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Use a bounded BV extension operator for the bounded Lipschitz domain. Extend $u_k$ to $w_k\in BV({\mathbb{R}}^n)$, supported in one fixed compact set, with $\Vert w_k{\Vert }_{BV({\mathbb{R}}^n)}\le C_{\Omega }\Vert u_k{\Vert }_{BV(\Omega )}$. Such an extension is obtained by reflection in Lipschitz boundary charts, a partition of unity and a fixed cutoff function. A naive zero extension must also charge its boundary-trace jump; it cannot discard that contribution.

Let $M$ bound these norms and let $w_{k,ϵ}=w_k\ast {\rho }_{ϵ}$ be their mollifications. The BV mollification error estimate givesThe variation measure here is on ${\mathbb{R}}^n$. The printed lemma's $|Dw|(\Omega )$ needs this correction unless it also assumes all the derivative mass lies in $\Omega$: a nonconstant bump supported outside $\overline{\Omega }$ disproves its literal wording. The correct estimate follows by averaging the BV translation estimate $\Vert w(\cdot -h)-w{\Vert }_1\le |h||Dw|({\mathbb{R}}^n)$ over a mollifier supported in $|h|\le ϵ$.

For each fixed $ϵ>0$, convolution gives uniform boundsThe mollifications have common compact support and are uniformly equicontinuous. The Arzelà-Ascoli theorem supplies a uniformly convergent subsequence at each scale $ϵ=1/m$. A diagonal subsequence converges at every one of those scales. For two late members of that subsequence,First choose large $m$, then late $k,\ell$; the sequence is a Cauchy sequence in $L^1$. Its limit $w$ belongs to the BV space because the total variation seminorm on a domain is sequentially lower semicontinuous. Uniform derivative bounds and integration by parts give $D{w}_k\stackrel{\ast }{\rightharpoonup }Dw$. Restriction to $\Omega$ proves the required weak-star convergence in BV.

The bounded-variation space carries the norm $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_{L^1(\Omega )}+|Du|(\Omega )$. Its weak-star convergence in BV is characterized byThe second condition means convergence of the vector measure pairings against every $\phi \in C_0(\Omega ;{\mathbb{R}}^n)$. Equivalently, strong convergence in $L^1$ together with ${\sup }_k|D{u}_k|(\Omega )<\infty$ suffices: integration by parts identifies the limit on smooth compactly supported tests, and uniform approximation extends this to $C_0$ tests.

The bounded-variation compactness theorem says thatguarantees a subsequence convergent in weak-star convergence in BV on a bounded Lipschitz domain. This is the uniform criterion for relative sequential compactness. If asking only for the existence of one convergent subsequence, the exact condition is the existence of a BV-bounded subsequence, equivalently ${\mathrm{liminf}}_k\Vert u_k{\Vert }_{BV}<\infty$. The entire sequence need not be bounded: alternating zero functions and constants tending to infinity gives a simple example. Conversely, a convergent subsequence has bounded $L^1$ norm and bounded total variation seminorm by the Uniform boundedness principle for its derivative-measure pairings.

The literal assumptions suffice for the scalar quadratic fidelity problem. One can avoid regularity of level-set boundaries by moving only a clipped part of the minimizer. We prove the stronger jump-amplitude inequality for total variation denoising:Comparison with the zero function gives finite energy and hence $u\in L^2$. Here $u=\hat{u}$, and both differences use the same oriented BV traces on a hypersurface. Reversing the normal reverses both differences and leaves the inequality unchanged. Outside $J_f$, the two traces of $f$ agree, so the inequality would read $-[u{]}^2\ge 0$ at a jump of $u$. It therefore gives the requested no-new-jumps property of total variation denoising in every dimension.

First use residual-preserving clipping of an ROF minimizer. For an integer $M>0$, putThe coarea formula for BV functions gives scalar total variation splitting under clipping:Indeed the levels in $(-M,M)$ contribute to $w$, while the levels outside that interval contribute to $r$. This uses the full signed coarea formula. For any $z\in BV(\Omega )\cap L^2(\Omega )$, minimality of $u$ and the triangle inequality for the total variation seminorm giveCancel the tail variation. Thus $w$ is a bounded ROF denoising minimizer for $g\in BV(\Omega )\cap L^2(\Omega )$. The data $g$ may still be unbounded. This is an exact reduction preserving $w-g=u-f$; it does not truncate the data and then pass to a limit of different reconstructions.

