Remark 6: exact use and zero-trajectory case of unique continuation

In the contradiction argument for the conditional observability estimate, the weak limit gives an adjoint trajectory annihilated by $\phi$. Lemma 4.5 makes that trajectory energy-continuous. Since $\phi$ is positive on $\omega$, local division of a compactly supported test function by $\phi$ shows that the trajectory vanishes on $(0,T)\times \omega$; (UCPw) then makes it identically zero. This is the application recorded in Proposition 4.6, so the hypothesis need only be formulated at the energy level for that step. The assertion that it is the only use in the article is retained as the authors' attribution in Remark 6.

Suppose $w=0$ and let $v$ be any distributional energy-level solution covered by (UCPw). Regard $V=H_0^1(\mathrm{\Omega };\mathbb{C})$ and $H=L^2(\mathrm{\Omega };\mathbb{C})$ as real Hilbert spaces, with inner products the real parts of the complex ones. The dense closed form gives completeness of $V$ and density of its continuous inclusion in $H$, hence the evolution triple of [F6]. For $u,\eta \in V$ define the real functional $(Bu)(\eta )=\mathrm{Re}(-i{\int }_{\mathrm{\Omega }}\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }\eta })$. Cauchy–Schwarz gives $\mathrm{\mid }(Bu)(\eta )\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }\eta {\mathrm{\parallel }}_{H^1}$, so $B:V\to V^{\ast }$ is bounded. Testing $i{v}_t+\mathrm{\Delta }v=0$ with products of a scalar time test and a conjugated smooth compactly supported spatial test gives ${\int }_0^T(v(t),\eta {)}_H{\alpha }^{\mathrm{\prime }}(t)dt=-{\int }_0^T(Bv(t))(\eta )\alpha (t)dt$. The defining $H^1$ density of spatial tests and the displayed bound extend this to every $\eta \in V$. Thus $v^{\mathrm{\prime }}=Bv$ in the real evolution-triple sense. Since $v$ and $Bv$ are norm-continuous on the compact time interval, finite sampled step approximations show strong measurability, and their bounded norms give $v\in L^2(0,T;V)$ and $Bv\in L^2(0,T;V^{\ast })$.

Put $h(t)=W(t)v(0)$. By [F4], $h\in C([0,T];V)$ and $h^{\mathrm{\prime }}=i{\mathrm{\Delta }}_{D}h$ in the complex pivot antidual. On the common spectral core, the represented form gives $⟨{\mathrm{\Delta }}_{D}u,\eta ⟩=-{\int }_{\mathrm{\Omega }}\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }\eta }$; continuity in both $H_D^1$ norms and core density extend the identity to all $u,\eta \in V$. Taking real parts therefore identifies the derivative of $h$ with $Bh$ in the same real dual as in step 1.2. Consequently $d=v-h$ belongs to the evolution space of [F6], satisfies $d^{\mathrm{\prime }}=Bd$, and is already continuous in $H$, with $d(0)=0$. The energy identity applies to this representative and gives $\mathrm{\parallel }d(t){\mathrm{\parallel }}_2^2=2{\int }_0^t\mathrm{Re}(-i\mathrm{\parallel }\mathrm{\nabla }d(r){\mathrm{\parallel }}_2^2)dr=0$. Hence $v(t)=W(t)v(0)$ for every $t\in [0,T]$. This proves the distributional-to-free-evolution bridge locally, rather than appealing to mild-class uniqueness for an arbitrary distributional solution.

Now impose the vanishing condition of (UCPw). Let $U$ be the representative in [F7] and set $N=\{(t,x)\in Q:x\in \omega ,U(t,x)\mathrm{\ne }0\}$. This is a measurable null set for the completed product measure: $U$ is jointly measurable, $\omega$ is open, and (UCPw)'s vanishing condition uses exactly this representative-independent space-time interpretation. The time factor has finite measure $T$, and the spatial factor is sigma-finite, as recorded in [F7]. Apply completed-product Tonelli in [F7] to ${\mathbf{1}}_N$. Outside a measurable null set of times its section integral is $I(t)=\mathrm{\mid }{N}_t\mathrm{\mid }$, where $N_t=\{x\in \mathrm{\Omega }:(t,x)\in N\}$ and $\mathrm{\mid }\cdot \mathrm{\mid }$ denotes spatial Lebesgue measure; assigning zero on the exceptional time set makes $I$ measurable, with ${\int }_0^{T}I(t)dt=0$. The zero-integral criterion in [F7] gives $\mathrm{\mid }{N}_t\mathrm{\mid }=0$ for almost every $t$. Intersecting this full-measure set with the one where $[U(t,\cdot )]=v(t)$ identifies these spatial sections with the path: $v(t)$ vanishes almost everywhere on $\omega$ for almost every time. Since $\phi$ vanishes outside $\omega$, this gives $\phi v(t)=0$ as an $L^2$ class for almost every time. Multiplication by $\phi$ has $L^2$ norm at most one, so $t\mapsto \phi v(t)$ is $L^2$-continuous and is therefore zero for every $t\in [0,T]$. By step 2.1 it equals $\phi W(t)v(0)$. The datum belongs to $H_D^1=H_0^1\subset H_D^0=L^2$, so Lemma 3.4 at index zero makes $v(0)=0$. Step 2.1 then makes the whole path zero. This establishes (UCPw) for $w=0$; the general-trajectory implication remains an explicit hypothesis.