Let $H$ be a complex Hilbert space. Denote $\mathcal{L}(H)$ the space of linear bounded operators on $H$ into $H.$ Let $N \in \mathcal{L}(H) $ be normal ($N^* N =NN^*$) such that $\sigma(N) \subset \Gamma := \{z \in \mathbb{C}\backslash \{0\}, -\pi /2<Arg(z) <\pi/2 \}.$ It is clear that there exists $C>0$ such that \begin{equation}\tag{1} (N+\lambda)^{-1} \in \mathcal{L}(H) \quad \text{and} \quad \|(N+\lambda)^{-1}\| < \frac{C}{Re(\lambda)} \forall \lambda \in \Gamma. \end{equation}
Is it possible to find $\delta >0$ such that (1) is still true for every $E \in \{A \in \mathcal{L}(H), \|N-A\| <\delta \}.$
Thank you for any hint.
Yes, this indeed holds for a normal operator. In fact, since $\sigma(N)$ is closed bounded, whence compact, the real part attains a minimum on $\sigma(N)\in \Gamma$, so we have: ${\rm Re\;} z \geq \alpha>0$ for all $z\in \sigma(N)$.
For a normal operator, its norm is the same as the spectral radius, so when ${\rm Re\;} \lambda \geq 0$, $(\lambda+N)$ is invertible and its inverse $A=(\lambda+N)^{-1}$ verifies: $$ \| A \| = \frac{1}{{\rm dist}(-\lambda,\sigma(N))} \leq \frac{1}{\alpha + {\rm Re\;} \lambda }\leq \frac{1}{\alpha} $$ Let $0<\delta<\alpha$. Whenever $\|U\|\leq \delta$ we see that $(\lambda+N+U)= (\lambda+N)(I+(\lambda+N)^{-1} U)$ is invertible by a von Neumann series and we have the wanted bound: $$\|(\lambda+N+U)^{-1} \| = \|(I - A \;U + (A\; U)^2 - \cdots) A \| \leq \left(\sum_{n\geq 0} \frac{\delta^n}{\alpha^n} \right) \frac{1}{\alpha+{\rm Re\; } \lambda} \leq \frac{\alpha}{\alpha-\delta} \frac{1}{{\rm Re\; } \lambda}$$