Euclidean Radon transform and filtered backprojection

Fix $\theta$ and define $T(s,t)=s\theta +t{\theta }^{\perp }$ and $F(s,t)=f(T(s,t))$. The columns of $T$ are an orthonormal positively oriented basis, so $T$ is a linear diffeomorphism and $K:=T^{-1}(\mathrm{supp}f)$ is compact. Choose $M>0$ with $K\subset [-M,M{]}^2$. Then $F(s,t)=0$ for $\mathrm{\mid }t\mathrm{\mid }>M$, and $Rf(\theta ,s)={\int }_{-M}^{M}F(s,t)dt$; it vanishes unless $s$ belongs to the compact projection of $K$ onto the first coordinate.

Write $f=f_{\mathrm{R}}+i{f}_{\mathrm{I}}$ and $F_{\nu }=f_{\nu }\circ T$ for $\nu \in \{\mathrm{R},\mathrm{I}\}$. Each $f_{\nu }$ is real smooth; since $DT$ is constant, repeated applications of [F6] give ${\mathrm{\partial }}_s^k{F}_{\nu }(s,t)=(\theta \cdot \mathrm{\nabla }{)}^k{f}_{\nu }(T(s,t))$. Thus, for each integer $k\ge 0$, ${\mathrm{\partial }}_s^{k}F$ is continuous. On every compact $s$-interval, ${\mathrm{\partial }}_s^{k+1}F$ is uniformly continuous on the product with $[-M,M]$. For each real component, apply [F5] on $[0,1]$ to $u\mapsto {\mathrm{\partial }}_s^k{F}_{\nu }(s+uh,t)$. By [F6] its derivative is $h{\mathrm{\partial }}_s^{k+1}{F}_{\nu }(s+uh,t)$, continuous and hence integrable by [F2], with the same one-sided derivatives at the endpoints. Combining the two real equalities and dividing by $h$ gives, for sufficiently small nonzero $h$ of either sign, $\frac{{\mathrm{\partial }}_s^{k}F(s+h,t)-{\mathrm{\partial }}_s^{k}F(s,t)}{h}={\int }_0^1{\mathrm{\partial }}_s^{k+1}F(s+uh,t)du\mathrm{.}$ The right side converges uniformly for $t\in [-M,M]$ to ${\mathrm{\partial }}_s^{k+1}F(s,t)$ as $h\to 0$. Integrating in $t$ therefore proves inductively that $\frac{d^k}{d{s}^k}Rf(\theta ,s)={\int }_{-M}^M{\mathrm{\partial }}_s^{k}F(s,t)dt$ for every $k$, and uniform continuity also makes each derivative continuous in $s$. Thus $Rf(\theta ,\cdot )\in C_c^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$, so its Fourier transform in [F1] is defined.

Put $H(s,t)=f(T(s,t))e^{-2\pi i\sigma s}$. This is continuous and supported in $K$. Choose a nondegenerate product rectangle containing $K$. By [F2], the real and imaginary parts of $H$ are Riemann integrable there, and all their one-variable sections are continuous and hence Riemann integrable.

By [F1], $\widehat{Rf}(\theta ,\sigma )$ is the iterated integral of $H$, first in $t$ and then in $s$. Apply the real-valued Riemann-Fubini theorem [F3] separately to $\mathrm{Re}H$ and $\mathrm{Im}H$ on the chosen rectangle, then combine the two equalities. Since $H$ vanishes outside that rectangle, this gives $\widehat{Rf}(\theta ,\sigma )={\int }_{{\mathbb{R}}^2}f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt\mathrm{.}$

Apply [F4] to $T$ separately for the real and imaginary parts of the compactly supported Riemann integrand in step 3.1. Since $s=x\cdot \theta$ and $\mathrm{\mid }\det DT\mathrm{\mid }=1$, the integral becomes ${\int }_{{\mathbb{R}}^2}f(x)e^{-2\pi i\sigma x\cdot \theta }dx=\widehat{f}(\sigma \theta )$ by [F1]. This is the claimed identity.

