]> 2.4 The Lyapunov operator equation

### 2.4 The Lyapunov operator equation

Let us consider an abstract observation system on a general Hilbert space $\text{H}$ with scalar product ${〈\cdot ,\cdot 〉}_{\text{H}}$
 $\left\{\begin{array}{cccc}\hfill ẋ\left(t\right)& \hfill =\hfill & \mathcal{𝒜}x\left(t\right),\hfill & \phantom{\rule{1em}{0ex}}t\ge 0\hfill \\ \hfill x\left(0\right)& \hfill =\hfill & {x}_{0}\hfill & \hfill \\ \hfill y& \hfill =\hfill & \mathcal{𝒞}x\hfill & \hfill \end{array}\right\}$ (2.19)

with the operator $\mathcal{𝒜}:\left(D\left(\mathcal{𝒜}\right)\subset \text{H}\right)\to \text{H}$ generating a linear ${\text{C}}_{0}$–semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ on $\text{H}$ and the operator $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$ where ${D}_{\mathcal{𝒜}}$ denotes the domain $D\left(\mathcal{𝒜}\right)$ equipped with the norm induced by the scalar product ${〈x,v〉}_{\mathcal{𝒜}}:={〈x,v〉}_{\text{H}}+{〈\mathcal{𝒜}x,\mathcal{𝒜}v〉}_{\text{H}}$. Y is another Hilbert space with the scalar product ${〈\cdot ,\cdot 〉}_{\text{Y}}$. Observe also that if ${\mathcal{𝒜}}^{-1}\in L\left(\text{H}\right)$ then the norm induced by ${〈x,v〉}_{\mathcal{𝒜}}$ is equivalent with the norm induced by the scalar product ${〈\mathcal{𝒜}x,\mathcal{𝒜}v〉}_{\text{H}}$. In this case without loss of generality one can assume that

 $\mathcal{𝒞}={\mathcal{𝒟}}^{\ast }\mathcal{𝒜}$ (2.20)

where $\mathcal{𝒟}\in L\left(\text{Y},\text{H}\right)$.

Deﬁnition 2.4.1. The observation operator $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$ is called admissible if there exists $\gamma >0$ such that

 ${\int }_{0}^{\infty }{∥\mathcal{𝒞}S\left(t\right){x}_{0}∥}_{\text{Y}}^{2}dt\le \gamma {∥{x}_{0}∥}_{\text{H}}^{2}\phantom{\rule{2em}{0ex}}\forall {x}_{0}\in D\left(\mathcal{𝒜}\right)$ (2.21)

i.e., the observability map

$P:\left(D\left(\mathcal{𝒜}\right)\subset \text{H}\right)\ni {x}_{0}↦\mathcal{𝒞}S\left(\cdot \right){x}_{0}\in {\text{L}}^{2}\left(0,\infty ;\text{Y}\right)$

is bounded.

In this section, we prove a few results concerning the existence, uniqueness, and regularity of solution of the Lyapunov operator equation.

Theorem 2.4.1. $\mathcal{𝒞}$ is admissible iff there exists $\mathcal{ℋ}={\mathcal{ℋ}}^{\ast }\in L\left(\text{H}\right)$, $\mathcal{ℋ}\ge 0$, and $\mathcal{ℋ}$ satisﬁes the Lyapunov operator equation

 ${〈\mathcal{𝒜}{x}_{1},\mathcal{ℋ}{x}_{2}〉}_{\text{H}}+{〈{x}_{1},\mathcal{ℋ}\mathcal{𝒜}{x}_{2}〉}_{\text{H}}=-{〈\mathcal{𝒞}{x}_{1},\mathcal{𝒞}{x}_{2}〉}_{\text{Y}}\phantom{\rule{2em}{0ex}}\forall {x}_{1},{x}_{2}\in D\left(\mathcal{𝒜}\right)$ (2.22)

Proof. Notice that, for $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$, the function $\left[0,\infty \right)\ni t↦\mathcal{𝒞}S\left(t\right){x}_{0}\in \text{Y}$ is continuous for every ﬁxed ${x}_{0}\in D\left(\mathcal{𝒜}\right)$.

