Limited-angle visibility and instability

Choose a nondegenerate closed rectangle $B$ containing $\mathrm{supp}f$. For every $z\in {\mathbb{C}}^2$, the integrand $x\mapsto f(x)e^{-2\pi ix\cdot z}$ is continuous on $B$, so [F2] makes its real and imaginary parts Riemann integrable. It vanishes outside $\mathrm{supp}f$; hence enlarging $B$ changes the integral only by adding regions on which the integrand is zero. This proves existence and independence of the containing rectangle.

Fix $j\in \{1,2\}$, fix the other complex coordinate, and let $h\in \mathbb{C}\setminus \{0\}$. By [F4], the coordinate difference quotient is the integral over $B$ of $f(x)e^{-2\pi ix\cdot z}(e^{-2\pi i{x}_{j}h}-1)\mathrm{/}h$. The coordinate $x_j$ is bounded on $B$; [F3] therefore makes $(e^{-2\pi i{x}_{j}h}-1)\mathrm{/}h$ converge uniformly on $B$ to $-2\pi i{x}_j$ as $h\to 0$. Multiplication by the fixed bounded continuous factor $f(x)e^{-2\pi ix\cdot z}$ preserves uniform convergence, and [F4] passes the real and imaginary integrals to their limits. Thus the coordinate derivative exists and is the displayed integral. Since the other coordinate and the point were arbitrary, every coordinate slice is complex differentiable everywhere.

By [F5], each coordinate slice is entire. At real $z=\xi$, the defining integral is exactly the compact-support Fourier integral of [F1], so $F(\xi )=\widehat{f}(\xi )$. This proves all claims without using a Lebesgue/Riemann identification theorem or Countable Choice.

Fix a derivative multi-index $\beta$ and an integer $N\ge 1$, and choose an integer $m\ge N$. Since $(1+\mathrm{\mid }x\mathrm{\mid }{)}^N\le (1+\mathrm{\mid }{x}_1\mathrm{\mid }{)}^m(1+\mathrm{\mid }{x}_2\mathrm{\mid }{)}^m$, expanding the right side and using the finite Schwartz seminorms in [F1] gives a finite $C_{\beta ,N}(f)$ with $\mathrm{\mid }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }\le C_{\beta ,N}(f)(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-N}$. For $x=s\theta +t{\theta }^{\perp }$, orthogonality gives $\mathrm{\mid }x\mathrm{\mid }=(s^2+t^2{)}^{1\mathrm{/}2}\ge (\mathrm{\mid }s\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid })\mathrm{/}\sqrt{2}$, so the line integrand is bounded by a constant times $(1+\mathrm{\mid }s\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid }{)}^{-N}$. Taking $N>1$ makes this majorant integrable in $t$, proving absolute convergence.

Let $D_{\theta }=\theta \cdot \mathrm{\nabla }$. The chain rule [F4] gives ${\mathrm{\partial }}_s^{k}f(s\theta +t{\theta }^{\perp })=(D_{\theta }^{k}f)(s\theta +t{\theta }^{\perp })$. Expanding $D_{\theta }^k$ gives finitely many partial derivatives with coefficients bounded uniformly for $\theta \in S^1$. For the target decay order $N$, apply step 1.1 with spatial decay order $N+2$ to each such derivative. Integrating $(1+\mathrm{\mid }s\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid }{)}^{-N-2}$ in $t$ gives $\mid {\int }_{\mathbb{R}}(D_{\theta }^{k}f)(s\theta +t{\theta }^{\perp })dt\mid \le C_{k,N}^{\mathrm{\prime }}(1+\mathrm{\mid }s\mathrm{\mid }{)}^{-N-1}\le C_{k,N}^{\mathrm{\prime }}(1+\mathrm{\mid }s\mathrm{\mid }{)}^{-N}$, uniformly in $\theta$ and $s$.

