Let $a\in A$ a stricly positive element (this means: for all states $\varphi$ of $A$ is $\varphi(a)>0$), let $u_n=a(\frac{1}{n}+a)^{-1}$, $n\in\mathbb{N}$ . Claim: for all $b\in A$, for all states $\varphi$ of $A$ is $$\lim\limits_{n\to\infty}\varphi(b^*(1-u_n)^2b)=0.$$
I have tried to prove it, but there must be a mistake because I didn't use that $a$ is strictly positive. But I can't find the mistake.
a remark: I still don't know that $(u_n)_{n\in\mathbb{N}}$ is an approximate unit.
My try: Let $\varphi$ a state of $A$, then there exist a Hilbert space $H_\varphi$, a $*$-representation $\pi:A\to L(H)$ with cyclic vector $x_{\pi}$, such that $\|x_{\pi}\|=1$ and $\varphi(a)=\langle \pi(a)x_{\pi},x_{\pi}\rangle$ for all $a\in A$.
Then it is: $$\varphi(b^*(1-u_n)^2b) \\ =\langle \pi(b^*(1-u_n)^2b)x_{\pi},x_{\pi}\rangle \\ =\langle \pi(1-u_n)\pi(b)x_{\pi},\pi(1-u_n)\pi(b)x_{\pi}\rangle \\ =\|\pi(1-u_n)\pi(b)x_{\pi}\|^2 =\|\pi(b-u_nb)x_{\pi}\|^2 \le \|b-u_nb\|^2$$
I think the argument has to be wrong in the following part, where I want to prove that $\|b-u_nb\|\to 0$.
I already know, that $\|a-u_na\|\to 0, \|a-au_n\|\to 0$ for positive $a\in A$. If $b\in A$ is arbitrary, there are positive elements $b_1,b_2,b_3,b_4\in A$ such that $b=(b_1-b_2)+i(b_3-b_4)$. Then it is:
$$\|b-u_nb\|=\|(b_1-b_2)+i(b_3-b_4)-u_n[(b_1-b_2)+i(b_3-b_4)]\| \\ =\|b_1-u_nb_1+u_nb_2-b_2+ib_3-iu_nb_3+iu_nb_4-ib_4\| \\ \le \|b_1-u_nb_1\|+\|u_nb_2-b_2\|+\|b_3-u_nb_3\|+\|u_nb_4-b_4\|\to 0$$
Here I nowhere use that $a$ is strictly positive. Could you help me to find the mistake?
Regards
No. You know that for your fixed $ a $ that happens. There is no reason why, for $a $ not invertible, $b-bu_n\to0$ for arbitrary $b $.
By the way, you are using that your algebra is unital. In that case, "strictly positive" is simply "positive and invertible".