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On a bounded Lipschitz domain, a sequence with bounded $L^1$ norms and bounded total variation seminorm has a subsequence converging strongly in $L^1$. The limit is in the BV space, and its variation does not exceed the limit inferior. This supplies the compactness step of the direct method in the calculus of variations for linearly growing gradient energies.

The calculus of variations studies functionals whose inputs are functions, curves or fields. A variational problem specifies an admissible class, including endpoint or other boundary conditions, and asks for stationary points or minimizers. The first variation gives an equation for stationary candidates; a second variation, convexity or a direct comparison is needed to decide minimality.

For example, minimizeA zero-endpoint variation $u+ϵ\eta$ gives $\delta E[u;\eta ]={\int }_0^{1}u\prime \eta \prime$. If $u$ is twice differentiable, integration by parts and the fundamental lemma of the calculus of variations yield $u\prime \prime =0$. The candidate is the straight line $\ell (x)=a+(b-a)x$.

It is also the unique global minimizer, not merely stationary. Write any admissible $u=\ell +w$ with zero endpoint values for $w$. The cross term is $(b-a){\int }_0^{1}w\prime =0$, soEquality forces $w=0$. This argument extends to Sobolev spaces with a meaningful endpoint trace and illustrates the Dirichlet principle. In more complicated problems, the direct method in the calculus of variations establishes existence without first solving the stationarity equation.

The direct method in the calculus of variations proves existence by passing to a limit of a minimizing sequence. A useful abstract version takes a nonempty weakly sequentially closed admissible subset $\mathcal{A}$ of a reflexive Banach space $X$ and a functional $E:X\to \mathbb{R}\cup \{+\infty \}$. Assume $E$ is finite at some admissible point, bounded below, coercive on $\mathcal{A}$, and has weak lower semicontinuity.

Choose $u_n\in \mathcal{A}$ with $E(u_n)\to m={\inf }_{\mathcal{A}}E$. Coercivity bounds $\Vert u_n{\Vert }_X$. Reflexivity supplies a weakly convergent subsequence, and closure puts its limit $u$ in $\mathcal{A}$. Finally,so $u$ attains the infimum. Each hypothesis has a distinct role: compactness must preserve the constraints, and lower semicontinuity must hold for the convergence actually obtained.

For the Dirichlet energy with linear forcing on $H_0^1(\Omega )$ over a bounded domain, the Poincaré inequality controls the forcing and gives coercivity, while the squared gradient norm is weakly lower semicontinuous. The resulting minimizer solves the weak Poisson equation by its first variation.^[ref] Strict convexity gives uniqueness when the admissible class is convex. The oscillating minimizing sequence with no minimizer shows what can fail without suitable lower semicontinuity.

On a finite-measure space, a bounded subset of $L^1$ is relatively weakly compact exactly when it is uniformly integrable. In particular every uniformly integrable bounded sequence has a weakly convergent subsequence. This is the compactness substitute for reflexivity at exponent one. A superlinear integral bound ${\sup }_j\int H(u_j)<\infty$, with $H\ge 0$ and $H(t)/|t|\to \infty$, gives the required uniform integrability: for $H(t)\ge A|t|$ outside $[-R_A,R_A]$,Choose $A$ large and then the measure of $E$ small. This is the direction used in the direct method in the calculus of variations; a mere bounded $L^1$ norm does not suffice, as concentrating spikes show.

Let $\Omega$ be a bounded connected Lipschitz domain, $\alpha >0$, $g\ge 0$ bounded, and $T:L^1\to L^1$ bounded and positivity preserving with $T1\ne 0$. Minimize $\alpha \mathrm{TV}(u)+D_g(u)$ over nonnegative BV space functions. The logarithmic Jensen inequality bounds $\Vert Tu{\Vert }_1$ and variation on sublevels; mean control for positive imaging operators then bounds the full $BV$ norm. Bounded-variation compactness gives strong $L^1$ convergence, including nonnegativity. The scalar fidelity has bounded derivative on $s\ge 0$, so continuity of $T$ makes its integral continuous in that limit. Lower semicontinuity of variation proves existence by the direct method in the calculus of variations. If $g>0$ almost everywhere and $T$ is injective, strict convexity of the fidelity gives uniqueness. For $0<g<1$, zero is the unique minimizer even without injectivity, since zero energy forces both $Tu=0$ and a constant nonnegative $u$.