We next prove the jump-amplitude inequality for a bounded ROF minimizer, allowing its data to be unbounded. Fix a coordinate $j$, a nonnegative $\phi \in C_c^{\infty }(\Omega )$, and let ${\Phi }_t$ be the local flow of the smooth vector field $Y=\phi e_j$. The flow is the identity near the domain boundary, preserves each line parallel to $e_j$, and satisfies ${\Phi }_{-t}={\Phi }_t^{-1}$. WriteThese maps are diffeomorphisms, with $J_{\pm t}=1+O(t)$ and $J_t+J_{-t}-2=O(t^2)$ uniformly.

The useful trace calculation is the BV jump-product limit with one bounded factor. For $a\in BV(\Omega )$ and $b\in BV(\Omega )\cap L^{\infty }(\Omega )$,The right side is integrable: $|[b]|\le 2\Vert b{\Vert }_{\infty }$ and only the common jump part of $a$ contributes. The same limit holds with both increments replaced by their negative-time increments, still dividing by positive $t$.

Here is why this calculation needs only one bounded factor. By the BV slicing theorem, almost every coordinate slice of $a$ and $b$ has one-sided representatives. On one such interval let $\mu =Da$ and use its right-continuous representative. Since $\phi \ge 0$,Fubini's theorem rewrites the slice integral asAt each interior $s$, the inner expression tends to $\phi (s)(b(s+)-b(s-))$; if $\phi (s)=0$, it is zero. Its absolute value is at most $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }$. Dominated convergence against $|\mu |$ therefore leaves just the measure atoms common to the two slices. The bound $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }|Da|$ is also integrable over the transverse coordinates, by the BV slicing theorem. Integrating the slice jump sums gives the surface integral and its factor $|{\nu }_b\cdot e_j|$. Negative-time increments give the same product, because both slice differences reverse sign. At no point is a uniform bound on $a$ across the slices required.

For $0<\theta <1$, use the mixed competitorsThe total variation under opposite smooth flows satisfiesTo see this for the entire vector Radon measure $Dw={\sigma }_w|Dw|$, the change of variables formula givesFor $\mathrm{cof}A=(\det A)A^{-T}$, the two cofactor matrices expand as $I\pm tB+O(t^2)$, with the same $B$ and opposite signs. Their norm expansions on $|{\sigma }_w|=1$ have cancelling linear terms. Integrating proves the estimate, including the absolutely continuous, jump and Cantor parts. The BV transformation formula is also given in Lemma 4.2 on differentiable regularizers; that lemma does not assume bounded data. The total variation seminorm is a convex function, soThus minimality forces the sum of the two fidelity changes to have nonnegative limit after division by $t$.

It remains to evaluate that sum without bounding $g$. Put $F_g(v)=\frac{1}{2}\Vert v-g{\Vert }_2^2$. Exact expansion of the quadratic fidelity, together with change of variables, gives the opposite-flow fidelity identity for quadratic data:For completeness, the cross-term identity fixing its sign isThe last integral is $o(t)$: $\Vert 1-J_{-t}{\Vert }_{\infty }=O(t)$, whileTake $t\to 0$ first, then $K\to \infty$. Here $g\in L^1$ and the local flow is strongly continuous in $L^1$. The remaining Jacobian mass term is $O(t^2)\Vert w{\Vert }_2^2$. This argument avoids multiplying an unbounded fidelity derivative by an uncontrolled derivative measure.

Apply the BV jump-product limit with one bounded factor with $(a,b)=(g,w)$ and $(w,w)$, and use minimality. For every nonnegative $\phi$ and every coordinate $j$,Let $\theta \downarrow 0$. These are inequalities for finite signed Radon measures, so arbitrary nonnegative smooth tests imply nonnegativity of their densities. Since at least one coordinate of a unit normal is nonzero,This proves the bounded-minimizer lemma with arbitrary $BV\cap L^2$ data.

Finally return to $w=T_{M}u$ and $g=f-u+w$. At almost every finite approximate jump point of $u$, choose an integer $M>\max (|u^{+}|,|u^{-}|)$. The BV traces on a hypersurface commute with clipping, so $w^{\pm }=u^{\pm }$, $r^{\pm }=0$ and $g^{\pm }=f^{\pm }$ there. It is consequently a jump point of $w$, and its inequality is exactly $[u]([f]-[u])\ge 0$. A countable union over $M$ removes all exceptional surface-null sets. Outside $J_f$, the BV traces on a hypersurface of $f$ agree almost everywhere. We concludeThe mechanism is exact scalar coarea splitting, paired smooth-flow variations, a one-bounded-factor BV trace limit, and localization through integer clipping levels. It works in every dimension under the printed hypotheses, without essential boundedness of $f$ or $u$ and without regularity of their level-set boundaries.