Fix $s_0\in \mathbb{R}$ and a bounded open interval $I$ containing $s_0$. For each $k\ge 0$, set $g_k(t,s)=(D_{\theta }^{k}f)(s\theta +t{\theta }^{\perp })$. Every slice $g_k(\cdot ,s)$ is integrable by step 1.1; it is differentiable in $s$ with derivative $g_{k+1}$ by [F4], and this derivative is measurable. Step 1.1 with spatial decay order $3$ bounds $\mathrm{\mid }{g}_{k+1}(t,s)\mathrm{\mid }\le C(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-3}$ uniformly for $s\in I$, an integrable majorant. Thus [F3] proves $\frac{d}{ds}\int g_k(t,s)dt=\int g_{k+1}(t,s)dt$. To see each resulting derivative is continuous, if $s=s^{\mathrm{\prime }}$ the difference is zero; otherwise apply [F5] to real and imaginary parts on the segment joining $s$ and $s^{\mathrm{\prime }}$. The same majorant gives $\mathrm{\mid }{g}_k(t,s)-g_k(t,s^{\mathrm{\prime }})\mathrm{\mid }\le C\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-3}$, and integration proves local continuity. Repeating for all $k$ and using the estimates in step 2.1 shows the offset function is Schwartz.

Choose $R>0$ with $\mathrm{supp}f\subseteq {\bar{B}}_R(0)$. If $\mathrm{\mid }t\mathrm{\mid }>R$, then $\mathrm{\mid }s\theta +t{\theta }^{\perp }\mathrm{\mid }\ge \mathrm{\mid }t\mathrm{\mid }>R$ and the integrand in [F1] is zero. On any compact set of $(\theta ,s)$, uniform continuity of $f$ on the compact image of $(s,\theta ,t)$ for $\mathrm{\mid }t\mathrm{\mid }\le R$ shows that the line integrals vary continuously: the difference of two such integrals is at most $2R$ times the uniform difference of their integrands. Hence the data function is continuous and measurable.

Put $D=\mathrm{diam}(\mathrm{\Omega })$ and $T_{\theta ,s}=\{t\in \mathbb{R}:s\theta +t{\theta }^{\perp }\in \mathrm{\Omega }\}$. Any two points of $T_{\theta ,s}$ differ by at most $D$, so this open set is contained in an interval of length at most $D$. Cauchy--Schwarz [F2] on $T_{\theta ,s}$, where the constant function $1$ has squared integral at most $D$, gives $\mathrm{\mid }{R}_{\mathcal{S}}f(\theta ,s){\mathrm{\mid }}^2\le D{\int }_{\mathbb{R}}\mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^{2}dt$.

The function $(\theta ,s,t)\mapsto \mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^2$ is bounded and vanishes unless $\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid }\le R$, so it is integrable on the finite-measure set $\mathrm{\Theta }\times [-R,R{]}^2$. Apply [F4] to exchange the iterated integrals, and for each $\theta$ apply [F3] to the orthogonal diffeomorphism $(s,t)\mapsto s\theta +t{\theta }^{\perp }$, whose absolute Jacobian is one. This gives ${\int }_{\mathrm{\Theta }}{\int }_{{\mathbb{R}}^2}\mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^{2}dsdtd\theta =\mathrm{\mid }\mathrm{\Theta }\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}^2$. Integrating step 1.2 yields the claimed bound and square-integrability.

The hypothesis and the compact-support Fourier-slice identity [F2] give $\widehat{f}(\sigma \theta )=0$ for every $\theta \in \mathrm{\Theta }$ and $\sigma \in \mathbb{R}$. By [F1], this includes the nonempty open positive-radius sector generated by $J$; choose a product of open real intervals $I_1\times I_2$ inside that sector. For each real ${\xi }_2\in I_2$, the entire function $z_1\mapsto \widehat{f}(z_1,{\xi }_2)$ vanishes on $I_1$, so [F5] makes it zero for every complex $z_1$. Then for every real ${\xi }_1$ the entire second-coordinate slice vanishes on $I_2$, and [F5] makes it zero everywhere. Thus $\widehat{f}$ vanishes on all of ${\mathbb{R}}^2$.