Necessity. Putting ${x}_{1}={x}_{2}=S\left(t\right){x}_{0}$, for ${x}_{0}\in D\left(\mathcal{𝒜}\right)$ and $t\ge 0$, in (2.22), we get

$\frac{d}{dt}{〈S\left(t\right){x}_{0},\mathcal{ℋ}S\left(t\right){x}_{0}〉}_{\text{H}}={〈\mathcal{𝒜}S\left(t\right){x}_{0},\mathcal{ℋ}S\left(t\right){x}_{0}〉}_{\text{H}}+{〈S\left(t\right){x}_{0},\mathcal{ℋ}\mathcal{𝒜}S\left(t\right){x}_{0}〉}_{\text{H}}=-{∥\mathcal{𝒞}S\left(t\right){x}_{0}∥}_{\text{Y}}^{2},$

and hence

${\int }_{0}^{t}{∥\mathcal{𝒞}S\left(\tau \right){x}_{0}∥}_{\text{Y}}^{2}d\tau ={〈{x}_{0},\mathcal{ℋ}{x}_{0}〉}_{\text{H}}-{〈S\left(t\right){x}_{0},\mathcal{ℋ}S\left(t\right){x}_{0}〉}_{\text{H}}\le {〈{x}_{0},\mathcal{ℋ}{x}_{0}〉}_{\text{H}}\le ∥\mathcal{ℋ}∥\phantom{\rule{0ex}{0ex}}{∥{x}_{0}∥}_{\text{H}}^{2}\phantom{\rule{0ex}{0ex}}.$

The integrand is nonnegative and continuous, thus the function

$t↦{\int }_{0}^{t}{∥\mathcal{𝒞}S\left(\tau \right){x}_{0}∥}_{\text{Y}}^{2}d\tau$

is C${}^{1}$, monotone increasing, and bounded from above. This implies the convergence of the improper integral, and (2.21) holds with $\gamma =∥\mathcal{ℋ}∥$.

Sufficiency. By (2.21), the operator $P:\phantom{\rule{0ex}{0ex}}\text{H}\to {\text{L}}^{2}\left(0,\infty ;\text{Y}\right)$ deﬁned as

 $\left(P{x}_{0}\right)\left(t\right)=\mathcal{𝒞}S\left(t\right){x}_{0},\phantom{\rule{2em}{0ex}}{x}_{0}\in D\left(P\right)=D\left(\mathcal{𝒜}\right)$ (2.23)

is linear, densely deﬁned, and bounded. Thus $P$ is closable, and its closure $\overline{P}$ is unique continuous extension of $P$ to the whole space $\text{H}$ [86, Theorem 4.5, p. 58, Theorem 5.2, p. 84, Theorem 5.3, p. 90]. Now, the bilinear form

 ${〈{x}_{1},\mathcal{ℋ}{x}_{2}〉}_{\text{H}}={〈\overline{P}{x}_{1},\overline{P}{x}_{2}〉}_{{\text{L}}^{2}\left(0,\infty ;\text{Y}\right)}$ (2.24)

uniquely determines an operator $\mathcal{ℋ}\in L\left(\text{H}\right)$, with $\mathcal{ℋ}={\mathcal{ℋ}}^{\ast }$ and $\mathcal{ℋ}\ge 0$. Actually $\mathcal{ℋ}={\left(\overline{P}\right)}^{\ast }\overline{P}={P}^{\ast }\overline{P}$. By virtue of (2.23), (2.24), and the semigroup axiom (i), for ${x}_{1},{x}_{2}\in D\left(\mathcal{𝒜}\right)$ we have

${〈S\left(t\right){x}_{1},\mathcal{ℋ}S\left(t\right){x}_{2}〉}_{\text{H}}={\int }_{t}^{\infty }{〈\mathcal{𝒞}S\left(\tau \right){x}_{2},\mathcal{𝒞}S\left(\tau \right){x}_{2}〉}_{\text{Y}}d\tau ,\phantom{\rule{2em}{0ex}}t\ge 0.$

