Is the set of positive elements in a $C^*$-algebra norm closed?

Question

Suppose, $H$ is a complex Hilbert space and consider the Banach space $B(H)$ with operator norm. If $B(H)^{+}$ denote the set of positive operators then is it closed subset in $B(H)$?

For this; let $(A_n)$ be a sequence in $B(H)^+$ such that $A_n \to A$ in $B(H)$. Will prove $A\in B(H)^+$: suppose $x \in H$ then we have $$|\langle A_nx, x\rangle - \langle Ax, x\rangle|=|\langle(A_n-A)x, x\rangle|\le ||A_n-A|| .||x||^2\to 0 ~\text{as} ~n\to \infty$$

Now since each $\langle A_nx, x\rangle \ge 0$ so is there limit $ \langle Ax, x\rangle$, i.e., $A \in B(H)^+$. This proves the claim.

My Question. Is this true for any $C^*$-Algebra too? i.e., if $\mathcal{A}$ is a $C^*$-algebra and $\mathcal{A}^+:=\{a \in \mathcal{A}: a^*=a, \sigma(a) \subset [0, \infty) \}$ then is $\mathcal{A}^+$ norm closed in $\mathcal{A}$?

I am not sure how to (dis)prove this. I know that $\mathcal{A}_{sa}$ is norm closed in $\mathcal{A}$ but is it true for the subset $\mathcal{A}^+$? Any suggestion is appreciated. Thanks

Answer 1

This is true for any $C^*$-algebra. I will edit and add an elementary proof soon, but here is a short explanation:

We know by the Gelfand-Naimark theorem that every $C^*$-algebra $A$ can be faithfully represented in $B(H)$ for some Hilbert space $H$, in other words, if $A$ is a $C^*$-algebra there exists an injective $*$-homomorphism $f:A\to B(H)$. So we can assume that $A\subset B(H)$. Since you already showed that $B(H)_+$ is closed, observe that $$A_+=A\cap B(H)_+$$ which is closed in $A$.

Warning! Usually the Gelfand-Naimark theorem comes much later than the proof of the fact that $A_+$ is closed, so there is a great chance there is cyclicity in the above argument. As I said, the standard proof of the fact that $A_+$ is closed is elementary and I will add it soon, but the above is just to convince you that this is indeed true.

Edit

Proposition: Let $A$ be a $C^*$-algebra. Then the set of positive elements $A_+$ is closed in $A$.

First we need to prove the following lemma:

Lemma: If $A$ is a unital $C^*$-algebra, $a\in A_{sa}$ and $t\in\mathbb{R}$ such that $\|a-t1\|\leq t$. Then $a\geq0$. Conversely, if $t\in\mathbb{R}$ is a number satisfying $t\geq\|a\|$ and $a\geq0$, then $\|a-t\|\leq t$.

Proof of lemma: Without loss of generality, we can pass to the $C^*$-algebra which is generated by $a$ and $1_A$. This is an abelian $C^*$-algebra and because of the continuous functional calculus it is $*$-isomorphic to $C(\sigma(a))$, the continuous functions over the spectrum of $a$. Therefore, the proof amounts to convince ourselves of the following: if $f$ is a continuous function over a compact set and $t\in\mathbb{R}$ satisfies $\|f-t\|_\infty\leq t$, then $f\geq0$ and, if $f\geq0$ and $t\geq\|f\|_\infty$, then $\|f-t\|_\infty\leq t$. But this is only a matter of verification.

Proof of proposition: Without loss of generality we can assume that $A$ is unital, otherwise we simply pass to the unitization. Let $(a_n)\subset A_+$ with $a_n\to a$. Since each $a_n$ is positive, by the above lemma we have that $$\bigg\|a_n-\|a_n\|1\bigg\|\leq\|a_n\|$$ so letting $n\to\infty$ we obtain $$\bigg\|a-\|a\|1\bigg\|\leq\|a\|$$ So, again by the above lemma, we obtain $a\geq0$.