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

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.

Use the value function of variational regularization in its selection-independent formBecause $J\ge 0$, each objective is non-decreasing with $\alpha$, hence so is its infimum. For $0\le t\le 1$, the affine function of the parameter at each fixed $u$ givesThus $\Psi$ is concave.

For continuity, on any interval $0<a\le \alpha <\beta \le b$, comparison using $u_{\alpha }$ givesIt follows that $\Psi$ is a locally Lipschitz function, hence continuous, on $(0,\infty )$. Therefore the value function of variational regularization is non-decreasing, concave and continuous. No uniqueness of minimizers is needed for these conclusions.

For $0<\alpha <\beta$, abbreviate $D_{\alpha }=D(u_{\alpha })$ and $J_{\alpha }=J(u_{\alpha })$. Optimality of the two variational regularization solutions givesAdding shows $(\beta -\alpha )(J_{\beta }-J_{\alpha })\le 0$, so $J_{\beta }\le J_{\alpha }$. Rearranging each comparison then gives the monotonicity of data fidelity and regularization penaltyHence the regulariser value is non-increasing and the data misfit is non-decreasing with the parameter. These inequalities hold for any choices of minimizers at the two distinct parameters, including when uniqueness fails.

Comparison with zero and nonnegativity of the two terms givesThereforeFor a concrete variational regularization example, take a Hilbert space and $J(u)=\Vert u{\Vert }^2$. The bound forces $u_{\alpha }\to 0$ in norm. Since $K$ is a bounded linear operator, $K{u}_{\alpha }\to 0$, and continuity of the squared norm givesThe same conclusion is not forced by $J(u_{\alpha })\to 0$ for a regulariser with a nontrivial zero set.

The small-parameter limit depends on data compatibility with the regularizer domain. SetFor every such $v$, optimality givesTaking the upper limit as $\alpha \downarrow 0$ and then the infimum over $v$ proves ${\Phi }_{\alpha ,f}(u_{\alpha })\to m$. Its two nonnegative contributions above $m$ must vanish, yielding the general resultIn particular, the requested zero-misfit limit holds when $f\in \overline{K(\mathrm{dom}J)}$. A sufficient condition is an exact solution $v$ with $Kv=f$ and $J(v)<\infty$; thenMembership of $f$ in $\mathcal{R}(K)$ alone is insufficient. Take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $K=I$, $f=1$, and $J$ the indicator functional of a constraint set $\{0\}$. All positive-parameter objectives are proper extended-real functions with coercivity and sequential lower semicontinuity, with the unique minimizer $u_{\alpha }=0$, yet $D(u_{\alpha })=1/2$ for every $\alpha$. Thus the first printed limit needs compatibility with the regulariser's effective domain; the second limit remains true under the given assumptions.

For a functional $E$ on a real Banach space, its convex conjugate is defined on the dual space byHere $⟨p,u⟩$ is the dual pairing. For complex spaces, use its real part, equivalently regard the space as real for convex analysis. The convex conjugate is a convex function, being the supremum of affine functions of $p$, even when $E$ itself is not convex.

Let $G=E\square F$. To prove convexity, take $u_1,u_2$ with finite $G$ values, $0<t<1$, and $\epsilon >0$. By the definition of infimum, choose $v_i$ withPut $u_t=t{u}_1+(1-t)u_2$ and $v_t=t{v}_1+(1-t)v_2$. Convexity of $E$ and $F$ givesLet $\epsilon \downarrow 0$. If either endpoint value is infinite, the convexity inequality is automatic, and $t=0,1$ gives equality. Thus the infimal convolution of two convex functions is convex under the stated properness assumption. The proof uses approximate minimizing splits, so no attainment of the inner infimum is assumed.

For $p\in {\mathcal{U}}^{\ast }$, the definition of the convex conjugate giveswhere the substitution $w=u-v$ is a bijection of the independent pairs. ThereforeThis infimal convolution identity does not need convexity of $E,F$ or attainment of the inner infimum. The assumptions ensure that both functions have nonempty effective domains; their conjugates never take $-\infty$, so the separated sum is well defined, allowing $+\infty$.

Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.