Let $K$ be a nonempty norm-closed convex set in a real Hilbert space, and let $F:K\to \mathbb{R}$ be continuous and a convex function. If $K$ is bounded, or $F(x)\to \infty$ as $x\in K$ tends to infinity in norm, then $F$ attains its infimum. Every sublevel set is bounded, norm-closed, and convex, hence contains the weak limits of its sequences by norm-closed convex set is weakly closed. A minimizing sequence in one fixed sublevel set has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Its limit lies in every sublevel set above the infimum. Since $F$ is real-valued, this also rules out infimum $-\infty$ and proves attainment. This is an instance of the direct method in the calculus of variations.

A minimizing sequence can converge uniformly while its derivatives keep oscillating between distinct energy wells. For the functional $I(u)={\int }_0^1[u^4+(u^{\prime 4}-16{)}^4]$, with zero endpoint values and Lipschitz admissible functions, the triangular waves displayed above have slopes $\pm 2$ almost everywhere and height $1/N$. Direct integration gives $I(u_N)=1/(5N^4)$. Nonnegativity therefore makes the infimum zero. Attainment would require both $u=0$ and $|u^{\prime }|=2$ almost everywhere, an impossibility. The uniform limit is zero but its energy is $16^4$; thus the energy fails to be lower semicontinuous for this convergence. The example explains why control of function values alone cannot replace suitable convexity in the derivative variable in the direct method in the calculus of variations.

Use the direct method in the calculus of variations. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a minimizing sequence $u_n$ with uniformly bounded energy. ThenThe fixed value $u_n(0)=1$ and the Cauchy-Schwarz inequality give $\Vert u_n{\Vert }_{\infty }\le 1+\Vert u_n^{\prime }{\Vert }_2$ and $|u_n(x)-u_n(y)|\le \Vert u_n^{\prime }{\Vert }_2|x-y{|}^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.

By reflexivity and the Arzelà-Ascoli theorem, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; uniform convergence also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, uniform convergence gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The derivative term is lower semicontinuous for weak convergence, soAdmissibility gives $I(u)\ge m$. Thus

We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

For each real $a$, define the sublevel set $K_a=\{u\in K:F(u)\le a\}$. Continuity of $F$ and closure of $K$ make $K_a$ a closed set in $X$; the convex function inequality makes it a convex set. The preceding result therefore makes every $K_a$ a sequentially weakly closed set. Equivalently, $F$ has weak lower semicontinuity along sequences in $K$.

Every $K_a$ is bounded. This is immediate if $K$ is bounded. Otherwise, an unbounded $K_a$ would contain $u_n$ with $\Vert u_n\Vert \ge n$, contradicting the assumed coercive function condition $F(u_n)\to \infty$.

Choose $z\in K$ and put $\ell ={\inf }_{K}F\in [-\infty ,F(z)]$. Choose a minimizing sequence $(u_n)$ entirely in $K_{F(z)}$, with $F(u_n)\to \ell$; when $\ell =-\infty$, require $F(u_n)\le -n$ for sufficiently large $n$. It is bounded by the previous paragraph.

The weak subsequence of a bounded Hilbert-space sequence theorem gives $u_{n_j}\rightharpoonup u_0$. This applies even if $X$ is not separable: the closed linear span of the sequence is a separable Hilbert space; successive extraction makes every coefficient in a countable Hilbertian basis converge, the common norm bound gives square-summable limiting coefficients by the Bessel inequality, and approximation by finite linear combinations proves weak convergence. Orthogonal decomposition extends that conclusion to all test vectors in $X$. Since $K$ is a sequentially weakly closed set, $u_0\in K$.

For any real $a>\ell$, a tail of $(u_{n_j})$ lies in $K_a$, so $u_0\in K_a$. If $\ell =-\infty$, this would give $F(u_0)\le a$ for every real $a$, impossible because $F$ is real-valued. Thus $\ell$ is finite; letting $a\downarrow \ell$ gives $F(u_0)\le \ell$, while the definition of infimum gives the reverse inequality. ThereforeThis proves minimum of a coercive convex functional on a Hilbert space by the direct method in the calculus of variations, with bounded sublevel sets supplying the required weak sequential compactness in a Hilbert space.

The required functional, with the sign appropriate to a positive divergence operator, isIt is differentiable on $W_0^{1,2}$, with $D\mathcal{F}(u)[v]=Q(u,v)+\int fv$. A critical point consequently satisfies the weak equation. We now prove existence by the direct method in the calculus of variations, rather than invoke the representation theorem again.

