Solution

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

For a bounded <linear operator> $A$ between <Hilbert spaces>, minimize $\|Au-f\|^2+\alpha\|u\|^4$, with $\alpha>0$. The objective is coercive and weakly lower semicontinuous, and its <strictly convex> norm penalty gives a unique minimizer. Its stationarity equation is
$$
A^*(Au-f)+2\alpha\|u\|^2u=0,
$$
so the map from data to solution is generally nonlinear. For admissible fixed $f$, comparison with $u^\dagger=A^\dagger f$ bounds the minimizer's norm by $\|u^\dagger\|$ and makes its squared residual converge to the minimum least-squares residual. Every weak cluster point is therefore a least-squares solution of norm at most $\|u^\dagger\|$, hence equals the <minimum-norm least-squares solution>. Weak lower semicontinuity then forces convergence of norms, and the <Radon-Riesz theorem> gives strong convergence. This supplies a nonlinear <regularization of an inverse problem>.

For completeness, the fixed-$\alpha$ data-to-solution map is continuous. For $g(u)=\|u\|^2u$, setting $m=(u+v)/2$ and $d=(u-v)/2$ gives
$$
\operatorname{Re}\langle g(u)-g(v),u-v\rangle
=4(\|m\|^2+\|d\|^2)\|d\|^2+8(\operatorname{Re}\langle m,d\rangle)^2
\geq\tfrac14\|u-v\|^4.
$$
Subtracting the stationarity equations for data $f,h$ and pairing with $u-v$ therefore gives $\frac\alpha2\|u-v\|^4\leq\|A\|\|f-h\|\|u-v\|$. Thus
$$
\|R_\alpha f-R_\alpha h\|\leq\left(\frac{2\|A\|}{\alpha}\|f-h\|\right)^{1/3},
$$
which proves continuity directly.

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 Incoming links: Quartic-norm variational regularization