Assumption 1. $T$ is a closed normal operator on a Hilbert space $H$ and $A$ is $T$-bounded with $T$-bound $\delta_A \lt 1$ and the constants $a\ge 0$ and $0\le b \leq \delta_A$ satisfy (1.1).

Assumption 2. $T$ is a closed normal operator on a Hilbert space $H$ and $A$ is $p$-subordinate to $T$ for some $p\in [0,1]$ and the constant $c \gt 0$ satisfies (1.2).

Proof. (i) is well known and follows from the Neumann series for the resolvent: If $z\in\rho(T)$, we can write $T+A-z = (I + A (T-z)^{-1})(T-z)$. Therefore, if $\| A (T-z)^{-1}\| \lt 1$, then $T+A-z$ is invertible with inverse

\begin{equation*} (T+A-z)^{-1} = (T-z)^{-1} \sum_{n=0}^\infty (-A (T-z)^{-1})^n, \end{equation*}

and

\begin{equation*}\Arrowvert(T+A-z)^{-1}\Arrowvert\leq\Arrowvert(T-z)^{-1}\Arrowvert\sum_{n=0}^\infty\Arrowvert A(T-z)^{-1}\Arrowvert^n=\frac{\Arrowvert(T-z)^{-1}\Arrowvert}{1-\Arrowvert A(T-z)^{-1}\Arrowvert}.\end{equation*}

(ii) Since $T$ is normal, we have $\||T|x\| = \|Tx\|$ for all $x\in \mathcal{D}(T) = \mathcal{D}(|T|)$ where $|T| := (T^*T)^{1/2}$. Using that $A$ is $T$-bounded, we obtain from (1.1) for all $x\in \mathcal{D}(|T|^2)$ and $z\in\rho(T)$ that

\begin{align*} \|Ax\|^2&\leq a^2\|x\|^2 +b^2\||T|x\|^2 =a^2\langle x,\, x\rangle +b^2\langle |T|x,\, |T|x\rangle \\ &= \langle (a^2+b^2|T|^2)x,\, x\rangle \\ &=\|\sqrt{a^2+b^2|T|^2}\, x\|^2. \end{align*}

Therefore, for all $x\in\mathcal{D}(T)$ with $\|x\|=1$,

(2.1)\begin{equation}\Arrowvert A(T-z)^{-1}x\Arrowvert\leq\Arrowvert\sqrt{a^2+b^2\vert T\vert^2}\, (T-z)^{-1}x\Arrowvert\leq\sup_{t\in\sigma(T)}\frac{\sqrt{a^2+b^2\vert t\vert^2}}{\vert t-z\vert}.\end{equation}

In the last step we used that, by the spectral theorem for normal operators, $\|g(T)x \| \le \sup_{t\in\sigma(T)} |g(t)| \|x\|$ for every bounded continuous function $g$. In our case, $g(t) = \sqrt{a^2+b^2t \overline t}\,(t-z)^{-1}$. Since $A(T-z)^{-1}$ is bounded on the dense subset $\mathcal{D}(T)$, it extends uniquely to a bounded operator on $H$ and (2.1) holds for all $x\in H$.

(iii) follows from (i) and (ii).

Proof. Since $\mathcal{D}(T)\subseteq \mathcal{D}(A)$, we can write $\mathcal{D}(T+A)=\mathcal{D}(T_1)\oplus \mathcal{D}(T_2) = \widetilde H \oplus \mathcal{D}_2$ and, if we define

\begin{equation*} A_{j1}:=P_j A|_{\mathcal{D}(T)\cap \widetilde H},\quad A_{j2}:=P_j A|_{\mathcal{D}(T)\cap \widetilde H^\perp}, \qquad j=1,2, \end{equation*}

we can represent both $T$ and $T+A$ as block operator matrices

\begin{align*} T = \begin{pmatrix} T_1 & 0 \\ 0 & T_2 \end{pmatrix} : \mathcal{D}(T_1)\oplus \mathcal{D}(T_2) \subseteq H \longrightarrow \widetilde H \oplus \widetilde H^\perp, \end{align*}

and

\begin{align*} T + A = \begin{pmatrix} T_1 + A_{11} & A_{12} \\ A_{21} & T_2 + A_{22} \end{pmatrix} : \mathcal{D}(T_1)\oplus \mathcal{D}(T_2) \subseteq H \longrightarrow \widetilde H \oplus \widetilde H^\perp. \end{align*}

