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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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    $\begingroup$ What is a stellar Banach algebra? $\endgroup$
    – Rasmus
    Jul 26, 2012 at 18:02
  • $\begingroup$ @Rasmus I think inolutive banach algebra with identity $\Vert a\Vert=\Vert a^*\Vert$ $\endgroup$
    – Norbert
    Jul 26, 2012 at 18:14
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    $\begingroup$ Cross-posted at MO: mathoverflow.net/questions/103229/… $\endgroup$
    – Philip Brooker
    Jul 26, 2012 at 20:00
  • $\begingroup$ That's exactly what I meant, Norbert. Thanks. $\endgroup$
    – Daniel
    Jul 26, 2012 at 21:02
  • $\begingroup$ 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. $\endgroup$
    – user16299
    Aug 26, 2012 at 1:46

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

Credits goes to Yemon Choi.

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  • $\begingroup$ No problem. I upvoted. $\endgroup$
    – Davide Giraudo
    Aug 19, 2013 at 20:50

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