Descent of source dynamics

Proof. (1)$\Rightarrow$(2). Let $n\in N$ and $t\ge0$. Then $Kn=0$, so by (5.1), $$K\,U(t)n=V(t)\,Kn=V(t)\,0=0,$$ i.e. $U(t)n\in\ker K=N$. As $n\in N$ and $t\ge0$ were arbitrary, (5.2) holds.

(2)$\Rightarrow$(1). Fix $t\ge0$ and suppose $K\Psi=K\Psi'$. Then $\Psi-\Psi'\in N$, so by (5.2) $U(t)(\Psi-\Psi')\in N$, hence $$K\,U(t)\Psi-K\,U(t)\Psi'=K\,U(t)(\Psi-\Psi')=0 .$$ Therefore the prescription (5.3) depends only on the value $K\Psi\in\operatorname{Ran}K$ and not on the representative $\Psi$; it defines a map $V(t):\operatorname{Ran}K\to\mathcal{H}_{\mathrm{obs}}$, whose image lies in $\operatorname{Ran}K$ because $V(t)K\Psi=K(U(t)\Psi)\in\operatorname{Ran}K$. Linearity of $V(t)$ follows from linearity of $K$ and $U(t)$ together with surjectivity of $K$ onto $\operatorname{Ran}K$. Equation (5.1) holds by construction.

Semigroup laws and uniqueness. $V(0)K\Psi=KU(0)\Psi=K\Psi$, so $V(0)=I$ on $\operatorname{Ran}K$. For $s,t\ge0$, $$V(t)V(s)K\Psi=V(t)K\bigl(U(s)\Psi\bigr)=K\,U(t)U(s)\Psi=K\,U(t+s)\Psi=V(t+s)K\Psi ,$$ and since $K$ is onto $\operatorname{Ran}K$ this gives $V(t)V(s)=V(t+s)$. Uniqueness is likewise forced: any $V'(t)$ satisfying (5.1) agrees with $V(t)$ at every point $K\Psi$ of $\operatorname{Ran}K$.

Quotient statement. Well-definedness of $\bar U(t)$ in (5.4) is exactly the implication just proved: $[\Psi]=[\Psi']$ means $\Psi-\Psi'\in N$, so $U(t)\Psi-U(t)\Psi'\in N$, so $[U(t)\Psi]=[U(t)\Psi']$. The semigroup laws for $\bar U$ follow from those of $U$ applied to representatives, and the intertwining $\widetilde{K}\bar U(t)[\Psi]=K U(t)\Psi=V(t)K\Psi=V(t)\widetilde{K}[\Psi]$ is immediate. Conversely, if $\bar U(t)$ is well defined by (5.4) for all $t$, then applying it to $[n]=0$ for $n\in N$ gives $[U(t)n]=0$, i.e. (5.2). ◻

Proof. (1) Boundedness: for any $n\in N$, (5.2) gives $U(t)n\in N$, hence $$\lVert\bar U(t)[\Psi]\rVert_{K}=\inf_{m\in N}\lVert U(t)\Psi+m\rVert \le\inf_{n\in N}\lVert U(t)(\Psi+n)\rVert \le\lVert U(t)\rVert\inf_{n\in N}\lVert\Psi+n\rVert =\lVert U(t)\rVert\,\lVert[\Psi]\rVert_{K},$$ where in the first inequality we restricted the infimum to the subfamily $m=U(t)n$, $n\in N$, which is legitimate precisely because $U(t)N\subseteq N$. Strong continuity: for fixed $\Psi$, $\lVert\bar U(t)[\Psi]-[\Psi]\rVert_{K}\le\lVert U(t)\Psi-\Psi\rVert_{\mathcal{H}_{\mathrm{src}}}\to0$ as $t\downarrow0$ by strong continuity of $U$, and the semigroup laws are Theorem 5.2. Strong continuity at $t=0$ plus local boundedness gives strong continuity on $[0,\infty)$ by the standard argument [3, Prop. I.5.3].

(2) By construction $\widetilde{K}:(\mathcal{H}_{K},\lVert\cdot\rVert_{K})\to(\operatorname{Ran}K,\lVert\cdot\rVert_{\sharp})$ is an isometric isomorphism, and conjugation by an isometric isomorphism carries $C_{0}$-semigroups to $C_{0}$-semigroups.

(3) If $\operatorname{Ran}K$ is closed, Theorem 2.3 makes $\widetilde{K}$ and $\widetilde{K}^{-1}$ bounded in the inherited norm, so $\lVert\cdot\rVert_{\sharp}$ and $\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}}$ are equivalent on $\operatorname{Ran}K$; equivalent norms have the same bounded operators and the same strongly convergent limits.

(4) Let $y=K\Psi\in\operatorname{Ran}K$ and $t,t_{0}\ge0$. By (5.3) and boundedness of $K$, $$\lVert V(t)y-V(t_{0})y\rVert_{\mathcal{H}_{\mathrm{obs}}} =\bigl\lVert K\bigl(U(t)\Psi-U(t_{0})\Psi\bigr)\bigr\rVert_{\mathcal{H}_{\mathrm{obs}}} \le\lVert K\rVert_{\mathrm{op}}\,\lVert U(t)\Psi-U(t_{0})\Psi\rVert_{\mathcal{H}_{\mathrm{src}}} \xrightarrow[\;t\to t_{0}\;]{}0$$ by strong continuity of $U$. Thus every orbit is continuous in the inherited norm, whatever $\operatorname{Ran}K$ may be. (Note that the estimate is made through the representative $\Psi$, which is legitimate because $V(t)y$ does not depend on the choice of representative, by Theorem 5.2.)

