How does the spectrum behave under WOT convergence? Suppose I have a sequence of normal operators $T_n$ on some Hilbert space (separable say) that converge in the weak operator topology to $T$ and also have the same essential spectrum and spectrum as we vary $n$. Is it true that
$\sigma_{ess}(T)\subset\sigma_{ess}(T_n)$
or
$$\sigma(T)\subset\sigma(T_n)$$
where $\sigma_{ess}$ denotes the essential spectrum (only one reasonable definition for normal operators)? (Is it even true that $T$ is normal?)
What happens in the case that all the $T_n$ are unitary equivalent?
 A: It is known that the set of unitary operators is wot dense in the unit ball of $B(H)$ when $H$ is infinite-dimensional. So, any non-normal element of the unit ball is a limit of unitaries (i.e. normals). 
Regarding an example: Let $T$ be any non-normal operator in the unit ball, with spectrum $\sigma(T)$ properly contained in $\mathbb D$, and $\sigma_{\rm ess}(T)\subsetneq\mathbb T$. Consider, on $B(H\oplus H)$, the operator
$$
X=\begin{bmatrix} T&0\\ 0& V\end{bmatrix},
$$
where $V$ is a unitary with  $\sigma(V)=\sigma_{\rm ess}(V)=\mathbb T$. Then $X$ is not normal, and $\sigma(X)=\sigma(T)\cup\mathbb T$. Now let $\{W_j\}$ be a net of unitaries such that $W_j\to T$ wot. Then 
$$
U_j=\begin{bmatrix}W_j&0\\0& V\end{bmatrix}
$$
is a unitary, $U_j\to X$ wot, and $\sigma(U_j)=\sigma_{\rm ess}(U_j)=\mathbb T$, while both $\sigma(X)$ and $\sigma_{\rm ess}(X)$ are strictly larger than $\mathbb T$. 
Finally, as $H$ is infinite-dimensional, there exists a unitary $H\oplus H\to H$, and so the above example can be cramped back into $B(H)$. 