Hence, taking the right derivative at $t=0$, we get (2.22). □

In , this theorem is proved in the case where $\mathcal{𝒞}\in L\left(\text{H},\text{Y}\right)$. If $\mathcal{𝒜}$ has an inverse ${\mathcal{𝒜}}^{-1}\in L\left(\text{H}\right)$, then (2.22) is equivalent to the following equation in $L\left(\text{H}\right)$:

 ${\left({\mathcal{𝒜}}^{\ast }\right)}^{-1}\mathcal{ℋ}+\mathcal{ℋ}\mathcal{𝒜}=-D{D}^{\ast }$ (2.25)

where ${D}^{\ast }=\mathcal{𝒞}{\mathcal{𝒜}}^{-1}\in L\left(\text{H},\text{Y}\right)$ by (2.20).

Theorem 2.4.2. Let $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$ be admissible. If the semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is AS then (2.22) has the unique solution.

Proof. Let ${\mathcal{ℋ}}_{1}$, ${\mathcal{ℋ}}_{2}\in L\left(\text{H}\right)$, with ${\mathcal{ℋ}}_{k}={\mathcal{ℋ}}_{k}^{\ast }\ge 0$ ($k=1,2$), be two solutions of (2.22). Then, taking $f$, $g\in D\left(\mathcal{𝒜}\right)$, and proceeding as in the proof of Theorem 2.4.1 (the "necessity" part), we obtain

$\frac{d}{dt}{〈S\left(t\right)f,\left({\mathcal{ℋ}}_{1}-{\mathcal{ℋ}}_{2}\right)S\left(t\right)g〉}_{\text{H}}=0,\phantom{\rule{2em}{0ex}}t\ge 0.$

Hence, by (2.12), and noting that $\overline{D\left(\mathcal{𝒜}\right)}=\text{H}$, we get ${〈f,\left({\mathcal{ℋ}}_{1}-{\mathcal{ℋ}}_{2}\right)g〉}_{\text{H}}=0$ for all $f,g\in \text{H}$, and the application of the Hahn–Banach theorem completes the proof. □

Theorem 2.4.1 and 2.4.2, in somewhat different versions, appeared in  and [34, Theorem 3, p. 322, Theorem 4, p. 323].

Exercise 2.4.1. Let $\text{H}={\text{L}}^{2}\left(0,\infty \right)$ with scalar product ${\int }_{0}^{\infty }{x}_{1}\left(\theta \right){x}_{2}\left(\theta \right)d\theta$. Assume that in (2.19) $\mathcal{𝒜}$ denotes the generator of the left–shift semigroup on $\text{H}$, i.e.,

$\mathcal{𝒜}x={x}^{\prime },\phantom{\rule{2em}{0ex}}D\left(\mathcal{𝒜}\right)={\text{W}}^{1,2}\left[0,\infty \right).$

Prove that the observation functional

$\mathcal{𝒞}x=x\left(0\right),\phantom{\rule{2em}{0ex}}D\left(\mathcal{𝒞}\right)\supset D\left(A\right)$

can be represented on $D\left(\mathcal{𝒜}\right)$ as $\mathcal{𝒞}x={〈{e}^{-\left(\cdot \right)},x〉}_{\mathcal{𝒜}}$. Indeed,

${〈{e}^{-\left(\cdot \right)},x〉}_{\mathcal{𝒜}}={〈{e}^{-\left(\cdot \right)},x〉}_{{\text{L}}^{2}\left(0,\infty \right)}-{〈{e}^{-\left(\cdot \right)},{x}^{\prime }〉}_{{\text{L}}^{2}\left(0,\infty \right)}=x\left(0\right)=\mathcal{𝒞}x.$

