Open subset of $\mathcal{L}(H)$?

Question

Let $H$ be a Hilbert space . Denote $\mathcal{L}(H)$ the vector space (on the field $\mathbb{C}$) of all bounded linear maps on $H$ into $H.$ Define the subset $K \subset \mathcal{L}(H) $ by $$K := \{A \in \mathcal{L}(H); \sigma(A) >0\},$$ where $\sigma(A)$ denotes the spectrum of $A.$

Is $K$ an open subset of $\mathcal{L}(H)$ ?

Answer 1

A famous result of Newburgh says:

If $U$ is an open set in $ \mathbb C$ and if $ \sigma(A) \subset U$ then there is $ \delta >0$ such that $||A-B|| < \delta$ implies $ \sigma(B) \subset U$.

(upper semicontnuity)

Newburgh: Variation of spectra. Duke. Math. J, $18 (1951), 165-167$

Answer 2

Consider the one-dimensional Hilbert space $\mathbb{C}^1$. The operators acting on it are represented by $1\times 1$ matrices $(z)$, and the topology of $L(H)$ is just the topology of $\mathbb{C}$. The spectrum of matrix $(z)$ is the point $z$ itself. So, the set of all operators with the spectrum contained in some set $A\subset\mathbb{C}$ cannot be open unless $A$ is open. In particular, for $A=(0,\infty)$ it is not open.

Conversely, if $A$ is open, then by the result cited by Fred the set of operators with spectrum contained in $A$ is open.

Answer 3

For $I=id_H$ we have $\sigma (I)=\{1\}\subset \mathbb R^+ .$ For $0\ne r\in \mathbb R$ we have $\sigma ((1+r\sqrt {-1}\;)I)=\{1+r\sqrt {-1}\;\}\not \subset \mathbb R^+$ and $\|I-(1+r\sqrt {-1}\;)I\|=|r|$. So $I\in K \cap \overline {K^c}.$