(5) The stated equivalence is the definition of boundedness of the linear map $V(t)$ from $(\operatorname{Ran}K,\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}})$ to itself, once one writes a general element of $\operatorname{Ran}K$ as $K\Psi$ and recalls $V(t)K\Psi=KU(t)\Psi$: the inequality $\lVert V(t)y\rVert\le C_{t}\lVert y\rVert$ for all $y\in\operatorname{Ran}K$ is exactly (5.5). It is not implied by (5.2): in the nonclosed case $\widetilde{K}^{-1}$ is unbounded in the inherited norm, so the conjugation argument of (2) is unavailable, and no such $C_{t}$ is produced.

Suppose now (5.5) holds for every $t\ge0$. Then each $V(t)$ is a bounded operator on the normed space $\operatorname{Ran}K$, the semigroup laws hold by Theorem 5.2, and by part (4) the map $t\mapsto V(t)y$ is continuous for every $y\in\operatorname{Ran}K$; so $\{V(t)\}$ is a strongly continuous semigroup of bounded operators on $\operatorname{Ran}K$. That is all that follows, and we stop there.

(6) We do not obtain (5.6) from (5). The usual route, pointwise boundedness of $\{V(t):0\le t\le T\}$ on each vector (which does hold, each orbit being the continuous image of a compact interval) plus Banach–Steinhaus, is unavailable, because Banach–Steinhaus requires the domain to be complete, and when $\operatorname{Ran}K$ is not closed it is an incomplete normed space. Nor may one evade this by passing to the completion $\overline{\operatorname{Ran}K}$ first: pointwise boundedness has been established only on the dense subspace $\operatorname{Ran}K$, not at every vector of $\overline{\operatorname{Ran}K}$, and a family of operators that is pointwise bounded on a dense subspace need not be pointwise bounded on the whole space absent a uniform bound, which is precisely what is at issue. So (5.6) is an extra hypothesis.

Assume it. Then for $0\le t\le T$ the operator $V(t)$ is bounded on the dense subspace $\operatorname{Ran}K$ of $\overline{\operatorname{Ran}K}$ with $\lVert V(t)\rVert_{\mathrm{op}}\le M_{T} :=\sup_{0\le t\le T}\lVert V(t)\rVert_{\mathrm{op}}<\infty$, so it extends uniquely to a bounded operator $\widehat V(t)$ on $\overline{\operatorname{Ran}K}$ with the same norm (uniformly continuous extension from a dense subspace of a complete space). The semigroup laws pass to the closure by density and continuity. For strong continuity, let $\hat y\in\overline{\operatorname{Ran}K}$ and $\varepsilon>0$; pick $y\in\operatorname{Ran}K$ with $\lVert\hat y-y\rVert<\varepsilon$; then for $0\le t\le T$, $$\lVert\widehat V(t)\hat y-\hat y\rVert \le\lVert\widehat V(t)(\hat y-y)\rVert+\lVert V(t)y-y\rVert+\lVert y-\hat y\rVert \le(M_{T}+1)\varepsilon+\lVert V(t)y-y\rVert ,$$ and the middle term tends to $0$ as $t\downarrow0$ by part (4). Hence $\widehat V(t)\hat y\to\hat y$, and $\{\widehat V(t)\}$ is a $C_{0}$-semigroup on $\overline{\operatorname{Ran}K}$. In the closed-range case $\operatorname{Ran}K=\overline{\operatorname{Ran}K}$ is complete, (5.6) follows from the standard local boundedness of the $C_{0}$-semigroup already produced in (3), and nothing new is needed. ◻

Proof. If such a $V(t_{0})$ existed, (5.1) applied to $n$ would give $KU(t_{0})n=V(t_{0})Kn=0$, contradicting (5.7). The general statement follows by linearity: $K(\Psi+n)=K\Psi$, while $KU(t_{0})(\Psi+n)-KU(t_{0})\Psi=KU(t_{0})n\neq0$. ◻

Proof. ($\Leftarrow$) If $LN\subseteq N$ then $L^{k}N\subseteq N$ for all $k\ge0$ by induction. For $n\in N$, the partial sums $\sum_{k=0}^{m}\tfrac{t^{k}}{k!}L^{k}n$ lie in $N$ and converge in norm to $e^{tL}n$; since $N$ is closed, $e^{tL}n\in N$.

($\Rightarrow$) If $e^{tL}N\subseteq N$ for all $t\ge0$ and $n\in N$, then $t^{-1}(e^{tL}n-n)\in N$ for $t>0$, and $N$ closed gives $Ln=\lim_{t\downarrow0}t^{-1}(e^{tL}n-n)\in N$, the limit existing in norm because $L$ is bounded. ◻