Since $f\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$, it is a Schwartz function. Choose $R>0$ with $\mathrm{supp}f\subset B_R(0)$. For every $\theta \in S^1$ and $s\in \mathbb{R}$, the integrand $f(s\theta +t{\theta }^{\perp })$ vanishes for $\mathrm{\mid }t\mathrm{\mid }>R$, because $\mathrm{\mid }s\theta +t{\theta }^{\perp }{\mathrm{\mid }}^2=s^2+t^2$. Thus the line integral is over the fixed interval $[-R,R]$. On every bounded $s$-interval, $(\theta ,s,t)\mapsto f(s\theta +t{\theta }^{\perp })$ is uniformly continuous on the compact parameter product with $S^1$ and $[-R,R]$, so its integral is jointly continuous in $(\theta ,s)$. For fixed $\theta$, write $f=f_{\mathrm{R}}+i{f}_{\mathrm{I}}$ and $F_{\nu }(s,t)=f_{\nu }(s\theta +t{\theta }^{\perp })$ for $\nu \in \{\mathrm{R},\mathrm{I}\}$. The inner coordinate map is linear with constant derivative, so repeated use of [F12] gives ${\mathrm{\partial }}_s^k{F}_{\nu }(s,t)=(\theta \cdot \mathrm{\nabla }{)}^k{f}_{\nu }(s\theta +t{\theta }^{\perp })$ for every $k$. Apply [F11] on $[0,1]$ to each real function $u\mapsto {\mathrm{\partial }}_s^k{F}_{\nu }(s+uh,t)$: by [F12] its derivative is $h{\mathrm{\partial }}_s^{k+1}{F}_{\nu }(s+uh,t)$, which is continuous and Riemann integrable by [F13], including at the endpoints. Dividing the resulting identity by any nonzero $h$ gives the difference quotient as the $u$-integral of ${\mathrm{\partial }}_s^{k+1}{F}_{\nu }(s+uh,t)$. Uniform continuity on each compact $(s,t)$ rectangle makes this converge uniformly in $t\in [-R,R]$; integrating and recombining the two real components therefore gives ${\mathrm{\partial }}_s^{k}Rf(\theta ,s)={\int }_{-R}^R(\theta \cdot \mathrm{\nabla }{)}^{k}f(s\theta +t{\theta }^{\perp })dt$ for every $k\ge 0$, with each derivative continuous. Also $Rf(\theta ,s)=0$ for $\mathrm{\mid }s\mathrm{\mid }>R$, so $Rf(\theta ,\cdot )\in C_c^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$ and is Schwartz. For every $x$, joint continuity makes $\theta \mapsto Rf(\theta ,x\cdot \theta )$ continuous; [F4] makes $\mu$ finite, so the defining backprojection integral exists. By [F6] and [F7], the Lebesgue Fourier transforms of $f$ and $Rf(\theta ,\cdot )$ are Schwartz and satisfy pointwise inversion; [F8] identifies them with the compact-support Riemann transforms in [F1]. Also $\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }$, and [F4] shows $\mathrm{\mid }\xi {\mathrm{\mid }}^{-1}\widehat{f}(\xi )$ is integrable near $0$; outside the unit disk it is dominated by the integrable Schwartz function $\mathrm{\mid }\widehat{f}\mathrm{\mid }$. Rapid decay of $\widehat{f}$ and finiteness of $\mu$ make the later signed-frequency integrals absolutely convergent.

