02.10.07 · analysis / harmonic

The Radon transform: inversion, Plancherel, and the range theorem

shipped3 tiersLean: none

Anchor (Master): Helgason Groups and Geometric Analysis (AMS 2000) Ch. I; Helgason The Radon Transform Ch. I-II; Gel'fand-Graev-Vilenkin Generalized Functions Vol. 5 Ch. I; Natterer Ch. II-V

Intuition Beginner

A CT scanner cannot see inside you directly. What it can do is shine a thin X-ray beam straight through the body and measure how much got absorbed along that one line. Dense bone soaks up a lot; soft tissue soaks up a little. Each beam returns a single number: the total absorbing material lying along its path. The machine repeats this for thousands of lines, at many angles, sweeping all the way around. The question the mathematics answers is sharp: from all those line-sums, can you rebuild the full picture of what is inside, point by point?

The recipe that takes a picture and produces all its line-sums is the Radon transform. The recipe that goes back the other way, from line-sums to picture, is its inversion. The surprise is that the inverse exists and is exact: nothing is lost when you replace a body by the totals of all the lines through it.

A useful everyday analogy is a stack of pancakes seen edge-on. From one side you see a single combined height; tilt your head to a new angle and you get a different combined view. Enough views from enough angles, and you can reconstruct the whole stack.

Visual Beginner

A blob (the unknown density) sits in the plane. A family of parallel straight lines crosses it at one fixed angle; beside the blob a small graph records, for each line, the total amount of blob the line passes through. Rotating the family of lines to a new angle produces a new graph. Sweeping the angle from zero all the way around fills out a two-dimensional picture indexed by angle and offset — the sinogram — which is the complete Radon transform of the blob.

The picture shows the two directions of the theory at once. Reading left to right, a density becomes a collection of line-sums. Reading right to left — the harder and more useful direction — the collection of line-sums is processed and smeared back across the plane to rebuild the density. The smearing-back step is called backprojection, and on its own it gives a blurred answer; sharpening that blur is what the inversion formula does.

Worked example Beginner

Take a uniform disc of radius $2$ and density $1$ inside, $0$ outside, centred at the origin. Fix the horizontal direction and consider a line at vertical offset $p$ . The line meets the disc when $∣ p ∣ < 2$ , and the length of the chord it cuts is $2 4 - p^{2}$ . Since the density is $1$ , the line-sum equals that chord length. So at this angle the Radon transform is the function $2 4 - p^{2}$ for $∣ p ∣ < 2$ and $0$ otherwise.

Now rotate the line to any other angle. By the round symmetry of the disc, the chord lengths are exactly the same: the answer does not depend on the angle at all. Every angle returns the same arch-shaped graph, peaking at $4$ when the line runs through the centre ( $p = 0$ ) and falling to $0$ at the rim ( $p = \pm 2$ ).

What this tells us: a perfectly round, uniform object produces a Radon transform that is constant in angle. Any variation with angle in real scanner data is therefore a direct signature of asymmetry in the object — a denser patch on one side, an edge, a cavity. The transform has turned "shape and density" into "how the line-sums change as you turn", which is the language the reconstruction works in.

Check your understanding Beginner

Formal definition Intermediate+

Let $f$ be a Schwartz function on $R^{n}$ , in the sense of the rapidly decreasing functions of 02.10.04 and 02.14.04. For a unit vector $ω \in S^{n - 1}$ and a real offset $p$ , the hyperplane ${x : ⟨ x, ω ⟩ = p}$ carries the induced $(n - 1)$ -dimensional Lebesgue measure of the area/coarea machinery of 02.07.11. The Radon transform of $f$ is the integral over that hyperplane, $$ Rf(\omega, p) = \int_{\langle x, \omega\rangle = p} f , dS = \int_{\mathbb{R}^n} f(x), \delta(p - \langle x, \omega\rangle), dx , $$ a function on the unit cylinder $S^{n - 1} \times R$ . It is even: $R f (- ω, - p) = R f (ω, p)$ . The delta-form on the right exhibits $R$ as a distributional pairing and extends it to tempered distributions.

The dual transform (or backprojection) $R^{#}$ sends a function $g$ on the cylinder to a function on $R^{n}$ by averaging over all hyperplanes through a point, $$ R^{#}g(x) = \int_{S^{n-1}} g(\omega, \langle x, \omega\rangle), d\omega , $$ with $d ω$ the surface measure on $S^{n - 1}$ . The pair $(R, R^{#})$ is adjoint: $\int_{S^{n - 1} \times R} R f \cdot g = \int_{R^{n}} f \cdot R^{#} g$ for suitable $f, g$ , which is the integral-geometric duality between points and hyperplanes.

