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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.

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 $u\in L^1(\Omega )$, the total variation seminorm on a domain isUsing $C_c^1$ instead gives the same definition. The zero test field shows nonnegativity, and $\mathrm{TV}(0)=0$, so this is a proper extended-real function. Each test field defines a linear functional of $u$; taking their supremum proves convexity, equivalentlyIt is not strictly convex: two distinct constant functions both have zero total variation, as does every convex combination of them.

The function of bounded variation on a domain space isTotal variation is not coercive on $BV(\Omega )$: for $u_k\equiv k$ on a domain of positive measure, $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$, whereas $\mathrm{TV}(u_k)=0$. This is noncoercivity of total variation on constants; a mean-zero constraint can remove the constant obstruction.

A proper convex function is both a convex function and a proper extended-real function. For example, the total variation seminorm on a domain is nonnegative, finite at zero, and a supremum of linear functionals, making it proper and convex.