Approximating the product of two positive elements in a $C^\star$ algebra by a positive element Let $\mathcal{A}$ be a (unital) $C^\star$ algebra, and $a,b$ be two positive elements of $\mathcal{A}$. We know that $ab$ is not positive in general (in fact, it is positive iff $a$ and $b$ commute), but it is always true that $\sigma(ab)\geq 0$. I was wondering if it would be possible to approximate $ab$ by a positive element in $\mathcal{A}$, that is, if for all $\epsilon>0$ there exists some $c>0$ in $\mathcal{A}$ such that $\lVert ab - c \rVert < \epsilon$ .
My obvious first attempt was with $c=\frac{ab+ba}{2}$ . But the noncommutativity of $a$ and $b$ blocks this from working. I was wondering if functional calculus techniques might work, but I am not sure. I know that the set of hermitian elements is not dense in $\mathcal{A}$, so the approximation might actually not work as well, but I don't have a counterexample at hand. I would be ineterested in both finite and infinite dimensional settings.
(Of course, the problem is trivial if $\mathcal{A}$ is commutative.)
 A: If $a,b$ are noncommuting positive elements in a $C^*$-algebra then $x:=ab$ is not self-adjoint.
Let $s$ be any self-adjont element.
Then
$$2\|x-s\|=\|x-s\|+\|x^*-s\|\ge \|x-x^*\|>0$$ Therefore $$\|x-s\|\ge {1\over 2}\|x-x^*\|$$
That means the distance from $x$ to self-adjoint operators is greater or equal $\|x-x^*\|/2.$  Hence $x=ab$ cannot be approximated by self-adjoint elements, positive in particular.
Remarks As $$\left\|x-{x+x^*\over 2}\right \|={1\over 2}\|x-x^*\|$$ the distance from $x$ to self-adjoint elements is equal $\|x-x^*\|/2.$ The proof shows that the set of self-adjoint elements in a $C^*$-algebra is closed.
A: Take
$$E_{11}=\begin{bmatrix}1 & 0 
\\ 0 & 0\end{bmatrix}\text{ and }J=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix},$$
both positive. Their product:
$$JE_{11}=\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}.$$
Can we approximate this using a sequence of hermitian matrices:
$$C_n=\begin{bmatrix}a_n & z_n \\ \overline{z_n} & b_n\end{bmatrix}?$$
No, because we cannot simultaneously approximate $z_n\to 0$ and $\overline{z_n}\to 1$.
Edit: In terms of the answer of Ryszard, $A:=JE_{11}$ gives $c=1/2$ as $$A-A^*=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix},$$
which has norm one.
A: If you had your approximation, you would have $ab$ as a limit of positive elements. But a limit of positive elements is positive.
This can be seen most easily if you work on a faithful representation $A\subset B(H)$: if $c_n\to ab$ with $c_n\geq0$, then
$$
\langle ab\xi,\xi\rangle=\lim\langle c_n\xi,\xi\rangle\geq0,\qquad\qquad \xi\in H.
$$
If you want to stay abstract, an element $a\in A$ is positive if and only if $\varphi(a)\geq0$ for all states $\varphi$ (proof below). And for any state you have
$$
\varphi(ab)=\lim\varphi(c_n)\geq0,
$$
so $ab\geq0$.

Proof that positivity can be tested on states.
Let $a\in A$ such that $\varphi(a)\geq0$ for all states $\varphi$. As $\varphi(\operatorname{Im} a)=\operatorname{Im} \varphi(a)=0$, we get $\operatorname{Im} a=0$ (by functional calculus, as any selfadjoint attains its norm at a state) and so $a=a^*$. Knowing that $a$ is selfadjoint we can do the following. If it were not positive, it has $\lambda<0$ in its spectrum; using functional calculus, we can get a state (a character, actually) $\varphi$ with $\varphi(a)=\lambda<0$.