The ellipticity and Poincare estimates implyThus the infimum is finite and any minimizing sequence is bounded in the Sobolev norm. A bounded sequence in a Hilbert space has a weakly convergent subsequence; write $u_j\rightharpoonup u$ in $W_0^{1,2}$. The equivalent $Q$ norm has the same continuous linear functionals and hence the same weak convergence. A Hilbert norm is Weak lower semicontinuity, giving $Q(u,u)\le {\mathrm{liminf}}_{j}Q(u_j,u_j)$. The term $\int f{u}_j$ converges to $\int fu$ because it is a bounded linear functional. ThereforeThe limit attains the infimum. Differentiating $t\mapsto \mathcal{F}(u+tv)$ at its minimum, for every $v$, gives $D\mathcal{F}(u)[v]=0$. This produces the desired weak solution. Strict positivity of $Q$ also makes the functional strictly convex, consistent with uniqueness.

Symmetry implies that all eigenvalues of $A(x)$ lie between $\lambda$ and $\Lambda$, so $\Vert A(x){\Vert }_{\mathrm{op}}\le \Lambda$. In particular the bilinear form $B$ from part (i) is bounded and symmetric under the printed assumptions. The symmetric elliptic Dirichlet energy with a nonpositive potential isAll its terms are finite. For an admissible variation $\phi \in H_0^1$, its first variation is $B(u,\phi )+\int f\phi$. Thus stationarity is exactly the weak formulation of $Lu=f$.

Here is the direct method in the calculus of variations. Put $u=\psi +w$. Expanding the symmetric quadratic form givesThe Poincaré inequality controls $\Vert w{\Vert }_2$ by $\Vert Dw{\Vert }_2$. Therefore $J$ is bounded below on the admissible class, and a minimizing sequence is bounded in $H^1$. By weak compactness in this reflexive Banach space, a subsequence converges weakly to $u$. The affine closed subspace $\psi +H_0^1$ is weakly closed, so $u$ remains admissible.

The symmetric square root $A^{1/2}(x)$ is measurable and bounded. Since $-q\ge 0$, the two bounded linear mapstake weak $H^1$ convergence to weak $L^2$ convergence. The weak lower semicontinuity of the Hilbert norm shows that $B(u,u)$ is weakly lower semicontinuous. The term $\int fu$ is weakly continuous. Hence $J(u)\le \mathrm{liminf}J(u_k)$ and $u$ attains the minimum. Varying in either sign yields the weak equation.

The sign $q\le 0$ was used both to make the potential energy nonnegative, hence weakly lower semicontinuous by this argument, and to obtain coercivity. Finally $J(u+z)-J(u)=B(z,z)/2\ge \lambda \Vert Dz{\Vert }_2^2/2$ for $z\in H_0^1$, since the first variation at $u$ vanishes. Therefore

The affine class $C_{\psi }=\psi +H_0^1(\Omega )$ is nonempty, and part (i) gives a finite comparison energy $\mathcal{F}(\psi )$. Since $F(p)\ge \alpha |p{|}^2$, its energy infimum $m$ satisfies $0\le m\le \mathcal{F}(\psi )$. Choose a minimizing sequence $u_k\in C_{\psi }$ with $\mathcal{F}(u_k)\le \mathcal{F}(\psi )+1$. ThenThe gradient bounds also control the full Sobolev norm. Indeed $w_k=u_k-\psi \in H_0^1$, so the Poincaré inequality givesThus $u_k$ is bounded in the Hilbert space $H^1(\Omega )$. Its weak compactness yields a weakly convergent subsequence $u_k\rightharpoonup u$. The closed linear subspace $H_0^1$ is weakly closed, hence $u-\psi \in H_0^1$ and $u\in C_{\psi }$. By part (ii),The direct method in the calculus of variations therefore provesNeither a smooth boundary nor a classical trace theorem is needed: the boundary condition is encoded by the closed zero-boundary Sobolev space.