We decompose the sum as

\begin{equation*} T+A=\mathcal{T}+\mathcal{A} \qquad\text{with }\quad \mathcal{T}:= \begin{pmatrix} T_1 & 0 \\ 0 & T_2+A_{22} \end{pmatrix}, \quad \mathcal{A}:= \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & 0 \end{pmatrix}. \end{equation*}

Note that for $j=1, 2$ and $x\in\mathcal{D}(T_j)$ we have that

\begin{equation*} \|A_{j1}x\|^2 + \|A_{j2}x\|^2 = \|Ax\|^2 \le a^2 \|x\|^2 + b^2 \|Tx\|^2 = a^2 \|x\|^2 + b^2 \|T_jx\|^2, \end{equation*}

hence $A_{j\ell}$ is $T_j$-bounded with relative bound less or equal to the $T$-bound of $A$. Since $\widetilde H$ is finite dimensional, an easy calculation shows that $\mathcal{A}$ is $\mathcal{T}$-compact. Thus, by [Reference Gohberg, Goldberg and Kaashoek11, Chapter XVII, Theorem 4.3],

(2.4)\begin{align} \sigma_{\mathrm{ess} }(T+A)=\sigma_{\mathrm{ess} }(\mathcal{T}+\mathcal{A})=\sigma_{\mathrm{ess} }(\mathcal{T}). \end{align}

Observe that for $\lambda \in \mathbb{C}$

\begin{align*} \ker(\mathcal{T}-\lambda) & =\ker(T_1-\lambda)\oplus \ker(T_2+A_{22} - \lambda), \\ \operatorname{Ran}(\mathcal{T}-\lambda) & =\operatorname{Ran}(T_1-\lambda)\oplus \operatorname{Ran}(T_2+A_{22} - \lambda). \end{align*}

Since $\ker(T_1-\lambda)$ and $\operatorname{Ran}(T_1-\lambda)$ are finite dimensional, we have that

\begin{align*} \dim(\ker(\mathcal{T}-\lambda)) \lt \infty & \quad \Longleftrightarrow \quad \dim(\ker(T_2+A_{22}-\lambda)) \lt \infty, \\ \operatorname{codim}(\operatorname{Ran}(\mathcal{T}-\lambda)) \lt \infty & \quad\Longleftrightarrow\quad \operatorname{codim}(\operatorname{Ran}(T_2+A_{22}-\lambda)) \lt \infty, \end{align*}

and therefore

(2.5)\begin{align} \sigma_{\mathrm{ess} }(\mathcal{T})=\sigma_{\mathrm{ess} }(T_2+A_{22}), \end{align}

which together with (2.4) shows (2.3). The claim about nonaccumulation of the spectrum follows from [Reference Gohberg, Goldberg and Kaashoek11, Chapter XVII, Theorem 2.1].

Proof. Let $z=\mu + \mathrm{i}\nu\in\mathbb{C}$ with $\nu \gt \gamma$. Then $z\in\rho(T)$ and it belongs to $\rho(T+A)$ if $\sup_{\operatorname{Im} t \le \gamma} H_z(t) \lt 1$. By Lemma A.1, $H_z$ has no local extrema, so we obtain

(3.5)\begin{equation} \sup_{\operatorname{Im} t \le \gamma} H_z(t) = \sup_{\operatorname{Im} t = \gamma} H_z(t). \end{equation}

By Lemma A.3, $\sup_{\operatorname{Im} t = \gamma} H_z(t) \lt 1$ for $\operatorname{Im} z \gt \gamma$ if and only if

\begin{equation*} a^2 + b^2\gamma^2 + \frac{b^2\mu^2}{1-b^2} \lt (\operatorname{Im} z -\gamma)^2, \end{equation*}

which is true if and only if $z\in\mathrm{i}\gamma + \mathrm{Hyp}_\gamma$. The resolvent estimate (3.4) follows from Proposition 2.1(iii) together with (A.3) and $\|(T-z)^{-1}\| = \frac{1}{\operatorname{dist}(z,\sigma(T))}\leq \frac{1}{\operatorname{Im} z -\gamma}$.

Proof. This is an immediate consequence of Lemma A.4(i). The resolvent estimate follows as in the previous proposition.

