Use the <singular system of a compact operator> convention $Ax_j=\sigma_jy_j$ and $A^*y_j=\sigma_jx_j$. Starting from zero gives no component in $\ker A$. Write $x(t)=\sum_jc_j(t)x_j$. The coefficient equations are
$$
c_j'=-\sigma_j^2c_j+\sigma_j\langle f,y_j\rangle,\qquad c_j(0)=0.
$$
Solving these scalar <linear ordinary differential equations> gives
$$
c_j(t)=\frac{1-e^{-\sigma_j^2t}}{\sigma_j}\langle f,y_j\rangle.
$$
Consequently <asymptotic regularization> has the <spectral regularization method> representation
$$
\boxed{u_\alpha=\sum_j\frac{1-e^{-\sigma_j^2/\alpha}}{\sigma_j}\langle f,y_j\rangle x_j.}
$$
The component of $f$ orthogonal to the closure of the range of $A$ is annihilated by $A^*$ and contributes nothing. The scalar filter satisfies $(1-e^{-\sigma^2/\alpha})/\sigma\leq\min(\sigma/\alpha,1/\sigma)\leq\alpha^{-1/2}$, so every fixed-$\alpha$ reconstruction is a bounded <linear operator>. Its long-time limit is the <Moore–Penrose inverse of an operator> on admissible data.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 326 / 1 / c / Solution

Asymptotic regularization stops the <gradient flow> $x'=-A^*(Ax-f)$ at $t=1/\alpha$, starting from zero. With the <singular system of a compact operator> convention $Av_j=\sigma_j u_j$, its spectral filter gives
$$
R_\alpha f=\sum_j\frac{1-e^{-\sigma_j^2/\alpha}}{\sigma_j}\langle f,u_j\rangle v_j.
$$
The scalar coefficient solves $c_j'=-\sigma_j^2c_j+\sigma_j\langle f,u_j\rangle$ with zero initial value. Since $1-e^{-s}\leq\min(s,1)$ for $s\geq0$, the <operator norm> is at most $\alpha^{-1/2}$. On the domain of the <Moore–Penrose inverse of an operator>, <dominated convergence theorem> of the squared spectral coefficients proves $R_\alpha f\to A^\dagger f$. The <noise-bias decomposition for linear regularization> then gives noisy-data convergence when $\delta/\sqrt\alpha\to0$.

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