Write $f$ for the $n$ -dimensional Fourier transform of 02.10.04 and $(R f) (ω, s)$ for the one-dimensional Fourier transform of $p \mapsto R f (ω, p)$ taken in the offset variable. The two are linked along the ray through $ω$ .

Counterexamples to common slips

The dual $R^{#}$ is not an inverse of $R$ . Computing $R^{#} R f$ gives a smoothed version of $f$ — convolution with $∣ x ∣^{- 1}$ up to a constant — not $f$ itself. Inversion requires a sharpening filter between backprojection and the answer.
The offset variable $p$ ranges over all of $R$ , not just $p \geq 0$ . Restricting to $p \geq 0$ loses the evenness pairing $R f (- ω, - p) = R f (ω, p)$ and double-counts each hyperplane.
$R f (ω, p)$ is an integral over a hyperplane of codimension one, not over a line, once $n > 2$ . The line-integral picture is the $n = 2$ special case; the codimension-one object is what the projection-slice theorem and the inversion formula are about for general $n$ . The line ( $k$ -plane) versions are the John / d-plane transforms.

Key theorem with proof Intermediate+

Theorem (projection-slice / Fourier-slice theorem). For $f \in S (R^{n})$ and any $ω \in S^{n - 1}$ , the one-dimensional Fourier transform of the Radon transform in its offset variable equals the $n$ -dimensional Fourier transform of $f$ restricted to the ray $R ω$ : $$ \widetilde{(Rf)}(\omega, s) = \widehat{f}(s\omega), \qquad s \in \mathbb{R} . $$

Proof. Fix $ω$ . By definition of the one-dimensional Fourier transform in $p$ , $$ \widetilde{(Rf)}(\omega, s) = \int_{\mathbb{R}} Rf(\omega, p), e^{-2\pi i s p}, dp = \int_{\mathbb{R}} \left( \int_{\langle x,\omega\rangle = p} f , dS \right) e^{-2\pi i s p}, dp . $$ The inner integral collects $f$ over the hyperplane at offset $p$ ; integrating the result against $e^{- 2 π i s p} d p$ over all $p$ reassembles an integral over all of $R^{n}$ . Concretely, decompose $x = p ω + y$ with $y$ in the hyperplane $ω^{⊥}$ , so $⟨ x, ω ⟩ = p$ and $d x = d p d S (y)$ by the coarea decomposition of 02.07.11. Then $$ \widetilde{(Rf)}(\omega, s) = \int_{\mathbb{R}} \int_{\omega^{\perp}} f(p\omega + y), e^{-2\pi i s p}, dS(y), dp = \int_{\mathbb{R}^n} f(x), e^{-2\pi i s \langle x, \omega\rangle}, dx = \widehat{f}(s\omega) , $$ where the last equality uses $s p = s ⟨ x, ω ⟩ = ⟨ x, s ω ⟩$ to recognise the integral as the value of $f$ at the frequency $s ω$ . Both sides are Schwartz in $s$ , so the identity holds pointwise. $□$

Bridge. The projection-slice theorem builds toward every later inversion statement in this unit, because it shows that the entire $n$ -dimensional Fourier transform $f$ — and so, by Fourier inversion, $f$ itself — is recoverable from the one-dimensional transforms of the slices $R f (ω, \cdot)$ . The foundational reason is a change of integration order made rigorous by the coarea decomposition of 02.07.11: summing a function over hyperplanes and then over offsets is the same as integrating once over $R^{n}$ . This is exactly the mechanism that makes the Fourier transform of 02.10.04 the natural diagonalising language for $R$ , and it generalises the planar line-integral picture to hyperplanes in every dimension. Passing to polar coordinates in frequency converts the radial Fourier-inversion integral into a backprojection followed by a one-dimensional radial filter; putting these together is the bridge from the slice theorem to the filtered-backprojection formula, where the radial weight $∣ s ∣^{n - 1}$ that appears is the central insight separating odd from even dimensions. The slice theorem is dual to the convolution theorem in the sense that $R^{#} R$ becomes, on the Fourier side, multiplication by $∣ s ∣^{- (n - 1)}$ , the symbol of a Riesz potential of 02.13.02.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Prove the adjoint relation $\int_{S^{n - 1} \times R} R f (ω, p) g (ω, p) d p d ω = \int_{R^{n}} f (x) R^{#} g (x) d x$ for $f, g$ Schwartz.

Hint

Write $R f$ with the delta-form, substitute, and exchange the order of integration so the $p$ -integral acts first.

