Limited-angle visibility and instability — Examples and Counterexamples

The ball $B_1(0)$ contains a smaller concentric ball, so [F2] supplies a bump with the stated support and plateau. It is nonzero because it equals one on the nonempty inner ball.

If $h$ has the same restricted data as $\psi$, linearity of the line integral gives $R(h-\psi )(\theta ,s)=0$ for every measured pair. By [F4], the Fourier transforms of $h$ and $\psi$ agree on $W_{\mathrm{\Theta }}$; [F3] identifies these values as restrictions of their coordinate-entire Fourier-Laplace transforms. Applying [F5] to the compactly supported difference $h-\psi$ gives $h-\psi =0$ everywhere.

If the data of $\psi$ were zero, [F5] would imply $\psi =0$, contradicting its value one on ${\bar{B}}_{1\mathrm{/}2}(0)$. Thus its restricted data are nonzero while uniquely determining it in the stated compact-support class.

The set $\mathrm{\Theta }$ is the inverse image of the open interval $(-1\mathrm{/}2,1\mathrm{/}2)$ under the continuous coordinate map $\theta \mapsto {\theta }_1$, so it is open; changing $\theta$ to $-\theta$ preserves $\mathrm{\mid }{\theta }_1\mathrm{\mid }$, so it is antipodally symmetric. Its complement contains arcs around $(1,0)$ and $(-1,0)$. For every $\theta \in \mathrm{\Theta }$, $\mathrm{\mid }{\theta }_2\mathrm{\mid }=\sqrt{1-{\theta }_1^2}>\sqrt{3}\mathrm{/}2$, hence the infimum is at least $\sqrt{3}\mathrm{/}2$; approaching either boundary $\mathrm{\mid }{\theta }_1\mathrm{\mid }=1\mathrm{/}2$ shows the infimum is exactly $\sqrt{3}\mathrm{/}2$. By [F1] this is ${\inf }_{\theta \in \mathrm{\Theta }}\mathrm{\mid }v\cdot {\theta }^{\perp }\mathrm{\mid }$.

A translated or rescaled bump from [F2] has compact support in $B_1(0)$ and equals one on a smaller ball. That inner ball contains a nondegenerate box, which has positive measure by [F3]; compact support makes its $L^2$ norm finite, and the plateau makes the norm positive. Divide by this norm to obtain the required real $\psi$ with $\mathrm{\parallel }\psi {\mathrm{\parallel }}_2=1$. Since $K={\bar{B}}_1(0)$ lies inside $\mathrm{\Omega }=B_2(0)$, its support is contained in the fixed set $K$.

The formula in the Example is the modulation family in [F4]. The chosen ${\lambda }_M$ satisfies ${\lambda }_M(1+\mathrm{\parallel }\mathrm{\nabla }\psi {\mathrm{\parallel }}_2+2\pi )=M\mathrm{/}2\le M$, and the geometric verification in step 1.1 supplies the positive lower bound required there. Thus the family has the stated fixed support, $H^1$ and $L^2$ bounds, and rapid restricted-data decay.

If a uniform Hölder estimate with exponent $\alpha >0$ and constant $C\ge 0$ held on the fixed-support class, choose an integer $m\ge 1$ with $\alpha (m+1)>1$. Step 2.1 would imply ${\lambda }_M{N}^{-1}\le C(C_m{N}^{-m-1}{)}^{\alpha }$, so ${\lambda }_M\le C{C}_m^{\alpha }{N}^{1-\alpha (m+1)}\to 0$, contradicting ${\lambda }_M>0$. Therefore no such estimate holds.

The sector $V$ is a nonempty open subset of ${\mathbb{R}}^2$ because $(r,\omega )\mapsto r\omega$ maps the open parameter set $(1,2)\times J$ to an open polar-coordinate sector. It therefore contains a ball; [F2] supplies a nonzero $q\in C_c^{\mathrm{\infty }}(V)$. This establishes the required bump choice.

By the Fourier definition [F5], the function in the Example is $f(x)=\widehat{q}(-x)$. Since $q$ is smooth with compact support, every Schwartz seminorm of $q$ is bounded, so $q\in \mathcal{S}$. By [F3] its transform is Schwartz, and reflection preserves each Schwartz seminorm up to sign, hence $f$ is Schwartz; [F7] gives $f\in L^2\cap L^1$. The reflection substitution [F6] and inversion [F4] give $\widehat{f}(\eta )=\int \widehat{q}(-x)e^{-2\pi ix\cdot \eta }dx=\int \widehat{q}(y)e^{2\pi iy\cdot \eta }dy=q(\eta )$. Since $q\mathrm{\ne }0$, this identity implies $f$ is nonzero as an $L^1$ and $L^2$ class.

For $\theta \in \mathrm{\Theta }$ and $\sigma \mathrm{\ne }0$, the point $\sigma \theta$ is outside the support of $q$ because every frequency in $V$ has direction in $J$ and $\bar{J}$ is disjoint from $\bar{\mathrm{\Theta }}$; also $q(0)=0$ since $V$ avoids the origin. Thus [F8] gives $\widehat{R_{\mathcal{S}}f}(\theta ,\sigma )=q(\sigma \theta )=0$ for all real $\sigma$. By [F9], the offset function is Schwartz, so Fourier inversion [F4] in one dimension gives $R_{\mathcal{S}}f(\theta ,s)=0$ for every $s$.

Let the parameter and target metric spaces both be $L^2({\mathbb{R}}^2)$ with distance $d(g,h)=\mathrm{\parallel }g-h{\mathrm{\parallel }}_2$, let the admissible truth class be $U=\{0,f\}$, and let the target map $q:U\to L^2({\mathbb{R}}^2)$ be inclusion. Give the data space $Y=L^2(\mathrm{\Theta }\times \mathbb{R})$ its norm metric and define $F:U\to Y$ by the restricted line integrals. The normed-space result [F10] and the metric axioms [F12] give the parameter, target, and data metrics used here. Theorem [F11] applies directly to this truth class and target map. Step 1.3 gives $F(0)=F(f)=0$, while step 1.2 gives $d(q(0),q(f))=\mathrm{\parallel }f{\mathrm{\parallel }}_2>0$. Apply [F11] with shared datum $0$ and radius $0$ to any deterministic estimator $\mathrm{\Phi }:Y\to L^2({\mathbb{R}}^2)$; its maximum target loss is at least $\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{/}2$. This proves the asserted lower bound on the stated two-truth class.