In total variation calibration notation, $\mathrm{div}z=(u-f)/\alpha$. Since this divergence is itself in the BV space, the proved inequality equivalently readsThus an upward output jump forces the appropriate nonpositive jump of the calibrated divergence. This is a consequence of the variational argument above, with common oriented BV traces on a hypersurface; no curvature of the rectifiable interface or differentiability of its normal is assumed.

For $t\in \mathbb{R}$, define the set functionalThe ROF level-set formulation states that the unique minimizer of $J(u)=\Vert u-f{\Vert }_2^2/2+\alpha \mathrm{TV}(u)$ has superlevel sets minimizing $F_t$ for almost every $t$. Conversely, an admissible $u$ with minimizing superlevel sets is the ROF minimizer. Thus the concise characterization is

The signed layer-cake identity for quadratic fidelity and the coarea formula for BV functions giveIndeed $u^2/2-fu={\int }_{\mathbb{R}}(t-f)({\chi }_{\{u>t\}}-{\chi }_{\{0>t\}})dt$ pointwise. Its absolute integral is bounded by $u^2/2+|fu|$, so Fubini's theorem applies for $u,f\in L^2$. The negative-level baseline is essential; integrating the unadjusted $F_t$ would generally diverge.

Each $F_t$ attains its minimum. A minimizing sequence of indicator functions has bounded $L^1$ norm, and comparison with the empty set bounds its perimeter by an $L^1$ forcing bound. Bounded-variation compactness supplies a limiting indicator function. The forcing integral converges by dominated convergence, while relative perimeter is sequentially lower semicontinuous.

For $s<t$, the supplied comparison lemma applied to $(f-t)/\alpha <(f-s)/\alpha$ makes every selected minimizer $A_t$ contained in $A_s$ up to a null set. The comparison of perimeter minimizers with ordered forcing also follows directly: compare $A_t$ with $A_t\cap A_s$, compare $A_s$ with $A_t\cup A_s$, add and use submodularity of relative perimeter to obtain $(t-s)|A_t\setminus A_s|\le 0$.

Select minimizers at rational levels, remove their countably many exceptional null sets, and reconstruct $v(x)=\sup \{q\in \mathbb{Q}:x\in A_q\}$. The strict superlevel set is $\{v>t\}={\bigcup }_{q>t}{A}_q$. This union also minimizes $F_t$: take $q\downarrow t$, use monotone $L^1$ convergence of its indicator functions and sequential lower semicontinuity, and note that $m(t)={\min }_E{F}_t(E)$ satisfies $|m(t)-m(s)|\le |\Omega ||t-s|$.

To justify finite energy before assuming it, clip this reconstruction to $v_M\in [-M,M]$. Comparison with the empty set gives $\alpha \mathrm{Per}(E_t;\Omega )\le {\int }_{\Omega }|t-f|$ uniformly on a bounded interval of levels. Since $v_M=-M+{\int }_{-M}^M{\chi }_{E_t}dt$, pairing with compactly supported test-field divergences bounds its total variation seminorm by ${\int }_{-M}^M\mathrm{Per}(E_t;\Omega )dt<\infty$. Thus $v_M$ belongs to the BV space before applying the coarea formula for BV functions. The layer-cake argument on the finite interval $[-M,M]$ shows $J(v_M)\le J(T_{M}w)$ for every finite-energy competitor $w$, where $T_M$ is clipping. In particular $w=0$ bounds the $L^2$ norms and variations uniformly. Fatou's lemma excludes infinite values of $v$ on a positive-measure set. Bounded-variation compactness and weak convergence in $L^2$ identify an admissible limit $v$. For any fixed competitor, $T_{M}w\to w$ in $L^2$ and its variation tends to that of $w$, by bounded-variation contraction under clipping and sequential lower semicontinuity. Thus $J(v)\le J(w)$.

The quadratic fidelity is strictly convex, so every ROF minimizer equals $v$ almost everywhere and has the selected minimizing superlevel sets. Conversely, for any $u$ whose levels minimize $F_t$, integrate the levelwise inequality against any competitor's levels in the displayed identity to obtain $J(u)\le J(w)$. This proves both directions for signed as well as nonnegative data.