Answer

Using $R f (ω, p) = \int_{R^{n}} f (x) δ (p - ⟨ x, ω ⟩) d x$ , $$ \int_{S^{n-1}}!\int_{\mathbb{R}} Rf(\omega,p),g(\omega,p),dp,d\omega = \int_{S^{n-1}}!\int_{\mathbb{R}^n} f(x) \left(\int_{\mathbb{R}} \delta(p-\langle x,\omega\rangle)g(\omega,p),dp\right) dx, d\omega . $$ The inner $p$ -integral evaluates $g$ at $p = ⟨ x, ω ⟩$ , giving $g (ω, ⟨ x, ω ⟩)$ . Exchanging the $x$ and $ω$ integrals (Fubini, justified by Schwartz decay) leaves $\int_{R^{n}} f (x) \int_{S^{n - 1}} g (ω, ⟨ x, ω ⟩) d ω d x = \int_{R^{n}} f (x) R^{#} g (x) d x$ . Rubric: full credit for the delta-collapse and the Fubini exchange.

Exercise 4 (medium, short-answer).

Using the slice theorem, show that on the Fourier side $R^{#} R$ acts as multiplication by a constant times $∣ ξ ∣^{- (n - 1)}$ , and identify it as a Riesz potential of 02.13.02.

Hint

Backprojection averages slices over $ω$ ; on the Fourier side each slice contributes along the ray $R ω$ , and integrating over $S^{n - 1}$ in frequency introduces the Jacobian $∣ ξ ∣^{- (n - 1)}$ from polar coordinates.

Answer

By the slice theorem $R f$ lives along rays as $f (s ω)$ . Backprojection $R^{#}$ integrates over $ω \in S^{n - 1}$ ; converting the resulting frequency-space integral to polar coordinates $ξ = s ω$ supplies the factor $∣ s ∣^{n - 1} = ∣ ξ ∣^{n - 1}$ in the volume element, so the inverse appears in the operator: $R^{#} R f (ξ) = c_{n} ∣ ξ ∣^{- (n - 1)} f (ξ)$ . Multiplication by $∣ ξ ∣^{- (n - 1)}$ is the Fourier symbol of the Riesz potential $I_{n - 1} = (- Δ)^{- (n - 1) /2}$ of 02.13.02. Hence $R^{#} R = c_{n} (- Δ)^{- (n - 1) /2}$ , a smoothing of order $n - 1$ . Rubric: full credit for the polar-coordinate Jacobian and the Riesz-potential identification.

Exercise 5 (medium, numeric).

In the inversion formula the density of $f$ is recovered by applying the operator $(- Δ)^{(n - 1) /2}$ after backprojection. For which dimensions $n$ is this operator a genuine (positive integer power) differential operator acting locally?

Hint

$(- Δ)^{(n - 1) /2}$ is a local differential operator exactly when the exponent $(n - 1) /2$ is a non-negative integer.

Answer

$(n - 1) /2$ is an integer precisely when $n$ is odd. For odd $n$ the inversion uses an integer power of the Laplacian, a local differential operator, so the value of $f$ at a point depends only on data near the hyperplanes through that point. For even $n$ the exponent is a half-integer, the operator is the nonlocal fractional Laplacian (a Hilbert-transform / Riesz-potential factor enters), and reconstruction at a point needs all the data. The boundary case asked for "positive integer power, local" therefore holds exactly for odd $n \geq 1$ . Rubric: full credit for "odd $n$ " with the locality reason.

Exercise 7 (hard, short-answer).

State and prove the lowest moment condition in the Helgason–Ludwig range theorem: for $f \in S$ , the zeroth moment $\int_{R} R f (ω, p) d p$ is independent of $ω$ .

Hint

Relate the zeroth moment to $f (0)$ via the slice theorem.

Answer

By the slice theorem at $s = 0$ , $\int_{R} R f (ω, p) d p = (R f) (ω, 0) = f (0) = \int_{R^{n}} f (x) d x$ , which does not involve $ω$ . So the total mass read off each projection is the same for every direction — the zeroth Helgason–Ludwig moment condition. More generally the $k$ -th moment $\int_{R} p^{k} R f (ω, p) d p$ is a homogeneous polynomial of degree $k$ in $ω$ (differentiate the slice identity $k$ times in $s$ at $s = 0$ ); a function on the cylinder lies in the range of $R$ exactly when it is even, Schwartz, and satisfies all these polynomial-in- $ω$ moment conditions. Rubric: full credit for the $s = 0$ slice evaluation and the angle-independence conclusion; bonus for the general moment form.

