Lemma 3.4: free unique continuation on the scale

The geometry and generalized-ray meaning of (H) are given by Exterior and truncated obstacle domains and Non-trapping obstacle. Choose $R>0$ with $\mathrm{\Theta }\subset B_R(0)$ and the real cutoff $\phi \in C^{\mathrm{\infty }}({\mathbb{R}}^n;[0,1])$ of Control region and nested real cutoffs, with $\phi (x)=0$ for $\mathrm{\mid }x\mathrm{\mid }\le R$, $\phi (x)>0$ for $\mathrm{\mid }x\mathrm{\mid }>R$, and $\phi (x)=1$ for $\mathrm{\mid }x\mathrm{\mid }\ge R+1$. Its derivatives through order two are bounded: the positive-order derivatives vanish outside the compact annulus $R\le \mathrm{\mid }x\mathrm{\mid }\le R+1$ and are continuous there, and $0\le \phi \le 1$.

Under Choice, $H_D^{\sigma }$ is the completed spectral scale, and $W(t)$ is the consistent free Dirichlet group with $W(0)=I$ Dirichlet spectral Sobolev scale and duality Scale isometries and free Schrödinger evolution The Axiom of Choice. For $T^{\mathrm{\prime }}\ge T$, $N_{T^{\mathrm{\prime }}}^{\sigma }\subseteq N_T^{\sigma }$; for $-2\le \sigma \le {\sigma }^{\mathrm{\prime }}\le 2$, $N_T^{{\sigma }^{\mathrm{\prime }}}\subseteq N_T^{\sigma }$ under the canonical injective scale embedding Invisible initial-data spaces.

For the fixed cutoff in [F1], for $T_0>0$ and $\sigma \in [-2,2)$, there are $T_0\mathrm{/}2\le a<b\le T_0$ and a finite-dimensional complex space $N=N_a^{\sigma }=N_b^{\sigma }$ such that ${\mathrm{\Delta }}_{D}N\subseteq N$, identifying $N$ with its image in $H_D^{\sigma -2}$. Here ${\mathrm{\Delta }}_D:H_D^{\sigma }\to H_D^{\sigma -2}$ is the consistent scale extension prop-se-invisible-invariant.

Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue. Every nonconstant complex polynomial has a complex root, so $\mathbb{C}$ is algebraically closed Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root, by the definition of an algebraically closed field An algebraically closed field: every nonconstant polynomial has a root in the field.

Under Choice, an element $v\in H_D^{\sigma }$ satisfying ${\mathrm{\Delta }}_{D}v=\mu v$ in $H_D^{\sigma -2}$ and invisible to $\phi$ in the test-pairing sense is zero, for any real $\sigma$ and complex $\mu$, on the connected exterior with the cutoff in [F1] Vanishing of an eigenfunction invisible outside a ball. Since the real cutoff in [F1] has bounded derivatives through order two, at order $-2$ multiplication by the real cutoff satisfies $⟨\phi v,\eta {⟩}_{-2,2}=⟨v,\phi \eta {⟩}_{-2,2}$; every compactly supported smooth test function and its product with $\phi$ belong to $H_D^2$ Endpoint multiplier bounds extended to negative spectral orders Dirichlet spectral Sobolev scale and duality.