We are reading a proof about the following limit \begin{equation}\tag{1} \lim_{n \to \infty} \sigma_1(T_n)= \sigma_1(T), \end{equation} where $T:D(T) \subseteq H \to H$ and $T_n:D(T_n) \subseteq H \to H$ are linear operators on a Hilbert space $H$ and $$\sigma_1(A)= \sigma(A) \cup \{ z \in \mathbb{C}: \|(z-A)^{-1} \|>1\}$$ for a linear operator $A$. In the proof it is said that it is enough to show the following:

(i) If $K \subseteq \sigma_1(T)$ is compact then there is $N \in \mathbb{N}$ such that $K \subseteq \sigma_1(T_n)$ for all $n \geq N$.

(ii) If $K$ is compact and $K \cap \overline{\sigma_1(T)}=\emptyset$ then there is $N \in \mathbb{N}$ such that $K \cap \overline{\sigma_1(T_n)}=\emptyset$ for all $n \geq N$.

But we don't know why (i) and (ii) implies (1)?.

The definition of convergence of sets which is used in (1) is the following: Let $\{X_n\}$ a sequence of subsets of $\mathbb{C}$. Then $x \in \limsup X_n$ iff there exists a subsequence $\{ X_{n_k}\}$ and a sequence $\{ x_k \}$ such that $x_k \in X_{n_k}$ and $\lim_{k \to \infty} x_k=x$. On the other hand, $x \in \liminf X_n$ iff there is sequence $\{x_n \}$ with $x_n \in X_n$ suc that $\lim_{n \to \infty} x_n=x$. The limit $\lim_{n \to \infty} X_n$ exists if $\limsup X_n=\liminf X_n$ and we define $\lim_{n \to \infty} X_n=\limsup X_n$.

Our attempt:

We think that (i) implies that $\sigma_1(T) \subseteq \liminf \sigma_1(T_n)$ because if $z \in \sigma_1(T)$ then $\{z \}$ is compact.

From (ii) we can get that $\limsup \overline{\sigma_1(T_n)} \subseteq \overline{\sigma_1(T)}$. Indeed if $z \notin \overline{\sigma_1(T)}$, then there exists $\varepsilon>0$ such that $\overline{B(z;\varepsilon )} \cap \overline{\sigma_1(T)}= \emptyset$ and from (ii) we get $\overline{B(z;\varepsilon )} \cap \overline{\sigma_1(T_n)}= \emptyset$ for all large enough $n$, so $z \notin \limsup \overline{\sigma_1(T_n)}$.

But how can we show that $\limsup \sigma_1(T_n) \subseteq \sigma_1(T)$?

Thank you for any help you can provide us.


1 Answer 1


You have hit the nail on the head and indeed there is a mistake with the proof you are reading. The argument already fails for matrices. If $T_n \equiv T \in \mathbb{C}^{n \times n}$ then $\sigma_1(T)$ is open by definition, but $$ \lim \sigma_1(T_n) = \overline{\sigma_1(T)}, $$ as any element in the closure of $\sigma_1(T)$ is the limit of a sequence lying in $\sigma_1(T)$. This sequence we can interpret as a sequence lying in the $\sigma_1(T_n)$.

The conditions (i) and (ii) provided are trivially satisfied as we have $\sigma_1(T_n) = \sigma_1(T)$. So (i) and (ii) do not imply the claimed equality. Indeed your definition of the limit of a sequence of sets implies that the limit is closed, if it exists. So the claim has to be modified accordingly.

  • $\begingroup$ Thank you so much. I had been coffused about that proof for weeks. I think that the claim is true if we change $\sigma_1(T)$ by $\overline{ \sigma_1(T)}$. You agree with me? $\endgroup$
    – Mainkit
    Jul 24, 2019 at 23:32
  • $\begingroup$ I do. This follows from your arguments if you note that from the definition already $\liminf$ and $\limsup$ are closed. $\endgroup$ Jul 24, 2019 at 23:59

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