# 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?

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 $$B(H)$$.
• 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. – Mathmo Jan 18 '17 at 17:47
• Consider the operators in matrix form - take the Hilbert space $l^2(\mathbb{N})$. Then matrix elements converge and hence $T$ must be normal. – Mathmo Jan 18 '17 at 17:57