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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?

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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 $B(H)$.

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  • $\begingroup$ That's a nice example, but here the spectrum of $T_n$ varies as we vary $n$. I'm wondering about the case when the spectrum and essential spectrum stays fixed. $\endgroup$ – Mathmo Jan 18 '17 at 17:47
  • $\begingroup$ My bad, I missed that. I'll edit soon with the full answer. $\endgroup$ – Martin Argerami Jan 18 '17 at 17:51
  • $\begingroup$ Consider the operators in matrix form - take the Hilbert space $l^2(\mathbb{N})$. Then matrix elements converge and hence $T$ must be normal. $\endgroup$ – Mathmo Jan 18 '17 at 17:57
  • $\begingroup$ @Mathmo: no. See the edit. $\endgroup$ – Martin Argerami Jan 18 '17 at 18:03
  • $\begingroup$ Ah, nice! I see, my argument only works in the self-adjoint case. $\endgroup$ – Mathmo Jan 18 '17 at 18:08

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