Proof. Write \[\mathcal{N}(h):=e^h \in L^1(\nu_0), \qquad Z(h):=\int_\Omega e^h\,\mathrm{d}\nu_0 \in (0,\infty).\] Then \[\mathcal{E}(h)=\frac{\mathcal{N}(h)}{Z(h)}.\]

We first show that \(\mathcal{N}:\mathcal{H}\to L^1(\nu_0)\) is Fréchet differentiable with \[D\mathcal{N}(h)[\xi]=e^h\xi.\] Indeed, \[\mathcal{N}(h+\xi)-\mathcal{N}(h)-e^h\xi = e^h\bigl(e^\xi-1-\xi\bigr).\] Using the elementary bound \[|e^r-1-r|\le \tfrac12 e^{|r|}r^2, \qquad r\in\mathbb{R},\] we obtain \[\|\mathcal{N}(h+\xi)-\mathcal{N}(h)-e^h\xi\|_{L^1(\nu_0)} \le \frac{1}{2} e^{\|h\|_{L^\infty}+\|\xi\|_{L^\infty}} \|\xi\|_{L^\infty}^2 = o(\|\xi\|_{L^\infty}).\] Here we used that \(\nu_0\) is a probability measure.

Similarly, \[Z(h+\xi)-Z(h)-\int_\Omega e^h\xi\,\mathrm{d}\nu_0 = \int_\Omega e^h\bigl(e^\xi-1-\xi\bigr)\,\mathrm{d}\nu_0,\] so \(Z:\mathcal{H}\to \mathbb{R}\) is Fréchet differentiable with \[DZ(h)[\xi]=\int_\Omega e^h\xi\,\mathrm{d}\nu_0.\]

Since \(Z(h)>0\) and inversion on \((0,\infty)\) is smooth, the quotient rule yields that \(\mathcal{E}=\mathcal{N}/Z\) is Fréchet differentiable, with \[D\mathcal{E}(h)[\xi] = \frac{e^h\xi}{Z(h)} - \frac{e^h}{Z(h)^2}\int_\Omega e^h\xi\,\mathrm{d}\nu_0 = \frac{e^h}{Z(h)} \left( \xi - \frac{1}{Z(h)}\int_\Omega e^h\xi\,\mathrm{d}\nu_0 \right) = \frac{\mathrm{d}\rho_h}{\mathrm{d}\nu_0} \bigl(\xi-\mathbb{E}_{\rho_h}[\xi]\bigr).\] ◻

Proof. Equation ?? follows from Proposition 4 and the Banach space chain rule. Equation ?? is obtained by dividing ?? by \(\mathrm{d}\rho_t/\mathrm{d}\nu_0\), which is positive \(\nu_0\)-a.e. ◻

Proof. Since \(\mathrm{d}\rho_h=w_h\,\mathrm{d}\nu_0\) and \(c\le w_h\le C\) \(\nu_0\)-a.e., we have \[c\int_\Omega |u|^2\,\mathrm{d}\nu_0 \le \int_\Omega |u|^2\,\mathrm{d}\rho_h \le C\int_\Omega |u|^2\,\mathrm{d}\nu_0,\] and similarly \[c\int_\Omega |\nabla u|^2\,\mathrm{d}\nu_0 \le \int_\Omega |\nabla u|^2\,\mathrm{d}\rho_h \le C\int_\Omega |\nabla u|^2\,\mathrm{d}\nu_0.\] Summing the two inequalities yields 5 . ◻

Proof. Fix \(h\in\mathcal{X}_{\mathrm{ad}}\) and \(u\in H^1(\Omega;\rho_h)\). Since \(u\mapsto \int |u-a|^2\,\mathrm{d}\rho_h\) is minimized at \(a=\mathbb{E}_{\rho_h}[u]\), for every constant \(a\in\mathbb{R}\), \[\int_\Omega \left|u-\mathbb{E}_{\rho_h}[u]\right|^2\,\mathrm{d}\rho_h \le \int_\Omega |u-a|^2\,\mathrm{d}\rho_h.\] Choosing \(a=\mathbb{E}_{\nu_0}[u]\) and using \(\mathrm{d}\rho_h=w_h\,\mathrm{d}\nu_0\), together with 3 , we obtain \[\begin{align} \int_\Omega \left|u-\mathbb{E}_{\rho_h}[u]\right|^2\,\mathrm{d}\rho_h & \le \int_\Omega \left|u-\mathbb{E}_{\nu_0}[u]\right|^2\,\mathrm{d}\rho_h \\ & \le C \int_\Omega \left|u-\mathbb{E}_{\nu_0}[u]\right|^2\,\mathrm{d}\nu_0 \\ & \le C C_{\nu_0}\int_\Omega |\nabla u|^2\,\mathrm{d}\nu_0 \\ & \le \frac{C}{c}\,C_{\nu_0}\int_\Omega |\nabla u|^2\,\mathrm{d}\rho_h, \end{align}\] which proves ?? . ◻

Proof. Define, on \(H^1_{\rho_h,\diamond}(\Omega)\), \[B_h(u,\eta):=\int_\Omega \nabla u\cdot \nabla\eta\,\mathrm{d}\rho_h, \qquad F_{h,\xi}(\eta) := \int_\Omega \bigl(\xi-\mathbb{E}_{\rho_h}[\xi]\bigr)\eta\,\mathrm{d}\rho_h.\] The bilinear form \(B_h\) is continuous. By Proposition 8, \[\|u\|_{L^2(\rho_h)}^2 \le \frac{C}{c}\,C_{\nu_0} \int_\Omega |\nabla u|^2\,\mathrm{d}\rho_h \qquad \forall u\in H^1_{\rho_h,\diamond}(\Omega),\] so \(B_h\) is coercive on \(H^1_{\rho_h,\diamond}(\Omega)\).

Next, \[\|\xi-\mathbb{E}_{\rho_h}[\xi]\|_{L^2(\rho_h)} \le 2\|\xi\|_{L^2(\rho_h)} \le 2\sqrt{C}\,\|\xi\|_{L^2(\nu_0)},\] and therefore, by Cauchy–Schwarz and Proposition 8, \[|F_{h,\xi}(\eta)| \le \|\xi-\mathbb{E}_{\rho_h}[\xi]\|_{L^2(\rho_h)} \|\eta\|_{L^2(\rho_h)} \le C'\|\xi\|_{L^2(\nu_0)}\|\eta\|_{H^1(\Omega;\rho_h)}.\] Thus \(F_{h,\xi}\) is continuous on \(H^1_{\rho_h,\diamond}(\Omega)\), and the conclusion follows from the Lax–Milgram theorem. The strong form is the usual divergence-form interpretation of 6 . ◻

Proof. Take \(\eta=\psi_{h,\xi}\) in 6 . Then \[\int_\Omega |\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h = \int_\Omega \bigl(\xi-\mathbb{E}_{\rho_h}[\xi]\bigr)\psi_{h,\xi}\,\mathrm{d}\rho_h.\] By Cauchy–Schwarz and Proposition 8, \[\int_\Omega |\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h \le \|\xi-\mathbb{E}_{\rho_h}[\xi]\|_{L^2(\rho_h)} \|\psi_{h,\xi}\|_{L^2(\rho_h)} \le M\|\xi\|_{L^2(\nu_0)} \|\nabla\psi_{h,\xi}\|_{L^2(\rho_h)}.\] Canceling the nonzero factor yields 9 . ◻

Proof. By Remark 10, \[\int_\Omega \nabla\psi_{h,\xi}\cdot\nabla\eta\,\mathrm{d}\rho_h = \int_\Omega \bigl(\xi-\mathbb{E}_{\rho_h}[\xi]\bigr)\eta\,\mathrm{d}\rho_h \qquad \forall \eta\in H^1(\Omega;\rho_h),\] so \(\nabla\psi_{h,\xi}\in\mathcal{A}_{h,\xi}\).

Now let \(v\in\mathcal{A}_{h,\xi}\). Taking \(\eta=\psi_{h,\xi}\) gives \[\int_\Omega \bigl(v-\nabla\psi_{h,\xi}\bigr)\cdot\nabla\psi_{h,\xi}\,\mathrm{d}\rho_h=0.\] Hence \[\begin{align} \int_\Omega |v|^2\,\mathrm{d}\rho_h & = \int_\Omega |\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h + \int_\Omega |v-\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h \\ & \ge \int_\Omega |\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h, \end{align}\] with equality if and only if \(v=\nabla\psi_{h,\xi}\) in \(L^2(\rho_h;\mathbb{R}^d)\). ◻

Proof. Bilinearity and symmetry follow from the linearity of \(\xi\mapsto \psi_{h,\xi}\) and the symmetry of the \(L^2(\rho_h)\) inner product. Nonnegativity is immediate from \[\mathfrak g_h(\xi,\xi)=\int_\Omega |\nabla\psi_{h,\xi}|^2\,\mathrm{d}\rho_h.\]

Assume \(\mathfrak g_h(\xi,\xi)=0\). Then \(\nabla\psi_{h,\xi}=0\) \(\rho_h\)-a.e., hence \(\psi_{h,\xi}=0\) in \(H^1_{\rho_h,\diamond}(\Omega)\). By 10 , \[\int_\Omega \bigl(\xi-\mathbb{E}_{\rho_h}[\xi]\bigr)\eta\,\mathrm{d}\rho_h =0 \qquad \forall \eta\in H^1(\Omega;\rho_h).\] In particular, the same holds for every \[\eta=\varphi-\mathbb{E}_{\rho_h}[\varphi], \qquad \varphi\in C_c^\infty(\Omega).\] Since \(C_c^\infty(\Omega)\) is dense in \(L^2(\nu_0)\) by Assumption 1, and since \(\mathrm{d}\rho_h=w_h\,\mathrm{d}\nu_0\) with \(w_h\) bounded above and below by Assumption 2, it is also dense in \(L^2(\rho_h)\). Hence the set \[\left\{ \varphi-\mathbb{E}_{\rho_h}[\varphi]:\varphi\in C_c^\infty(\Omega) \right\}\] is dense in \(L^2_0(\rho_h)\). Therefore \[\xi-\mathbb{E}_{\rho_h}[\xi]=0 \qquad \rho_h\text{-a.e.}\] The converse implication is immediate from 11 .

