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Let $A \colon \mathcal{D}(A) \subset \mathbb{H} \to \mathbb{H}$ be a closed linear operator in a Hilbert space $\mathbb{H}$, which generates a $C_{0}$-semigroup. Suppose that in a $\varepsilon$-neighbourhood of the imaginary axis the operator $A$ does not have spectrum. Is it true that the resolvents $(A-i\omega I)^{-1}$ are bounded in $\mathcal{L}(\mathbb{H})$ uniformly for $\omega \in \mathbb{R}$? Can the same be said for the boundedness in the norm of $\mathcal{L}(\mathbb{H};\mathcal{D}(A))$, where $\mathcal{D}(A)$ is endowed with the graph norm. If not, what are additional conditions to ensure the boundedness.

For example, if $A$ generates an exponentially stable $C_{0}$-semigroup then the boundedness of resolvents follows from the resolvent estimate in the Hille-Yosida theorem.

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    $\begingroup$ While the answer is, in general, negative (see Michael Renardy's answer), a sufficient condition for a positive answer is that $\mathbb{H}$ is an $L^2$-space and the semigroup is positive; see for instance Corollary C-III-1.3 on page 294 in "Arendt et. al.: One-parameter Semigroups of Positive Operators (1986)". $\endgroup$ – Jochen Glueck May 14 at 7:50
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    $\begingroup$ By the way, a rather comprehensive overview about the relation between spectral bounds, pseudo-spectral bounds and growth bounds of $C_0$-semigroups can be found in Secctions 5.1 and 5.2 of "Arendt, Batty, Hieber, Neubrander: Vector-valued Laplace Transforms and Cauchy Problems (2011)". Using the notation from this book, your question is whether $s(A) = s_0(A)$, and an explicit counterexample to this is given in Example 5.1.10 (according to the notes at the end of Chapter 5, this example is a modification, due to Wrobel, of Zabczyk's counterexample mentioned in Michael Renardy's answer. $\endgroup$ – Jochen Glueck May 14 at 8:02
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This is essentially equivalent to asking whether the spectrum determines the growth bound. This is well known to be false. A pioneering counterexample is due to Zabczyk.

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    $\begingroup$ It might be worthwhile to add that, in the general Banach space setting, the question is equivalent to whether the spectral bound equals the pseudo-spectral bound; since the OP restricted the question to the Hilbert space setting, the pseudo-spectral bound always coincides with the growth bound (by the Gearhart-Prüss theorem), which leads to the statement in the answer. $\endgroup$ – Jochen Glueck May 14 at 7:42

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