Positivity means $u>0$ almost everywhere, since the real logarithm must belong to $L^1$. A positivity-preserving operator gives $s=Tu\ge 0$. For fixed $x$, the scalar shifted Poisson data fidelity is $f_x(s)=s-g(x)\log (1+s)$, withComposition with the linear operator $T$ proves convexity in $u$, and strict convexity holds along pairs whose forward images differ on a set of positive measure. The admissible class itself is convex: the concavity of the logarithm gives $\log (\theta u+(1-\theta )v)\ge \theta \log u+(1-\theta )\log v$, controlling its negative part, while ${\log }^{+}w\le w$ controls its positive part.

Because $|\Omega |=1$ and $\log (1+s)\ge 0$, the Jensen inequality gives the requested bound:The final equality uses nonnegativity and the unit area. Since $t-G\log (1+t)\to \infty$, this controls the forward-image norm on energy sublevels.

The printed strictly positive problem has no minimizer. This is nonattainment under strict positivity for shifted Poisson fidelity, rather than a failure of the coercivity calculation. Indeed, with $0<g<1$ and $s>0$, $\log (1+s)<s$ implies $f_x(s)>0$. Also $Tu$ cannot vanish identically for a strictly positive $u$. To see this, let $E_n=\{u\ge 1/n\}$. If $Tu=0$, positivity and $0\le {\chi }_{E_n}\le nu$ give $T{\chi }_{E_n}=0$. But ${\chi }_{E_n}\to {\chi }_{\Omega }$ in $L^1$ and continuity of $T$ would imply $T{\chi }_{\Omega }=0$, contradicting the hypothesis. Thus every admissible $u$ has positive fidelity and hence positive total energy.

Conversely, the constants $u_{\epsilon }=\epsilon {\chi }_{\Omega }$ are admissible, have zero total variation, and satisfyTheir logarithms are integrable for each $\epsilon >0$, but the limit is excluded. ThereforeA bounded minimizing sequence and bounded-variation compactness do not repair a nonclosed positivity/logarithm constraint. In particular, a literal existence or uniqueness proof for that domain is impossible.

The natural correction is to minimize over $BV(\Omega )$ with $u\ge 0$, omitting the unnecessary $\log u\in L^1$ condition: the fidelity only contains $\log (1+Tu)$, which is already integrable. Here is the full existence for nonnegative shifted Poisson regularization argument, also valid for any bounded nonnegative data $g$. Let $G=\Vert g{\Vert }_{\infty }$, $m_G={\inf }_{t\ge 0}\{t-G\log (1+t)\}>-\infty$, and take a minimizing sequence of energy at most $C$. The displayed bound gives $\Vert T{u}_n{\Vert }_1\le C_1$ and $\alpha \mathrm{TV}(u_n)\le C-m_G$. Write $c_n={\int }_{\Omega }{u}_n\ge 0$. The Poincaré inequality for total variation and mean control for positive imaging operators yieldThe denominator is nonzero, so the full $BV$ norm is bounded. Bounded-variation compactness gives $u_n\to u$ in $L^1$ along a subsequence, with $u\ge 0$. Continuity gives $T{u}_n\to Tu$ in $L^1$. On $s\ge 0$, $|f_x^{\prime }(s)|\le 1+G$, so the fidelity converges in the integral; lower semicontinuity of variation completes the direct method in the calculus of variations.

For the actual printed data $0<g<1$, the corrected problem has the unique minimizer $u=0$, even if $T$ is not injective. Zero attains energy zero. Any other zero-energy candidate would have both $Tu=0$ and $\mathrm{TV}(u)=0$; on the connected square, zero variation makes $u$ a nonnegative constant, and $T{\chi }_{\Omega }\ne 0$ forces that constant to vanish. For general positive bounded data, an injective $T$ is a sufficient uniqueness condition, because its fidelity is strictly convex; injectivity is not a necessary condition in every instance.

In the finite-dimensional interpretation, let ${\lambda }_{ij}=1+(Tu{)}_{ij}$. Independent Poisson observations with these intensities have negative log-likelihood ${\sum }_{ij}[{\lambda }_{ij}-g_{ij}\log {\lambda }_{ij}]+C(g)$. Removing the constant $\sum 1$ gives precisely the stated fidelity. Thus the model is Poisson counting noise with a unit background intensity, or an approximate version of it for rescaled/continuous grey values. Literal Poisson counts are integers; the constraint $0<g<1$ is a grey-value normalization, not a literal unscaled count sample.