Finally, since \(\rho_h\) and \(\nu_0\) are equivalent and \(\xi\in L^2_0(\nu_0)\), the identity \(\xi=\mathbb{E}_{\rho_h}[\xi]\) \(\rho_h\)-a.e.implies that \(\xi\) is almost everywhere constant, hence zero by the \(\nu_0\)-mean-zero constraint. ◻

Proof. Fix \(h_1,h_2\in\mathcal{X}_{\mathrm{ad}}\) and \(\xi_1,\xi_2\in L^2_0(\nu_0)\). Write \[w_i:=\frac{\mathrm{d}\rho_{h_i}}{\mathrm{d}\nu_0}, \qquad q_i:=\xi_i-\mathbb{E}_{\rho_{h_i}}[\xi_i], \qquad \psi_i:=\psi_{h_i,\xi_i}, \qquad i=1,2.\] By 4 , \[\|w_1-w_2\|_{L^\infty(\nu_0)} \le M_{\mathcal{X}_{\mathrm{ad}}}\|h_1-h_2\|_{\mathcal{Z}}.\] Also, \[\|q_1-q_2\|_{L^2(\nu_0)} \le \|\xi_1-\xi_2\|_{L^2(\nu_0)} + \left| \mathbb{E}_{\rho_{h_1}}[\xi_1]-\mathbb{E}_{\rho_{h_2}}[\xi_2] \right|.\] The second term is bounded by \[\begin{align} \left| \int_\Omega \xi_1(w_1-w_2)\,\mathrm{d}\nu_0 \right| + \left| \int_\Omega (\xi_1-\xi_2)w_2\,\mathrm{d}\nu_0 \right| & \le \|\xi_1\|_{L^2(\nu_0)}\|w_1-w_2\|_{L^2(\nu_0)} + \sqrt{C}\,\|\xi_1-\xi_2\|_{L^2(\nu_0)} \\ & \le M\Big( \|\xi_1-\xi_2\|_{L^2(\nu_0)} + \|h_1-h_2\|_{\mathcal{Z}}\|\xi_1\|_{L^2(\nu_0)} \Big). \end{align}\] Hence \[\label{eq:ambient-centered-forcing-stability} \|q_1-q_2\|_{L^2(\nu_0)} \le M\Big( \|\xi_1-\xi_2\|_{L^2(\nu_0)} + \|h_1-h_2\|_{\mathcal{Z}} \bigl(\|\xi_1\|_{L^2(\nu_0)}+\|\xi_2\|_{L^2(\nu_0)}\bigr) \Big).\tag{12}\]

Let \(\delta\psi:=\psi_1-\psi_2\). Subtract the weak formulations: \[\int_\Omega w_1\nabla\delta\psi\cdot\nabla\eta\,\mathrm{d}\nu_0 = \int_\Omega (w_1q_1-w_2q_2)\eta\,\mathrm{d}\nu_0 - \int_\Omega (w_1-w_2)\nabla\psi_2\cdot\nabla\eta\,\mathrm{d}\nu_0\] for all \(\eta\in H^1(\Omega;\nu_0)\). Taking \[\eta:=\delta\psi-\mathbb{E}_{\nu_0}[\delta\psi]\] and using 2 , 9 , 12 , and the uniform bounds on the weights yields \[\|\nabla\delta\psi\|_{L^2(\nu_0)} \le M\Big( \|\xi_1-\xi_2\|_{L^2(\nu_0)} + \|h_1-h_2\|_{\mathcal{Z}} \bigl(\|\xi_1\|_{L^2(\nu_0)}+\|\xi_2\|_{L^2(\nu_0)}\bigr) \Big),\] which proves ?? . Subtracting the \(\nu_0\)-mean and using 2 gives ?? . ◻

Proof. Apply Proposition 13 with \(\xi_1=\xi_2=\xi\), then take the supremum over \(\|\xi\|_{L^2(\nu_0)}\le 1\). ◻

Proof. Write \[Z(h):=\int_\Omega e^h\,\mathrm{d}\nu_0, \qquad w_h=\frac{e^h}{Z(h)}.\] Since \(h,\eta\in L^\infty(\nu_0)\) and \(h+\varepsilon\eta\) remains in the uniformly bounded admissible set for \(|\varepsilon|<\varepsilon_0\), the map \[\varepsilon\longmapsto \frac{e^{h+\varepsilon\eta}}{Z(h+\varepsilon\eta)}\] is differentiable in \(L^\infty(\nu_0)\), with derivative \[w_h\bigl(\eta-\mathbb{E}_{\rho_h}[\eta]\bigr) = w_h\,a_h(\eta).\] This proves 13 . Then, for \(\xi\in L^2(\nu_0)\), \[\mathbb{E}_{\rho_{h+\varepsilon\eta}}[\xi] = \int_\Omega \xi\,w_{h+\varepsilon\eta}\,\mathrm{d}\nu_0,\] so 13 implies \[\begin{align} \frac{\mathbb{E}_{\rho_{h+\varepsilon\eta}}[\xi]-\mathbb{E}_{\rho_h}[\xi]}{\varepsilon} & \to \int_\Omega \xi\,w_h a_h(\eta)\,\mathrm{d}\nu_0 \\ & = \int_\Omega \xi\,a_h(\eta)\,\mathrm{d}\rho_h = \operatorname{Cov}_{\rho_h}(\eta,\xi), \end{align}\] which is 14 . ◻

Proof. The right-hand side of ?? is continuous on \(H^1(\Omega;\rho_h)\), so existence and uniqueness of \(\chi_{h;\eta,\xi}\) follow from Lax–Milgram exactly as in Theorem 9.

Let \[w_\varepsilon:=w_{h+\varepsilon\eta}, \qquad q_\varepsilon:=\xi-\mathbb{E}_{\rho_{h+\varepsilon\eta}}[\xi], \qquad \psi_\varepsilon:=\psi_{h+\varepsilon\eta,\xi}.\] Define the difference quotient \[\delta_\varepsilon := \frac{\psi_\varepsilon-\psi_{h,\xi}}{\varepsilon}.\] Subtract the weak formulations for \(\psi_\varepsilon\) and \(\psi_{h,\xi}\), divide by \(\varepsilon\), and compare with the weak equation for \(\chi_{h;\eta,\xi}\). Using Lemma 2, 14 , the boundedness of the weights, and the energy bound 9 , one obtains an error equation for \[r_\varepsilon:=\delta_\varepsilon-\chi_{h;\eta,\xi}\] of the form \[\int_\Omega \nabla r_\varepsilon\cdot\nabla\varphi\,\mathrm{d}\rho_{h+\varepsilon\eta} = R_\varepsilon(\varphi), \qquad \forall \varphi\in H^1(\Omega;\rho_{h+\varepsilon\eta}),\] where \(R_\varepsilon(\varphi)\to 0\) uniformly for \(\|\varphi\|_{H^1(\Omega;\rho_{h+\varepsilon\eta})}\le 1\). Taking \(\varphi=r_\varepsilon-\mathbb{E}_{\rho_{h+\varepsilon\eta}}[r_\varepsilon]\), applying Proposition 8, and using the equivalence of the \(\rho_{h+\varepsilon\eta}\)- and \(\nu_0\)-norms, we conclude that \[\|\nabla r_\varepsilon\|_{L^2(\nu_0)}\to 0.\] Subtracting an irrelevant constant and using 2 then yields \[\inf_{a\in\mathbb{R}} \|r_\varepsilon-a\|_{H^1(\Omega;\nu_0)} \to 0.\] In particular, \(\nabla r_\varepsilon\to 0\) in \(L^2(\nu_0;\mathbb{R}^d)\), which proves the gradient differentiability claim and the equivalent differentiability statement in \(H^1(\Omega;\nu_0)/\mathbb{R}\). ◻

Proof. By definition, \[\mathfrak g_h(\xi,\zeta) = \int_\Omega \nabla\psi_{h,\xi}\cdot \nabla\psi_{h,\zeta}\, \mathrm{d}\rho_h.\] Differentiate this identity with respect to \(h\) in the direction \(\eta\). The derivative of \(\mathrm{d}\rho_h=w_h\,\mathrm{d}\nu_0\) is given by Lemma 2, while the derivatives of \(\psi_{h,\xi}\) and \(\psi_{h,\zeta}\) are given by Proposition 14. Passing to the limit in the resulting difference quotient yields 15 . ◻

Proof. Since \(h\in C^1([0,T];\mathcal{Z})\) and \(\mathcal{Z}\hookrightarrow L^\infty(\nu_0)\cap L^2_0(\nu_0)\) continuously, \(h\) is a regular coordinate path in the sense of Section 2. By Proposition 6 from Section 2, \[\partial_t \frac{\mathrm{d}\rho_t}{\mathrm{d}\nu_0} = \frac{\mathrm{d}\rho_t}{\mathrm{d}\nu_0} \Bigl(\dot{h}(t)-\mathbb{E}_{\rho_t}[\dot{h}(t)]\Bigr).\] On the other hand, by definition of \(v_t\) and Remark 10, \[\int_\Omega \nabla\eta\cdot v_t\,\mathrm{d}\rho_t = \int_\Omega \Bigl(\dot{h}(t)-\mathbb{E}_{\rho_t}[\dot{h}(t)]\Bigr)\eta\,\mathrm{d}\rho_t\] for every \(\eta\in H^1(\Omega;\rho_t)\). Therefore \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\int_\Omega \eta\,\mathrm{d}\rho_t & = \int_\Omega \eta\, \partial_t\!\left(\frac{\mathrm{d}\rho_t}{\mathrm{d}\nu_0}\right)\,\mathrm{d}\nu_0 \\ & = \int_\Omega \eta \Bigl(\dot{h}(t)-\mathbb{E}_{\rho_t}[\dot{h}(t)]\Bigr)\,\mathrm{d}\rho_t \\ & = \int_\Omega \nabla\eta\cdot v_t\,\mathrm{d}\rho_t, \end{align}\] which is 17 . ◻

Proof. The Sobolev embedding \(H^{s'}(\Omega)\hookrightarrow L^\infty(\Omega)\) holds because \(s'>d/2\), and the Poincaré inequality on bounded connected Lipschitz domains is classical. Since \(s>s'\), we also have the continuous embedding \(\mathcal{X}\hookrightarrow \mathcal{Z}\). Thus Assumption 1 holds with the chosen \(\mathcal{Z}\). Together with Assumption 2, this yields the abstract forward theory in the bounded-domain setting. The final statement is immediate from the local stability hypothesis. ◻