Hence

$\left|\mathcal{𝒞}x\right|\le {∥{e}^{\left(\cdot \right)}∥}_{\mathcal{𝒜}}{∥x∥}_{\mathcal{𝒜}}={∥x∥}_{\mathcal{𝒜}}$

and $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\mathbb{ℝ}\right)$. It follows from (2.15) that for ${x}_{0}\in D\left(\mathcal{𝒜}\right)$

$\left(P{x}_{0}\right)\left(t\right)=\mathcal{𝒞}S\left(t\right){x}_{0}={{x}_{0}\left(t+\theta \right)|}_{\theta =1}={x}_{0}\left(t\right).$

Hence (2.21) holds with equality sign and with $\gamma =1$. We have proved that $\mathcal{𝒞}$ is admissible. Verify that $\mathcal{ℋ}=I$ is a unique solution of (2.22).

Exercise 2.4.2. The operator $\mathcal{𝒜}$,

$\mathcal{𝒜}x={x}^{\prime }-x,\phantom{\rule{2em}{0ex}}D\left(\mathcal{𝒜}\right)={\text{W}}^{1,2}\left[0,\infty \right)$

generates on $\text{H}={\text{L}}^{2}\left(0,\infty \right)$ the semigroup,

$\left(S\left(t\right){x}_{0}\right)\left(\theta \right)={e}^{-t}{x}_{0}\left(t+\theta \right),\phantom{\rule{2em}{0ex}}t,\theta \ge 0,\phantom{\rule{1em}{0ex}}{x}_{0}\in \text{H},$

which is clearly EXS. Show that the observation functional

 $\mathcal{𝒞}x={〈\mathcal{𝒜}x,h〉}_{\text{H}},\phantom{\rule{2em}{0ex}}h\in {\text{L}}^{2}\left(0,\infty \right),\phantom{\rule{1em}{0ex}}h\left(\theta \right)=\frac{{e}^{-\theta }}{\sqrt{\theta }}$ (2.26)

is not admissible. Indeed, observe that for $x\in D\left(\mathcal{𝒜}\right)$ we have

$\left(P{x}_{0}\right)\left(t\right)=\mathcal{𝒞}S\left(t\right){x}_{0}={〈\mathcal{𝒜}S\left(t\right){x}_{0},h〉}_{\text{H}}={\int }_{0}^{\infty }\left[{e}^{-t}{x}_{0}^{\prime }\left(t+\theta \right)-{e}^{-t}{x}_{0}\left(t+\theta \right)\right]\frac{{e}^{-\theta }}{\sqrt{\theta }}d\theta .$

It is sufficient to show that $P$ is unbounded. If we take the sequence ${\left\{{x}_{n}\right\}}_{n\in \mathbb{ℕ}}\subset D\left(\mathcal{𝒜}\right)$, ${x}_{n}\left(\theta \right)=\sqrt{2n}{e}^{-n\theta }$, ${∥{x}_{n}∥}_{\text{H}}=1$ then

$\left(P{x}_{n}\right)\left(t\right)={\int }_{0}^{\infty }\left[-{e}^{-t}n\sqrt{2n}{e}^{-n\left(\theta +t\right)}-{e}^{-t}\sqrt{2n}{e}^{-n\left(\theta +t\right)}\right]\frac{{e}^{-\theta }}{\sqrt{\theta }}d\theta =$

$=-{e}^{-t\left(n+1\right)}\left(n+1\right)\sqrt{2n}{\int }_{0}^{\infty }\frac{{e}^{-\left(n+1\right)}\theta }{\sqrt{\theta }}d\theta =-{e}^{-t\left(n+1\right)}\sqrt{2n}\sqrt{n+1}\Gamma \left(\frac{3}{4}\right).$

Thus

${∥P{x}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}=\frac{n}{\sqrt{n+1}}{\left[\Gamma \left(\frac{3}{4}\right)\right]}^{2}↗\infty$

and $P$ maps the bounded sequence ${\left\{{x}_{n}\right\}}_{n\in \mathbb{ℕ}}$ into an unbounded one, which means that $P$ is unbounded. Repeating the arguments above one can prove that the observation functional (2.26) with $h$ replaced by $h\left(\theta \right)={\theta }^{\nu }{e}^{-\theta }$ for $\nu \in \left(-\frac{1}{2},0\right)$ is also not admissible.

