Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 2

For $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe corresponding bounded-variation space and its zero-mean subspace areIf $u_{\Omega }=|\Omega {|}^{-1}{\int }_{\Omega }udx$, the Poincaré inequality for total variation isIn particular, $\Vert u{\Vert }_{L^1}\le C_{\Omega }\mathrm{TV}(u)$ for $u\in B{V}_0(\Omega )$.

A $J$-minimizing solution is an exact solution $u_J^{†}$ satisfyingThe source condition in variational regularization says that some $p^{†}\in Y$ satisfieswhere $\partial J$ is the subdifferential.

Fix $\alpha >0$. The subgradient optimality condition forat $u_J^{†}$ isIf the source condition holds, choose $g=f+\alpha p^{†}$. Since $A{u}_J^{†}=f$, the displayed inclusion holds. Because the objective is a convex function, this condition is necessary and sufficient for global minimality.

Conversely, if the stated range condition holds for some $g$, optimality supplies $\xi \in \partial J(u_J^{†})$ withThus $\xi =A^{\ast }((g-f)/\alpha )$, which is precisely the source condition. This proves the equivalence.

For an absolutely one-homogeneous functional $J$, the subdifferential has the characterizationIf $c>0$, then $J(cu)=cJ(u)$, soThe second condition does not involve the base point. Therefore

Because $\lambda f\in \partial J(f)$ and $1-\lambda \alpha >0$ when $\alpha <1/\lambda$, part a givesSet $u=(1-\lambda \alpha )f$. ThenThis is the subgradient optimality condition forThe squared norm is strictly convex, so the objective has at most one minimizer. Henceis its unique minimizer.