Proof. For each \(j=1,\dots,r\), define \[\mathcal{G}_{t,j}^{y}(h):=\int_\Omega \kappa(x-y_j(t))\,\mathrm{d}\rho_h(x).\] Using the differential of the exponential-normalization map from Proposition 4, we have \[\begin{align} D\mathcal{G}_{t,j}^{y}(h)[\xi] & = \int_\Omega \kappa(x-y_j(t)) \bigl(\xi(x)-\mathbb{E}_{\rho_h}[\xi]\bigr)\,\mathrm{d}\rho_h(x) \\ & = \operatorname{Cov}_{\rho_h}(\kappa(\cdot-y_j(t)),\xi). \end{align}\] This proves ?? . The formulas ?? and ?? follow immediately from Definition 10. ◻

Proof. This is immediate from ?? . ◻

Proof. By definition of the transport form and the weighted Neumann solve, \[\mathfrak g_h(\zeta,\zeta) = \int_\Omega |\nabla\psi_{h,\zeta}|^2\,\mathrm{d}\rho_h.\] By the energy estimate from Theorem 9, together with the uniform density bounds on \(\mathcal{X}_{\mathrm{ad}}\), there exists \(C_{\mathrm{tr}}>0\) such that \[\mathfrak g_h(\zeta,\zeta) \le C_{\mathrm{tr}}\|\zeta\|_{L^2(\nu_0)}^2 \qquad \text{for all } h\in\mathcal{X}_{\mathrm{ad}},\;\zeta\in L^2_0(\nu_0).\] Applying this pointwise in time with \(\zeta=\dot{\xi}(t)\) and integrating over \([0,T]\) gives the claim. ◻

Proof. Fix \(h\in \mathcal{A}_{\mathrm{ad}}\), and let \[\xi\in L^2(0,T;H^{s'}(\Omega)\cap L^2_0(\nu_0)) \cap H^1(0,T;L^2_0(\nu_0)).\] Decompose \(\xi\) pointwise in time into its visible and invisible parts: \[\xi(t)=\xi_\perp(t)+\xi_0(t), \qquad \xi_\perp(t):=(I-\Pi_{y,h}(t))\xi(t), \qquad \xi_0(t):=\Pi_{y,h}(t)\xi(t).\] Since \(\Pi_{y,h}(t)\) is the \(L^2(\nu_0)\)-orthogonal projection onto \(K_{y,h}(t)\), we have \[\xi_\perp(t)\perp \xi_0(t) \qquad\text{in }L^2(\nu_0)\] for a.e.\(t\in[0,T]\), and therefore \[\label{eq:orthogonal-splitting} \|\xi\|_{L^2(0,T;L^2(\nu_0))}^2 = \int_0^T \|\xi_\perp(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t + \int_0^T \|\xi_0(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t.\tag{21}\]

Now apply the two hypotheses. By ?? , \[\int_0^T \mathfrak j_{t,h(t)}^{y}(\xi(t),\xi(t))\,\mathrm{d}t \ge c_{\mathrm{obs}} \int_0^T \|\xi_\perp(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t.\] By ?? , \[\int_0^T \mathfrak g_{h(t)}(\dot{\xi}(t),\dot{\xi}(t))\,\mathrm{d}t \ge c_{\mathrm{dyn}} \int_0^T \|\xi_0(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t.\] Adding the two inequalities yields \[\begin{align} Q_y[h](\xi,\xi) & = \int_0^T \mathfrak j_{t,h(t)}^{y}(\xi(t),\xi(t))\,\mathrm{d}t + \int_0^T \mathfrak g_{h(t)}(\dot{\xi}(t),\dot{\xi}(t))\,\mathrm{d}t \\ & \ge c_{\mathrm{obs}} \int_0^T \|\xi_\perp(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t + c_{\mathrm{dyn}} \int_0^T \|\xi_0(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t \\ & \ge \min\{c_{\mathrm{obs}},c_{\mathrm{dyn}}\} \left( \int_0^T \|\xi_\perp(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t + \int_0^T \|\xi_0(t)\|_{L^2(\nu_0)}^2\,\mathrm{d}t \right). \end{align}\] Using 21 , we conclude that \[Q_y[h](\xi,\xi) \ge \min\{c_{\mathrm{obs}},c_{\mathrm{dyn}}\} \|\xi\|_{L^2(0,T;L^2(\nu_0))}^2,\] which is exactly Assumption 5. ◻

Proof. For each \(h\in\mathcal{X}_{\mathrm{ad}}\), define \[\mathcal{S}_h : L^2_0(\nu_0)\to L^2(\nu_0;\mathbb{R}^d), \qquad \mathcal{S}_h \xi := w_h^{1/2}\,\mathcal{T}_h\xi,\] where \(w_h:=\mathrm{d}\rho_h/\mathrm{d}\nu_0\). Then \[\mathfrak g_h(\xi,\xi) = \|\mathcal{S}_h \xi\|_{L^2(\nu_0;\mathbb{R}^d)}^2.\]

We claim that \[\sup_{t\in[0,T]} \|\mathcal{S}_{h_n(t)}-\mathcal{S}_{h(t)}\|_{\mathcal{L}(L^2_0,L^2)} \to 0.\] Indeed, writing \[\mathcal{S}_{h_n}-\mathcal{S}_h = w_{h_n}^{1/2}(\mathcal{T}_{h_n}-\mathcal{T}_h) + (w_{h_n}^{1/2}-w_h^{1/2})\mathcal{T}_h,\] the first term vanishes by Corollary 2 and the second by the Sobolev embedding \(H^{s'}(\Omega)\hookrightarrow L^\infty(\Omega)\) and the Lipschitz dependence of the square-root weight on the state.

Now define \[u_n(t):=\mathcal{S}_{h_n(t)}\dot{h}_n(t), \qquad u(t):=\mathcal{S}_{h(t)}\dot{h}(t).\] The uniform operator convergence of \(\mathcal{S}_{h_n}\) to \(\mathcal{S}_h\), together with the weak convergence \(\dot{h}_n\rightharpoonup \dot{h}\) in \(L^2(0,T;L^2_0(\nu_0))\), implies \[u_n\rightharpoonup u \quad\text{weakly in } L^2(0,T;L^2(\nu_0;\mathbb{R}^d)).\] By weak lower semicontinuity of the norm, \[\|u\|_{L^2_tL^2_x}^2 \le \liminf_{n\to\infty} \|u_n\|_{L^2_tL^2_x}^2,\] which is the desired inequality. ◻

Proof. The assumed bound gives boundedness in \[L^2(0,T;H^s(\Omega))\cap H^1(0,T;L^2_0(\nu_0)).\] By the Lions–Magenes interpolation theorem, applied to the Hilbert triple \(H^s(\Omega)\hookrightarrow H^{s/2}(\Omega)\hookrightarrow L^2(\Omega)\), this space embeds continuously into \[C([0,T];H^{s/2}(\Omega)).\] Thus \((h_n)\) is bounded in \(C([0,T];H^{s/2}(\Omega))\). Since \(s'<s/2\), the embedding \(H^{s/2}(\Omega)\hookrightarrow H^{s'}(\Omega)\) is compact.

The \(H^1(0,T;L^2_0(\nu_0))\) bound gives equicontinuity in \(L^2(\nu_0)\): for all \(t_0,t_1\in[0,T]\), \[\|h_n(t_1)-h_n(t_0)\|_{L^2(\nu_0)} \le |t_1-t_0|^{1/2} \|\dot{h}_n\|_{L^2(0,T;L^2(\nu_0))}.\] Let \(\theta=2s'/s\in(0,1)\). Since \(\nu_0=|\Omega|^{-1}\mathrm{d}x\), the \(L^2(\nu_0)\) and Lebesgue \(L^2\) norms are equivalent, and interpolation gives \[\|f\|_{H^{s'}(\Omega)} \le C\|f\|_{L^2(\nu_0)}^{1-\theta} \|f\|_{H^{s/2}(\Omega)}^\theta.\] Applying this to \(f=h_n(t_1)-h_n(t_0)\), and using the uniform \(C([0,T];H^{s/2}(\Omega))\) bound, gives equicontinuity in \(H^{s'}(\Omega)\).

The Arzela–Ascoli theorem now gives relative compactness in \[C([0,T];H^{s'}(\Omega)),\] because pointwise relative compactness follows from \(H^{s/2}(\Omega)\hookrightarrow H^{s'}(\Omega)\) compactly, and equicontinuity was established above. The weak convergences follow from Banach–Alaoglu.

Finally, take the continuous representatives. Since \(h_n(t)\in\mathcal{X}_{\mathrm{ad}}\) for a.e.\(t\), continuity and the \(\mathcal{Z}\)-closedness of \(\mathcal{X}_{\mathrm{ad}}\) imply \(h_n(t)\in\mathcal{X}_{\mathrm{ad}}\) for all \(t\in[0,T]\). Passing to the uniform \(H^{s'}\)-limit and using closedness again gives \(h(t)\in\mathcal{X}_{\mathrm{ad}}\) for all \(t\in[0,T]\). ◻

Proof. Let \((h_n)\subset \mathcal{A}_{\mathrm{ad}}\) be a minimizing sequence. Since the first two terms in 22 are nonnegative, \[\mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h_n] \ge \frac{\mu}{2}\int_0^T \Big( \|h_n(t)\|_{L^2(\nu_0)}^2+\|\dot{h}_n(t)\|_{L^2(\nu_0)}^2 \Big)\,\mathrm{d}t + \frac{\gamma}{2}\int_0^T \|h_n(t)\|_{H^s(\Omega)}^2\,\mathrm{d}t.\] Thus \((h_n)\) is bounded in \[L^2(0,T;H^s(\Omega))\cap H^1(0,T;L^2_0(\nu_0)).\] By Proposition 29, after passing to a subsequence we obtain \[h_n\to h \quad\text{in } C([0,T];H^{s'}(\Omega)),\] \[h_n\rightharpoonup h \quad\text{weakly in } L^2(0,T;H^s(\Omega)),\] and \[\dot{h}_n\rightharpoonup \dot{h} \quad\text{weakly in } L^2(0,T;L^2_0(\nu_0))\] for some \(h\in \mathcal{A}_{\mathrm{ad}}\).

