On the Hautus test for exact observability of normal semigroups

Let $T$ be an exponentially stable strongly continuous semigroup on a Hilbert space $X$, $A$ its generator, $X_1$ the domain of $A$ with the norm $\mathrm{\parallel }v{\mathrm{\parallel }}_1:=\mathrm{\parallel }Av\mathrm{\parallel }$, and $C$ an admissible observation operator taking values in a Hilbert space $Y$. Russell and Weiss conjectured that if $(A,C)$ satisfies the infinite-dimensional Hautus test, then $(A,C)$ is exactly observable. After a counterexample was found their conjecture was modified to include the additional assumption that $T$ is similar to a contraction semigroup. We disprove this modified conjecture by constructing a counterexample in which $A$ is normal, $X$ has an orthonormal basis of eigenvectors of $A$ and $C:X_1\to Y$ is Hilbert-Schmidt. Since $A$ is normal and $T$ is exponentially stable, it follows that $T$ is a contraction semigroup. In contrast, we prove that the Hautus test implies exact observability for self-adjoint $A$. Finally, we show that if $A$ is normal and $C:X_1\to Y$ is compact, then the Hautus test implies that $A$ has compact resolvent. Consequently, for normal $A$ and finite-dimensional $Y$ the Hautus test implies exact observability.