a question on hereditary $C^*$- subalgebras Let $A$ be a $C^*$-algebra and $a\in A$ be positive. It is known that $\overline{aAa}$ is the hereditary subalgebra generated by $a$. Now, let $f$ be a continuous function on $[0,\|a\|]$ such that $f(0)=0$ and $f(x)>0$ whenever $x>0$.
My question is whether $\overline{f(a)Af(a)}=\overline{aAa}$? Note that the inclusion $\overline{f(a)Af(a)}\subset\overline{aAa}$ is trivial. So how does one prove the converse?
Thanks for all helps!
 A: What follows is an incomplete solution, but perhaps has some merit. One can apply Proposition 2.5 of this paper here, but there appears to be a problem I describe below.

Lemma: Let $X$ be a compact Hausdorff space, $f,g\in C(X)_+$ be two positive functions such that $\text{supp}(g) \subset \text{supp}(f)$. Then, for any $\epsilon >0, \exists h\in C(X)$ such that $\|g-hfg\| < \epsilon$

Now apply the above lemma to the function $f$ and $g(t) =t$. Let $h$ as in the lemma, and let $b = h(a)f(a)a$. Then for any $x\in A, bxb \in f(a)Af(a)$, and
$$
\|axa - bxb\| \leq \|axa-axb\| + \|axb - bxb\| < \epsilon\|x\|(\|a\| + \|b\|)
$$
Now 
suppose one can control $\|b\|$ in terms of $\|a\|$ and $\|f(a)\|$,
then this would imply $axa \in \overline{f(a)Af(a)}$ proving that $\overline{aAa} \subset \overline{f(a)Af(a)}$

The problem then is to control $\|b\|$, which amounts to controlling $\|h\|$ in the above lemma. Going through the proof, one sees that
$$
\|h\| \leq \frac{1}{\delta}
$$
where $\delta > 0$ is obtained by the continuity of $f$. I am not sure if there is a way to control this quantity. However, for certain functions $f$, it is possible (for instance if $f(t) \geq t$ for all $t\in [0,\|a\|]$).
Hope this helps.