For the data term, fix \(j\in\{1,\dots,r\}\). Since \(\kappa\in L^\infty\), \[\left| \int_\Omega \kappa(x-y_j(t))\,\mathrm{d}\rho_{h_n(t)}(x) \right| \le \|\kappa\|_{L^\infty(\mathbb{R}^d)}\] for all \(n\) and a.e.\(t\). Moreover, \(h_n(t)\to h(t)\) in \(H^{s'}(\Omega)\) for each \(t\), hence \[\mathcal{G}_t^y(h_n(t))\to \mathcal{G}_t^y(h(t)) \qquad\text{for a.e.\;}t\in[0,T].\] By dominated convergence, \[\mathcal{G}_t^y(h_n(t))\to \mathcal{G}_t^y(h(t)) \quad\text{in } L^2(0,T;\mathbb{R}^r),\] so the data term is continuous. Proposition 28 gives lower semicontinuity of the transport action. The final two regularization terms are weakly lower semicontinuous by convexity. Therefore \[\mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h] \le \liminf_{n\to\infty} \mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h_n].\] Thus \(h\) is a minimizer. ◻

Proof. Since \(h\) minimizes \(\mathcal{I}_{\lambda,\mu,\gamma}^{\,y}\) and \(h^\dagger\in\mathcal{A}_{\mathrm{ad}}\), \[\mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h] \le \mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h^\dagger] = \tfrac12\|\eta\|_{L^2(0,T;\mathbb{R}^r)}^2+R[h^\dagger] \le \tfrac12\delta^2+R[h^\dagger].\] The bounds 26 and 27 follow because each term in \(\mathcal{I}_{\lambda,\mu,\gamma}^{\,y}[h]\) is nonnegative. ◻

Proof. Set \(\xi:=h-h^\dagger\). We estimate \(\|\xi\|_{L^2(0,T;L^2(\nu_0))}^2\) by applying the joint coercivity of \(Q_y[h^\dagger]\) and bounding the resulting terms using minimality and the proximity condition.

Step 1: joint coercivity. By Assumption 5 applied at the path \(h^\dagger\), \[\label{eq:stab-joint-coercivity} \kappa\|\xi\|_{L^2_tL^2_x}^2 \le \underbrace{ \int_0^T \|J_{t,h^\dagger(t)}^{y}\xi(t)\|_{\mathbb{R}^r}^2\,\mathrm{d}t }_{\mathrm{Term\;I}} + \underbrace{ \int_0^T \mathfrak g_{h^\dagger(t)}(\dot{\xi}(t),\dot{\xi}(t))\,\mathrm{d}t }_{\mathrm{Term\;II}}.\tag{30}\]

Step 2: control of Term I. Since \(h^\dagger\in \mathcal{A}_{\mathrm{ad}}^{s'}\), the image \(h^\dagger([0,T])\subset \mathcal{Z}\) is compact. Because \(U\subset\mathcal{Z}\) is an open neighborhood of \(\mathcal{X}_{\mathrm{ad}}\) and \(h^\dagger(t)\in\mathcal{X}_{\mathrm{ad}}\subset U\) for all \(t\), there exists \(r_U>0\) such that \[\bigcup_{t\in[0,T]} B_{\mathcal{Z}}(h^\dagger(t),r_U)\subset U.\] We choose \(r\) so that \[r\le \min\left\{r_0,\;r_U,\;\sqrt{\frac{\kappa}{L^2}}\right\}.\] Then for every \(t\in[0,T]\) and every \(\theta\in[0,1]\), \[h^\dagger(t)+\theta\xi(t)\in U,\] so the fundamental-theorem-of-calculus expansion and the Lipschitz estimate 23 may be applied along the segment between \(h^\dagger(t)\) and \(h(t)\).

By the fundamental theorem of calculus in Banach spaces, \[\mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t)) = J_{t,h^\dagger(t)}^{y}\xi(t) + \int_0^1 \bigl(J_{t,h^\dagger(t)+\theta\xi(t)}^{y}-J_{t,h^\dagger(t)}^{y}\bigr)\xi(t)\,\mathrm{d}\theta.\] Denote the remainder by \[r(t) := \int_0^1 \bigl(J_{t,h^\dagger(t)+\theta\xi(t)}^{y}-J_{t,h^\dagger(t)}^{y}\bigr)\xi(t)\,\mathrm{d}\theta.\] By the Lipschitz condition 23 and the proximity condition 28 , \[\|r(t)\|_{\mathbb{R}^r} \le \int_0^1 L\theta\|\xi(t)\|_{H^{s'}(\Omega)}\|\xi(t)\|_{L^2(\nu_0)}\,\mathrm{d}\theta \le \frac{Lr}{2}\|\xi(t)\|_{L^2(\nu_0)}.\] Since \[J_{t,h^\dagger(t)}^{y}\xi(t) = \mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t))-r(t),\] we obtain \[\|J_{t,h^\dagger(t)}^{y}\xi(t)\|_{\mathbb{R}^r}^2 \le 2\|\mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t))\|_{\mathbb{R}^r}^2 + 2\|r(t)\|_{\mathbb{R}^r}^2.\] For the first piece, write \[\mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t)) = \bigl(\mathcal{G}_t^y(h(t))-\mathsf{d}(t)\bigr)+\eta(t),\] so that \[\|\mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t))\|_{\mathbb{R}^r}^2 \le 2\|\mathcal{G}_t^y(h(t))-\mathsf{d}(t)\|_{\mathbb{R}^r}^2+2\|\eta(t)\|_{\mathbb{R}^r}^2.\] Integrating in time and applying Lemma 3, \[\int_0^T \|\mathcal{G}_t^y(h(t))-\mathcal{G}_t^y(h^\dagger(t))\|_{\mathbb{R}^r}^2\,\mathrm{d}t \le 4\bigl(\tfrac12\delta^2+R[h^\dagger]\bigr)+2\delta^2 \le C_1\bigl(\delta^2+R[h^\dagger]\bigr).\] For the remainder, \[\int_0^T \|r(t)\|_{\mathbb{R}^r}^2\,\mathrm{d}t \le \frac{L^2 r^2}{4}\|\xi\|_{L^2_tL^2_x}^2.\] Therefore, \[\label{eq:stab-term-I-bound} \mathrm{Term\;I} \le C_1\bigl(\delta^2+R[h^\dagger]\bigr) + \frac{L^2 r^2}{2}\|\xi\|_{L^2_tL^2_x}^2.\tag{31}\]

Step 3: control of Term II. By Proposition 23, \[\mathrm{Term\;II} = \int_0^T \mathfrak g_{h^\dagger(t)}(\dot{\xi}(t),\dot{\xi}(t))\,\mathrm{d}t \le C_{\mathrm{tr}}\|\dot{\xi}\|_{L^2_tL^2_x}^2.\] Since \(\xi=h-h^\dagger\), we have \[\dot{\xi}=\dot{h}-\dot{h}^\dagger,\] and therefore \[\|\dot{\xi}\|_{L^2_tL^2_x}^2 \le 2\|\dot{h}\|_{L^2_tL^2_x}^2 + 2\|\dot{h}^\dagger\|_{L^2_tL^2_x}^2.\] By the a priori bound 27 , \[\frac{\mu}{2}\|\dot{h}\|_{L^2_tL^2_x}^2 \le \frac{1}{2}\delta^2+R[h^\dagger],\] so \[\|\dot{h}\|_{L^2_tL^2_x}^2 \le \frac{1}{\mu}\bigl(\delta^2+2R[h^\dagger]\bigr).\] Also, by the definition of \(R[h^\dagger]\), \[R[h^\dagger] \ge \frac{\mu}{2}\|\dot{h}^\dagger\|_{L^2_tL^2_x}^2,\] hence \[\|\dot{h}^\dagger\|_{L^2_tL^2_x}^2 \le \frac{2}{\mu}R[h^\dagger].\] Combining these estimates gives \[\label{eq:stab-term-II-bound} \mathrm{Term\;II} \le \frac{C_3}{\mu}\bigl(\delta^2+R[h^\dagger]\bigr)\tag{32}\] for a constant \(C_3>0\) depending only on \(C_{\mathrm{tr}}\).

Step 4: combine and absorb. Substituting 31 and 32 into 30 gives \[\kappa\|\xi\|_{L^2_tL^2_x}^2 \le \frac{L^2 r^2}{2}\|\xi\|_{L^2_tL^2_x}^2 + C_1\bigl(\delta^2+R[h^\dagger]\bigr) + \frac{C_3}{\mu}\bigl(\delta^2+R[h^\dagger]\bigr).\] Choose \[r \le \min\left\{r_0,\;r_U,\;\sqrt{\frac{\kappa}{L^2}}\right\},\] so that \(\frac{L^2 r^2}{2}\le \frac{\kappa}{2}\). Then \[\frac{\kappa}{2}\|\xi\|_{L^2_tL^2_x}^2 \le C(1+\mu^{-1})\bigl(\delta^2+R[h^\dagger]\bigr),\] and dividing by \(\kappa/2\) gives 29 . ◻

Proof. Since \(h_n\) minimizes \(\mathcal{I}_n\) and \(h^\dagger\in\mathcal{A}_{\mathrm{ad}}\), we have \[\mathcal{I}_n[h_n] \le \mathcal{I}_n[h^\dagger] = \frac{1}{2}\|\eta_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \mathcal{R}_{\lambda,\mu,\gamma}[h^\dagger] \le \frac{1}{2}\delta_n^2+\mathcal{R}_{\lambda,\mu,\gamma}[h^\dagger].\] Since the data-misfit term and the transport term are nonnegative, this yields the uniform bound \[\sup_n \left( \|h_n\|_{L^2(0,T;H^s(\Omega))} + \|h_n\|_{H^1(0,T;L^2_0(\nu_0))} \right) <\infty.\] By Proposition 29, after passing to a subsequence, \[h_n\to h_\ast \quad\text{in } C([0,T];H^{s'}(\Omega)),\] \[h_n\rightharpoonup h_\ast \quad\text{weakly in } L^2(0,T;H^s(\Omega)),\] and \[\dot{h}_n\rightharpoonup \dot{h}_\ast \quad\text{weakly in } L^2(0,T;L^2_0(\nu_0))\] for some \(h_\ast\in\mathcal{A}_{\mathrm{ad}}\).