Advanced results Master

The slice theorem turns Fourier inversion into a reconstruction formula. Writing Fourier inversion in polar frequency coordinates $ξ = s ω$ , $s \in R$ , $ω \in S^{n - 1}$ , with volume element $∣ s ∣^{n - 1} d s d ω$ , and substituting $f (s ω) = (R f) (ω, s)$ gives the filtered-backprojection inversion formula $$ f(x) = c_n \int_{S^{n-1}} \left( \Lambda^{n-1} Rf \right)(\omega, \langle x,\omega\rangle), d\omega = c_n, R^{#}!\left(\Lambda^{n-1} Rf\right)(x) , $$ where $Λ = (- \partial_{p}^{2})^{1/2}$ is the one-dimensional operator with Fourier multiplier $∣ s ∣$ acting in the offset variable, and $c_{n} = \frac{1}{2} (2 π)^{1 - n}$ . The filter $Λ^{n - 1}$ is the radial weight $∣ s ∣^{n - 1}$ promoted to an operator; backprojection $R^{#}$ does the angular averaging. Equivalently, in terms of the fractional Laplacian on $R^{n}$ of 02.13.02, $$ f = c_n, (-\Delta)^{(n-1)/2} R^{#} Rf , $$ which exhibits the inversion as the sharpening of the smoothing $R^{#} R = c_{n}^{- 1} (- Δ)^{- (n - 1) /2}$ established in Exercise 4. The two faces — filter the projections then backproject, or backproject then sharpen — are conjugate by the slice theorem.

The parity of $n$ governs the character of $Λ^{n - 1}$ , and this is the odd/even dichotomy that ties the Radon transform to the wave equation of 02.13.04. When $n$ is odd, $n - 1$ is even and $Λ^{n - 1} = (- \partial_{p}^{2})^{(n - 1) /2} = (- 1)^{(n - 1) /2} \partial_{p}^{n - 1}$ is a local differential operator; reconstruction of $f (x)$ then needs $R f$ only on hyperplanes through an arbitrarily small neighbourhood of $x$ . This locality is the Radon-transform shadow of the strong Huygens principle: in odd space dimensions a sharp wavefront leaves no tail, and the fundamental solution of the wave operator is supported exactly on the light cone (the spherical-means analysis of 02.13.04). When $n$ is even, $n - 1$ is odd and $Λ^{n - 1}$ carries a factor of the Hilbert transform $H$ (multiplier $- i sgn (s)$ ), which is nonlocal; reconstruction at a point requires data from all hyperplanes, and the corresponding wave propagation has a residual tail. The same parity that splits $(- Δ)^{(n - 1) /2}$ into local versus nonlocal is the parity that splits Huygens' principle into sharp versus diffuse.

The $L^{2}$ theory is the Plancherel identity for $R$ . From the slice theorem and the polar-coordinate Plancherel theorem of 02.10.04, $$ | f |{L^2(\mathbb{R}^n)}^2 = c_n' \int{S^{n-1}}!\int_{\mathbb{R}} \big| ,\widehat{|D_p|^{(n-1)/2} Rf},(\omega,s)\big|^2, ds, d\omega = c_n', \big| |D_p|^{(n-1)/2} Rf \big|_{L^2(S^{n-1}\times\mathbb{R})}^2 , $$ so $R$ becomes an isometry (up to constant) from $L^{2} (R^{n})$ onto a subspace of $L^{2} (S^{n - 1} \times R)$ after the half-order filter $∣ D_{p} ∣^{(n - 1) /2}$ . The smoothing order of $R$ — gaining $(n - 1) /2$ derivatives — is read directly off this identity and quantifies the ill-posedness of tomographic inversion: high-frequency components of $f$ are attenuated by $∣ s ∣^{- (n - 1) /2}$ in the data, so noise at high frequency is amplified on inversion, the analytic origin of the regularisation built into every practical reconstruction (Natterer).

The range of $R$ is characterised by the Helgason–Ludwig conditions: an even Schwartz function $g (ω, p)$ on the cylinder is $R f$ for some $f \in S$ if and only if, for each non-negative integer $k$ , the $k$ -th moment $\int_{R} p^{k} g (ω, p) d p$ extends to a homogeneous polynomial of degree $k$ in $ω$ . These polynomial-moment conditions are the integral-geometry analogue of a compatibility (integrability) condition: the data of all hyperplane integrals is enormously redundant, and the moment conditions pin down exactly which cylinder functions are consistent. In Gel'fand's formulation on the affine Grassmannian, the redundancy is encoded by a differential equation. For the John / X-ray transform $P f (ℓ) = \int_{ℓ} f$ over lines $ℓ$ in $R^{3}$ — the four-parameter family of lines exceeding the three dimensions of $R^{3}$ by one — the range is cut out by the ultrahyperbolic John equation $$ \frac{\partial^2 u}{\partial x_1,\partial y_2} - \frac{\partial^2 u}{\partial x_2,\partial y_1} = 0 $$ on the line coordinates $(x, y)$ , the prototype of a Gel'fand–Graev range condition. The Helgason support theorem sharpens injectivity into a localisation statement: if $f \in S (R^{n})$ decays faster than any polynomial and $R f (ω, p) = 0$ for every hyperplane that misses a fixed closed ball $B$ , then $f$ vanishes outside $B$ . Knowing the line-sums only of hyperplanes avoiding a region forces the function to be supported there — no mass can hide where every probing hyperplane reports zero.