An image segmentation separates an observed image signal into regions that are smooth within themselves and separated by meaningful image edges. Pure image smoothing blurs the very transitions that ought to define these regions. The Mumford–Shah segmentation model instead chooses the reconstruction and its discontinuity set together. For a bounded planar Lipschitz domain and bounded grey-value data $g$, one standard normalization isHere $K$ is the relatively closed edge set and $u\in H^1(\Omega \setminus K)$ can have different traces on its two sides. The first term is quadratic fidelity, the second penalizes variation within regions, and the Hausdorff measure term charges total edge length. It balances fitting, denoising and economical region geometry. The model was developed by David Mumford and Jayant Shah; their 1989 paper formulates this joint variational approach.

All three terms matter. Without fidelity, a constant reconstruction with no edge has zero energy. Without the gradient term, smooth data can be fitted exactly with no edge penalty. Without the length term, fine partitions with nearly constant region means can drive fidelity arbitrarily low while creating excessive boundaries. This is segmentation overfitting without an edge penalty. Increasing $\alpha$ favors flatter regions; increasing $\beta$ makes extra boundaries more expensive and can remove small features. These parameter effects describe a balance, not a guaranteed monotone evolution of every individual boundary.

For fixed $K$, varying $u$ gives the fixed-edge Euler-Lagrange equation for Mumford–Shah:The boundary condition is a separate one-sided Neumann boundary condition on each side of an edge, not continuity of $u$ across it. The fidelity makes this fixed-edge problem strictly convex, so its weak solution is unique by the Lax-Milgram theorem. Optimizing the edge set remains a different geometric problem. The nonconvexity of Mumford–Shah segmentation prevents a general uniqueness assertion or a guarantee that a numerical stationary point is globally optimal.

The piecewise-constant Mumford–Shah problem imposes $u=c_i$ on regions $E_i$. It minimizesThe factor $1/2$ counts each shared internal boundary once. The region means in piecewise-constant segmentation give $c_i=|E_i{|}^{-1}{\int }_{E_i}g$, provided $|E_i|>0$. Thus the remaining optimization concerns the partition. For a smooth interface between $i$ and $j$, moving it in the normal pointing out of $i$ changes fidelity by $[(c_i-g{)}^2-(c_j-g{)}^2]$ per unit displacement, while length changes by its curvature $\kappa$. The resulting segmentation interface curvature balance isWith equal isotropic interface costs, three freely meeting smooth edges satisfy the triple-junction angle in isotropic segmentation: force balance of their unit tangents gives angles of $120$ degrees. These are local stationarity conditions on regular interfaces, not a description of every singular edge configuration.

A simple piecewise-constant segmentation contrast threshold explains why small objects can disappear. On a domain of area $A$, suppose two constant intensities differ by $d$, occupying areas $a$ and $A-a$ with internal boundary length $L$. Keeping that boundary fits the data exactly and costs $\beta L$. Merging both regions costs the within-region squared error $a(A-a)d^2/A$. Among these two candidates, splitting wins precisely when $\beta L<a(A-a)d^2/A$. Other partitions may beat either candidate, so this is not a universal global segmentation formula.

This original synthetic example compares those two candidate geometries using their least-squares means. It illustrates the edge cost; it does not claim to compute the globally optimal Mumford–Shah segmentation.

Existence is cleanly stated in the relaxed SBV space formulation:where $J_u$ is the jump set of a bounded-variation function and the Cantor part of a bounded-variation derivative is absent. Clip a minimizing sequence to the bounded data range: this cannot increase fidelity, gradient energy or jump length. Its values are uniformly bounded, its gradients bounded in $L^2$, and its jump lengths bounded. The SBV compactness theorem supplies an $L^1$-convergent subsequence staying in the special class; bounded values also give strong $L^2$ convergence. The fidelity then converges, while gradient energy and jump length are lower semicontinuous. The direct method in the calculus of variations produces a minimizer. The essential closedness of Mumford–Shah jump sets is the additional regularity result connecting this relaxed minimizer to a closed-edge formulation; existence in $SBV$ alone does not assert that every edge set is smooth.

A practical continuous approximation is the Ambrosio–Tortorelli approximation, introduced in Ambrosio and Tortorelli's 1990 paper. An auxiliary field $v\in [0,1]$ is near one in regions and near zero at edges. One normalized energy isThe edge field weakens smoothing across a narrow transition, and its own energy approximates interface length. The Gamma-convergence result, together with the required compactness, relates global minimizing sequences to the limiting segmentation energy; it does not make the finite-parameter problem jointly convex. Alternating the two fields solves quadratic elliptic subproblems, but initialization and stopping can affect which local stationary configuration is found. The strengths of the model are joint denoising and segmentation, sharp transitions and a geometric cost. Limitations include competing local minima, sensitivity to scale parameters, loss of fine texture, and boundaries driven by intensity rather than semantic object identity.