Because \(\mathsf{d}_n\to \mathcal{G}_\cdot^y(h^\dagger(\cdot))\) in \(L^2(0,T;\mathbb{R}^r)\), for each fixed \(h\in\mathcal{A}_{\mathrm{ad}}\) we have \[\mathcal{I}_n[h]\to \mathcal{I}^0[h].\] Also, by dominated convergence, \[\mathcal{G}_\cdot^y(h_n(\cdot))\to \mathcal{G}_\cdot^y(h_\ast(\cdot)) \quad\text{in } L^2(0,T;\mathbb{R}^r).\] Together with Proposition 28 and weak lower semicontinuity of the \(\mu\)- and \(\gamma\)-terms, this gives \[\mathcal{I}^0[h_\ast] \le \liminf_{n\to\infty}\mathcal{I}_n[h_n].\]

Now let \(h\in\mathcal{A}_{\mathrm{ad}}\) be arbitrary. Since \(h_n\) minimizes \(\mathcal{I}_n\), \[\mathcal{I}_n[h_n]\le \mathcal{I}_n[h].\] Passing to the limit superior on the right and combining with the previous liminf inequality yields \[\mathcal{I}^0[h_\ast]\le \mathcal{I}^0[h].\] Thus \(h_\ast\) is a minimizer of \(\mathcal{I}^0\).

If \(\mathcal{I}^0\) has a unique minimizer \(h^0_{\lambda,\mu,\gamma}\), then every convergent subsequence of \((h_n)\) has the same limit, so the whole sequence converges to \(h^0_{\lambda,\mu,\gamma}\) in \(C([0,T];H^{s'}(\Omega))\).

For the final claim, assume \[\sup_{t\in[0,T]} \|h^0_{\lambda,\mu,\gamma}(t)-h^\dagger(t)\|_{H^{s'}(\Omega)}<r.\] Since \(h_n\to h^0_{\lambda,\mu,\gamma}\) in \(C([0,T];H^{s'}(\Omega))\), it follows that for all sufficiently large \(n\), \[\sup_{t\in[0,T]} \|h_n(t)-h^\dagger(t)\|_{H^{s'}(\Omega)}<r.\] Hence the proximity condition 28 holds for \(h_n\), and Theorem 32 gives \[\|h_n-h^\dagger\|_{L^2(0,T;L^2(\nu_0))}^2 \le \frac{C}{\kappa}(1+\mu^{-1}) \bigl(\delta_n^2+R[h^\dagger]\bigr)\] for all sufficiently large \(n\). ◻

Proof. Let \(\tilde{h}\in \mathcal{S}(\mathsf{d}^\dagger)\) be arbitrary. Since \(h_n\) minimizes \(\mathcal{I}_n\), \[\frac{1}{2} \|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[h_n] \le \frac{1}{2} \|\mathcal{G}_\cdot^y(\tilde{h}(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[\tilde{h}].\] Because \(\mathcal{G}_\cdot^y(\tilde{h}(\cdot))=\mathsf{d}^\dagger\), this gives \[\frac{1}{2} \|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[h_n] \le \frac{1}{2} \|\mathsf{d}^\dagger-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[\tilde{h}] \le \frac{1}{2} \delta_n^2 + \alpha_n \mathcal{R}[\tilde{h}].\] Hence \[\label{eq:vanishing-regularization-bound} \mathcal{R}[h_n] \le \mathcal{R}[\tilde{h}] + \frac{\delta_n^2}{2\alpha_n}.\tag{35}\] Since \(\delta_n^2/\alpha_n \to 0\), the sequence \((\mathcal{R}[h_n])\) is bounded.

Because \(\bar\mu,\bar\gamma>0\), boundedness of \(\mathcal{R}[h_n]\) implies boundedness of \[\|h_n\|_{L^2(0,T;H^s(\Omega))} \qquad\text{and}\qquad \|h_n\|_{H^1(0,T;L^2_0(\nu_0))}.\] Therefore Proposition 29 applies, and after passing to a subsequence, \[h_n \rightharpoonup h_\ast \quad\text{weakly in } L^2(0,T;H^s(\Omega)),\] \[h_n \rightharpoonup h_\ast \quad\text{weakly in } H^1(0,T;L^2_0(\nu_0)),\] and \[h_n \to h_\ast \quad\text{in } C([0,T];H^{s'}(\Omega))\] for some \(h_\ast\in \mathcal{A}_{\mathrm{ad}}\).

Next, from the minimizing inequality above we also have \[\frac{1}{2} \|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 \le \frac{1}{2} \delta_n^2 + \alpha_n \mathcal{R}[\tilde{h}].\] Since \(\delta_n\to 0\) and \(\alpha_n\to 0\), it follows that \[\|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)} \to 0.\] Therefore \[\|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}^\dagger\|_{L^2(0,T;\mathbb{R}^r)} \le \|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)} + \|\mathsf{d}_n-\mathsf{d}^\dagger\|_{L^2(0,T;\mathbb{R}^r)} \to 0.\] Since \(h_n\to h_\ast\) in \(C([0,T];H^{s'}(\Omega))\), dominated convergence yields \[\mathcal{G}_\cdot^y(h_\ast(\cdot))=\mathsf{d}^\dagger.\] Thus \(h_\ast\in \mathcal{S}(\mathsf{d}^\dagger)\).

It remains to show that \(h_\ast\) minimizes \(\mathcal{R}\) over \(\mathcal{S}(\mathsf{d}^\dagger)\). Let \(z\in \mathcal{S}(\mathsf{d}^\dagger)\) be arbitrary. By minimality of \(h_n\), \[\frac{1}{2} \|\mathcal{G}_\cdot^y(h_n(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[h_n] \le \frac{1}{2} \|\mathcal{G}_\cdot^y(z(\cdot))-\mathsf{d}_n\|_{L^2(0,T;\mathbb{R}^r)}^2 + \alpha_n \mathcal{R}[z].\] Using \(\mathcal{G}_\cdot^y(z(\cdot))=\mathsf{d}^\dagger\), we obtain \[\alpha_n \mathcal{R}[h_n] \le \frac{1}{2} \delta_n^2 + \alpha_n \mathcal{R}[z].\] Dividing by \(\alpha_n\) yields \[\mathcal{R}[h_n] \le \mathcal{R}[z] + \frac{\delta_n^2}{2\alpha_n}.\] Taking the limit superior and using \(\delta_n^2/\alpha_n\to 0\), we obtain \[\limsup_{n\to\infty}\mathcal{R}[h_n]\le \mathcal{R}[z].\]

On the other hand, Proposition 28 gives lower semicontinuity of the transport part of \(\mathcal{R}\), while the remaining two terms are weakly lower semicontinuous. Hence \[\mathcal{R}[h_\ast] \le \liminf_{n\to\infty}\mathcal{R}[h_n].\] Combining the last two inequalities gives \[\mathcal{R}[h_\ast]\le \mathcal{R}[z].\] Since \(z\in \mathcal{S}(\mathsf{d}^\dagger)\) was arbitrary, \(h_\ast\) minimizes \(\mathcal{R}\) over \(\mathcal{S}(\mathsf{d}^\dagger)\).

Finally, if the \(\mathcal{R}\)-minimizing exact solution is unique, then every convergent subsequence of \((h_n)\) has the same limit \(h_\ast\). Therefore the whole sequence converges to \(h_\ast\) in \(C([0,T];H^{s'}(\Omega))\). ◻

Proof. If \(Q_y[h](\xi,\xi)=0\), then both terms in the definition of \(Q_y[h]\) vanish: \[\int_0^T \|J_{t,h(t)}^{y}\xi(t)\|_{\mathbb{R}^r}^2\,\mathrm{d}t=0, \qquad \int_0^T \mathfrak g_{h(t)}(\dot{\xi}(t),\dot{\xi}(t))\,\mathrm{d}t=0.\] The first identity gives \(\xi(t)\in\ker J_{t,h(t)}^{y}\) for a.e.\(t\). By Proposition 12, the second identity gives \(\dot{\xi}(t)=0\) for a.e.\(t\). Hence \(\xi\) is time-constant. The converse is immediate. ◻

Proof. Define \[\mathcal{K}_y[h]:L^2_0(\nu_0)\to L^2(0,T;\mathbb{R}^r)\] by \[(\mathcal{K}_y[h]\xi_0)(t) := J_{t,h(t)}^{y}\xi_0.\] Then \[\bar{\mathfrak j}_y[h](\xi_0,\eta_0) = \langle \mathcal{K}_y[h]\xi_0,\mathcal{K}_y[h]\eta_0\rangle_{L^2(0,T;\mathbb{R}^r)},\] so \(\bar J_y[h]=\mathcal{K}_y[h]^\ast \mathcal{K}_y[h]\).

It remains to show that \(\mathcal{K}_y[h]\) is compact. For \(j=1,\dots,r\), write \[(\mathcal{K}_y[h]\xi_0)_j(t) = \int_\Omega b_j(t,x)\,\xi_0(x)\,\mathrm{d}\nu_0(x),\] where \[b_j(t,x) := \Bigl( \kappa(x-y_j(t)) - \mathbb{E}_{\rho_{h(t)}}[\kappa(\cdot-y_j(t))] \Bigr) \,w_{h(t)}(x), \qquad w_{h(t)}:=\frac{\mathrm{d}\rho_{h(t)}}{\mathrm{d}\nu_0}.\] Since \(\kappa\in L^\infty\) and the admissibility bounds give \[0<c\le w_{h(t)}(x)\le C<\infty \qquad\text{for a.e.\;}(t,x)\in[0,T]\times\Omega,\] we have \[|b_j(t,x)|\le 2C\|\kappa\|_{L^\infty(\mathbb{R}^d)},\] hence \(b_j\in L^2((0,T)\times\Omega)\). Therefore each component \[\xi_0\longmapsto (\mathcal{K}_y[h]\xi_0)_j\] is a Hilbert–Schmidt operator from \(L^2_0(\nu_0)\) to \(L^2(0,T)\), and so \(\mathcal{K}_y[h]\) itself is Hilbert–Schmidt, hence compact.

Thus \(\bar J_y[h]=\mathcal{K}_y[h]^\ast\mathcal{K}_y[h]\) is compact, self-adjoint, and nonnegative.

