Suppose $A$ and $B$ are two matrices or bounded operators such that $A=\lim_{n\to\infty} A_n$ and $B=\lim_{n\to\infty} B_n$ for a sequence of matrices or operators $A_n$ and $B_n$. (The limit is to be understood as the matrix or operator norm.) Suppose in addition that for each $n$, $A_n$ is similar (i.e. conjugate) to $B_n$.
Does it follow that $A$ is similar to $B$? I think the answer is positive at least in some cases, for instance when $A_n$ and $B_n$ are matrices and are orthogonally similar.
I wonder what general statements can be proved (even by modifying the conditions, modes of convergence, etc) for this setting and in what cases counterexamples can be constructed. Any help is appreciated.
The answer is no. As a counterexample, consider sequences defined by $$ A_n = A = \pmatrix{0&1\\0&0}\\ B_n = \pmatrix{0&1/n\\0&0}, \quad B = 0 $$ Although each $A_n$ is similar to each $B_n$, the limit $A$ is not similar to the limit $B$.
Another counterexample: $$ A_n = \pmatrix{1/n & 1\\0&0}, \quad B_n = \pmatrix{1/n & 0\\0&0} $$
For orthogonal similarity: indeed, it is true (for matrices) that if each $A_n$ is orthogonal to each $B_n$, then the limits $A$ and $B$ are also orthogonally similar. To show this, it suffices to apply the following fact:
Now, it suffices to note that for any such $w$, we have $$ \operatorname{trace}(w(A_n,A_n^*)) \to \operatorname{trace}(w(A,A^*)) $$ which means that since each $A_n$ is orthogonally similar to the corresponding $B_n$, $A$ must be orthogonally similar to $B$.