Proof. Let $z=\mu + \mathrm{i}\nu$ with $\mu \in (\alpha_{T},\beta_{T})$ and $\nu \in\mathbb{R}\setminus [\gamma_1, \gamma_2]$. Then

(3.11)\begin{align} \sup_{t\in\sigma(T)} \frac{a^2+b^2|t|^2}{|t-z|^2} &\le \sup_{\sigma\in \{\gamma_1, \gamma_2\}\; } \sup_{t\in(\mathbb{R}\setminus(\alpha_T, \beta_T)) + \mathrm{i}\sigma} \frac{a^2+b^2|t|^2}{|t-z|^2}\cr &\le \sup_{\sigma\in \{\gamma_1, \gamma_2\} } \sup_{\tau\in \mathbb{R}\setminus(\alpha_T, \beta_T)} \frac{a^2+b^2(\tau^2+\sigma^2)}{(\tau-\mu)^2}\cr &= \sup_{\tau\in \mathbb{R}\setminus(\alpha_T, \beta_T)} \frac{a^2 + b^2\widetilde \gamma^2 + b^2 \tau^2}{(\tau-\mu)^2}\cr & \leq \max\left\{b^2,~\frac{{a^2+b^2\widetilde \gamma^2+b^2\alpha_T^2}}{(\mu-\alpha_T)^2},~\frac{{a^2+b^2\widetilde \gamma^2+b^2\beta_T^2}}{(\beta_T-\mu)^2} \right\} \end{align}

where the last inequality follows from Lemma A.6 with $a^2+b^2\widetilde \gamma^2$ instead of $a^2$. Note that the expression in (3.11) is less than 1 if $\alpha_{T+A}' \lt \operatorname{Re} z = \mu \lt \beta_{T+A}'$. Therefore (3.9) follows from Proposition 2.1(iii).

The resolvent estimate (3.10) follows again from Proposition 2.1(iii) together with (3.11) and

\begin{align*} \operatorname{dist}(z, \sigma(T)) \ge \sqrt{(\min\{\operatorname{Re} z-\alpha_T,\beta_T-\operatorname{Re} z\})^2+(\min\{0, \gamma_2-\operatorname{Im} z, \operatorname{Im} z - \gamma_1 \})^2}. \\[-40pt] \end{align*}

Proof. We proof only (i). Let $\beta_{T+A}'$ be as in (3.8). Since the spectrum of $T$ is bounded from below, we can choose $\alpha_T$, and hence also $\alpha_{T+A}'$ from (3.8) arbitrarily small because $0\le b \lt 1$ and therefore

\begin{equation*} \alpha_{T+A}' = \alpha_T+\sqrt{a^2+b^2\widetilde \gamma^2+b^2\alpha_{T}^2} \ \longrightarrow -\infty \quad\text{for }\ \alpha_T\to -\infty. \end{equation*}

Hence (3.9) shows that

\begin{equation*} (-\infty,~\beta'_{T+A})+\mathrm{i} \mathbb{R} \subseteq \rho(T+A).\\[-28pt] \end{equation*}

Proof. For $n\in\mathbb{N}$ define the sets

\begin{equation*} S_n := \left\{z\in\mathbb{C} : \alpha+\frac{1}{n} \le \operatorname{Re} z \le \beta-\frac{1}{n},\ \gamma_1 - \frac{1}{n} \le \operatorname{Im} z \le \gamma_2 + \frac{1}{n} \right\} \end{equation*}

and $U_n := \mathbb{C}\setminus S_n$. Then $\sigma_{\mathrm{ess} }(T)\subseteq S_n$ for every $n\in\mathbb{N}$ and $U_n\cap \sigma(T)$ is finite and contains only finite eigenvalues of finite multiplicity. Let $P_1$ be the spectral projection of $T$ for the set $U_n$ and set $\widetilde H_n := \operatorname{Ran}(P_1) = \bigoplus_{\lambda\in U_n\cap \sigma(T)} \ker(T-\lambda)$. Set $\mathcal{D}_2 = \mathcal{D}(T)\cap \widetilde H_n^\perp$. Then $\dim \widetilde H_n \lt \infty$, $\mathcal{D}(T) = \widetilde H_n \oplus \mathcal{D}_2$ and both $\widetilde H_n$ and $\widetilde H_n^\perp$ are $T$-invariant, see [Reference Schmüdgen20, pag. 83]). Let us define $T_j^{(n)}$ and $A_{ij}^{(n)}$ as in Theorem 2.2. Then $A_{22}^{(n)}$ is $T_2^{(n)}$-bounded with the same constants $a,b$ that work for $T$ and $A$ and, by the spectral theorem, $\sigma(T_2^{(n)})\subseteq S_n$. If we apply Proposition 3.2 to $T_2^{(n)}$ and $A_{22}^{(n)}$, we obtain that

\begin{align*} F_n & := \textstyle (\mathrm{i}(\gamma_1 - \frac{1}{n} ) - \mathrm{Hyp}_{\widetilde \gamma + \frac{1}{n} }) \cup (\mathrm{i}(\gamma_2 + \frac{1}{n} ) + \mathrm{Hyp}_{\widetilde \gamma + \frac{1}{n} }) \subseteq \rho(T_2^{(n)} + A_{22}^{(n)}). \end{align*}