For the coercivity claim, suppose by contradiction that there exists \(\kappa_{\mathrm{obs}}>0\) such that \[Q_y[h](\xi,\xi) \ge \kappa_{\mathrm{obs}} \|\xi\|_{L^2(0,T;L^2(\nu_0))}^2\] for every admissible perturbation \(\xi\). Let \(\xi_0\in H^{s'}(\Omega)\cap L^2_0(\nu_0)\), and consider the time-constant path \(\xi(t)\equiv\xi_0\). Then \(\dot{\xi}=0\), so \[Q_y[h](\xi,\xi) = \bar{\mathfrak j}_y[h](\xi_0,\xi_0), \qquad \|\xi\|_{L^2(0,T;L^2(\nu_0))}^2 = T\|\xi_0\|_{L^2(\nu_0)}^2.\] Hence \[\bar{\mathfrak j}_y[h](\xi_0,\xi_0) \ge \kappa_{\mathrm{obs}}T\|\xi_0\|_{L^2(\nu_0)}^2 \qquad \forall \xi_0\in H^{s'}(\Omega)\cap L^2_0(\nu_0).\] Because \(H^{s'}(\Omega)\cap L^2_0(\nu_0)\) is dense in \(L^2_0(\nu_0)\) and \(\bar{\mathfrak j}_y[h]\) is continuous on \(L^2_0(\nu_0)\), this extends to all \(\xi_0\in L^2_0(\nu_0)\). Thus the compact operator \(\bar J_y[h]\) would be bounded below by a positive multiple of the identity on an infinite-dimensional Hilbert space, which is impossible. ◻

Proof. This is the spectral theorem for compact self-adjoint operators applied to \(\bar J_y[h]\). ◻

Proof. By bilinearity of \(\mathfrak g_{h(a)}\), \[\mathfrak g_{h(a)}(\alpha_h,\beta_h) = \sum_{k,\ell=1}^m \alpha_k\beta_\ell\,\mathfrak g_{h(a)}(\phi_k,\phi_\ell) = \sum_{k,\ell=1}^m \alpha_k\beta_\ell\,H_{k\ell}(a) = \alpha^\top H(a)\beta.\] Taking \[\dot{h}(t)=\sum_{k=1}^m \dot{a}_k(t)\phi_k\] gives ?? . ◻

Proof. By Proposition 17, \[J_{t,h(a)}^y\xi = \Bigl( \operatorname{Cov}_{\rho_{h(a)}}(\kappa(\cdot-y_1(t)),\xi),\dots, \operatorname{Cov}_{\rho_{h(a)}}(\kappa(\cdot-y_r(t)),\xi) \Bigr).\] Substituting \(\xi=\sum_{k=1}^m \alpha_k\phi_k\) and using bilinearity of covariance gives \[\operatorname{Cov}_{\rho_{h(a)}}(\kappa(\cdot-y_j(t)),\xi) = \sum_{k=1}^m \alpha_k \operatorname{Cov}_{\rho_{h(a)}}(\kappa(\cdot-y_j(t)),\phi_k) = \sum_{k=1}^m [M_y(t,a)]_{jk}\alpha_k,\] which proves ?? . The identity ?? follows from Definition 10. ◻

Proof. By Proposition 44, \[\mathfrak g_{h(a(t))}\bigl(\dot{\xi}_\alpha(t),\dot{\xi}_\alpha(t)\bigr) = \dot{\alpha}(t)^\top H(a(t))\dot{\alpha}(t).\] By Proposition 45, \[\mathfrak j_{t,h(a(t))}^{y}\bigl(\xi_\alpha(t),\xi_\alpha(t)\bigr) = \alpha(t)^\top M_y(t,a(t))^\top M_y(t,a(t))\alpha(t).\] Substituting these identities into Definition 12 gives ?? .

If ?? holds, then \[Q_y[h(a)](\xi_\alpha,\xi_\alpha) \ge \kappa_m \int_0^T |\alpha(t)|^2\,\mathrm{d}t.\] On the other hand, ?? gives \[\|\xi_\alpha(t)\|_{L^2(\nu_0)}^2 = \left\|\sum_{k=1}^m \alpha_k(t)\phi_k\right\|_{L^2(\nu_0)}^2 \le C_m |\alpha(t)|^2\] for a.e.\(t\in[0,T]\). Integrating in time yields \[\|\xi_\alpha\|_{L^2(0,T;L^2(\nu_0))}^2 \le C_m \int_0^T |\alpha(t)|^2\,\mathrm{d}t.\] Therefore \[Q_y[h(a)](\xi_\alpha,\xi_\alpha) \ge \frac{\kappa_m}{C_m} \|\xi_\alpha\|_{L^2(0,T;L^2(\nu_0))}^2,\] which is exactly Assumption 5 on the reduced class. ◻

Proof. Substitute \(h(t)=h(a(t))\) into Definition 13. The data term becomes \[\|\mathcal{G}_t^y(h(a(t)))-d(t)\|_{\mathbb{R}^r}^2 = \|\mathcal{G}_{m,t}^{y}(a(t))-d(t)\|_{\mathbb{R}^r}^2,\] and the transport term is identified by Proposition 44. Since \[h(a(t))=\sum_{k=1}^m a_k(t)\phi_k, \qquad \dot{h}(t)=\sum_{k=1}^m \dot{a}_k(t)\phi_k,\] the remaining terms are simply the pullbacks of ambient norms to the finite-dimensional space \(V_m\), hence are equivalent to Euclidean quadratic forms in \(a(t)\) and \(\dot{a}(t)\). ◻

Proof. Since \(\int \kappa_\delta(\cdot-y)\,\mathrm{d}x=1\), one has \[\operatorname{Cov}_{\rho_h}\!\bigl(\kappa_\delta(\cdot-y),\xi\bigr) = \int_\Omega \kappa_\delta(x-y)\,\xi(x)\,\mathrm{d}\rho_h(x) - \Bigl(\int_\Omega \kappa_\delta(x-y)\,\mathrm{d}\rho_h(x)\Bigr)\,m_h(\xi).\] Using \(\mathrm{d}\rho_h(x)=r_h(x)\,\mathrm{d}x\), this becomes \[\operatorname{Cov}_{\rho_h}\!\bigl(\kappa_\delta(\cdot-y),\xi\bigr) = \int_\Omega \kappa_\delta(x-y)\,r_h(x)\bigl(\xi(x)-m_h(\xi)\bigr)\,\mathrm{d}x = \int_{\mathbb{R}^d}\kappa_\delta(y-x)\,g_{h,\xi}(x)\,\mathrm{d}x,\] which is exactly ?? .

Because \(y\) is a Lebesgue point of \(g_{h,\xi}\), the approximate-identity theorem implies \[(\kappa_\delta*g_{h,\xi})(y)\longrightarrow g_{h,\xi}(y) \qquad\text{as }\delta\downarrow 0,\] which proves ?? . If \(g_{h,\xi}(y)\neq 0\), then for all sufficiently small \(\delta\) the quantity \((\kappa_\delta*g_{h,\xi})(y)\) has the same sign as \(g_{h,\xi}(y)\), and in particular is nonzero. ◻

Proof. Since \(r_h\in L^\infty(\Omega)\) and \(\xi\in L^2(\nu_0)\), one has \(g_{h,\xi}\in L^1(\Omega)\). If \(g_{h,\xi}=0\) almost everywhere, then \[r_h(x)\bigl(\xi(x)-m_h(\xi)\bigr)=0 \qquad\text{for a.e. }x.\] Because \(r_h>0\) a.e., it follows that \(\xi=m_h(\xi)\) almost everywhere. Since \(\xi\in L^2_0(\nu_0)\), this forces \(\xi=0\) almost everywhere, a contradiction.

Hence \(g_{h,\xi}\) is not zero a.e. Its set of nonzero points has positive Lebesgue measure. By the Lebesgue differentiation theorem, almost every point of that set is a Lebesgue point of \(g_{h,\xi}\). Choose such a point \(y\in\Omega\) with \(g_{h,\xi}(y)\neq 0\) and \(\mathop{\mathrm{dist}}(y,\partial\Omega)>0\). Proposition 48 now yields the claim. ◻

Proof. Because \(\xi\) is continuous and nonzero with \(\xi\in L^2_0(\nu_0)\), it cannot be constant on \(\Omega\). Hence \(\xi-\mathbb{E}_{\rho_h}[\xi]\) is not identically zero, so there exists \(y\in\Omega\) such that \[\xi(y)\neq \mathbb{E}_{\rho_h}[\xi].\] For such \(y\), \[g_{h,\xi}(y)=r_h(y)\bigl(\xi(y)-\mathbb{E}_{\rho_h}[\xi]\bigr)\neq 0.\] Since both \(r_h\) and \(\xi\) are continuous, \(g_{h,\xi}\) is continuous, hence every point is a Lebesgue point of \(g_{h,\xi}\). The conclusion follows from Proposition 48. ◻

Proof. For \(y\in K\), define the evaluation vector \[v(y):=\bigl(f_0(y),\dots,f_m(y)\bigr)\in\mathbb{R}^{m+1}.\] If the span of \(\{v(y):y\in K\}\) had dimension strictly smaller than \(m+1\), then there would exist \(c=(c_0,\dots,c_m)\neq 0\) such that \[c\cdot v(y)=\sum_{\ell=0}^m c_\ell f_\ell(y)=0 \qquad\forall y\in K,\] contradicting the linear independence of \(f_0,\dots,f_m\). Hence \(\{v(y):y\in K\}\) spans \(\mathbb{R}^{m+1}\), so one can choose \(y_0,\dots,y_m\in K\) such that \(v(y_0),\dots,v(y_m)\) form a basis of \(\mathbb{R}^{m+1}\). Equivalently, the matrix \(A_Y\) is invertible. ◻

Proof. Apply Lemma 4 to the functions \[f_0:=1,\qquad f_k:=\phi_k,\quad k=1,\dots,m,\] on the compact space \(K_{\mathrm{sens}}\). This yields points \(y_0,\dots,y_m\in K_{\mathrm{sens}}\) such that the augmented evaluation matrix \[A_Y := \begin{pmatrix} 1 & \phi_1(y_0) & \cdots & \phi_m(y_0) \\ \vdots & \vdots & & \vdots \\ 1 & \phi_1(y_m) & \cdots & \phi_m(y_m) \end{pmatrix} \in\mathbb{R}^{(m+1)\times(m+1)}\] is invertible.