The whole apparatus is one instance of Gel'fand's integral geometry on a homogeneous space. Replace the incidence "point $\in$ hyperplane" by a double fibration $X \leftarrow Z \to Ξ$ with $X = G / K$ a space of points and $Ξ = G / H$ a space of submanifolds (" $ξ$ -planes"), both homogeneous under a common group $G$ ; the transform $f \mapsto \int_{ξ} f$ and its dual $g \mapsto \int_{x \in ξ} g$ are then the two legs of the fibration, and inversion is an intertwining operator commuting with $G$ . The Euclidean hyperplane transform is the case $G = M (n)$ , the rigid-motion group; the Funk transform integrating over great circles of $S^{2}$ is the case $G = O (3)$ (where the kernel is the even functions, so antipodally-even data is the natural range); the d-plane / Grassmann transform integrates over affine $d$ -planes and is inverted by a power of the Laplacian depending on $d$ and $n$ . Helgason's programme is the systematic development of this picture for symmetric spaces, with the horocycle transform on $G / K$ as its non-Euclidean centrepiece.

Synthesis. The projection-slice theorem is the foundational reason every statement above coheres: it is the single identity from which inversion, Plancherel, injectivity, and the range conditions all descend, because it identifies the Radon data with the Fourier transform read in polar coordinates. The central insight is that backprojection and inversion differ by exactly the Riesz factor $(- Δ)^{(n - 1) /2}$ , so the smoothing $R^{#} R$ and the sharpening inversion are dual operations conjugate through the Fourier transform of 02.10.04; this is exactly the structure that makes the filtered-backprojection algorithm both correct and ill-posed, the half-order smoothing measured by the Plancherel identity. The odd/even-dimension split in the locality of $(- Δ)^{(n - 1) /2}$ generalises, and is dual to, the sharp-versus-diffuse split of Huygens' principle for the wave equation of 02.13.04: the same parity that makes the inversion filter a local differential operator makes the wave's fundamental solution live on the light cone. Putting these together, the Helgason–Ludwig moment conditions and the John ultrahyperbolic equation are the redundancy of an over-determined family of integrals made into a differential constraint, and the support theorem turns injectivity into localisation; the bridge from Euclidean tomography to Gel'fand's general integral geometry is the replacement of "point and hyperplane" by a double fibration of two homogeneous spaces, where the entire theory recurs with the rigid-motion group swapped for an arbitrary $G$ .

Full proof set Master

The projection-slice theorem and its corollaries are proved in the Key theorem and Exercises sections. The remaining Master claims are recorded here.

Proposition (filtered-backprojection inversion). For $f \in S (R^{n})$ , $$ f = c_n,(-\Delta)^{(n-1)/2} R^{#} R f, \qquad c_n = \tfrac{1}{2}(2\pi)^{1-n}, $$ equivalently $f = c_{n} R^{#} (Λ^{n - 1} R f)$ with $Λ = (- \partial_{p}^{2})^{1/2}$ .

Proof. Start from Fourier inversion on $R^{n}$ in polar frequency coordinates. Writing $ξ = s ω$ with $s \in [0, \infty)$ and $ω \in S^{n - 1}$ , the Lebesgue measure is $d ξ = s^{n - 1} d s d ω$ , and since $f (s ω)$ and $f (- s ω) = f (s (- ω))$ both occur, symmetrising over $ω \in S^{n - 1}$ extends the radial variable to all of $R$ with weight $\frac{1}{2} ∣ s ∣^{n - 1}$ : $$ f(x) = \int_{\mathbb{R}^n} \widehat f(\xi), e^{2\pi i \langle x,\xi\rangle}, d\xi = \tfrac12 \int_{S^{n-1}}!\int_{\mathbb{R}} \widehat f(s\omega), e^{2\pi i s\langle x,\omega\rangle},|s|^{n-1}, ds, d\omega . $$ Insert the slice theorem $f (s ω) = (R f) (ω, s)$ : $$ f(x) = \tfrac12 \int_{S^{n-1}} \left( \int_{\mathbb{R}} |s|^{n-1}, \widetilde{(Rf)}(\omega,s), e^{2\pi i s\langle x,\omega\rangle},ds \right) d\omega . $$ The inner integral is the inverse one-dimensional Fourier transform, in the variable conjugate to $s$ , of $∣ s ∣^{n - 1} (R f) (ω, s)$ evaluated at $p = ⟨ x, ω ⟩$ . Multiplication by $∣ s ∣^{n - 1}$ on the Fourier side is the operator $Λ^{n - 1}$ with $Λ = (- \partial_{p}^{2})^{1/2}$ acting in $p$ , so the inner integral equals $(Λ^{n - 1} R f) (ω, ⟨ x, ω ⟩)$ . Hence $$ f(x) = \tfrac12 \int_{S^{n-1}} \big(\Lambda^{n-1} Rf\big)(\omega, \langle x,\omega\rangle), d\omega = \tfrac12, R^{#}\big(\Lambda^{n-1} Rf\big)(x). $$ Absorbing the Fourier normalisation constants into $c_{n} = \frac{1}{2} (2 π)^{1 - n}$ and recognising, by Exercise 4, that the angular average of the radially-filtered slices realises $(- Δ)^{(n - 1) /2} R^{#} R$ gives the stated equivalent forms. $□$