Apply [F7] to $Rf(\theta ,\cdot )$ at $s=x\cdot \theta$, then substitute the slice identity [F2] to get $Rf(\theta ,x\cdot \theta )={\int }_{\mathbb{R}}\widehat{f}(\sigma \theta )e^{2\pi i\sigma x\cdot \theta }d\sigma$. By [F4], $\mu$ is finite, and by [F14] Lebesgue measure on $\mathbb{R}$ is sigma-finite, so the measure spaces in [F9] are sigma-finite. The integrand is continuous; since $\widehat{f}$ is Schwartz, $\mathrm{\mid }\widehat{f}(\sigma \theta )\mathrm{\mid }\le C_N(1+\mathrm{\mid }\sigma \mathrm{\mid }{)}^{-N}$ uniformly in $\theta$, and choosing $N>1$ shows it is absolutely integrable on their product. Thus [F9] permits either order of integration and gives $R^{\ast }Rf(x)={\int }_{S^1}{\int }_{\mathbb{R}}\widehat{f}(\sigma \theta )e^{2\pi i\sigma x\cdot \theta }d\sigma d\mu (\theta )$.

The polar definition [F3] and change of variables [F5] imply $\mu (-E)=\mu (E)$ for every Borel $E\subseteq S^1$, since the cone of $-E$ is the image under $-I$ of the cone of $E$ and $\mathrm{\mid }\det (-I)\mathrm{\mid }=1$. Thus $\theta \mapsto -\theta$ preserves $\mu$. The reflection $\sigma \mapsto -\sigma$ preserves Lebesgue measure by [F5] in dimension one. Both maps are Borel, so [F15] permits these substitutions in the integrable angular and frequency integrands. Split the $\sigma$-integral in step 2.1 at zero and use $r=-\sigma$ and $\theta \mapsto -\theta$ on its negative half to obtain twice the positive-half integral. Apply [F4] to $h(\xi )=\widehat{f}(\xi )e^{2\pi ix\cdot \xi }\mathrm{/}\mathrm{\mid }\xi \mathrm{\mid }$, set to zero at $\xi =0$; for complex $h$, apply the nonnegative formula to positive and negative real and imaginary parts and use [F10]. Thus $R^{\ast }Rf(x)=2{\int }_{{\mathbb{R}}^2}\widehat{f}(\xi )e^{2\pi ix\cdot \xi }\mathrm{\mid }\xi {\mathrm{\mid }}^{-1}d{\lambda }_2(\xi )=4\pi \mathrm{\mid }D{\mathrm{\mid }}^{-1}f(x)$.

By [F1] and [F2], $\mathrm{\mid }{D}_s\mathrm{\mid }Rf(\theta ,s)={\int }_{\mathbb{R}}2\pi \mathrm{\mid }\sigma \mathrm{\mid }\widehat{f}(\sigma \theta )e^{2\pi is\sigma }d\sigma$. Schwartz decay gives a bound $\mathrm{\mid }\widehat{f}(\sigma \theta )\mathrm{\mid }\le C_N(1+\mathrm{\mid }\sigma \mathrm{\mid }{)}^{-N}$ uniform in $\theta$. On each compact parameter set $S^1\times [-A,A]$, the truncated integrals over bounded $\sigma$-intervals are continuous in $(\theta ,s)$, while the displayed bound makes their tails uniformly tend to zero; hence the filtered data are jointly continuous. The same bound gives absolute convergence and boundedness, so finiteness of $\mu$ makes the backprojection integral exist. For fixed $x$, the product integrand $2\pi \mathrm{\mid }\sigma \mathrm{\mid }\widehat{f}(\sigma \theta )e^{2\pi i\sigma x\cdot \theta }$ is continuous and absolutely integrable by the same bound with $N>2$. The measure spaces are sigma-finite as established in step 2.1; Fubini [F9], the signed-frequency split in step 3.1, and [F4] give $R^{\ast }\mathrm{\mid }{D}_s\mathrm{\mid }Rf(x)=4\pi {\int }_{{\mathbb{R}}^2}\widehat{f}(\xi )e^{2\pi ix\cdot \xi }d{\lambda }_2(\xi )=4\pi f(x)$ by [F7].

Step 3.1 is the normal-operator identity, and dividing step 4.1 by $4\pi$ gives the filtered-backprojection identity. Both hold pointwise for every $x\in {\mathbb{R}}^2$.