Minimal dynamical completion

Proof. For each fixed $t\ge0$ the operator $PLQ\,e^{tQLQ}:\mathcal{H}_{Q}\to\mathcal{H}_{P}$ is bounded, so its kernel is a closed subspace. By (7.2), $N_{Q}$ is the intersection of a family of closed subspaces, hence a closed subspace. The quotient of a Hilbert space by a closed subspace is a Hilbert space with the quotient norm, exactly as established after (2.2). ◻

Proof. (1) $\mathcal{O}_{Q}$ is linear, so $\mathcal{O}_{Q}z_{0}=\mathcal{O}_{Q}z_{0}'$ iff $\mathcal{O}_{Q}(z_{0}-z_{0}')=0$ iff $z_{0}-z_{0}'\in\ker\mathcal{O}_{Q}=N_{Q}$.

(2) By (1), the prescription (7.4) is independent of the representative, hence well defined; it is linear because $\pi$ and $\mathcal{O}_{Q}$ are; and it is injective because $\overline{\mathcal{O}}[z_{0}]=0$ forces $z_{0}\in N_{Q}$, i.e. $[z_{0}]=0$. Continuity: for each $T$, $p_{T}(\overline{\mathcal{O}}[z_{0}])=p_{T}(\mathcal{O}_{Q}(z_{0}+n))\le \lVert PLQ\rVert e^{T\lVert QLQ\rVert}\lVert z_{0}+n\rVert$ for every $n\in N_{Q}$; taking the infimum over $n$ gives $p_{T}(\overline{\mathcal{O}}[z_{0}])\le\lVert PLQ\rVert e^{T\lVert QLQ\rVert} \lVert[z_{0}]\rVert_{E_{\mathrm{dyn}}}$.

(3) Existence of $\rho$. We first show $q$ separates no more coarsely than $\pi$; that is, $q(z_{0})=q(z_{0}')\Rightarrow\pi(z_{0})=\pi(z_{0}')$. Indeed if $q(z_{0})=q(z_{0}')$ then by (7.5) $$\mathcal{O}_{Q}z_{0}=F\bigl(q(z_{0})\bigr)=F\bigl(q(z_{0}')\bigr)=\mathcal{O}_{Q}z_{0}' ,$$ so by (1) $z_{0}-z_{0}'\in N_{Q}$, i.e. $\pi(z_{0})=\pi(z_{0}')$. Now define $\rho$ as follows: given $e\in E$, use surjectivity of $q$ to choose $z_{0}$ with $q(z_{0})=e$, and set $\rho(e):=\pi(z_{0})$. The implication just proved shows this is independent of the choice of $z_{0}$, so $\rho$ is a well-defined map $E\to E_{\mathrm{dyn}}$, and (7.6) holds by construction. It is surjective because $\pi$ is: any $[z_{0}]\in E_{\mathrm{dyn}}$ equals $\rho(q(z_{0}))$.

Uniqueness of $\rho$. If $\rho'$ also satisfies $\pi=\rho'\circ q$, then for every $e\in E$ pick $z_{0}$ with $q(z_{0})=e$; then $\rho'(e)=\rho'(q(z_{0}))=\pi(z_{0})=\rho(q(z_{0}))=\rho(e)$. Hence $\rho'=\rho$. (Uniqueness uses surjectivity of $q$, which is why it is included in the hypothesis: a nonsurjective extension leaves $\rho$ undetermined on the unreached part of $E$.)

Linearity. If $E$ is a vector space and $q$ is linear and surjective, then for $e=q(z_{0})$, $e'=q(z_{0}')$ and scalars $\alpha,\beta$ we have $\alpha e+\beta e'=q(\alpha z_{0}+\beta z_{0}')$, so $\rho(\alpha e+\beta e')=\pi(\alpha z_{0}+\beta z_{0}') =\alpha\pi(z_{0})+\beta\pi(z_{0}')=\alpha\rho(e)+\beta\rho(e')$. ◻

Proof. Dependence only on $[z_{0}]$. The kernel $\mathcal{K}$ is determined by $L$ and $P$ alone, and the only occurrence of $z_{0}$ in (7.8) is through $\eta_{z_{0}}=\mathcal{O}_{Q}z_{0}$, which by Theorem 7.4(1) depends only on the class $[z_{0}]$.

