Let $\mathcal{A}=\operatorname{M}_k(\mathbb{R})$ be the Banach algebra of $k\times k$ real matrices and let $(U_n)_{n\in\mathbb{N}}\subset\operatorname{GL}_k(\mathbb{R})$ be a sequence of invertible elements such that $U_n\to 0$ as $n\to\infty$. Define $\sigma_n\in\operatorname{Aut}(\mathcal{A})$ via $X\mapsto U_nXU_n^{-1}$. Suppose I have a sequence $(W_n)_{n\in\mathbb{N}}\subset\mathcal{A}$ such that $W_n\to W\in\mathcal{A}$ as $n\to\infty$. I would like to determine $\lim_{n\to\infty}\sigma_n(W_n)$.
My question is how can I approach such a problem? It looks like something that should have a general answer (for $\mathcal{A}$ not necessarily finite-dimensional Banach algebra over the reals) in the theory of operator algebras, but I have a rather poor background there. If it is something relatively easy, I would rather appreciate a hint or reference than a full answer, so I can work it out further on my own (I am just trying to get back on the math track after some time of troubles).
Thanks in advance for any help!
This limit need not exist. For example, let's work in $M_2(\mathbb R)$.
If $$ U_n= \left( \begin{array}{cc} \frac{1}{n} & 0 \\ 0 & \frac{1}{n^2} \end{array} \right), $$ then $\Vert U_n \Vert \to 0$ as $n \to \infty$, and $$ U_n^{-1}= \left( \begin{array}{cc} {n} & 0 \\ 0 & {n^2} \end{array} \right). $$ If we now let $$ W_n= \left( \begin{array}{cc} 0 & \frac{1}{\sqrt{n}} \\ 0 & 0 \end{array} \right), $$ then $W_n \to 0$ as $n \to \infty$, but $$ \sigma_n (W_n) = U_n W_n U_n^{-1}= \left( \begin{array}{cc} 0 & \sqrt{n} \\ 0 & 0 \end{array} \right), $$ which does not converge as $n \to \infty$.