# strictly positive elements in $C^*$-algebra

Let $A$ be a $C^*\text{-algebra}$ and $A_+$ denote the positive elements. An element $a\in A_+$ is called

strictly positive if $\overline{aAa}=A$. Want to find the following:

a)What are the strictly positive elements of $C_0(\Omega)$ where $\Omega$ is locally compact Hausdorff space, and $C_0(\Omega)$ is the space of continuous complex valued functions that vanishes at $\infty$.

b)If $A$ is unital, then $a\in A_+$ is strictly positive iff $a$ is invertible.

c)if $(e_n)$ is an approximate identity of $A$, then $a:=\sum_{n=}^{\infty}\frac{1}{2^n}e_n$is strictly positive.

What I tried and know:

a) I know that positive elements in this space are functions with positive images, and thus, strictly positive elements are functions with positive images and now real root. is this correct? and how can I show that the positive elements are these kind of functions.

b) I figures out that if $a$ in invertible then $a$ is strictly positive, but no idea how to do the other direction.

c)I did this but using another characterization of strictly positive elements but want to do it with this one, and I don't know how.

• @julien, Yes, Now I need to show what are the positive functions you described satisfies $\overline{aAa}=A$., my intuition says it should be functions with range in $(0,\infty)$ but I can't really work the details. – i.a.m Mar 24 '13 at 20:52
• @julien, Yes there are no non zero constant functions here, can you tell me how to fix this? – i.a.m Mar 24 '13 at 21:35
• @julien : But that only shows that the set of strictly positive elements in $C_0(\Omega)$ lies in the set $\{ f \in C_0(\Omega) | Image (f) \subset (0, \infty)\}$. How to derive the exact set of strictly positive elements ? – user44349 Mar 25 '13 at 4:33
• I just wrote an answer for part (b) in this question – user56706 Mar 25 '13 at 15:07
• @VahidShirbisheh, Thank you for the answer, any idea how to do part c? – i.a.m Mar 25 '13 at 15:13

Question a) Denoting $\widetilde{\Omega}$ the one point compactification of $\Omega$, and denoting $\infty$ the extra point, we have a natural isometric embedding of $C_0(\Omega)$ into $C(\widetilde{\Omega})$. The range of this embedding is $$A=\{f\in C(\widetilde{\Omega})\;;\;f(\infty)=0\}.$$ This is clearly a sub $C^*$ algebra of $C(\widetilde{\Omega})$. We claim that its strictly positive elements are $$A_{++}=\{f\in A\;;\; f(x)\neq 0\;\forall x\in \Omega\}.$$

The key tool is Urysohn's lemma for $C(\widetilde{\Omega})$, which holds because $\widetilde{\Omega}$ is Hausdorff compact, hence normal.

First, assume that $f\in A$ vanishes at some $x_0$ in $\Omega$. By Urysohn, there exists $g\in C(\widetilde{\Omega})$ such that $g(x_0)=1$ and $g(\infty)=0$. Now every function $h$ in $fAf$ has $f(x_0)=0$, hence also every function in $\overline{fAf}$. So $g$ can't belong to the latter, and $f$ is not strictly positive.

Assume now that $f\in A$ does not vanish on $\Omega$. For every function $g\in A$ with support contained in $\Omega$, we can write $g=fhf$ with $h=g/f^2$ on the support of $g$ and $0$ elsewhere. So $g$ belongs to $fAf$. Now every function in $A$ is the uniform limit of functions with support in $\Omega$, using Urysohn again. So $\overline{fAf}=A$, i.e. $f$ is strictly positive.

Back to $C_0(\Omega)$, the strictly positive elements are the positive elements which do not vanish on $\Omega$. That is the positive functions with limit $0$ at $\infty$.

Question b) If $a$ is invertible, then every $x\in A$ can be written $x=aya$ with $y=a^{-1}xa^{-1}$ in $A$. So $A=aAa$. Conversely (the argument has laready been given here by Vahid Shirbisheh), if $A=\overline{aAa}$, then a fortiori $1$ belongs to $\overline{aAa}$. So there exists $x\in A$ such that $\|1-axa\|<1$. As is well-known, in a Banach algebra, this implies that $axa$ is invertible (cf. Neumann series). It follows that $a$ is both left and right invertible, hence invertible. This proves the equivalence.

Question c) I assume by approximate identity you mean every $e_n$ is positive, has $\|e_n\|\leq 1$, and $\lim xe_n=\lim e_nx=x$ for every $x\in A$. The series defining $a$ is absolutely convergent, so $a$ is well-defined.

The only proof I can think of uses the following characterization: $a\geq 0$ such that $\|a\|\leq 1$ is strictly positive if and only if $a^\frac{1}{n}$ is an approximate identity of $A$.

It only remains to observe that $\|a\|\leq 1$ and that, indeed $a^\frac{1}{n}$ is an approximate identity. This follows from the inequalities $$h^\frac{1}{m}\geq \frac{1}{2^\frac{n}{m}}h_n^\frac{1}{m}\geq \frac{1}{2^\frac{n}{m}}h_n$$ for all $n,m$.

• Thank you for this answer, and yes please do more, when you have the time its much appreciated. Thank you – i.a.m Mar 25 '13 at 13:28
• Hi @julien I have a question. But I cannot produce any idea. There exist an answer.but not enough in order for me to understand..i think. Please can you teach me the solution more precisely. Thank you. – user315 Mar 25 '13 at 17:28