Use
$$
(\mathcal Fg)(\mathbf q)=\int_{\mathbb R^3}
g(\mathbf r)e^{-i\mathbf q\cdot\mathbf r}d\mathbf r,
\qquad
\mathcal F^{-1}h=\frac1{(2\pi)^3}\int h(\mathbf q)e^{i\mathbf q\cdot\mathbf r}d\mathbf q.
$$
The <adjoint operator> is consequently
$$
\boxed{\mathcal F^*=(2\pi)^3\mathcal F^{-1}.}
$$
With a unitary Fourier normalization, this is simply $\mathcal F^*=\mathcal F^{-1}$.

At fixed incident direction and wavenumber, define
$$
(TV)(\widehat{\mathbf r})
=-\frac1{4\pi}\int_D
e^{-ik_0(\widehat{\mathbf r}-\widehat{\mathbf x}_0)\cdot\mathbf r}
V(\mathbf r)d\mathbf r.
$$
Taking the complex conjugate of the kernel gives
$$
\boxed{
(T^*g)(\mathbf r)
=-\frac1{4\pi}\int_{S^2}
e^{ik_0(\widehat{\mathbf r}-\widehat{\mathbf x}_0)\cdot\mathbf r}
g(\widehat{\mathbf r})dS(\widehat{\mathbf r}),
\qquad \mathbf r\in D.}
$$
The least-squares minimizer of $\|TV-f_\infty\|^2$ satisfies the <normal equation for a linear inverse problem> $T^*TV=T^*f_\infty$. Fourier inversion on the measured transfer-vector set is exactly the corresponding Moore--Penrose reconstruction $T^\dagger f_\infty$; hence the formal solution in part (i) is the minimum-norm least-squares solution when the data are incomplete or inconsistent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 335 / 2 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-335/2/ii/solution