On strong convergence versus weak in operator topology and semi-continuity of the spectrum

Question

Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the Banach algebra of bounded operators on $\mathbb H$. Let $(A_k)_{k\ge 1}$ be a sequence in $\mathcal B(\mathbb H)$.

$\bullet$ If $(A_k)_{k\ge 1}$ is uniformly convergent (i.e. convergent for the norm-topology of $\mathcal B(\mathbb H)$), I guess that there is a continuity result.

$\bullet$ If $(A_k)_{k\ge 1}$ is converging strongly towards $A$ in $\mathcal B(\mathbb H)$ (i.e. $\forall u\in \mathbb H$, $\lim_{\mathbb H} A_k u = A u$), what could be said about the respective spectra of $A_k$ and $A$? Some sort of semi-continuity?

$\bullet$ I guess that if the convergence of $(A_k)_{k\ge 1}$ is weak (i.e. for all $u,v\in \mathbb H$, $\lim\langle A_k u, v\rangle=\langle A u, v\rangle$), nothing or not much could be said.

In fact, I feel that uniform convergence is too much to ask and weak convergence is too little, so my interest goes primarily to strong convergence.

Answer 1

The spectrum (as a multivalued map) is upper semicontinuous with respect to the operator norm. With respect to strong convergence no analogous assertion is true, as can be seen in $\ell_2$ by the operator sequence $A_n(\xi_1,\xi_2,\ldots)=(0,\ldots,0,\xi_n,\xi_{n+1})$ which converges strongly to $0$ and whose spectrum contains $1$.