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I know that if there are enough Hermitian elements in a Banach algebra, then the Banach algebra is stellar. In particular, I'm interested in the two spaces $B(L^1(S^1,\Sigma,\mu))$ the space of bounded linear operators on Lebesgue integrable functions of the circle and $B(ba(\Sigma))$ the space of bounded linear operators on finite, finitely-additive Borel measures. I know about the results that having enough Hermitian elements is sufficient, but I'm not quite sure how to apply them.

The issue comes up because I am trying to bound the inverse of a Hermitian element in terms of its spectral radius. From my reading, we have an equality for $C^\star$ algebras and an inequality for Banach algebras.

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What is a stellar Banach algebra? –  Rasmus Jul 26 '12 at 18:02
    
@Rasmus I think inolutive banach algebra with identity $\Vert a\Vert=\Vert a^*\Vert$ –  no identity Jul 26 '12 at 18:14
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Cross-posted at MO: mathoverflow.net/questions/103229/… –  Philip Brooker Jul 26 '12 at 20:00
    
That's exactly what I meant, Norbert. Thanks. –  Daniel Jul 26 '12 at 21:02
    
Daniel and @Norbert: the first sentence appears to allude to the Vidav-Palmer theorem, but the conclusion of that theorem is stronger than being involutive with isometric involution: the conclusion is that we actually get a $C^*$-algebra. So I am not sure whether this mention of stellar Banach algebras is precisely what is meant. –  user16299 Aug 26 '12 at 1:46

1 Answer 1

The question was cross-posted at mathoverflow and answered there.

Credits goes to Yemon Choi.

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No problem. I upvoted. –  Davide Giraudo Aug 19 '13 at 20:50

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