Apply the direct method in the calculus of variations to $J(u)=\Vert f-Ku{\Vert }_1+\alpha |Du|(\Omega )$. Since $J(0)=\Vert f{\Vert }_1$, take a minimizing sequence with $J(u_k)\le C:=\Vert f{\Vert }_1+1$. ThenThus the sequence is bounded in the bounded-variation space. The bounded-variation compactness gives a subsequence converging in the strong convergence sense in $L^1$ to $u\in BV(\Omega )$. Boundedness of the linear operator $K$ gives $K{u}_k\to Ku$ strongly in $L^1$, so the residual norm converges. The variation term is sequentially lower semicontinuous, henceTherefore $u$ attains the infimum. The useful inverse hypothesis is the lower bound $\Vert u{\Vert }_1\le C_K\Vert Ku{\Vert }_1$; a bounded inverse on the range already suffices for this proof. No uniqueness follows from the nonsquared $L^1$ fidelity.

A one-dimensional Sobolev representative is absolutely continuous. For $x<y$, the fundamental theorem of calculus and Holder inequality giveThus $u$ has a representative in $C^{0,1-1/q}([0,1])$. The qualification about representatives matters because a Sobolev space element is an almost-everywhere equivalence class.

In two dimensions, the Sobolev fundamental theorem of calculus on lines and Fubini's theorem imply that almost every horizontal and vertical slice belongs to $W^{1,q}(0,1)$ and has this one-dimensional Hölder continuity. The slice seminorm depends on the slice; this does not give one uniform pointwise estimate on the square. For $q>2$, Morrey's inequality additionally gives a globally Hölder continuous representative of exponent $1-2/q$. For $1<q\le 2$, global continuity need not hold. For example, with a smooth cutoff around an interior point, $u(x)=|x-x_0{|}^{-a}$ lies in $W^{1,q}$ when $0<a<2/q-1$, but is unbounded. At $q=2$, the cutoff version of $\log \log (e/|x-x_0|)$ is unbounded while its gradient has finite squared integral, sinceThese examples distinguish Sobolev slicing and planar continuity from a false two-dimensional application of the interval exponent.

Put $\Omega =(0,1{)}^2$. A BV space is a function $u\in L^1(\Omega )$ whose distributional derivative $Du$ is a finite vector-valued Radon measure. Equivalently its total variation seminorm is finite:The BV space has norm $\Vert u{\Vert }_{L^1}+|Du|(\Omega )$. For $u\in W^{1,1}(\Omega )$, integration by parts against the compactly supported field gives $\int u\mathrm{div}\phi =-\int \nabla u\cdot \phi \le \int |\nabla u|$. Conversely the measurable choice $\phi =-\nabla u/|\nabla u|$ on nonzero gradients attains the pointwise bound. Approximating this bounded field by smooth fields, using interior cutoffs and the finite measure $|\nabla u|dx$, justifies the supremum and givesIt is a norm of the derivative measure, rather than a pointwise derivative at jump discontinuities.

There is a genuine mismatch in the printed definition of the next functional. Its constraints on ${\phi }_0$ and $\phi$ are independent. Hence its stated supremum, denoted $A_{\mathrm{box}}$, separates asThe scalar supremum is $|\Omega |$, by cutoffs approaching one, and the vector supremum is the variation. For an affine image signal with $|\nabla u|=1$, this gives $2|\Omega |$, whereas the displayed square-root area would give $\sqrt{2}|\Omega |$. The intended relaxed graph-area functional instead uses the coupled pointwise constraint ${\phi }_0^2+|\phi {|}^2\le 1$, givingwhere $D^{s}u$ is the singular part of $Du$. Both readings have a minimizer, but their equations are different.

