Proposition 34.2.3.label Let $A$ be a unital $C^{*}$-algebra, $B \subset A$ be a $C^{*}$-subalgebra containing $1$, then $G(B) = G(A) \cap B$.
Proof. Let $x \in G(A) \cap B$, then $x^{*}x \in G(A) \cap B$ as well. In particular, $0 \not\in \sigma_{A}(x^{*}x)$. Since $x^{*}x \in A_{sa}$, $\sigma_{A}(x^{*}x) \subset \real$ by Proposition 34.4.6. By Proposition 33.5.12, $\partial \sigma_{B}(x^{*}x) \subset \sigma_{A}(x^{*}x) \subset \real$. Thus $\sigma_{B}(x^{*}x) \subset \real$ as well, which means that $\partial \sigma_{B}(x^{*}x) = \sigma_{B}(x^{*}x) = \sigma_{A}(x^{*}x)$. Therefore $0 \not\in \sigma_{A}(x^{*}x) = \sigma_{B}(x^{*}x)$, $x^{*}x \in G(B)$, and $x \in G(B)$.$\square$
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