# When is a Banach Algebra stellar?

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.

• What is a stellar Banach algebra? – Rasmus Jul 26 '12 at 18:02
• @Rasmus I think inolutive banach algebra with identity $\Vert a\Vert=\Vert a^*\Vert$ – Norbert Jul 26 '12 at 18:14
• 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