Moreover, Theorem 3.4 applied to $T_2^{(n)}$ and $A_{22}^{(n)}$ shows that

\begin{align*} G_n := \left\{z\in\mathbb{C} : \alpha_{n}' \lt \operatorname{Re} z \lt \beta_{n}' \right\} \subseteq \rho(T_2^{(n)} + A_{22}^{(n)}), \end{align*}

where

\begin{align*} \alpha_{n}' & := \alpha + \frac{1}{n} + \sqrt{\textstyle a^2 + b^2(\widetilde \gamma+\frac{1}{n})^2 + b^2 (\alpha + \frac{1}{n})^2}, \\ \beta_{n}' & := \beta - \frac{1}{n} + \sqrt{\textstyle a^2 + b^2(\widetilde \gamma+\frac{1}{n})^2 + b^2 (\beta - \frac{1}{n})^2}. \end{align*}

In summary, $\sigma_{\mathrm{ess} }(T_2^{(n)}+A_{22}^{(n)}) \subseteq \sigma(T_2^{(n)}+A_{22}^{(n)})\subseteq \mathbb{C}\setminus (F_n\cup G_n)$. Since by Theorem 2.2 the essential spectra of $T+A$ and $T_2^{(n)} + A_{22}^{(n)}$ coincide for every $n\in\mathbb{N}$, we obtain that

\begin{equation*} \sigma_{\mathrm{ess} }(T+A) \subseteq \bigcap_{n\in\mathbb{N}} \mathbb{C}\setminus (F_n\cup G_n). \end{equation*}

Since $\bigcup_{n\in\mathbb{N}} (F_n\cup G_n)$ is equal to the right hand side of (3.16), that inclusion is proved. The claim about nonaccumulation of the spectrum follows from [Reference Gohberg, Goldberg and Kaashoek11, Chapter XVII, Theorem 2.1] because by Proposition 3.2 the exterior of certain hyperbolas belongs to the resolvent set of $T+A$, hence the complement of the left hand side of (3.16) contains points in the resolvent set of $T+A$.

Proof. In order to prove (i), we start with $z\in\rho(T)$. As in the proofs of Proposition 3.2 and Theorem 3.4, we have to show that $\sup_{t\in\sigma(T)} H_z(t) \lt 1$ for $z$ in the set on the left side of (3.17). Note that

\begin{equation*} \sup_{t\in\sigma(T)} H_z(t) = \max\left\{\sup_{t\in\Sigma_0} H_z(t),\ \sup_{t\in\Sigma_1} H_z(t) \right\} \le \max\left\{\sup_{t\in\mathrm{Str}} H_z(t),\ \sup_{t\in\Sigma_1} H_z(t) \right\}. \end{equation*}

By Proposition 3.2 and Theorem 3.4 we know that $\sup_{t\in\mathrm{Str}} H_z(t) \lt 1$ if $z$ belongs to the union of the first three sets on the left hand side in (3.17). Moreover,

\begin{equation*} \sup_{t\in\Sigma_1} H_z(t) = \max_{j=1}^N H_z(\lambda_j) = \max_{j=1}^N \frac{a^2 + b^2 |\lambda_j|^2}{|\lambda_j - z |^2 } \end{equation*}

which is smaller than $1$ if $z$ is outside $\bigcup_{j=1}^N K_{r_j}(\lambda_j)$.

We proceed to prove (ii). For $s\in [0,1]$ define $A_s := sA$. Then all $A_s$ are $T$-bounded and if $a,b$ are the constants from (1.1), then we have the corresponding inequality for $A_s$ if we use $a_s:=\sqrt{s}a$ and $b_s:=\sqrt{s}b$ instead of $a$ and $b$. Note that the numbers $\alpha'_{T+A_s}, \beta'_{T+A_s}$ and $r_{j,s} := \sqrt{a_s^2 + b_s^2|\lambda_j|^2}$ are non-decreasing in $s$. In particular we have (3.17) for all $A_s$ with the corresponding $\alpha'_{T+A_s}, \beta'_{T+A_s}$. Therefore, if for the moment we denote the set (3.18) by $U(A)$, we have that $U(sA)\supset U(s'A)$ for all $0\le s \le s' \le 1$ and for any $s,s'\in [0,1]$ we have that $K_{r_{k,s}+2\epsilon} \subseteq U(s'A)$. Therefore we can define the Riesz projections