Exercise 2.4.3. The following result is proved in [44, Theorem 4.1].

Theorem 2.4.3. Assume that $\mathcal{𝒜}$ generated an EXS semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ on $\text{H}$ and $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$, $\mathcal{𝒞}x={D}^{\ast }\mathcal{𝒜}x$ for $x\in D\left(\mathcal{𝒜}\right)$ and some $D\in L\left(\text{Y},\text{H}\right)$, is an addmissible observation operator. Suppose that $L$ and $R={L}^{\ast }$ denote the generators of the semigroups of left– and right–shifts on ${\text{L}}^{2}\left(0,\infty \right)$, viz.

$\left\{\begin{array}{cc}\hfill Lf={f}^{\prime },& D\left(L\right)={\text{W}}^{1,2}\left(0,\infty \right)\hfill \\ \hfill Rf=-{f}^{\prime },& D\left(R\right)=\left\{f\in {\text{W}}^{1,2}\left(0,\infty \right):\phantom{\rule{0ex}{0ex}}f\left(0\right)=0\right\}\hfill \end{array}\right\}.$

Let $V\in L\left(\text{H},{\text{L}}^{2}\left(0,\infty \right)\right)$,

$\left(Vx\right)\left(t\right)={h}^{\ast }S\left(t\right)x⇔{V}^{\ast }f={\int }_{0}^{\infty }{S}^{\ast }\left(t\right)hf\left(t\right)dt.$

Then we have

$\left\{\begin{array}{c}R\left(V\right)\subset D\left(L\right),\phantom{\rule{2em}{0ex}}\overline{P}=LV⇔{{P}^{\ast }|}_{D\left(R\right)}={V}^{\ast }R\hfill \\ P=V\mathcal{𝒜}⇔R\left({V}^{\ast }\right)\subset D\left({\mathcal{𝒜}}^{\ast }\right),\phantom{\rule{2em}{0ex}}{P}^{\ast }={\mathcal{𝒜}}^{\ast }{V}^{\ast }\hfill \end{array}\right\}.$

Write down the explicit forms of $\overline{P}$, ${{P}^{\ast }|}_{D\left(R\right)}$, ${P}^{\ast }$ and $\mathcal{ℋ}={P}^{\ast }\overline{P}$. The last representation holds due to (2.24). Derive a formula expressing the Laplace transform $\stackrel{̂}{\overline{P}x}$ of $\overline{P}x$, $x\in \text{H}$.

The next result [34, Theorem 5, p. 324] relates to the existence of a special type of solution to (2.22).

Deﬁnition 2.4.2. Assume that ${\text{H}}_{1}$ and ${\text{H}}_{2}$ are Hilbert spaces with scalar products ${〈\cdot ,\cdot 〉}_{{\text{H}}_{1}}$, ${〈\cdot ,\cdot 〉}_{{\text{H}}_{2}}$, respectively. An operator $\mathcal{𝒯}\in L\left({\text{H}}_{1},{\text{H}}_{2}\right)$ is called a Hilbert–Schmidt operator (HS) if there exists an orthonormal basis ${\left\{{e}_{i}\right\}}_{i\in J}$ in ${\text{H}}_{1}$ such that

$\sum _{i\in J}{∥\mathcal{𝒯}{e}_{i}∥}_{{\text{H}}_{2}}^{2}<\infty .$

Theorem 2.4.4. Suppose that $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\mathbb{ℝ}\right)$ is such that the estimate

 (2.27)

holds for some $k\in {\text{L}}^{2}\left(0,\infty \right)$. Then (2.22) has a solution $\mathcal{ℋ}\in L\left(\text{H}\right)$, with $\mathcal{ℋ}={\mathcal{ℋ}}^{\ast }$ and $\mathcal{ℋ}\ge 0$, and $\mathcal{ℋ}$ is a HS operator.