Here is the direct method in the calculus of variations for either reading. Let $A_{\bullet }$ be the literal $A_{\mathrm{box}}$ or the corrected $A$, and define the energy on $BV(\Omega )\cap L^2(\Omega )$, assigning infinity elsewhere. A minimizing sequence has bounded energy by comparison with $u=0$. Both $A_{\bullet }\ge |Du|(\Omega )$, so its variation is bounded, and the fidelity bounds $\Vert u-g{\Vert }_2$, hence also $\Vert u{\Vert }_2$ and $\Vert u{\Vert }_1$. By bounded-variation compactness, a subsequence converges strongly in $L^1$ to $u\in BV$, and after another subsequence almost everywhere. Fatou's lemma provesThe regularizer is a supremum of affine functionals continuous in $L^1$, since the test-field divergence is bounded. It is therefore lower semicontinuous. Combining the two lower bounds proves existence of a minimizer. In fact the convex regularizer and the strictly convex squared fidelity make the minimizer unique up to null sets. This does not assert that the minimizer must belong to $W^{1,1}$.

For the intended graph area, conditionally assume that the minimizer is in $W^{1,1}$. For $\eta \in C_c^{\infty }(\Omega )$, differentiate at $u+t\eta$. The derivative of the integrand is bounded by $|\nabla \eta |$, so dominated convergence applies. The weak equation isthat is,This is the graph-area Euler-Lagrange equation. Compactly supported variations impose no boundary condition in this statement.

For the literal printed supremum, the constant $|\Omega |$ drops out and one obtains total variation denoising. Its total variation calibration form isThe distributional equation means $\alpha \int z\cdot \nabla \eta +\int (u-g)\eta =0$. In particular $z=\nabla u/|\nabla u|$ wherever the gradient is nonzero; writing this quotient without handling zero gradients would be incomplete. Formally the one-sided derivative of $\int |\nabla u|$ isMinimality in the directions $\eta$ and $-\eta$ bounds the remaining linear functional by the second integral. The Hahn-Banach theorem extends it on that zero-gradient set to a bounded vector field of magnitude at most one, furnishing $z$ and the displayed weak equation. Thus the literal definition has a nonsmooth subgradient equation, not the square-root equation above.

Work with real-valued functions. The screened sine-Gordon energy is well defined: $0\le 1-\cos u\le u^2/2$, and $fu$ is integrable by the Cauchy-Schwarz inequality. It is coercive, becauseA minimizing sequence is bounded in the Hilbert space $H^1$. Weak sequential compactness of bounded sequences in a reflexive Banach space supplies a weakly convergent subsequence. The squared H1 space norm is weakly lower semicontinuous, the source pairing is weakly continuous, and the nonlinear term is covered by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. The direct method in the calculus of variations therefore gives a minimizer $\varphi$.

Taking its first variation in any $v\in H^1$ givesThus the Euler-Lagrange equation isas a weak solution, equivalently in distributions when tested against smooth compactly supported functions. The derivative of the nonlinear term is justified by $|\sin \varphi |\le |\varphi |$ and the second-order remainder bound $|\cos |\le 1$.

Now $G=f-\sin \varphi \in L^2$, so the supplied elliptic regularity estimate puts $\varphi$ in $H^2$. Applying the Sobolev inequality to $\varphi$ and each first weak derivative gives $\varphi \in W^{1,6}({\mathbb{R}}^3)$. Morrey's inequality and uniformly continuous integrable functions vanish at infinity prove that its continuous representative tends to zero.

The final printed supremum estimate is false in general. The maximum bound for a monotone reaction term involves $h(s)=s+\sin s$, which is odd and strictly increasing: $h^{\prime }=1+\cos s\ge 0$, and its zeros are isolated. If $M=\Vert f{\Vert }_{\infty }<\infty$, a positive maximum $m$ of $\varphi$ satisfies $h(m)\le f\le M$; apply the same argument to $-\varphi$. The valid general estimate isThe bound by $M$ is valid if $M\le \pi$, but $h(s)$ can be smaller than $s$ for larger positive $s$.

For an explicit failure of the source-size bound for the screened sine-Gordon equation, take $a=3\pi /2$, $L=10$, $\varphi (x)=a{e}^{-|x{|}^2/L^2}$ and $f=-\Delta \varphi +\varphi +\sin \varphi$. These are smooth functions in the required spaces. With $q=|x{|}^2/L^2$,Hence $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert \varphi {\Vert }_{\infty }$. Moreover $V(s)=s^2/2+1-\cos s$ has $V^{\prime \prime }=1+\cos s\ge 0$, so the energy is convex and this critical point is a minimizer. The example therefore satisfies even the minimizing and $C^2$ hypotheses of the printed claim.

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.