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If $A$ is a unital Banach algebra and $B$ is a closed subalgebra and $\sigma$ denotes the spectra then the following inclusion holds:

$$ \partial \sigma_B (b) \subseteq \partial \sigma_A (b)$$

for every $b \in B$. Consider the following proof of the statement:

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where condition (1) and (2) are

(1) The set $\mathrm{Inv}(B)$ is a clopen subset of $B \cap \mathrm{Inv}(A)$

(2) For each $b \in B$: $ \sigma_A (b) \subseteq \sigma_B (b) $ and $ \partial \sigma_B (b) \subseteq \partial \sigma_A (b)$

Where in the proof is it used that $b-\lambda \notin \mathrm{Inv}(A)$? It holds that $\mathrm{Inv}(B) \subseteq \mathrm{Inv}(A)$ so that if $b-\lambda_n \in \mathrm{Inv}(B)$ then also $b-\lambda_n \in \mathrm{Inv}(A)$ and hence $\lambda \in \partial \sigma_A(b)$ since $(b-\lambda_n) \to (b-\lambda)$.

Student
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    Well, if you had $b-\lambda \in \operatorname{Inv}(A)$, then you couldn't conclude $\lambda \in \sigma_A(b)$, which is necessary for $\lambda \in \partial \sigma_A(b)$. – Daniel Fischer Mar 25 '14 at 13:02
  • @DanielFischer But the proof goes into the opposite direction: because $\lambda \in \partial \sigma_A(b) $ it follows that $\lambda \in \sigma_A(b)$ (due to the fact that the spectrum is a closed set). If this is flawed reasoning I do not understand where the flaw is. Please could you elaborate? – Student Mar 25 '14 at 14:36

1 Answers1

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The first thing in the proof concerning $A$ is $b - \lambda \notin \operatorname{Inv}(A)$, which just means $\lambda \in \sigma_A(b)$.

Next, it is observed that the terms of the sequence $(\lambda_n)$ do not belong to $\sigma_A(b)$ - namely, it is stated that $b - \lambda_n \in \operatorname{Inv}(A)$ - which says that $\lambda$ is not an interior point of $\sigma_A(b)$. Since $\lambda \in \sigma_A(b)\setminus \sigma_A(b)^{\Large\circ}$, it follows that $\lambda \in \partial \sigma_A(b)$.

$b-\lambda \notin \operatorname{Inv}(A)$, or equivalently $\lambda\in\sigma_A(b)$ is essential for the argument. If we had $b-\lambda\in \operatorname{Inv}(A)$, or, equivalently, $\lambda\notin\sigma_A(b)$, then we could not have $\lambda\in\partial\sigma_A(b)$, since spectra are closed.

Daniel Fischer
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  • Actually one more thing is not clear to me (I hope you can help me): To derive $b-\lambda \notin \mathrm{Inv}(A)$ am I right that one must argue as follows: – Student Mar 25 '14 at 16:04
  • One has $\lambda_n - b \in \mathrm{Inv}(B)$ and $(\lambda_n - b) \to (\lambda -b)$. Since $\mathrm{Inv}(B)$ is closed in $B \cap \mathrm{Inv}(A)$, if $\lambda -b$ was in $B \cap \mathrm{Inv}(A)$ it would follow that $\lambda -b \in \sigma_B^c (b)$.This would be a contradiction hence $\lambda -b$ must lie in $(B \cap \mathrm{Inv}(A))^c = B^c \cup \mathrm{Inv}(A)^c$. The crucial step is this: the author assumes that $B$ contains the unit of $A$ because only then does it hold that $b-\lambda \in B$. Right? – Student Mar 25 '14 at 16:05
  • Yes, that's right. – Daniel Fischer Mar 25 '14 at 16:11
  • Maybe the definition of $\sigma_B$ doesn't make sense if $1 \notin B$. But I think it does make sense and then $\sigma_B(b) = \mathbb C - {0}$ because for all $\lambda$, $\lambda -b \notin B$. – Student Mar 25 '14 at 20:47
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    If $1\notin B$, then you would have $\sigma_B(b) = \mathbb{C}$, since $\mathbb{C} \setminus \sigma_B(b) = \left{ \lambda \in\mathbb{C} : (\exists c\in B) \bigl(c\cdot(\lambda-b) = 1\bigr)\right}$, and if $1\notin B$, such a $c$ cannot exist for any $\lambda$. – Daniel Fischer Mar 25 '14 at 21:01
  • Great, thanks!. – Student Mar 26 '14 at 07:39