Let $A$ be a unital $C^*$-algebra in $\mathcal{B}(H)$ with unit the identity operator $I$.
Assume that $\mathcal{A}\subset A$ is a $*$-subalgebra of $A$ that contains $I$.
Moreover, assume that $\mathcal{A}$ is closed under holomorphic functional calculus; that is, for every $a\in \mathcal{A}$ and every function $f$ holomorphic in a neighbourhood of the spectrum $\sigma_A(a)=\sigma_{\mathcal{}B(H)}(a)$, we have $f(a)\in \mathcal{A}$.
Question: Let $p\in \mathcal{A}$ be a non-zero projection. Is it true that the $*$-subalgebra $p\mathcal{A}p\subset pAp$ is also closed under holomorphic functional calculus?
Issues:
- The unit of $pAp$ is now $p$.
- Is the spectrum of an element in $pAp$ related to the spectrum in $A$?
Would it help if $p$ was a very big projection? Like a full projection; that is, $ApA$ is dense in $A$.
Thanks!
Best Answer
Before tackling the question itself, it is perhaps useful to discuss a minor point regarding the fact that the unit of $p A p$ is $p$, rather than $I$. To highlight this difference, whenever we are given an element $b\in p A p$, we will write $\sigma _p(b)$ for the spectrum of $b$ relative to $p A p$, reserving the notation $\sigma (a)$ for the spectrum of any element $a\in A$ relative to $A$ (or, equivalently, to $B(H)$).
Leaving aside the trivial case in which $p=1$, observe that no element $b\in pAp$ is invertible relative to $A$, so 0 is always in $\sigma (b)$. In fact it is easy to show that, for every such $b$, one has $$ \sigma (b) = \sigma _p(b)\cup \{0\}. \tag{*} $$
Likewise, if $b\in p A p$, and $f$ is a holomorphic function on a neighborhood of $\sigma _p(b)$, we will denote by $f_p(b)$ the outcome of the holomorphic functional calculus computed relative to $pAp$. As before, we will reserve the undecorated expression $f(a)$ for the holomorphic functional calculus relative to $A$.
In the event that $f$ is holomorphic on the larger set $\sigma _p(b)\cup \{0\}$, one may easily prove that $$ f(b)=f_p(b)+f(0)(1-p), \tag{**} $$ for every $b\in p A p$.
This said, let $b\in \mathscr A$ and let $f$ be a holomorphic function on an open set $U$ such that $\sigma _p(b)\subseteq U$.
CASE 1: Assuming first that $0\in U$, we have by ($*$) that $f$ is also holomorphic on a neighborhood of $\sigma (b)$ and we have by hypothesis that $f(b)\in \mathscr A$. Applying ($**$) it then follows that $$ f_p(b)=pf(b)p \in p\mathscr A p, $$ as desired.
CASE 2: $0\notin U$.
In this case it is clear that $0\notin \sigma _p(b)$, and since $\sigma _p(b)$ is compact, one has that $r:=\text{dist}(0,\sigma _p(b))>0$. So $$ \sigma _p(b)\subseteq V:= \mathbb C\setminus \overline {B_{r/2}(0)}. $$
By restricting $f$ to $U\cap V$, we may assume that $U\subseteq V$. The open ball $B_{r/3}(0)$ is therefore disjoint from $U$, so we may extend $f$ to $U\cup B_{r/3}(0)$ by declaring it to be identically zero on $B_{r/3}(0)$. Consequently the extended $f$ is now defined on a neighborhood of $\sigma _p(b)\cup \{0\}$.
Since the extension process did not change the values of $f$ on points of $\sigma _p(b)$, the outcome of $f_p(b)$ remains unchanged and hence the conclusion follows as in case 1.