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A global minimizer is a point attaining the infimum of an objective. For an extended-real functional, it is usual to require a finite objective value; this excludes points outside its effective domain. A convention allowing $+\infty$ as the attained value gives every point as a formal minimizer of the identically infinite function.

A minimizing sequence satisfies $E(u_n)\to \inf E$. Coercivity bounds an appropriate sublevel set, Compactness supplies a convergent subsequence, and sequential lower semicontinuity turns its limit into a global minimizer.

For $0<\alpha <\beta$, choosing global minimizers of $D+\alpha J$ and $D+\beta J$ gives $J(u_{\beta })\le J(u_{\alpha })$ and $D(u_{\beta })\ge D(u_{\alpha })$. The decrease in penalty and increase in data misfit satisfy $\alpha \Delta J\le \Delta D\le \beta \Delta J$.

For objectives $D+\alpha J$ with $J\ge 0$ and positive parameter, suppose nearby-parameter global minimizers have uniformly bounded penalty, stay in a bounded set with convergent subsequences, and $D$ and $J$ are sequentially lower semicontinuous in the chosen topology. Their optimality inequalities and lower semicontinuity show that every subsequential limit minimizes the limiting objective. Uniqueness makes the regularized solution continuous in the parameter. Existence alone does not make an arbitrary selection among multiple global minimizers continuous.

A finite-valued global minimizer of the functional $E$ is an element $\bar{u}\in \mathrm{dom}E$ satisfyingHere $\mathrm{dom}E$ is its effective domain. The finite-value convention avoids treating an infeasible problem as solved.

The functional is a proper extended-real function if its effective domain is nonempty; the specified codomain already excludes $-\infty$. It has coercivity if $\Vert u_n\Vert \to \infty$ always implies $E(u_n)\to +\infty$, equivalently every finite sublevel set is bounded. It is $\tau$-sequentially lower semicontinuous ifThe topology in this definition matters: norm sequential lower semicontinuity and weak sequential lower semicontinuity need not coincide for a nonconvex functional.

On $\mathcal{U}=\mathbb{R}$ with its usual topology, the following examples isolate the three failures.

A nonproper functional is $E(x)=+\infty$ for every $x$. Its effective domain is empty, so it has no finite-valued global minimizer. It nevertheless has coercivity and is sequentially lower semicontinuous. If a minimizer is instead defined only by $E(x)=\inf E$ without requiring finiteness, every point formally minimizes this function; under that convention the requested nonproper counterexample is impossible with the given codomain.

A noncoercive functional is $E(x)=e^x$. It is a proper extended-real function and is continuous, but its infimum zero is approached as $x\to -\infty$ and is never attained. Hence it has no global minimizer.

A failure of sequential lower semicontinuity isIt is a proper extended-real function with coercivity, but $E(1/n)\to 0<E(0)$. Its infimum is zero, while every function value is positive. Thus it has no global minimizer.

The direct method in the calculus of variations gives the following existence theorem. Suppose bounded sequences in the Banach space $\mathcal{U}$ have $\tau$-convergent subsequences, and $E$ is a proper extended-real function with coercivity and is $\tau$-sequentially lower semicontinuous. Then $E$ has a finite-valued global minimizer.

To prove this, properness makes $m=\inf E<+\infty$. First rule out $m=-\infty$: a sequence with $E(u_n)\to -\infty$ eventually lies in a fixed sublevel set, which is bounded by coercivity. A $\tau$-convergent subsequence would then have a limit $u$ with $E(u)\le -\infty$, contradicting the codomain. Hence $m$ is finite.

Choose a minimizing sequence with $E(u_n)\to m$. It eventually belongs to the bounded sublevel set $\{E\le m+1\}$. Extract $u_{n_j}\stackrel{\tau }{\to }\bar{u}$. By sequential lower semicontinuity,Thus $E(\bar{u})=m$. In particular, a reflexive Banach space with the weak topology supplies the required subsequence property by weak sequential compactness of bounded sequences in a reflexive Banach space. A strictly convex function has at most one minimizer; this is an additional property, not part of the existence theorem.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

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.

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.