(3.19)\begin{equation} P_s := \frac{1}{2\pi \mathrm{i}} \int_{\partial K_{r_k+\epsilon}(\lambda_k)} (T+A_s-\lambda)^{-1}\,\mathrm d\lambda, \qquad s\in[0,1]. \end{equation}

The dimension of the range of $P_s$ is equal to the total algebraic multiplicity of the eigenvalues of $A_s$ contained in $B_{r_k+\epsilon}(\lambda_k)$, hence in $K_{r_k}(\lambda_k)$. So we have to show that $\dim\operatorname{Ran} P_0 = \dim\operatorname{Ran} P_1$. To this end, observe that for $\lambda\in\partial K_{r_k+\epsilon}(\lambda_k)$

\begin{equation*} (T+A_s-\lambda)^{-1} = (T-\lambda)^{-1} \sum_{n=0}^\infty s^n(-A(T-\lambda)^{-1})^n \end{equation*}

which is continuous in $s$, uniformly with respect to $\lambda$. Therefore $P_s$ is continuous in $s$, hence $\dim\operatorname{Ran} P_s$ is constant in $s$ by [Reference Kato13, Ch.I, Lemma 4.10].

The claim in (iii) follows as in (ii) if we replace the contour of integration in (3.19) by $\Gamma = \bigcup_{j=1}^N \partial K_{r_j+\epsilon}(\lambda_j)$.

Proof. As in the proof of (iii) in Theorem 3.8 it can be shown that the vertical parts of $\Gamma$ belong to $\rho(T+sA)$ for all $s\in [0,1]$ by Theorem 3.4, applied to the spectral gaps $\{\alpha \lt \operatorname{Re} z \lt \operatorname{Re}\lambda_1\}$ and $\{\operatorname{Re}\lambda_N \lt \operatorname{Re} z \lt \beta\}$. The same is true for the horizontal lines from $\widetilde R_1\pm\mathrm{i}\eta$ if we choose $\eta$ large enough such that they lie in $\mathrm{i}\gamma_2 + \mathrm{Hyp}_{\widetilde \gamma}$ and $\mathrm{i}\gamma_1 - \mathrm{Hyp}_{\widetilde \gamma}$ respectively. Now the same continuity argument shows that the rank of the Riesz projections in (3.19) with contour $\Gamma$ is constant in $s$.

Proof of Theorem 4.3

Let $z = \mu + \mathrm{i}\nu \in\mathbb{C}\setminus S_{\theta}(\beta_T)$ and define the function $H_z$ as in (2.2). Note that

\begin{align*} \sup_{t\in \sigma(T)} H_z(t) \le \sup_{t\in S_{\theta}(\beta_T)} H_z(t), \end{align*}

and the right hand side was calculated in (A.6). Therefore we obtain the inclusion (4.4) from Lemma A.4(ii).

Now let us assume that $\mu = \operatorname{Re} z \lt \beta_T$. Then (A.12) shows that

(4.8)\begin{align} \sup_{t\in \partial S_{\theta}(\beta_T)} H_z(t) \le \max\left\{b^2,\ \frac{a^2+b^2\beta_T^2}{(\beta_T-\mu)^2} \right\} + b^2\tan^2\theta. \end{align}

Under the additional assumption $\beta \lt \cos\theta$ we have that $b^2(1+\tan^2\theta) \lt 1$. So the right hand side in (4.8) is smaller than 1 if $\frac{{a^2+b^2\beta_T^2}}{(\beta_T-\mu)^2}+b^2\tan^2\theta \lt 1$. This is the case if and only if $\mu \lt \beta_T - \frac{\sqrt{a^2+b^2\beta_T^2}}{1-b^2\tan^2\theta} = \beta_{T+A}$. It follows that $\{z\in\mathbb{C} : \operatorname{Re} z \lt \beta_{T+A} \}\subseteq\rho(T+A)$ and (4.5) is proved.

Proof. Let $z = \mu + \mathrm{i}\nu \in\mathbb{C}\setminus\Omega(\alpha_T, \beta_T, \theta)$ and define the function $H_z$ as in (2.2). Proposition 2.1(iii) shows that $z\in\rho(T+A)$ if $\sup_{\sigma(T)} H_z(t) \lt 1$. Observe that

(4.11)\begin{align} \sup_{t\in \sigma(T)} H_z(t) \le \max\left\{\sup_{t\in -S_{\theta}(-\alpha_T)} H_z(t),\ \sup_{t\in S_{\theta}(\beta_T)} H_z(t) \right\}. \end{align}

From the proof of Theorem 4.3, we know that $\sup_{t\in S_{\theta}(\alpha_T)} H_z(t) \lt 1$ if $z$ belongs to the left hand side of (4.4) and that $\sup_{t\in -S_{\theta}(-\alpha_T)} H_z(t) \lt 1$ if $z$ belongs to the left hand side of (4.6). Therefore the right hand side of (4.11) is less than 1 if $z$ belongs to both sets which proves (4.9).

