Let $A$ be a $C^*$ algebra. Let $a\in A$ be such that $a^*a-aa^*\geq 0$. Doe this imply that the spectral radius of $a$ is equal to $\|a\|$?
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1$\begingroup$ It may help, in looking for proofs or counter-examples, to know that if we realise $A$ as a unital star-subalgebra of $B(H)$, then the condition on $a$ is the definition of being a hyponormal operator $\endgroup$– Yemon ChoiCommented Apr 28, 2018 at 14:37
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1$\begingroup$ yes. I know for a Hyponormal operator $r(T)=\|T\|$. I am not able to prov e it for C* algebra case. $\endgroup$– user31459Commented Apr 28, 2018 at 14:40
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3$\begingroup$ Why not? If you take any unital embedding $A \to B(H)$ then this embedding preserves norm (by definition) and preserves spectral radius. In any case, why not look at the proof for the case of hyponormal operators and modify it if necessary? $\endgroup$– Yemon ChoiCommented Apr 28, 2018 at 14:43
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$\begingroup$ Actually, my approach was that since $a^*a-aa^*\geq 0$, then $\|a^*a\|\geq\|aa^*\|$, Then using Corollary 15.6 in Complete normed algebras by Bonsal and Duncan, $a^*a=aa^*$. But that will mean that every Hyponormal operator is normal in the case of B(H). Can you tell me where I am wrong? $\endgroup$– user31459Commented Apr 28, 2018 at 14:50
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2$\begingroup$ In a $C^*$ algebra you always have $ \Vert a a^* \Vert = \Vert a^* a \Vert$ for any $a $. I can't go check what this corollary 15.6 is, but you are probably misusing it. $\endgroup$– Simon HenryCommented Apr 28, 2018 at 15:22
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Yes, see T. Ando, On hyponormal operators, Proc. Amer. Math. Soc. 14 (1963), 290-291. His main result states that $\|T^n\| = \|T\|^n$ for any hyponormal operator, which implies the conclusion by Gelfand's spectral radius formula.
answered Apr 28, 2018 at 18:06