Proof. Since (2.27) implies (2.21), it is sufficient to show that a solution $\mathcal{ℋ}$, deﬁned by (2.23), (2.24), is an HS operator [86, Theorem 6.12, p. 140, Theorem 6.10, p. 137]). Hence there exists an orthonormal basis ${\left\{{e}_{j}\right\}}_{i\in J}$ in $\text{H}$ for which

$\sum _{i\in J}{∥\overline{P}{e}_{i}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}=\sum _{i\in J}{〈{e}_{i},\mathcal{ℋ}e〉}_{\text{H}}=\sum _{i\in J}{∥{\mathcal{ℋ}}^{\frac{1}{2}}{e}_{i}∥}_{\text{H}}^{2}<\infty ,$

Thus ${\mathcal{ℋ}}^{\frac{1}{2}}$ is an HS operator, which implies that $\mathcal{ℋ}$ is an HS operator. □

The assertion of this theorem remains true for a ﬁnite–rank operator $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$ satisfying (2.27). The assumption $dim\text{Y}<\infty$ is essential. The implication (2.21) $⇒$ (2.27) does not hold generally.

Remark 2.4.1. Assume that $\mathcal{𝒜}$ is the inﬁnitesimal generator of an EXS linear analytic semigroup, and that $\mathcal{𝒞}$ is a linear functional bounded on ${\text{H}}^{\alpha }$, where ${\text{H}}^{\alpha }$ ($0\le \alpha \le 1$) denotes the Banach space equal to $D\left({\mathcal{𝒜}}^{\alpha }\right)$, the domain of a fractional power of $\mathcal{𝒜}$, equipped with the topology induced by the norm ${∥x∥}_{{\text{H}}^{\alpha }}={∥{\mathcal{𝒜}}^{\alpha }x∥}_{\text{H}}$. Then (2.27) holds for $\alpha <\frac{1}{2}$. Indeed, there exists $\delta >0$ such that

$\left|\mathcal{𝒞}S\left(t\right)x\right|\le {c}_{\alpha }{∥S\left(t\right)x∥}_{{\text{H}}^{\alpha }}={c}_{\alpha }{∥{\mathcal{𝒜}}^{\alpha }S\left(t\right)x∥}_{\text{H}}\le {\overline{c}}_{\alpha }{t}^{-\alpha }{e}^{-\delta t}{∥x∥}_{\text{H}}$

[69, p. 74]. For $\alpha <\frac{1}{2}$ the function $k:t↦{t}^{-\alpha }{e}^{-\delta t}$ belongs to ${\text{L}}^{2}\left(0,\infty \right)$ [22, 860.19].

The last theorem of this section enables us to express the solution to the Lyapunov operator equation in the Riesz basis of eigenvectors of the semigroup generator.

Theorem 2.4.5. Suppose that $\text{H}$ and $\text{Y}$ are Hilbert spaces with inner products ${〈\cdot ,\cdot 〉}_{\text{H}}$ and ${〈\cdot ,\cdot 〉}_{\text{Y}}$ respectively. Let $\mathcal{𝒜}:\phantom{\rule{0ex}{0ex}}\left(D\left(\mathcal{𝒜}\right)\subset \text{H}\right)\to \text{H}$ be a linear operator which is similar to a discrete normal operator and generating EXS linear C${}_{0}$–semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ on $\text{H}$. Also, let $\mathcal{𝒞}\in L\left({D}_{\mathcal{𝒜}},\text{Y}\right)$ be admissible. Then the unique solution of the Lyapunov operator equation (2.22) has the spectral representation