If $b \lt \cos\theta$, then the proofs of (4.5) and (4.7) show that the right hand side of (4.11) is less than 1 if $\alpha_{T+A} \lt \operatorname{Re} z \lt \beta_{T+A}$. Hence (4.10) holds.

Proof. Let $z= \mu + \mathrm{i}\nu$ with $\alpha_{T+A} \lt \mu \lt \beta_{T+A}$. Then, by (4.8) and the analogous estimate for $-S_{\theta}(-\alpha)$, we obtain

(4.12)\begin{equation} \sup_{t\in\sigma(T)} H_z(t) \le \max\left\{b^2,\ \frac{{a^2+b^2\beta_T^2}}{(\beta_T-\operatorname{Re} z)^2},\ \frac{{a^2+b^2\alpha_T^2}}{(\operatorname{Re} z-\alpha_T)^2} \right\} + b^2\tan^2\theta. \end{equation}

Note that

(4.13)\begin{align} \|(T-z)^{-1}\| &\leq \frac{1}{\operatorname{dist}(z,\sigma(T))} \leq \frac{1}{\operatorname{dist}(z,\Omega(\alpha_T,\beta_T,\theta))}. \end{align}

Now the resolvent estimate follows from Proposition 2.1(iii), (4.12) and (4.13).

Proof. The proof is analogous to that of Theorem 4.6. We define the function $H_z$ as in (2.2). In order to localise $\rho(T+A)$ we need to find those $z\in\mathbb{C}$ for which $\sup_{t\in\sigma(T)} H_z(t) \lt 1$. As before, we use the estimate

\begin{align*} \sup_{t\in\sigma(T)} H_z(t) & \le \sup \left\{H_z(t) : t\in \Omega(\alpha_T,\beta_T, \theta) \cup (\mathbb{R} + \mathrm{i} [-\gamma_T,\gamma_T]) \right\} \\ & \le \max\left\{ \sup_{t\in \Omega(\alpha_T,\beta_T, \theta)} H_z(t),\ \sup_{t\in \mathbb{R} + \mathrm{i} [-\gamma_T,\gamma_T]} H_z(t) \right\}. \end{align*}

We know from the proof of Theorem 4.6 that $\sup\limits_{t\in \Omega(\alpha_T,\beta_T, \theta)} H_z(t) \lt 1$ if $z\in \Big( \mathrm{e}^{\mathrm{i}\theta}(\beta_T + \widetilde{\mathrm{Hyp}}_{\beta_T}) \cup \mathrm{e}^{-\mathrm{i}\theta}(\beta_T - \widetilde{\mathrm{Hyp}}_{\beta_T}) \Big) \cap \Big( \mathrm{e}^{-\mathrm{i}\theta}(\alpha_T + \widetilde{\mathrm{Hyp}}_{\alpha_T}) \cup \mathrm{e}^{\mathrm{i}\theta}(\alpha_T - \widetilde{\mathrm{Hyp}}_{\alpha_T}) \Big)$ while $\sup\limits_{t\in \mathbb{R} + \mathrm{i} [-\gamma_T,\gamma_T]} H_z(t) \lt 1$ if $z\in (\mathrm{i}\gamma + \mathrm{Hyp}_{\gamma_T}) \cup (-\mathrm{i}\gamma - \mathrm{Hyp}_{\gamma_T})$ by the proof of Proposition 3.2.

Proof. The proof is analogous to that of Theorem 3.7 if we use for instance the sets $S_n = \Omega(\alpha+\frac{1}{n}, \beta-\frac{1}{n}, \theta)$.

Proof. We write $\sigma(T) \subseteq \{\operatorname{Re} z \le \gamma_1 \} \cup \{\operatorname{Re} z \ge \gamma_2 \}$. We know that a sufficient condition for $z\in \rho(T+A)$ is that $\sup_{t\in\partial M} H_z(t) \lt 1$ for $M = \{\operatorname{Re} z \le \gamma_2 \}$ and $M = \{\operatorname{Re} z \ge \gamma_1 \}$. The claim then follows as in Proposition 3.1. See Figure 6.

Figure 6.

