Let $\mathfrak{A}$ be a unital $C^*$-Algebra ($\mathfrak{A} \neq \mathbb{C}$) and $A$ a positive element with norm smaller or equal to $1$. Knowing that $A \mathfrak{A} A$ is the hereditary subalgebra generated by $A$:
- (Answered) Is there a sufficient (and maybe even necessary) condition (either on $\mathfrak{A}$ or $A$) to tell $A \mathfrak{A} A$ has a projection?
- (Open) Given $c \in [0, 1) \backslash \sigma(A)$, define a function $f$ in $[0, 1]$ such that $f \vert_{[0, c]} =0$ and $f \vert_{[c, 1]}$ is a straight line joining $(c, 0)$ and $(1, 1)$. If $A \mathfrak{A} A$ has a projection, how to show $P \leq E_{A}[c, 1]$? Here $\chi_{[c, 1]}$ is a characteristic function of a component and hence $E_A{[c, 1]} \in \mathfrak{A}$.
- (Answered) Is there a continuous function $g$ defined on $[0, 1]$ such that $g(A) \mathfrak{A} g(A) = A \mathfrak{A} A$?
(Edit) For question 1, if the given $\mathfrak{A}$ is simple and purely infinite (for equivalent definitions of purely infinite please refer to this post1). If $\mathfrak{A}$ is finite-dimensional, then the answer to question 1 will be yes. In general, when $\mathfrak{A}$ is real rank zero, $A \mathfrak{A} A$ will also have a proper projection (but this might overkill ... Hopefully for a general $\mathfrak{A}$ there will be some restriction on $A$ such that $A \mathfrak{A} A$ will have a proper projection).
(Edit) For question 3, in this post2 there is a sufficient condition on the function $f$ such that $f(A) \mathfrak{A} f(A) = A \mathfrak{A} A$. The answer in this post proved that whenever $f(0) = 0$ and $f$ strictly positive in $[0, 1]$ then $f(A) \mathfrak{A} f(A) = A \mathfrak{A} A$. The direction here is the same the one mentioned by Prahlad.
2 is not true. Let $X=\{\frac1n:n\in\mathbb N\}\cup\{0\}$ with the topology inherited from being a subspace of $\mathbb R$, let $\mathfrak A=C(X)$, and let $f\in C(X)$ be the function $f(x)=x$. Then the projection $E_f[\frac34,1]$ (i.e. the characteristic function of the set $\{1\}$) is a minimal projection, while the hereditary subalgebra $fC(X)f$ contains plenty of projections.