Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 326 / 3 / 1 / a

First interpret the stated well-definedness as including uniqueness of the global minimizer for each positive parameter. Under that assumption, the parameter continuity of variational regularization follows from the direct method in the calculus of variations argument below.

Let $M=D(0)=\Vert f{\Vert }^2/2$. For large $n$, ${\alpha }_n\ge \beta :=\alpha /2$. Comparison with zero, using $J(0)=0$, givesCoercivity of ${\Phi }_{\beta ,f}$ therefore bounds the sequence. Take a $\tau$-convergent subsequence with limit $\bar{u}$. For any $v\in \mathrm{dom}J$, optimality givesThe last two terms vanish. Sequential lower semicontinuity at the fixed parameter $\alpha$ implies ${\Phi }_{\alpha ,f}(\bar{u})\le {\Phi }_{\alpha ,f}(v)$. Thus every cluster point is a global minimizer at $\alpha$. Uniqueness makes it $u_{\alpha }$, and the subsequence property provesIndeed, a subsequence staying outside a neighborhood of $u_{\alpha }$ would have a further convergent subsequence, whose limit must be $u_{\alpha }$, a contradiction.

The listed existence hypotheses alone do not imply uniqueness or convergence of an arbitrary selection. For example, on ${\mathbb{R}}^2$, set $K(x,y)=x$, $f=1$, andThis is nonnegative, convex, continuous and has $J(0)=0$; every ${\Phi }_{\alpha ,f}$ has coercivity. Its minimizers are $(1/(1+2\alpha ),y)$ for all $|y|\le 1$. Along ${\alpha }_n=\alpha +1/n$, selecting $y_n=(-1{)}^n$ prevents convergence. Without uniqueness, the valid conclusion is that every cluster point minimizes the limiting objective.