Proposition (the dual-composition intertwining $R^{#} R$ ). For $f \in S (R^{n})$ , $$ (R^{#} R f)(x) = \frac{2,(2\pi)^{n-1}}{;}, \big(f * |x|^{-1}\big)(x)\ \text{up to a dimensional constant}, \qquad \widehat{R^{#}Rf}(\xi) = c_n^{-1},|\xi|^{-(n-1)},\widehat f(\xi). $$

Proof. By the slice theorem, $R f$ is carried on rays. Computing $R^{#} R$ on the Fourier side: $R^{#}$ averages the slices over $ω \in S^{n - 1}$ , and each slice $(R f) (ω, s) = f (s ω)$ contributes to the frequency $ξ = s ω$ . Reassembling the angular average as a frequency integral and passing to polar coordinates $ξ = s ω$ with element $s^{n - 1} d s d ω$ , the absence of the radial weight in the plain backprojection (which integrates $d ω$ only) leaves a deficit of $∣ s ∣^{n - 1} = ∣ ξ ∣^{n - 1}$ relative to full Fourier inversion. Thus $R^{#} R f (ξ) = c_{n}^{- 1} ∣ ξ ∣^{- (n - 1)} f (ξ)$ . The multiplier $∣ ξ ∣^{- (n - 1)}$ is, by 02.13.02, the Fourier symbol of the Riesz potential $I_{n - 1}$ , whose kernel is a constant times $∣ x ∣^{- (n - (n - 1))} = ∣ x ∣^{- 1}$ ; convolving $f$ with that kernel gives the stated spatial form up to the dimensional Riesz constant. $□$

Proposition (Helgason support theorem). Let $f \in C (R^{n})$ decay faster than every power $∣ x ∣^{- k}$ , and let $B$ be a closed ball. If $R f (ω, p) = 0$ for every hyperplane ${⟨ x, ω ⟩ = p}$ disjoint from $B$ , then $f (x) = 0$ for all $x \in / B$ .

Proof (Helgason's reduction to the support theorem on spheres). Translate so $B$ is centred at the origin with radius $ρ$ . Fix a point $x_{0}$ with $∣ x_{0} ∣ > ρ$ ; the claim is $f (x_{0}) = 0$ . After a rotation assume $x_{0}$ lies on a coordinate axis. Consider the family of spheres through $x_{0}$ enclosing $B$ ; Helgason shows that the hypothesis " $R f = 0$ on hyperplanes missing $B$ " forces the spherical means of $f$ over all sufficiently large spheres centred near $x_{0}$ to vanish, by writing a hyperplane integral as a limit of integrals over large spheres and using the decay of $f$ to control the limit. The vanishing of these spherical means, together with the rapid decay, is exactly the hypothesis of the classical spherical support theorem (a consequence of the injectivity of the spherical-mean transform on rapidly decreasing functions), which yields $f \equiv 0$ outside the convex hull of $B$ . Since $B$ is already convex, $f (x_{0}) = 0$ . The decay hypothesis cannot be dropped: there exist non-zero $f$ decaying like a fixed power for which $R f$ vanishes on hyperplanes missing a ball, so the support theorem is genuinely a statement about the Schwartz/rapidly-decreasing class. $□$

Proposition (John's equation as a range condition for the X-ray transform). Parametrise an oriented line in $R^{3}$ by a point $y$ on the plane $x_{3} = 0$ and a direction; in suitable affine coordinates $(x_{1}, x_{2}, y_{1}, y_{2})$ for the four-dimensional line manifold, the X-ray data $u = P f$ satisfies $u_{x_{1} y_{2}} - u_{x_{2} y_{1}} = 0$ . Conversely a function on an open set of lines satisfying this equation and the decay conditions is locally $P f$ for some $f$ .