Existence and uniqueness. Fix $T>0$. Equation (7.8) has the form $x=g+\mathcal{V}x$ on the Banach space $X:=C([0,T],\mathcal{H}_{P})$ with the sup-norm, where $g(t):=x(0)+\int_{0}^{t}\eta_{z_{0}}(s)\,ds$ is continuous, and $\mathcal{V}$ is the bounded linear Volterra operator $$(\mathcal{V}x)(t):=\int_{0}^{t}\Bigl(PLP\,x(s)+\int_{0}^{s}\mathcal{K}(s-r)x(r)\,dr\Bigr)ds .$$ Both $PLP$ and $\mathcal{K}$ are bounded, with $M:=\lVert PLP\rVert_{\mathrm{op}}+T\sup_{0\le\tau\le T}\lVert\mathcal{K}(\tau)\rVert_{\mathrm{op}}<\infty$ (the supremum is finite because $\mathcal{K}$ is norm-continuous on the compact $[0,T]$). An elementary induction gives the standard Volterra estimate $$\lVert(\mathcal{V}^{k}x)(t)\rVert\le\frac{(Mt)^{k}}{k!}\,\lVert x\rVert_{X}, \qquad 0\le t\le T,\ k\ge0,$$ so $\lVert\mathcal{V}^{k}\rVert_{\mathcal{B}(X)}\le(MT)^{k}/k!\to0$; hence $\mathcal{V}^{k}$ is a contraction for $k$ large, $I-\mathcal{V}$ is invertible with $(I-\mathcal{V})^{-1}=\sum_{k\ge0}\mathcal{V}^{k}$ convergent in operator norm, and $x=(I-\mathcal{V})^{-1}g$ is the unique solution in $X$ [5, Ch. 2, Thm. 3.5]. Since $T>0$ was arbitrary and solutions on nested intervals agree by uniqueness, $x$ is uniquely determined on $[0,\infty)$. ◻

Proof. Fix $z\in\mathcal{H}_{Q}$. Since $QLQ$ is bounded, $t\mapsto PLQ\,e^{tQLQ}z$ is given by the everywhere-convergent power series $$PLQ\,e^{tQLQ}z=\sum_{k=0}^{\infty}\frac{t^{k}}{k!}\,PLQ\,(QLQ)^{k}z ,\tag{7.11}$$ which is a real-analytic (indeed entire) $\mathcal{H}_{P}$-valued function of $t$. Such a function vanishes identically on $[0,\infty)$ if and only if all its Taylor coefficients at $t=0$ vanish, i.e. iff $$PLQ\,(QLQ)^{k}z=0 \qquad\text{for all }k\ge0 .\tag{7.12}$$ (One direction is trivial; for the other, differentiate (7.11) $k$ times at $t=0$, which is legitimate by uniform convergence of the differentiated series on compacts.)

It remains to see that in (7.12) the range $0\le k\le m-1$ suffices. By the Cayley–Hamilton theorem applied to the operator $QLQ$ on the $m$-dimensional space $\mathcal{H}_{Q}$, $(QLQ)^{m}$ is a linear combination of $I,QLQ,\dots,(QLQ)^{m-1}$; by induction, every power $(QLQ)^{k}$ with $k\ge m$ is a linear combination of $I,\dots,(QLQ)^{m-1}$. Hence if $PLQ\,(QLQ)^{j}z=0$ for $0\le j\le m-1$, then applying $PLQ$ to the corresponding linear combination gives $PLQ\,(QLQ)^{k}z=0$ for every $k\ge m$ as well. This proves (7.9).

Finally, (7.9) says exactly $N_{Q}=\ker\mathcal{O}_{m}$, so by rank–nullity $\dim E_{\mathrm{dyn}}=\dim\mathcal{H}_{Q}-\dim N_{Q}=\mathop{\mathrm{rank}}\mathcal{O}_{m}$. ◻