For \(a\in U\), define \[c(a):= \bigl(c_1(a),\dots,c_m(a)\bigr)^\top, \qquad c_k(a):=\mathbb{E}_{\rho_{h(a)}}[\phi_k].\] Also define \[X_Y:=\bigl(\phi_k(y_j)\bigr)_{j=0,\dots,m}^{k=1,\dots,m}\in\mathbb{R}^{(m+1)\times m}, \qquad \mathbf{1}:=(1,\dots,1)^\top\in\mathbb{R}^{m+1},\] and \[D_Y(a):=\operatorname{diag}(r_a(y_0),\dots,r_a(y_m)).\] Consider the ideal point-sensor matrix \[\widetilde{M}_Y(a):=D_Y(a)\bigl(X_Y-\mathbf{1}\,c(a)^\top\bigr).\]

We first show that \(\widetilde{M}_Y(a)\) has full column rank \(m\) for every \(a\in U\). Suppose \[\widetilde{M}_Y(a)\alpha=0 \qquad\text{for some }\alpha\in\mathbb{R}^m.\] Since \(D_Y(a)\) is invertible by the uniform positivity of \(r_a\) on \(K_{\mathrm{sens}}\), this implies \[\bigl(X_Y-\mathbf{1}\,c(a)^\top\bigr)\alpha=0,\] hence \[X_Y\alpha=(c(a)^\top\alpha)\mathbf{1}.\] Therefore \[A_Y \binom{-c(a)^\top\alpha}{\alpha} = 0.\] Because \(A_Y\) is invertible, we conclude \(\alpha=0\). Thus \(\operatorname{rank}\widetilde{M}_Y(a)=m\).

Next, the map \(a\mapsto \widetilde{M}_Y(a)\) is continuous on \(U\): the values \(r_a(y_j)\) are continuous in \(a\), and the averages \(c_k(a)=\int \phi_k\,\mathrm{d}\rho_{h(a)}\) are continuous by dominated convergence. Hence the smallest singular value of \(\widetilde{M}_Y(a)\) is a positive continuous function on the compact set \(U\), so there exists \(\sigma_*>0\) such that \[\widetilde{M}_Y(a)^\top \widetilde{M}_Y(a)\succeq 2\sigma_* I_m \qquad \forall a\in U.\]

It remains to compare the actual bump-sensor matrix \(M_Y^\delta(a)\) with \(\widetilde{M}_Y(a)\). Because \(K_{\mathrm{sens}}\subset\Omega\) is compact, there exists \[d_*:=\mathop{\mathrm{dist}}(K_{\mathrm{sens}},\partial\Omega)>0.\] For \(0<\delta<d_*\), the support of \(\kappa_\delta(\cdot-y_j)\) stays inside \(\Omega\) for every \(j\). For each \(j,k\), \[[M_Y^\delta(a)]_{jk} = \int_\Omega \kappa_\delta(x-y_j)\,r_a(x)\bigl(\phi_k(x)-c_k(a)\bigr)\,\mathrm{d}x.\] Define \[g_{a,k}(x):=r_a(x)\bigl(\phi_k(x)-c_k(a)\bigr).\] Since \((a,x)\mapsto r_a(x)\) is continuous on \(U\times\overline{\Omega}\), each \(\phi_k\) is continuous on \(K_{\mathrm{sens}}\), and \(a\mapsto c_k(a)\) is continuous, the family \(\{g_{a,k}:a\in U\}\) is uniformly continuous in a neighborhood of \(K_{\mathrm{sens}}\). Hence the approximate-identity property yields \[\sup_{a\in U}\bigl|[M_Y^\delta(a)]_{jk}-g_{a,k}(y_j)\bigr|\longrightarrow 0 \qquad\text{as }\delta\downarrow 0.\] But \(g_{a,k}(y_j)=[\widetilde{M}_Y(a)]_{jk}\), so \[\sup_{a\in U}\|M_Y^\delta(a)-\widetilde{M}_Y(a)\|\longrightarrow 0 \qquad\text{as }\delta\downarrow 0.\] Choosing \(\delta_0>0\) sufficiently small, we may arrange \[M_Y^\delta(a)^\top M_Y^\delta(a)\succeq \sigma_* I_m \qquad \forall a\in U,\;\forall 0<\delta<\delta_0.\] This proves both full column rank and 41 . ◻

Proof. Let \[B_m:=\bigl(\langle \phi_k,\phi_\ell\rangle_{L^2(\nu_0)}\bigr)_{k,\ell=1}^m \in \mathbb{R}^{m\times m}.\] Since \(\{\phi_1,\dots,\phi_m\}\) is a basis of \(V_m\), \(B_m\) is symmetric positive definite.

Write \[\xi_0=\sum_{k=1}^m c_k\phi_k, \qquad c=(c_1,\dots,c_m)^\top\in\mathbb{R}^m.\] Then \[\label{eq:xi0-norm-gram} \|\xi_0\|_{L^2(\nu_0)}^2 = c^\top B_m c.\tag{42}\] By Proposition 45, \[J_{t,h(a(t))}^{y}\xi_0 = M_y(t,a(t))\,c,\] and therefore \[\begin{align} \int_0^T \|J_{t,h(a(t))}^{y}\xi_0\|_{\mathbb{R}^r}^2\,\mathrm{d}t & = \int_0^T c^\top M_y(t,a(t))^\top M_y(t,a(t))\,c\,\mathrm{d}t \notag \\ & = c^\top \overline{G}_y[a]\,c. \label{eq:obs-energy-gram} \end{align}\tag{43}\]

Assume \((1)\). Since both \(B_m\) and \(\overline{G}_y[a]\) are symmetric positive definite, the matrix \[A_m:=B_m^{-1/2}\,\overline{G}_y[a]\,B_m^{-1/2}\] is symmetric positive definite. Hence \[\lambda_*:=\lambda_{\min}(A_m)>0.\] Using 42 and 43 , \[c^\top \overline{G}_y[a]\,c = (B_m^{1/2}c)^\top A_m (B_m^{1/2}c) \ge \lambda_*\,|B_m^{1/2}c|^2 = \lambda_*\,\|\xi_0\|_{L^2(\nu_0)}^2.\] Thus \((2)\) holds.

Conversely, if \((2)\) holds and \(\overline{G}_y[a]\) were not positive definite, there would exist \(c\neq 0\) such that \[c^\top \overline{G}_y[a]\,c=0.\] Then the associated \(\xi_0=\sum_{k=1}^m c_k\phi_k\) is nonzero, while 43 contradicts \((2)\). Hence \((1)\) follows.

Finally, a time-constant perturbation \(\xi(t)\equiv \xi_0\in V_m\) has \(\dot{\xi}=0\), so its transport contribution vanishes. Therefore \[Q_y[h(a)](\xi_0,\xi_0) = \int_0^T \|J_{t,h(a(t))}^{y}\xi_0\|_{\mathbb{R}^r}^2\,\mathrm{d}t.\] Thus \[\ker Q_y[h(a)]\cap V_m = \left\{ \xi_0\in V_m: \int_0^T \|J_{t,h(a(t))}^{y}\xi_0\|_{\mathbb{R}^r}^2\,\mathrm{d}t=0 \right\},\] which shows \((2)\Longleftrightarrow(3)\). ◻

Proof. Choose once and for all a kernel width \(0<\delta<\delta_0\), where \(\delta_0\) is the radius given by Theorem 49. We suppress the dependence on \(\delta\) in the notation.

By Theorem 49, there exist points \[z_0,\dots,z_m\in K_{\mathrm{sens}}\] and a constant \(\sigma_*>0\) such that for every \(b\in U\), \[\label{eq:static-uniform-coercivity-proof} \sum_{j=0}^m M_{z_j}(b)^\top M_{z_j}(b) \succeq \sigma_* I_m,\tag{44}\] where \(M_{z_j}(b)\in \mathbb{R}^{1\times m}\) denotes the reduced sensing row for a single static sensor at the location \(z_j\).

For each \(j=0,\dots,m-1\), choose an \(H^1\)-curve in \(K_{\mathrm{sens}}\) connecting \(z_j\) to \(z_{j+1}\), and choose an \(H^1\)-curve connecting \(z_m\) back to \(z_0\). Because the set of curves is finite, we may regard these as fixed transition templates.

For the fixed path \(a\in \mathcal{A}_m(U)\), define \[G_j(t):=M_{z_j}(a(t))^\top M_{z_j}(a(t)) \in \mathbb{R}^{m\times m}, \qquad G_\Sigma(t):=\sum_{j=0}^m G_j(t).\] Since \(a\in H^1(0,T;\mathbb{R}^m)\subset C([0,T];\mathbb{R}^m)\) and \(b\mapsto M_{z_j}(b)\) is continuous on \(U\), each \(G_j\) is continuous on \([0,T]\), hence uniformly continuous. Also, by 44 , \[\label{eq:Gsigma-lower-bound} G_\Sigma(t)\succeq \sigma_* I_m \qquad\forall t\in[0,T].\tag{45}\]

Let \(q:=m+1\), and choose \(\eta>0\) so small that \[q\eta \le \frac{\sigma_*}{2}.\] By uniform continuity, there exists a partition \[0=t_0<t_1<\cdots<t_N=T\] such that on each interval \(I_n:=[t_{n-1},t_n]\), \[\label{eq:Gj-small-variation} \|G_j(t)-G_j(s)\|\le \eta \qquad \forall s,t\in I_n,\;\forall j=0,\dots,m.\tag{46}\]

We now construct \(y[a]\) interval by interval. On each \(I_n\), first let the sensor dwell at \(z_0,\dots,z_m\) in that order, each for a time interval of length \[\frac{|I_n|}{2q}.\] Use the remaining half of \(I_n\) to traverse the chosen transition curves between successive locations, scaled in time so as to fit. This gives a piecewise \(H^1\) path on each \(I_n\), and by construction the endpoint of the last transition on \(I_n\) matches the initial position of the first dwell on \(I_{n+1}\). Hence the resulting global path \(y[a]\) belongs to \(\mathcal{Y}_{\mathrm{ad}}^{(1)}\).

