( M ) . As linear map from ( B, k k ) to C ( M , k k ) it is continuous with operator norm exactly 1 . We know that the homomorphism x 7→ ˆ x need not be injective Exercise 91. Justify this statement by considering the Banach algebra of Exercise 88. The following simple observation is the key to the question of when we have isomorphism. Lemma 92. Suppose x is an element of a commutative Banach algebra with unit. Then the complement of the resolvent R ( x ) is the range of ˆ x . That is to say, { ˆ x ( M ) : M ∈ M} = { λ C : ( x - λe ) is not invertible } . There are two immediate corollaries. Lemma 93. If x is an element of a commutative Banach algebra with unit, then k ˆ x k = ρ ( x ) . Lemma 94. If x is an element of a commutative Banach algebra with unit, then ρ ( x ) = 0 if and only if x is contained in every maximal ideal. We make the following definitions. Definition 95. If B is a commutative Banach algebra with unit we define the radical of B to be the set of all elements contained in every maximal ideal. Thus x radical( B ) if and only if ρ ( x ) = 0. 25
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Definition 96. We say that a commutative Banach algebra with unit is semi-simple if and only if its radical consists of 0 alone. Theorem 97. Let B be a commutative Banach algebra with unit. The map- ping x 7→ ˆ x is injective if and only if B is semi-simple. Exercise 98. Consider the Banach algebra X of continuous linear maps T : l l . Let S be the map given by S ( a 1 , a 2 , . . . ) = (0 , c 1 a 1 , c 2 a 2 , . . . ) , (with the sequence c j bounded. Explain why the closed Banach subalgebra generated by I and S is a commutative Banach algebra. Show that with an appropriate choice of c j we can have S n 6 = 0 for all n but ρ ( S ) = 0 . Theorem 99. Let B be a commutative Banach algebra with unit. If there exists a K > 0 such that k x k 2 K k x 2 k for all x B , then ρ is a norm equivalent to the original norm on B . 12 Finding the Gelfand representation Suppose we are given a commutative Banach algebra B and we wish to find its Gelfand representation. It is not enough to find its maximal ideals (or, equivalently its multiplicative linear functionals). We must also find the correct topology on the space of maximal ideals. The following simple remarks resolve the problem in all the cases that we shall consider. Exercise 100. Write out the proof that if ( X, τ ) and ( Y, σ ) are topological spaces with ( X, τ ) compact and ( Y, σ ) Hausdorff then, if f : ( X, τ ) ( Y, σ ) is a continuous bijection, f is a homeomorphism. Lemma 101. Suppose τ is a compact topology on the space M of maxi- mal ideals of commutative Banach space B with identity. If the maps ˆ x : ( M , τ ) C are continuous for each x B then τ is the weak star topology on M .
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