 ${〈{x}_{1},\mathcal{ℋ}{x}_{2}〉}_{\text{H}}=\sum _{i\in J}{Res}_{\lambda ={\lambda }_{i}}{〈\mathcal{𝒞}{\left(-\overline{\lambda }I-\mathcal{𝒜}\right)}^{-1}{x}_{1},\mathcal{𝒞}{\left(\lambda I-\mathcal{𝒜}\right)}^{-1}{x}_{2}〉}_{\text{Y}}\phantom{\rule{2em}{0ex}}\forall {x}_{1},{x}_{2}\in \text{H}$ (2.28)

Proof. It follows from Theorems 2.4.1 and 2.4.2 that (2.22) has a nique solution $\mathcal{ℋ}\in L\left(\text{H}\right)$ with $\mathcal{ℋ}={\mathcal{ℋ}}^{\ast }$ nd $\mathcal{ℋ}\ge 0$. As in Corollary 2.3.1, denote by ${\left\{{\phi }_{i}\right\}}_{i\in j}$ the Riesz basis built from eigenvectors of $\mathcal{𝒜}$, nd by ${\left\{{\psi }_{i}\right\}}_{i\in j}$ its biorthogonal system. Putting ${x}_{2}={\phi }_{i}$ for $i\in J$ in (2.22), we obtain

 ${〈\left(\mathcal{𝒜}+{\overline{\lambda }}_{i}I\right){x}_{1},\mathcal{ℋ}{\phi }_{i}〉}_{\text{H}}=-{〈\mathcal{𝒞}{x}_{1},\mathcal{𝒞}{\phi }_{i}〉}_{\text{Y}}\phantom{\rule{2em}{0ex}}\forall {x}_{1}\in D\left(\mathcal{𝒜}\right)\phantom{\rule{2em}{0ex}}\forall i\in J$ (2.29)

By EXS, $-{\overline{\lambda }}_{i}\in \rho \left(\mathcal{𝒜}\right)$, and 2.29) is equivalent to

 ${〈f,\mathcal{ℋ}{\phi }_{i}〉}_{\text{H}}=-{〈\mathcal{𝒞}{\left({\overline{\lambda }}_{i}I+\mathcal{𝒜}\right)}^{-1}f,\mathcal{𝒞}{\phi }_{i}〉}_{\text{Y}}\phantom{\rule{2em}{0ex}}\forall f\in \text{H}\phantom{\rule{1em}{0ex}}\forall i\in J$ (2.30)

Taking into account the fact that ${\left\{{\phi }_{i}\right\}}_{i\in J}$ is a Riesz basis nd $\mathcal{ℋ}\in L\left(\text{H}\right)$, we get from (2.30) that

$\begin{array}{cc}\hfill {〈{x}_{1},\mathcal{ℋ}{x}_{2}〉}_{\text{H}}=& {〈{x}_{1},\sum _{i\in J}{〈{x}_{2},{\psi }_{i}〉}_{\text{H}}\mathcal{ℋ}{\phi }_{i}〉}_{\text{H}}=\sum _{i\in J}{〈{x}_{1},\mathcal{ℋ}{\phi }_{i}〉}_{\text{H}}\overline{{〈{x}_{2},{\psi }_{i}〉}_{\text{H}}}=\hfill \\ \hfill =& \sum _{i\in J}{〈\mathcal{𝒞}{\left(-{\overline{\lambda }}_{i}I-\mathcal{𝒜}\right)}^{-1}{x}_{1},\mathcal{𝒞}\left({〈{x}_{2},{\psi }_{i}〉}_{\text{H}}{\phi }_{i}\right)〉}_{\text{Y}}=\hfill \\ \hfill =& \sum _{i\in J}{〈\mathcal{𝒞}{\left(-{\overline{\lambda }}_{i}I-\mathcal{𝒜}\right)}^{-1}{x}_{1},\mathcal{𝒞}{Res}_{\lambda ={\lambda }_{i}}{\left(\lambda I-\mathcal{𝒜}\right)}^{-1}{x}_{2}〉}_{\text{Y}}\phantom{\rule{2em}{0ex}}\forall {x}_{1},{x}_{2}\in \text{H}.\hfill \end{array}$

In the next four sections, we give a complete solution of Problem 2.2.1.