Illustration of Proposition 5.1 when $\sigma(T)\subseteq \mathbb{C}\setminus \{\gamma_{1}\le \operatorname{Re}(z) \le \gamma_{2} \}$. In the upper row we have the special case $\gamma_1 = -\gamma_2$. The dashed red lines show the boundaries of the corresponding hyperbolas. The spectrum of $T+A$ is contained in coloured regions.

Proof. We write $\sigma(T)$ as union of the half-planes $\{\operatorname{Im} z \ge \eta_2 \},\, \{\operatorname{Im} z \le \eta_2 \},\, \{\operatorname{Re} z \le \gamma_1 \}$ and $\{\operatorname{Re} z \ge \gamma_2 \}$. A sufficient condition for $z\in \rho(T+A)$ is that $\sup_{t\in\partial M} H_z(t) \lt 1$ for all of the four half-planes $M$. The claim then follows as in Proposition 3.1. See Figure 7.

Figure 7.

Illustration of Proposition 5.2 when $\sigma(T)\subseteq \mathbb{C}\setminus \{\gamma_{1}\le \operatorname{Re}(z) \le \gamma_{2},\ \eta_{1} \le \operatorname{Im}(z) \le \eta_{2} \}$. In the upper row we have the special case $\gamma_1 = -\gamma_2, \eta_1 = -\eta_2$. The dashed red lines show the boundaries of the corresponding hyperbolas. The white area is contained in $\rho(T+A)$.

Proof. Again we will use that $\sup_{t\in\sigma(T)} H_z(t) \lt 1$ is a sufficient condition for $z\in \rho(T+A)$. We estimate

\begin{equation*} \sup_{t\in\sigma(T)} H_z(t) \le \sup_{|t|\ge R} H_z(t) = \max\left\{b^2, \sup_{|t|= R} H_z(t) \right\}. \end{equation*}

Note that $\sup_{|t|= R} H_z(t) = \sup_{|t|= R} \frac{a^2 + b^2|t|^2}{|z-t|^2} = \sup_{|t|= R} \frac{a^2 + b^2R^2}{|z-t|^2}$. Clearly, the supremum is attained when $\arg t = \arg z$, so $\sup_{|t|= R} H_z(t) = \frac{a^2 + b^2R^2}{(R - |z|)^2}$. This is smaller than 1 if and only if $|z| \lt R - \sqrt{a^2 + b^2 R^2} =: r$. Therefore, such $z$ belong to $\rho(T+A)$ by Proposition 2.1. See Figure 8.

Proof. Let $z\in \mathbb{C}$ with $ \alpha_{T} \lt \operatorname{Re} z \lt \beta_{T}$. We set $\mu = \operatorname{Re} z$ and $\nu=\operatorname{Im} z$. Then $z\in \rho(T)$ and

\begin{equation*} \|(T-z)^{-1}\|=\frac{1}{\operatorname{dist} (z,\sigma (T))}\leq \max\left\{\frac{1}{\mu-\alpha_T}, \frac{1}{\beta_T-\mu}\right\}. \end{equation*}

If we set $\tau =\operatorname{Re} t$ and $\sigma = \operatorname{Im} t$ for $t\in\sigma(T)$, then we obtain

\begin{align*} \|T(T-z)^{-1}\|^2 &= \sup_{t\in\sigma(T)} \frac{|t|^2}{|t-z|^2} =\sup_{\tau + \mathrm{i}\sigma \in\sigma(T)} \frac{\tau^2+\sigma^2}{(\tau-\mu)^2+(\sigma-\nu)^2} \leq \sup_{\tau + \mathrm{i}\sigma \in\sigma(T)} \frac{\tau^2+\gamma_T^2}{(\tau-\mu)^2} \\ & \le \sup_{\tau\in\mathbb{R}\setminus\ (\alpha_T,\beta_T)} \frac{\tau^2+\gamma_T^2}{(\tau-\mu)^2} = \max \left\{\frac{\alpha_T^2+\gamma_T^2}{(\alpha_T-\mu)^2},\, \frac{\beta_T^2+\gamma_T^2}{(\beta_T-\mu)^2} \right\} \end{align*}

where the last equality is a straightforward calculation. Therefore, if $\alpha_{T,p}' \lt \operatorname{Re} z \lt \beta_{T,p}'$, then

