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

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.

• There is huge literature on the fact the the spectrum is in general not continuous with respect to the operator norm. Which result do you mean in your first bullet point? – András Bátkai Feb 5 at 18:57
• @András Bátkai, the spectrum is upper semicontinuous with respect to the operator norm. – Martin Väth Feb 5 at 19:48
• Yes, I was aware of that general result of semi-continuity but uniform convergence is way too much for me. I guess for instance that semi-continuity is not true under weak convergence, but I am not quite sure of what could happen for strong convergence. – Bazin Feb 5 at 19:54
• @MartinVäth this has been dicussed here many times before, see for example mathoverflow.net/a/231222/12898 or mathoverflow.net/q/262510/12898 – András Bátkai Feb 6 at 8:10
• For selfadjoint operators, there are prositive results. See for example in Section 9.3 in Weidmann: Linear Operators in Hilbert space book for strong convergence.Theorem 9.19 states a detailed convergence result. – András Bátkai Feb 6 at 8:17

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 $$1$$.