Let $K=\mathrm{supp}f$. If $K=\mathrm{\varnothing }$, then $f=0$. Otherwise, fix a rectangle $B$ whose interior contains $K$ and interpret all integrals below as compact-support Riemann integrals over $B$. Repeated use of the coordinate derivative formula in [F3] shows ${\int }_B{x}^{\alpha }f(x)dx=0$ for every multi-index $\alpha$, since all real-frequency derivatives of $\widehat{f}$ vanish at the origin. The real polynomials restricted to $K$ form a unital subalgebra separating points, because the coordinate functions distinguish distinct points. For an arbitrary $\epsilon >0$, apply [F6] separately to $\mathrm{Re}\bar{f}$ and $\mathrm{Im}\bar{f}$ with error $\epsilon \mathrm{/}2$ and combine the two real polynomials into one complex polynomial $p$ with $\mathrm{\parallel }p-\bar{f}{\mathrm{\parallel }}_{C(K)}<\epsilon$. Every moment is zero, so ${\int }_{B}fp=0$. Since $f=0$ outside $K$, the real Riemann integral bound [F4] on both components gives $0\le {\int }_B\mathrm{\mid }f{\mathrm{\mid }}^2=\mid {\int }_{B}f(\bar{f}-p)\mid \le \sqrt{2}\mathrm{vol}(B)\mathrm{\parallel }f{\mathrm{\parallel }}_{C(B)}\epsilon$. As this holds for every $\epsilon >0$, ${\int }_B\mathrm{\mid }f{\mathrm{\mid }}^2=0$; no sequence of approximants is selected. If $f$ were nonzero at a point, continuity would bound $\mathrm{\mid }f{\mathrm{\mid }}^2$ below by a positive number on a nondegenerate box inside $B$, forcing its nonnegative Riemann integral to be positive. Thus $f=0$ everywhere. This proof uses neither Fourier inversion nor Countable Choice.

Choose $0<r<R$ and apply [F3] to obtain a bump supported in $B_R(x_0)$ and equal to one on a smaller ball; translation of the concentric bump places it inside $K$. Its $L^2$ norm is finite and positive because it is bounded with compact support and equals one on a ball containing a nondegenerate box, whose measure is positive by [F10]. Rescale it to a real-valued $\psi$ with $\mathrm{\parallel }\psi {\mathrm{\parallel }}_2=1$. Choose $\lambda >0$ so that $\lambda (1+\mathrm{\parallel }\mathrm{\nabla }\psi {\mathrm{\parallel }}_2+2\pi )\le M$, and put $u_N(x)=\lambda N^{-1}{e}^{2\pi iNv\cdot x}\psi (x)$. Then all supports equal $\mathrm{supp}\psi ⋐\mathrm{int}K$ and $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_2=\lambda N^{-1}$. By [F7]–[F9], ${\mathrm{\partial }}_j{u}_N=\lambda N^{-1}{e}^{2\pi iNv\cdot x}({\mathrm{\partial }}_j\psi +2\pi iN{v}_j\psi )$, so $\mathrm{\parallel }\mathrm{\nabla }{u}_N{\mathrm{\parallel }}_2\le \lambda N^{-1}\mathrm{\parallel }\mathrm{\nabla }\psi {\mathrm{\parallel }}_2+2\pi \lambda \mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }\psi {\mathrm{\parallel }}_2\le \lambda (\mathrm{\parallel }\mathrm{\nabla }\psi {\mathrm{\parallel }}_2+2\pi )$. Therefore $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^1}\le M$; set $a_M=\lambda$.

Put $a=v\cdot {\theta }^{\perp }$, so $\mathrm{\mid }a\mathrm{\mid }\ge c>0$. The line formula gives $R_{\mathcal{S}}{u}_N(\theta ,s)=\lambda N^{-1}{e}^{2\pi iNs(v\cdot \theta )}{\int }_{\mathbb{R}}{e}^{2\pi iNat}\psi (s\theta +t{\theta }^{\perp })dt$. For each integer $m\ge 1$, integrate the last integral by parts $m$ times over an interval containing the support of its integrand. The boundary terms vanish because $\psi$ and all its derivatives vanish near the endpoints. Applying [F6] repeatedly gives $\mid {\int }_{\mathbb{R}}{e}^{2\pi iNat}\psi (s\theta +t{\theta }^{\perp })dt\mid \le (2\pi N\mathrm{\mid }a\mathrm{\mid }{)}^{-m}{\int }_{\mathbb{R}}\mathrm{\mid }({\theta }^{\perp }\cdot \mathrm{\nabla }{)}^m\psi (s\theta +t{\theta }^{\perp })\mathrm{\mid }dt$.

Let $D=\mathrm{diam}(\mathrm{supp}\psi )$. The set of $t$ values on a fixed line meeting this support has measure at most $D$. Cauchy--Schwarz [F4] bounds the square of the last integral in step 1.2 by $D{\int }_{\mathbb{R}}\mathrm{\mid }({\theta }^{\perp }\cdot \mathrm{\nabla }{)}^m\psi (s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^{2}dt$. Integrate in $s$ and apply [F5] to the orthogonal change of variables $(s,t)\mapsto s\theta +t{\theta }^{\perp }$, whose absolute Jacobian is one. Expanding $({\theta }^{\perp }\cdot \mathrm{\nabla }{)}^m$ gives an $L^2$ bound independent of $\theta$, since there are finitely many derivatives and all coefficients are bounded by one. Since $\mathrm{\mid }a\mathrm{\mid }\ge c$ and $\mathrm{\mid }\mathrm{\Theta }\mathrm{\mid }\le 2\pi$, step 1.2 yields a finite constant $C_m>0$, independent of $N$, with $\mathrm{\parallel }{R}_{\mathrm{\Theta }}{u}_N{\mathrm{\parallel }}_D\le C_m{N}^{-m-1}$.

