Let $B$ be a unital Banach algebra and $A$ be a Banach subalgebra of $B$ containing the identity of $B.$ Let $a \in A$ be such that there exists a sequence $\{a_n\}_{n \geq 1}$ in $\text {GL} (A)$ such that $\lim\limits a_n = a.$ If $a \in \text {GL} (B),$ then show that $a \in \text {GL} (A).$
$\textbf {My Attempt} :$ I am trying to show that $a_n^{-1} \to a^{-1}.$ For that I used the following identity $$a_n^{-1} - a^{-1} = a_{n}^{-1} (a - a_n) a^{-1}.$$ So if we can show that $\{\|a_n^{-1}\|\}_{n \geq 1}$ is a bounded sequence of non-negative real numbers then we are through.
But this is where I got stuck. Any small hint in this regard would be warmly appreciated.
Thanks a bunch.
Since $a$ is invertible, you may assume without loss of generality that $a=1$. So we assume that $a_n\to 1$. Then \begin{align} \|1-a_n^{-1}\|&=\|a_n(1-a_n^{-1})+1-a_n^{-1}-a_n(1-a_n^{-1})\|\\[0.3cm] &=\|(a_n-1) + (1-a_n)(1-a_n^{-1})\|\\[0.3cm] &\leq\|1-a_n\|+\|1-a_n\|\,\|1-a_n^{-1}\|. \end{align} Regrouping, and as long as $\|1-a_n\|<1$ (which is guaranteed for $n$ big enough), $$ \|1-a_n^{-1}\|\leq \frac{\|1-a_n\|}{1-\|1-a_n\|} $$