\begin{align*} \|A(T-z)^{-1}\|&\leq c\|(T-z)^{-1}\|^{1-p}\|T(T-z)^{-1}\|^{p}\\ &\leq c\cdot \max\left\{\frac{1}{(\mu-\alpha_T)^{1-p}}, \frac{1}{(\beta_T-\mu)^{1-p}}\right\} \max\left\{\frac{|\alpha_T+\mathrm{i}\gamma_T|^p}{(\mu-\alpha_T)^p},\ \frac{|\beta_T+\mathrm{i}\gamma_T|^p}{(\beta_T-\mu)^p} \right\} \\ &\leq c\cdot \max\left\{\frac{1}{\mu-\alpha_T}, \frac{1}{\beta_T-\mu}\right\} \max\left\{|\alpha_T+\mathrm{i}\gamma_T|^p,\ |\beta_T+\mathrm{i}\gamma_T|^p \right\} \\ & \lt c\cdot \max\left\{\frac{1}{\alpha_{T,p}'-\alpha_T}, \frac{1}{\beta_T-\beta_{T,p}'}\right\} \max\left\{|\alpha_T+\mathrm{i}\gamma_T|^p,\ |\beta_T+\mathrm{i}\gamma_T|^p \right\} = 1 \end{align*}

and (6.2) follows from Proposition 2.1.

Proof. Let $z\in \rho(T)$ with $\mu=\operatorname{Re} z$ and $\nu=\operatorname{Im} z$ and let $d(z) :=\operatorname{dist}(z,\sigma(T)$. Since $T$ is normal, we have that

\begin{equation*} \|(T-z)^{-1}\| = \frac{1}{d(z)}, \qquad \|T(T-z)^{-1}\| = \|I+z(T-z)^{-1}\| \leq 1+\frac{|z|}{d(z)}. \end{equation*}

Together with

\begin{equation*} \|Ax\|\leq c\|x\|^{1-p}\|Tx\|^{p},\qquad x\in \mathcal{D}(T), \end{equation*}

we obtain for $\alpha_{T+A}^{(p)} \lt \operatorname{Re} z \lt \beta_{T+A}^{(p)}$

\begin{equation*}\begin{aligned} \|A(T-z)^{-1}\| \leq c\|(T-z)^{-1}\|^{1-p}\|T(T-z)^{-1}\|^{p} \leq c\frac{1}{d(z)^{1-p}} \left(1+\frac{|z|}{d(z)}\right)^p \\ \leq \begin{cases} \frac{c\left(1+\frac{|z|}{\sqrt{\min\{\mu-\alpha_T, \beta_T-\mu\}^2+\nu^2}}\right)^p}{\left(\sqrt{\min\{\mu-\alpha_T, \beta_T-\mu\}^2+\nu^2}\right)^{1-p}}, &\text{if } |\nu| \leq 1/2, \\[4ex] \frac{c\left(1+\frac{|z|}{\sqrt{\min\{\mu-\alpha_T, \beta_T-\mu\}^2+|\nu|-\frac{1}{4}} }\right)^p}{\left(\sqrt{\min\{\mu-\alpha_T, \beta_T-\mu\}^2+|\nu|-\frac{1}{4}}\right)^{1-p}}, &\text{if } |\nu| \gt 1/2 \end{cases} \end{aligned} \end{equation*}

where in the last step we used Lemma A.8. If $|\nu|\leq \nu_0$, then

\begin{align*} \|A(T-z)^{-1}\|&\leq c\left(1+\frac{|z|}{\min\{\mu-\alpha_T, \beta_T-\mu\}}\right)^p\frac{1}{\left(\min\{\mu-\alpha_T, \beta_T-\mu\}\right)^{1-p}}\\ &\leq c\left(1+\frac{\sqrt{\zeta_T^2+\nu_0^2}}{M}\right)^p\frac{1}{M^{1-p}}=\frac{c\left(M+\sqrt{\zeta_T^2+\nu_0^2}\right)^p}{M} \lt 1 \end{align*}

whereas if $|\nu| \gt \nu_0\geq 1/2$, then

\begin{align*} \|A(T-z)^{-1}\|&\leq c\left(1+\frac{|z|}{\sqrt{|\nu|-\frac{1}{4}} }\right)^p\frac{1}{\left(\sqrt{|\nu|-\frac{1}{4}}\right)^{1-p}}= \frac{c\left(\sqrt{|\nu|-\frac{1}{4}}+|z|\right)^p}{\sqrt{|\nu|-\frac{1}{4}} }\\ &\leq \frac{c\left(\sqrt{|\nu|-\frac{1}{4}}+\sqrt{\zeta_T^2+\nu^2}\right)^p}{\sqrt{|\nu|-\frac{1}{4}} } \lt 1 \end{align*}

by (6.5). We conclude that, in any case, $T+A-z$ is bijective and so $z\in \rho(T+A)$.