Spectrum of elements in $C^*$-subalgebras Assume $\mathcal{A}$ is a $C^*$-algebra with unit $1$ and $\mathcal{B}\subset\mathcal{A}$ is a $C^*$-subalgebra (i.e. a closed $*$-subalgebra) such that $1\in\mathcal{B}$. It is said that under these assumptions, for any $a\in\mathcal{B}$ the spectrum $\sigma_\mathcal{B}(a)$ of $a$ in $\mathcal{B}$ coincides with $\sigma_\mathcal{A}(a)$, that is:
If $a-\lambda 1$ has an inverse $b\in\mathcal{A}$, then $b\in\mathcal{B}$.
Now, my questions are the following:


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*Do you know a nice proof of the above statement? I have found a proof that goes like this: If $b=(a-\lambda)^{-1}$ exists in $\mathcal{A}$, it can be expressed as a convergent power series, i.e. it is the norm limit of partial sums each belonging to $\mathcal{B}$, hene also $b\in\mathcal{B}$. Although this argument looks really nice and I'm aware of Neumann series, I do not see why $b$ can be expressed as a power series. Do you?

*Under the above assumptions, is the more general statement $\mathcal{B}^\times=\mathcal{A}^\times\cap B$ true? (here, we denote by $\mathcal{A}^\times$ and $\mathcal{B}^\times$ the set of invertible elements in $\mathcal{A}$ and $\mathcal{B}$, respectively)

 A: We essentially want to prove that $a \in B$ is invertible in $A$ iff it is invertible in $B$. Let us assume that $a$ is selfadjoint; after all, if $a^*a$ has an inverse in $B$, then so does $a$.
I think one has to use the Stone-Weierstrass theorem in some form. The argument you indicate basically does this in the algebra generated by $a$, which is commutative and actually of the form $C(Spec(a))$. I am slightly worried about the non-circularity of such a proof, however: That the spectrum of an element does not change when passing to a subalgebra is quite deeply ingrained into the continuous function calculus.
I would prove it as follows: Let $X$ be the $C^*$-algebra generated by $1,a,a^{-1}$ inside $A$ and $Y$ the algebra generated by $1,a$ inside $A$ and hence inside $B$ since $1, a \in B$. Since $a$ is self-adjoint, both $X$ and $Y$ are commutative and we have $Y \subset X$. We seek to prove $X = Y$.
Since $X$ is commutative, the Gelfand transform yields an isomorphism $X \rightarrow C(Spec(X))$. Let $Y' \subset C(Spec(X))$ be the image of $Y$ under this map. $Y'$ is a closed $*$-subalgebra of $C(Spec(X))$. Furthermore, if $l,k \in Spec(X)$ are different, we must have $l(a) \neq k(a)$ since otherwise $l, k$ agree on $1, a , a^{-1}$ and then everywhere. But since $a \in Y$, the evaluation at $a$-map $Spec(X) \rightarrow \mathbb{C}$ lies in $Y'$. In other words, $Y'$ separates the points of $Spec(X)$ and is hence dense in $C(Spec(X))$ by Stone-Weierstrass.  Since $Y'$ is also closed, it follows $Y' = C(Spec(X))$ and hence $Y = X$ as desired.