Proof (necessity). Write a line as $t \mapsto (y_{1} + t x_{1}, y_{2} + t x_{2}, t)$ and $u (x, y) = \int_{R} f (y_{1} + t x_{1}, y_{2} + t x_{2}, t) d t$ . Differentiating under the integral, $u_{y_{1}} = \int \partial_{1} f d t$ and $u_{x_{2} y_{1}} = \int t \partial_{2} \partial_{1} f d t$ , while $u_{x_{1} y_{2}} = \int t \partial_{1} \partial_{2} f d t$ . Since mixed partials of $f$ commute, $\partial_{1} \partial_{2} f = \partial_{2} \partial_{1} f$ , the two second derivatives of $u$ coincide, giving $u_{x_{1} y_{2}} - u_{x_{2} y_{1}} = 0$ . The ultrahyperbolic operator on the left is John's; the converse (local surjectivity onto its kernel) is John's theorem, proved by exhibiting the inverse via the dual transform and verifying the moment/compatibility conditions, and is stated here with the construction referred to Gel'fand–Graev–Vilenkin. $□$

Connections Master

The Fourier transform and Plancherel theorem 02.10.04 are the engine of this entire unit. The projection-slice theorem says the Radon transform is the Fourier transform read along rays, so Fourier inversion is the inversion formula once written in polar frequency coordinates, and the $L^{2}$ Plancherel identity for $R$ is the Plancherel theorem of 02.10.04 pushed through that change of variables. Every smoothing-order and ill-posedness statement here is a frequency-attenuation statement there.

The area and coarea formulas 02.07.11 supply the measure on each hyperplane and the decomposition $d x = d p d S (y)$ that makes the projection-slice proof a single change of order of integration. The Radon transform is, by definition, an integral over level sets of the linear function $x \mapsto ⟨ x, ω ⟩$ , and the coarea formula is exactly the tool for integrating over level sets; without it the slice theorem would be a formal manipulation rather than a proved identity.

The Poisson equation, Newtonian potential, and Riesz potentials 02.13.02 furnish the operator $(- Δ)^{(n - 1) /2}$ that converts backprojection into inversion. The composition $R^{#} R$ is a Riesz potential of order $n - 1$ — the Fourier multiplier $∣ ξ ∣^{- (n - 1)}$ that 02.13.02 identifies with convolution against a power of $∣ x ∣$ — and inverting it is precisely applying the fractional Laplacian developed there. The Newtonian potential of 02.13.02 is the $n - 1 = 2$ instance, linking tomographic backprojection to classical potential theory.

The wave equation, spherical means, and Huygens' principle 02.13.04 share the Radon transform's odd/even-dimension dichotomy at a structural level. The decomposition of a solution into plane waves is a Radon synthesis: $f = c_{n} (- Δ)^{(n - 1) /2} R^{#} R f$ rebuilds an initial datum from its plane-wave (hyperplane) components, and the parity that makes the inversion filter local in odd dimensions is the parity that makes the wave's fundamental solution supported on the light cone there. The method of plane waves solves the Cauchy problem of 02.13.04 by Radon-transforming the data, solving a one-dimensional wave equation per direction, and backprojecting.

The theory of distributions and the Schwartz kernel theorem 02.14.04 is the setting in which $R$ extends beyond Schwartz functions. The delta-form $R f (ω, p) = ⟨ f, δ (p - ⟨ \cdot, ω ⟩)⟩$ defines $R$ on tempered distributions, the dual transform is the adjoint in the distributional pairing, and the inversion and support theorems are most cleanly stated and proved in the distributional category developed there. The John ultrahyperbolic equation is a statement about a distribution on the line manifold.

Historical & philosophical context Master

Johann Radon solved the planar reconstruction problem in 1917 (Über die Bestimmung von Funktionen durch ihre Integralwerte längs gewisser Mannigfaltigkeiten, Berichte über die Verhandlungen der Sächsischen Akademie der Wissenschaften zu Leipzig 69, 262–277) ^{[Radon 1917]}, giving the inversion of the line-integral transform in $R^{2}$ and noting the extension to hyperplanes in higher dimensions; the work sat largely unread by applied scientists for half a century. Fritz John developed the higher-dimensional and line-complex theory in the 1930s–1940s, isolating the ultrahyperbolic equation that bears his name as the compatibility condition for the over-determined X-ray data in $R^{3}$ . The Soviet school of Israel Gel'fand, Mark Graev, and Naum Vilenkin recast the subject as integral geometry on homogeneous spaces in Volume 5 of Generalized Functions (1962, English translation 1966) ^{[Gel'fand-Graev-Vilenkin 1966]}, tying the inversion problem to the representation theory of the underlying transformation group and treating the line and plane complexes through the Plücker incidence relations. Sigurður Helgason then built the systematic theory of the Radon transform on symmetric spaces, proving the support theorem and developing the horocycle transform as the non-Euclidean analogue, collected in The Radon Transform (1980; 2nd ed. 1999) ^{[Helgason 1999]}.

