Limited-angle inversion has no Hölder modulus on a fixed-support H1 ball

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.