Let \(D_{n,j}\subset I_n\) denote the dwell interval on which \(y[a](t)=z_j\). Since the transition contributions are positive semidefinite, we have \[\overline{G}_{y[a]}[a] \succeq \sum_{n=1}^N \sum_{j=0}^m \int_{D_{n,j}} G_j(t)\,\mathrm{d}t.\] Fix \(n\in\{1,\dots,N\}\) and choose any \(s_n\in I_n\). By 46 , \[G_j(t)\succeq G_j(s_n)-\eta I_m \qquad \forall t\in I_n,\;\forall j=0,\dots,m.\] Therefore \[\begin{align} \sum_{j=0}^m \int_{D_{n,j}} G_j(t)\,\mathrm{d}t & \succeq \sum_{j=0}^m \frac{|I_n|}{2q}\bigl(G_j(s_n)-\eta I_m\bigr) \\ & = \frac{|I_n|}{2q} \left( G_\Sigma(s_n)-q\eta I_m \right). \end{align}\] Using 45 and the choice of \(\eta\), \[G_\Sigma(s_n)-q\eta I_m \succeq \frac{\sigma_*}{2} I_m.\] Hence \[\sum_{j=0}^m \int_{D_{n,j}} G_j(t)\,\mathrm{d}t \succeq \frac{|I_n|\,\sigma_*}{4q} I_m.\] Summing over \(n\) gives \[\overline{G}_{y[a]}[a] \succeq \sum_{n=1}^N \frac{|I_n|\,\sigma_*}{4q} I_m = \frac{T\,\sigma_*}{4(m+1)} I_m.\] Thus \(\overline{G}_{y[a]}[a]\) is positive definite. ◻

Proof. Suppose the conclusion fails. Then there exists a sequence \[\alpha_n\in H^1(0,T;\mathbb{R}^m), \qquad \|\alpha_n\|_{L^2(0,T;\mathbb{R}^m)}=1,\] such that \[\label{eq:fixed-path-coercivity-contradiction} \int_0^T \alpha_n^\top M_{y^\dagger}(t,a^\dagger(t))^\top M_{y^\dagger}(t,a^\dagger(t)) \alpha_n\,\mathrm{d}t + \int_0^T \dot{\alpha}_n^\top H(a^\dagger(t))\,\dot{\alpha}_n\,\mathrm{d}t \longrightarrow 0.\tag{50}\]

By 47 , \[c_{\mathrm{tr}}\int_0^T |\dot{\alpha}_n(t)|^2\,\mathrm{d}t \le \int_0^T \dot{\alpha}_n(t)^\top H(a^\dagger(t))\,\dot{\alpha}_n(t)\,\mathrm{d}t \longrightarrow 0,\] so \[\|\dot{\alpha}_n\|_{L^2(0,T;\mathbb{R}^m)}\to 0.\] Let \[\bar\alpha_n:=\frac{1}{T}\int_0^T \alpha_n(t)\,\mathrm{d}t.\] By the Poincaré–Wirtinger inequality, \[\|\alpha_n-\bar\alpha_n\|_{L^2(0,T;\mathbb{R}^m)} \le C_P \|\dot{\alpha}_n\|_{L^2(0,T;\mathbb{R}^m)} \to 0.\] Since \((\bar\alpha_n)\) is bounded in \(\mathbb{R}^m\), after passing to a subsequence we may assume \[\bar\alpha_n\to c\in\mathbb{R}^m.\] Therefore \[\alpha_n\to c \qquad\text{strongly in }L^2(0,T;\mathbb{R}^m),\] where \(c\) is identified with the corresponding time-constant function on \([0,T]\). Since \(\|\alpha_n\|_{L^2}=1\), we have \[\|c\|_{L^2(0,T;\mathbb{R}^m)}=1,\] so \(c\neq 0\).

Set \[M^\dagger(t):=M_{y^\dagger}(t,a^\dagger(t)).\] Because \(M^\dagger\in L^\infty(0,T;\mathbb{R}^{1\times m})\), \[M^\dagger \alpha_n \to M^\dagger c \qquad\text{strongly in }L^2(0,T;\mathbb{R}).\] Passing to the limit in the first term of 50 gives \[0 = \int_0^T |M^\dagger(t)c|^2\,\mathrm{d}t = c^\top \overline{G}_{y^\dagger}[a^\dagger]\,c.\] This contradicts 48 , since \(\overline{G}_{y^\dagger}[a^\dagger]\succ 0\) and \(c\neq 0\). Hence 49 holds. ◻

Proof. The idealized Gramian computed above is \(\frac{9T}{128}I_2\), hence positive definite. By continuity of the reduced sensing map with respect to the state, sensor location, and kernel width, the actual averaged Gramian is a small perturbation of this ideal matrix whenever \(|\bar a|\), \(\varepsilon\), \(\delta\), and \(\tau\) are sufficiently small. Positive definiteness is stable under small perturbations, so \(\overline{G}_y[a^\dagger]\succ 0\). The final claim then follows from Theorem 55. ◻

Proof. Let \((a_n)\subset \mathcal{A}_m(U)\) be a minimizing sequence. Since \(\mu>0\), the definition of \(\mathcal{J}_{\lambda,\mu}^{(m),y}\) and Remark 58 imply that \((a_n)\) is bounded in \(H^1(0,T;\mathbb{R}^m)\). Passing to a subsequence if necessary, \[a_n \rightharpoonup a \quad\text{weakly in }H^1(0,T;\mathbb{R}^m), \qquad a_n\to a \quad\text{strongly in }C([0,T];\mathbb{R}^m).\] Since \(U\) is compact, \(a(t)\in U\) for all \(t\in[0,T]\), so \(a\in \mathcal{A}_m(U)\).

By continuity of the reduced forward map under strong \(C([0,T];\mathbb{R}^m)\)-convergence, \[\mathcal{G}_m^y[a_n]\to \mathcal{G}_m^y[a] \qquad\text{in }\mathcal{Y}_T.\] The transport term in \(\mathcal{R}_{\lambda,\mu}^{(m)}\) is weakly lower semicontinuous because \(H(\cdot)\) is continuous on the compact set \(U\), hence bounded and uniformly positive semidefinite there. The remaining two regularization terms are convex quadratic forms in \(a\) and \(\dot{a}\), and therefore also weakly lower semicontinuous. Consequently, \[\mathcal{J}_{\lambda,\mu}^{(m),y}[a;d] \le \liminf_{n\to\infty}\mathcal{J}_{\lambda,\mu}^{(m),y}[a_n;d],\] so \(a\) is a minimizer. ◻

Proof. By the minimizing property of \(\hat{a}_m^\delta\), \[\frac{1}{2}\|\mathcal{G}_m^{y^\dagger}[\hat{a}_m^\delta]-d^\delta\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[\hat{a}_m^\delta] \le \frac{1}{2}\|\mathcal{G}_m^{y^\dagger}[a^\dagger]-d^\delta\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger].\] Hence \[\label{eq:reduced-minimizer-basic-bound} \mathcal{R}_{\lambda,\mu}^{(m)}[\hat{a}_m^\delta] \le \frac{1}{2}\|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger]\tag{55}\] and \[\label{eq:reduced-data-residual-basic-bound} \|\mathcal{G}_m^{y^\dagger}[\hat{a}_m^\delta]-d^\delta\|_{\mathcal{Y}_T}^2 \le \|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + 2\,\mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger].\tag{56}\]

Now \[\begin{align} \|\mathcal{G}_m^{y^\dagger}[\hat{a}_m^\delta]-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 & \le 2\|\mathcal{G}_m^{y^\dagger}[\hat{a}_m^\delta]-d^\delta\|_{\mathcal{Y}_T}^2 + 2\|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 \\ & \le 4\|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + 4\,\mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger], \end{align}\] where in the last step we used 56 .

Applying the global reduced stability estimate 53 with \(a=\hat{a}_m^\delta\), and then using the last inequality together with 55 , we obtain \[\begin{align} \|h(\hat{a}_m^\delta)-h(a^\dagger)\|_{L^2(0,T;L^2(\nu_0))}^2 & \le C_m^{\mathrm{stab}} \left( \|\mathcal{G}_m^{y^\dagger}[\hat{a}_m^\delta]-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[\hat{a}_m^\delta] \right) \\ & \le C_m^{\mathrm{stab}} \left( 4\|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + 4\,\mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger] + \frac{1}{2}\|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger] \right) \\ & \le 5\,C_m^{\mathrm{stab}} \left( \|d^\delta-\mathcal{G}_m^{y^\dagger}[a^\dagger]\|_{\mathcal{Y}_T}^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[a^\dagger] \right), \end{align}\] which is 54 . ◻

Proof. Set \[d_m^\dagger:=\mathcal{G}_m^{y_m^\dagger}[a_m^\dagger] =\mathcal{G}^{y_m^\dagger}[z_m^\dagger].\] By the observation Lipschitz estimate 57 , \[\|\mathcal{G}^{y_m^\dagger}[h^\dagger]-d_m^\dagger\|_{\mathcal{Y}_T} = \|\mathcal{G}^{y_m^\dagger}[h^\dagger]-\mathcal{G}^{y_m^\dagger}[z_m^\dagger]\|_{\mathcal{Y}_T} \le L_{\mathcal{K},m,y_m^\dagger}\|h^\dagger-z_m^\dagger\|_{\mathfrak H} = L_{\mathcal{K},m,y_m^\dagger}\|(I-P_m)h^\dagger\|_{\mathfrak H}.\] Therefore \[\label{eq:effective-reduced-data-error-global} \|d_m^\delta-d_m^\dagger\|_{\mathcal{Y}_T} \le \|d_m^\delta-\mathcal{G}^{y_m^\dagger}[h^\dagger]\|_{\mathcal{Y}_T} + \|\mathcal{G}^{y_m^\dagger}[h^\dagger]-d_m^\dagger\|_{\mathcal{Y}_T} \le \delta + L_{\mathcal{K},m,y_m^\dagger}\|(I-P_m)h^\dagger\|_{\mathfrak H}.\tag{59}\]

Now apply Theorem 60 with reference path \(a_m^\dagger\), chosen experiment \(y_m^\dagger\), and reduced data \(d_m^\delta\). Using 59 , we obtain \[\|\hat{h}_m^\delta-z_m^\dagger\|_{L^2(0,T;L^2(\nu_0))}^2 \le 5\,C_m^{\mathrm{stab}} \Bigl( \bigl(\delta + L_{\mathcal{K},m,y_m^\dagger}\|(I-P_m)h^\dagger\|_{\mathfrak H}\bigr)^2 + \mathcal{R}_{\lambda,\mu}^{(m)}[a_m^\dagger] \Bigr).\] Finally, use the triangle inequality: \[\|\hat{h}_m^\delta-h^\dagger\|_{L^2(0,T;L^2(\nu_0))} \le \|\hat{h}_m^\delta-z_m^\dagger\|_{L^2(0,T;L^2(\nu_0))} + \|z_m^\dagger-h^\dagger\|_{L^2(0,T;L^2(\nu_0))},\] and note that \(z_m^\dagger=P_m h^\dagger\). This gives 58 . ◻