The applied vindication came with Allan Cormack's independent rederivation of the inversion formula in 1963–1964 in the context of medical imaging and Godfrey Hounsfield's engineering of the first computed-tomography scanner in 1971; the two shared the 1979 Nobel Prize in Physiology or Medicine, neither initially aware of Radon's paper. The mathematical content that a body is determined by the totals of all lines through it, settled in principle in 1917, became a clinical instrument only once the half-order ill-posedness measured by the Plancherel identity could be tamed by the sampling and regularisation theory that Frank Natterer and others systematised ^{[Natterer 1986]}. Gel'fand's reframing made the parochial fact about hyperplanes in $R^{n}$ a single point in a representation-theoretic landscape: the Funk transform on the sphere, the horocycle transform on hyperbolic space, and the Penrose transform of twistor theory are siblings under the double-fibration formalism, each an instance of recovering a function from its integrals over a homogeneous family of submanifolds.

Bibliography Master

@article{radon1917,
  author  = {Radon, Johann},
  title   = {{\"U}ber die Bestimmung von Funktionen durch ihre Integralwerte l{\"a}ngs gewisser Mannigfaltigkeiten},
  journal = {Berichte {\"u}ber die Verhandlungen der S{\"a}chsischen Akademie der Wissenschaften zu Leipzig, Mathematisch-Physische Klasse},
  volume  = {69},
  pages   = {262--277},
  year    = {1917}
}

@book{helgason1999radon,
  author    = {Helgason, Sigurdur},
  title     = {The Radon Transform},
  edition   = {2nd},
  series    = {Progress in Mathematics},
  volume    = {5},
  publisher = {Birkh\"auser, Boston},
  year      = {1999}
}

@book{gelfandgraevvilenkin1966,
  author    = {Gel'fand, Israel M. and Graev, Mark I. and Vilenkin, Naum Ya.},
  title     = {Generalized Functions, Volume 5: Integral Geometry and Representation Theory},
  publisher = {Academic Press, New York},
  year      = {1966},
  note      = {Translated by E. Saletan}
}

@book{natterer1986,
  author    = {Natterer, Frank},
  title     = {The Mathematics of Computerized Tomography},
  publisher = {Wiley/B.G. Teubner, Stuttgart},
  year      = {1986},
  note      = {SIAM Classics in Applied Mathematics reprint, 2001}
}

@article{john1938,
  author  = {John, Fritz},
  title   = {The ultrahyperbolic differential equation with four independent variables},
  journal = {Duke Mathematical Journal},
  volume  = {4},
  number  = {2},
  pages   = {300--322},
  year    = {1938}
}

@article{cormack1963,
  author  = {Cormack, Allan M.},
  title   = {Representation of a function by its line integrals, with some radiological applications},
  journal = {Journal of Applied Physics},
  volume  = {34},
  number  = {9},
  pages   = {2722--2727},
  year    = {1963}
}

Prerequisites

02.10.04
02.07.11
02.13.02
02.14.04

Tier anchors

beginner: Strang Introduction to Linear Algebra / CT-scan motivation; Epstein Introduction to the Mathematics of Medical Imaging Ch. 1 (informal X-ray picture)
intermediate: Helgason The Radon Transform (2nd ed., Birkhäuser 1999) Ch. I §§1-3; Natterer The Mathematics of Computerized Tomography Ch. II §§1-2
master: Helgason Groups and Geometric Analysis (AMS 2000) Ch. I; Helgason The Radon Transform Ch. I-II; Gel'fand-Graev-Vilenkin Generalized Functions Vol. 5 Ch. I; Natterer Ch. II-V

References

Helgason — The Radon Transform (2nd ed., Birkhäuser, 1999) · Ch. I §§1-4 (inversion, dual transform, support theorem); Ch. II (Grassmann / d-plane transform)
Gel'fand, Graev, Vilenkin — Generalized Functions, Vol. 5: Integral Geometry and Representation Theory (Academic Press, 1966) · Ch. I (the line/plane complex, inversion); the John ultrahyperbolic equation as a range condition
Natterer — The Mathematics of Computerized Tomography (Wiley/Teubner, 1986; SIAM reprint 2001) · Ch. II §1 (Radon transform, projection-slice), §2 (filtered backprojection), Ch. V (sampling)
Radon — Über die Bestimmung von Funktionen durch ihre Integralwerte längs gewisser Mannigfaltigkeiten (1917) · Berichte Sächs. Akad. Wiss. Leipzig 69, pp. 262-277; the original inversion formula in the plane

Estimated time

beginner: 18m
intermediate: 45m
master: 80m