The following question was raised when I read a papaer "MF actions and K-theoretic dynamics". In one of the proofs in that paper, the author utilized a so called "elementary perturbation result in C*-algebra":
Let $\varepsilon>0$, there is a $0<\delta<min\{1, \frac{\varepsilon}{4}\}$ with the following property: if $D$ is any C*-algebra with $d\in D$ satisfying $||d^{*}d-1||<\delta$ and $||dd^{*}-1||<\delta$, there is a $v\in U(D)$ with $||d-v||<\frac{\varepsilon}{4K}$ for some fix number $K$. (Here, the U(D) denotes the set of all unitaries in $D$.)
I think it is an elementary exercise in operator algebra, but I do know how to find such a $v$ in the exercise, could someone give me some hints?
Something like this should work : Given $\epsilon > 0$, consider $f(x) = x^{-1/2}$. This is uniformly continuous on $[1/2,2]$, so $\exists 1/2>\delta > 0$ such that for any positive invertible $a \in A$ such that $\sigma(a) \subset [1/2,2]$, we have $$ \|a - 1\| < \delta \Rightarrow \|\sqrt{a}^{-1} - 1\| < \frac{\epsilon}{\sqrt{2}} \qquad\qquad (\ast) $$ Now suppose $\|dd^{\ast} - 1\| < \delta$ and $\|d^{\ast}d - 1\| < \delta$, then $dd^{\ast}$ and $d^{\ast}d$ are both invertible. Hence, $d$ is left and right invertible - and so it is invertible. Therefore $|d|$ is invertible. Also $$ \sigma(d^{\ast}d) \subset [1-\delta,1+\delta] \subset [1/2,2] $$ So we may apply $(\ast)$ $$ \||d|^{-1} - 1\| < \frac{\epsilon}{\sqrt{2}} $$ Now $\|d\| = \sqrt{\|d^{\ast}d\|} \leq \sqrt{1+\delta} \leq \sqrt{2}$. Hence $$ \|d|d|^{-1} - d\| < \epsilon $$ Now check that if $u = d|d|^{-1}$ is a unitary.