Since Proposition
4.1 tells that
\(\sigma ({{\mathbb {A}}}) \cap \mathrm{i}{\mathbb {R}}= \emptyset\), we only need to prove (
5.1). To this end, assuming by contradiction that (
5.1) fails to hold, it is readily seen that there exist a real sequence
\(\lambda _n\) with
\(|\lambda _n|\rightarrow \infty\) and a sequence of vectors
\({\varvec{u}}_n =(u_n,v_n,w_n,\theta _n)\in \mathrm{dom}({{\mathbb {A}}})\) of unit norm such that
$$\begin{aligned} \mathrm{i}{\varvec{u}}_n - \lambda _{n}^{-1}{{\mathbb {A}}}{\varvec{u}}_n \rightarrow 0 \quad \text {in } {\mathcal H}. \end{aligned}$$
(5.2)
Our aim is to reach a contradiction by proving that every component of
\({\varvec{u}}_n\) goes to zero in its own norm. In what follows,
\(c>0\) will denote a generic constant, whose value might change from line to line, or even within the same line. We will also tacitly employ the boundedness of
\(u_n\) and
\(v_n\) in
\(H^\varrho\) (and consequently in
\(H^1\), as
\(\varrho \ge 1\)) as well as the boundedness of
\(w_n\) and
\(\theta _n\) in
H.
First, we multiply relation (
5.2) by
\({\varvec{u}}_n\) in
\({\mathcal H}\). Exploiting (
4.2), we infer that
$$\begin{aligned} |\lambda _{n}|^{-1/2}\theta _n \rightarrow 0 \quad \text {in } H^1. \end{aligned}$$
(5.3)
Next, writing (
5.2) componentwise, we obtain the system
$$\begin{aligned}&\mathrm{i}u_n - \lambda _{n}^{-1} v_n \rightarrow 0 \quad \text {in } H^\varrho , \end{aligned}$$
(5.4)
$$\begin{aligned}&\mathrm{i}v_n - \lambda _{n}^{-1} w_n \rightarrow 0 \quad \text {in } H^\varrho , \end{aligned}$$
(5.5)
$$\begin{aligned}&\mathrm{i}w_n + \lambda _{n}^{-1} (\alpha w_n - A(\eta \theta _n - A^{\varrho -1}(\beta v_n + \gamma u_n))) \rightarrow 0 \quad \text {in } H, \end{aligned}$$
(5.6)
$$\begin{aligned}&\mathrm{i}\theta _n + \lambda _{n}^{-1} A (\kappa \theta _n +\eta (w_n + \alpha v_n)) \rightarrow 0 \quad \text {in } H. \end{aligned}$$
(5.7)
Recalling that
\(|\lambda _n|\rightarrow \infty\), it follows immediately from (
5.4) that
$$\begin{aligned} u_n \rightarrow 0 \quad \text {in } H^\varrho . \end{aligned}$$
(5.8)
We now define
$$\begin{aligned} \varepsilon _n= \mathrm{i}\langle \theta _n , w_n \rangle + \kappa \lambda _{n}^{-1}\langle \theta _n, w_n \rangle _1 +\eta \lambda _{n}^{-1} \Vert w_n\Vert _1^2 + \eta \alpha \lambda _{n}^{-1} \langle v_n , w_n\rangle _1. \end{aligned}$$
Multiplying (
5.7) by
\(w_n\) in
H we deduce that
\(\varepsilon _n\rightarrow 0\), hence
\(|\varepsilon _n|\le c\). Since (
5.3) ensures that
\(|\lambda _{n}|^{-1/2}\Vert \theta _n\Vert _1\le c\), we can estimate
$$\begin{aligned} |\lambda _{n}|^{-1} \Vert w_n\Vert _1^2&\le c |\varepsilon _n| + c\Vert \theta _n\Vert \Vert w_n\Vert + c \big [|\lambda _{n}|^{-1/2}\Vert w_n \Vert _1\big ] \big [|\lambda _{n}|^{-1/2} \Vert \theta _n\Vert _{1} + |\lambda _{n}|^{-1/2} \Vert v_n\Vert _1\big ]\\&\le \frac{1}{2} |\lambda _{n}|^{-1} \Vert w_n\Vert _1^2 + c. \end{aligned}$$
Accordingly, we are led to the bound
$$\begin{aligned} |\lambda _{n}|^{-1/2} \Vert w_n\Vert _1\le c. \end{aligned}$$
(5.9)
Next, we take the inner product in
H of (
5.7) and
\(\theta _n\) to get
$$\begin{aligned} \mathrm{i}\Vert \theta _n\Vert ^2 + \kappa \lambda _n^{-1}\Vert \theta _n\Vert _1^2 + \eta \lambda _n^{-1}\langle w_n +\alpha v_n, \theta _n \rangle _1 \rightarrow 0. \end{aligned}$$
The second term converges to zero due to (
5.3). Recalling (
5.9) and exploiting once more (
5.3), we also infer that
$$\begin{aligned} |\eta \lambda _n^{-1}\langle w_n +\alpha v_n, \theta _n \rangle _1 | \le c \big [|\lambda _{n}|^{-1/2} \Vert \theta _n\Vert _{1}\big ]\big [|\lambda _{n}|^{-1/2}\Vert w_n\Vert _1 + |\lambda _{n}|^{-1/2}\Vert v_n \Vert _1\big ] \rightarrow 0. \end{aligned}$$
