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For a Schwartz function $f$, the Hilbert transform satisfies $\mathcal{H}f(x)=(\int f)/(\pi x)+o(|x{|}^{-1})$ at both infinities. Splitting the inverse Fourier transform at the sign multiplier's jump and integrating each half-line by parts extracts the zero-frequency value $\hat{f}(0)$; the remainder vanishes by the Riemann-Lebesgue lemma. A nonzero zeroth moment therefore obstructs Schwartz decay.

Assume for this spectral construction that the source and known attenuation are sufficiently smooth and decaying, for instance compactly supported, and that $\mu \ge 0$ for the physical interpretation. The following steps identify both the forward attenuated Radon transform and the route to its inversion.

First write the complex transport operator asFor $k=e^{i\theta }$ this is the real directional derivative ${\mathbf{d}}_{\theta }\cdot \nabla$, with ${\mathbf{d}}_{\theta }=(\cos \theta ,\sin \theta )$. Put ${\mathbf{n}}_{\theta }=(-\sin \theta ,\cos \theta )$ and $x=p{\mathbf{n}}_{\theta }+s{\mathbf{d}}_{\theta }$. The spectral equation on the unit circle becomesIts minus sign fixes the appropriate endpoint condition: use $F(p,+\infty )=0$. The integrating factor givesThus the measured quantity at the opposite end of the line isThe exponent is the attenuation accumulated between the source point and the detector at the negative end. The common convention with detector at the positive end is the same transform after reversing the direction, ${\mathcal{R}}_{\mu }^{+}f(p,\theta )={\mathcal{R}}_{\mu }^{-}f(-p,\theta +\pi )$. Using an incoming zero condition at the negative end while retaining the printed minus sign would instead give a growing integrating factor, not physical attenuation.

Next, for $|k|\ne 1$, the operator $L_k$ is a complex elliptic first-order operator. Its decaying whole-plane Green function isThis is obtained by a real-linear change of variables in the Cauchy-Green operator; its change of orientation explains the sign. Write $G_{k}a$ for convolution with this kernel. Solve $L_{k}h=\mu$, and set $F=e^h\Psi$. This removes the attenuation:The normalized spectral solution is analytic separately inside and outside the unit circle. As $k$ approaches that circle, the Green function's characteristic singularity produces two limiting values. Their relation is computed from the weighted line integrals above together with transverse Hilbert transforms; the known attenuation determines the integrating factor weights. The two spectral limits are not individually just the incoming and outgoing real characteristic solutions: the singular-kernel prescription matters.

Finally formulate the resulting additive Riemann-Hilbert problem on the unit circle. Its jump is determined by the measured attenuated Radon transform and known $\mu$. With the unit circle oriented counterclockwise and jump $J=F_{\mathrm{inside}}-F_{\mathrm{outside}}$, a Cauchy integral formula reconstructs the normalized spectral solution:The normalization at zero supplies the compatibility condition $\int J/\zeta d\zeta =0$. Recover $f$ from $L_{k}F-\mu F$, or from its small-$k$ coefficient: if $F=k{F}_1+O(k^2)$, then $f={\partial }_{\bar{z}}{F}_1$. Equivalently the large-$k$ coefficient gives $f={\partial }_z{F}_{-1}$ when $F=k^{-1}{F}_{-1}+O(k^{-2})$. This spectral reconstruction of an attenuated Radon transform is the analogue of recovering $q$ from a coefficient in (ii). Known attenuation and full directed line data are inputs; one does not determine an arbitrary unknown attenuation and source simultaneously from this argument.

We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The Schwartz space consists of smooth functions with finite seminormsfor every pair of multi-indices. Convergence means convergence in each seminorm. The tempered distribution space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual space, with weak convergence tested against every Schwartz function. A continuous functional satisfies a bound by finitely many of these seminorms, equivalently by one sufficiently large weighted derivative seminorm.

Use the Fourier transform conventionDifferentiation under the integral and integration by parts express ${\xi }^{\alpha }{\partial }_{\xi }^{\beta }\hat{f}$ as a constant of modulus one times the Fourier transform of ${\partial }_x^{\alpha }(x^{\beta }f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz seminorms, using the integrable weight $(1+|x|{)}^{-s}$. Thus $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

For completeness, Fourier inversion follows by inserting $e^{-\epsilon |\xi {|}^2}$ in the inverse integral and using Fubini's theorem. The result is convolution with the Gaussian approximate identityIt tends to $f$, while integrability of $\hat{f}$ allows the damping factor to be removed by dominated convergence theorem. Consequently ${\mathcal{F}}^{2}f=(2\pi {)}^{n}f(-\cdot )$. Reflection preserves every Schwartz seminorm, so the inverse transform is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the Fourier transform of a tempered distribution by transposition,The Schwartz-space continuity just proved makes this a tempered distribution. Its inverse is $(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$, where $⟨\mathcal{R}u,\psi ⟩=⟨u,\psi (-\cdot )⟩$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the strong dual topology, because the Schwartz-space maps take bounded sets to bounded sets.

The convolution of a tempered distribution with a Schwartz function isSmooth dependence of translated Schwartz functions gives ${\partial }^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$. The finite-seminorm estimate and $1+|y|\le (1+|x|)(1+|x-y|)$ show that each derivative has at most polynomial growth. In particular, $u\ast \phi$ is a smooth function defining a tempered distribution. It need not itself be a Schwartz function; for example $1\ast \phi =\int \phi$.

Writing $\check{\phi }(y)=\phi (-y)$, its distributional pairing is $⟨u\ast \phi ,\psi ⟩=⟨u,\check{\phi }\ast \psi ⟩$. The inner convolution is a Schwartz function, and this identity follows by integration in the Schwartz topology, justified by the weighted seminorm estimates. A direct Fubini's theorem calculation gives $\check{\phi }\ast \hat{\psi }=\mathcal{F}(\hat{\phi }\psi )$. HenceMultiplication is well defined because multiplication by $\hat{\phi }$ acts continuously on $\mathcal{S}$.

Now write the Hilbert transform as convolution with $K={\pi }^{-1}\mathrm{pv}(1/x)$, the principal-value reciprocal distribution. The given Heaviside function transform, together with ${\mathcal{F}}^{2}H=2\pi H(-\cdot )$, yieldsThe value of the sign function at zero is irrelevant to its regular distribution. Thus the Hilbert-transform Fourier multiplier isApplying Plancherel theorem, whose normalization here is $\Vert f{\Vert }_2^2=(2\pi {)}^{-1}\Vert \hat{f}{\Vert }_2^2$, proves the isometry:The principal-value integral agrees with this convolution: near its singular point subtract $\phi (x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The Hilbert transform has domain $\mathcal{S}$, but generally does not take values in $\mathcal{S}$, as the tail in part (c) demonstrates.

With the stated sign convention, write $t=y-x$ to obtainThe Cauchy principal value of the cosine part vanishes by oddness, while the Dirichlet integral givesThus the Hilbert transform has Fourier multiplier $-i\mathrm{sgn}(\omega )$. Taking the real part yields

Using the convention adopted here, the Hilbert transform isThe convention with denominator $x-y$ differs by a minus sign.

Complexifying the directional transport equation yields spectral solutions analytic inside and outside the unit circle. An integrating factor built from known attenuation removes its multiplication term. The limiting Green kernels encode weighted line measurements and transverse Hilbert transforms. Their jump gives a Riemann-Hilbert problem; its solution recovers the source from an asymptotic coefficient. A general primary treatment of this spectral construction is www.stat.uchicago.edu/~guillaumebal/PAPERS/AttRadTh.pdf .