Fix $\alpha >0$ and choose an integer $m\ge 1$ with $\alpha (m+1)>1$. The square-integrability bound [F11] makes the data norm finite for every member of the fixed-support class. If a uniform Hölder estimate with constant $C\ge 0$ held on the class in the Statement, steps 1.1 and 2.1 would give $\lambda N^{-1}\le C(C_m{N}^{-m-1}{)}^{\alpha }$ for every $N\ge 1$. Multiplication by $N$ makes the right side $C{C}_m^{\alpha }{N}^{1-\alpha (m+1)}$, which tends to zero as $N\to \mathrm{\infty }$, contradicting $\lambda >0$. Therefore no such estimate exists.

The Schwartz seminorms and the estimate proved in The Radon transform maps Schwartz functions to Schwartz offset data give $\mathrm{\mid }f(s\theta +t{\theta }^{\perp })\mathrm{\mid }\le C_N(1+\mathrm{\mid }s\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid }{)}^{-N}$ for every $N$, after changing the constant. For $N>2$ this is integrable on ${\mathbb{R}}^2$ in $(s,t)$. The offset integral is absolutely convergent, and [F6] applied to the absolutely integrable function $f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}$ yields $\widehat{R_{\mathcal{S}}f}(\theta ,\sigma )={\int }_{{\mathbb{R}}^2}f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt$. The Lebesgue Fourier-transform definitions in [F3] are applicable by [F8].

Choose a smooth radial cutoff ${\chi }_R$ equal to one on the radius-$R$ ball and supported in the radius-$2R$ ball, using [F4], and put $f_R={\chi }_{R}f$. Then $f_R\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$. The matrix with columns $\theta ,{\theta }^{\perp }$ is a positively oriented orthogonal matrix, so its determinant is one. Apply the Lebesgue change-of-variables formula [F7] to the integrable compactly supported function $x\mapsto f_R(x)e^{-2\pi i\sigma x\cdot \theta }$ and the map $(s,t)\mapsto s\theta +t{\theta }^{\perp }$; its pullback integral is ${\int }_{{\mathbb{R}}^2}{f}_R(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt=\widehat{f_R}(\sigma \theta )$.

As $R\to \mathrm{\infty }$, $f_R(x)\to f(x)$ and $\mathrm{\mid }{f}_R(x)\mathrm{\mid }\le \mathrm{\mid }f(x)\mathrm{\mid }$. The function $f$ is integrable by [F8], and the same Schwartz estimate in $(s,t)$ gives an integrable majorant for $f_R(s\theta +t{\theta }^{\perp })$. Dominated convergence [F5] passes both integrals in step 1.2 to their limits; step 1.1 identifies the data-side limit with $\widehat{R_{\mathcal{S}}f}(\theta ,\sigma )$, while the spatial-side limit is $\widehat{f}(\sigma \theta )$. This proves the formula.

For each $\theta \in \mathrm{\Theta }$ and $\sigma \in \mathbb{R}$, [F3] gives the Fourier-slice identity. If the restricted data vanish, their one-dimensional Fourier transforms vanish, so $\widehat{f}(\sigma \theta )=0$. Every point of $W_{\mathrm{\Theta }}$ has this form, hence $\widehat{f}$ vanishes on $W_{\mathrm{\Theta }}$.

Now assume $f\in C_c^{\mathrm{\infty }}$ and its restricted data are zero. By [F2] and the stated Countable Choice assumption, those data use the compact-support Riemann line-integral convention. The compact-support Fourier-slice identity [F4] gives $\widehat{f}(\sigma \theta )=0$ for every $\theta \in \mathrm{\Theta }$ and $\sigma \in \mathbb{R}$, so the transform vanishes on the measured wedge. The compact-support uniqueness result [F5] then yields $f=0$ everywhere through its separate choice-free analytic-continuation proof.