Therefore, we end up with the convergence
$$\begin{aligned} \theta _n \rightarrow 0 \quad \text {in } H. \end{aligned}$$
(5.10)
In turn, as
\(\varrho \le 2\), we learn from (
5.7) together with the Poincaré inequality that
$$\begin{aligned} \lambda _{n}^{-1} (\kappa \theta _n +\eta (w_n + \alpha v_n) )\rightarrow 0 \quad \text {in } H^\varrho . \end{aligned}$$
Invoking now (
5.4)-(
5.5) together with (
5.8), the relation above implies that
$$\begin{aligned} \lambda _{n}^{-1} \beta \kappa \theta _n +\mathrm{i}\eta (\beta v_n + \gamma u_n) \rightarrow 0 \quad \text {in } H^\varrho . \end{aligned}$$
(5.11)
In particular, we find
$$\begin{aligned} |\lambda _n|^{-1} \Vert \theta _n\Vert _{\varrho }\le c. \end{aligned}$$
On the other hand, we readily learn from (
5.6) that the sequence
$$\begin{aligned} \lambda _{n}^{-1}(\eta \theta _n - A^{\varrho -1}(\beta v_n + \gamma u_n)) \end{aligned}$$
is bounded in
\(H^2\), hence in
\(H^\varrho\) appealing again on
\(\varrho \le 2\). Thus,
$$\begin{aligned} |\lambda _{n}|^{-1} \Vert A^{\varrho -1}(\beta v_n + \gamma u_n)\Vert _{\varrho } =|\lambda _n|^{-1} \Vert \beta v_n + \gamma u_n\Vert _{3\varrho -2} \le c. \end{aligned}$$
By interpolation, since
\(\varrho \ge 1\), we now obtain the bound
$$\begin{aligned} |\lambda _n|^{-1/2} \Vert \beta v_n + \gamma u_n\Vert _{2\varrho -1} \le |\lambda _n|^{-1/2} \Vert \beta v_n + \gamma u_n\Vert _{3\varrho -2}^{1/2} \, \Vert \beta v_n + \gamma u_n\Vert _\varrho ^{1/2} \le c. \end{aligned}$$
Next, a multiplication of (
5.11) by
\(\beta v_n + \gamma u_n\) in
\(H^\varrho\) gives
$$\begin{aligned} \mathrm{i}\eta \Vert \beta v_n + \gamma u_n\Vert _{\varrho }^2 + \lambda _{n}^{-1} \beta \kappa \langle \theta _n , \beta v_n + \gamma u_n\rangle _\varrho \rightarrow 0. \end{aligned}$$
Relation (
5.3) together with the boundedness of
\(|\lambda _n|^{-1/2} \Vert \beta v_n + \gamma u_n\Vert _{2\varrho -1}\) ensure that
$$\begin{aligned} |\lambda _{n}^{-1} \beta \kappa \langle \theta _n , \beta v_n + \gamma u_n\rangle _\varrho | \le c\big [|\lambda _{n}|^{-1/2} \Vert \theta _n\Vert _{1}\big ] \big [|\lambda _{n}|^{-1/2} \Vert \beta v_n + \gamma u_n\Vert _{2\varrho -1}\big ]\rightarrow 0, \end{aligned}$$
yielding the convergence
\(\beta v_n + \gamma u_n\rightarrow 0\) in
\(H^\varrho\). In the light of (
5.8), the latter entails
$$\begin{aligned} v_n \rightarrow 0 \quad \text {in } H^\varrho . \end{aligned}$$
(5.12)
At this point, we take the inner product in
H of (
5.6) and
\(w_n\) to get
$$\begin{aligned} \mathrm{i}\Vert w_n\Vert ^2 + \alpha \lambda _{n}^{-1} \Vert w_n\Vert ^2 - \eta \lambda _{n}^{-1} \langle \theta _n , w_n\rangle _1 + \lambda _{n}^{-1} \langle \beta v_n +\gamma u_n, w_n\rangle _\varrho \rightarrow 0. \end{aligned}$$
Since
\(|\lambda _n|\rightarrow \infty\), the second term converges to zero. Owing to (
5.3) and (
5.9), we also see that
$$\begin{aligned} |\eta \lambda _{n}^{-1} \langle \theta _n , w_n\rangle _1| \le c \big [|\lambda _{n}|^{-1/2} \Vert \theta _n\Vert _{1}\big ]\big [|\lambda _{n}|^{-1/2}\Vert w_n\Vert _1\big ]\rightarrow 0. \end{aligned}$$
Finally, collecting (
5.5) and (
5.12), we learn that
\(\lambda _{n}^{-1} w_n\rightarrow 0\) in
\(H^\varrho\), which leads to
$$\begin{aligned} \lambda _{n}^{-1} \langle \beta v_n +\gamma u_n, w_n\rangle _\varrho \rightarrow 0. \end{aligned}$$
In conclusion, we have proved the convergence
\(w_n\rightarrow 0\) in
H. Together with (
5.8), (
5.10) and (
5.12), the latter leads to the desired contradiction.
\(\square\)