The structure theorem for functions of bounded variation is the measure representation of the distributional derivative: for $u\in BV(\Omega )$ there is a unique finite vector Radon measure $Du$ withfor every $\phi \in C_c^1(\Omega ;{\mathbb{R}}^n)$. It also has a polar decomposition of a vector measure $Du={\sigma }_u|Du|$ with $|{\sigma }_u|=1$ for $|Du|$-almost every point.

To prove it, let $L(\phi )=-\int u\mathrm{div}\phi$. The test-function definition of total variation seminorm, applied to both signs, gives $|L(\phi )|\le \mathrm{TV}(u)\Vert \phi {\Vert }_{\infty }$. Compactly supported smooth vector fields are uniformly dense in $C_0(\Omega ;{\mathbb{R}}^n)$, so $L$ extends uniquely to a bounded functional there. The Riesz-Markov-Kakutani representation theorem, applied componentwise, gives the unique vector measure $Du$. The operator norm of $L$ is exactly the defining variation supremum; the vector-measure dual norm is $|Du|(\Omega )$, proving equality. Conversely any finite measure satisfying the identity bounds that supremum, so this also characterizes membership in the BV space.

The Radon-Nikodym theorem applied to the components relative to $|Du|$ gives ${\sigma }_u$; the definition of the variation measure forces $|{\sigma }_u|=1$ almost everywhere. If $u$ is smooth, ordinary integration by parts gives $Du=\nabla u{\mathcal{L}}^n$. Compact support of the test field removes any boundary contribution; no boundary regularity is needed for this representation statement.

A character of an algebra is a nonzero complex linear multiplicative map $\varphi :A\to \mathbb{C}$. It necessarily satisfies $\varphi (1)=1$. If $\lambda =\varphi (x)$, then $x-\lambda 1$ cannot be invertible, since applying $\varphi$ to an inverse identity would give $0=1$. Therefore $\varphi (x)\in \sigma (x)$. The Neumann series bound $\sigma (x)\subseteq \{z:|z|\le \Vert x\Vert \}$ givesproving automatic continuity of characters. In particular every algebra character belongs to the closed unit ball of $A^{\ast }$; no prior continuity was needed.

The character space of an algebra is ${\Phi }_A$, the set of all algebra characters. Its Gelfand topology is the weakest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous, equivalently the weak-star topology inherited from $A^{\ast }$. Within the weak-star compact ball $B_{A^{\ast }}$, it is the closed set specified byThese equations are closed conditions on finitely many evaluations at a time. By the Banach-Alaoglu theorem, ${\Phi }_A$ is compact, and it is Hausdorff because distinct characters differ on an evaluation.

To obtain every spectral value from a algebra character, take $\lambda \in \sigma (x)$. Commutativity makes the ideal $(x-\lambda 1)A$ proper, so it is contained in an algebraic maximal ideal $\mathfrak{m}$. Every such maximal ideal is norm closed: its closure is an ideal, and cannot be all of $A$, because an element of $\mathfrak{m}$ sufficiently close to $1$ would be invertible by the Neumann series. Maximality then makes its closure equal to itself.

The quotient $A/\mathfrak{m}$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. Composing the quotient map with this scalar identification gives a algebra character with $\varphi (x)=\lambda$. Together with the first inclusion,The same maximal-ideal argument shows that the algebra character space of a nonzero algebra is nonempty.

Now suppose $A$ is a commutative C-star algebra. All its elements are normal. The C-star identity givessince the self-adjoint element $h=x^{\ast }x$ satisfies $\Vert h^2\Vert =\Vert h{\Vert }^2$. Iterating yields $\Vert x^{2^n}\Vert =\Vert x{\Vert }^{2^n}$. The spectral radius formula along this subsequence provesThis is spectral radius norm equality for normal elements, applied to the commutative case.

For a self-adjoint $h$, the element $v_t=\exp (ith)$ is unitary for every real $t$, so $\Vert v_t\Vert =1$. Algebra character continuity and multiplicativity giveIts modulus is $\exp (-t\mathrm{Im}\varphi (h))$, forcing $\varphi (h)\in \mathbb{R}$. Write $x=h+ik$ with $h,k$ self-adjoint. ThenThis proves that characters of a C-star algebra respect the involution. The conjugation bar is present in the original PDF and must be retained; the TeX aid omits it.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra is isometrically star-isomorphic to $C(K)$ for a compact Hausdorff space, canonically $K={\Phi }_A$. Its Gelfand transform isIt is a unital algebra homomorphism; the conjugation identity just proved makes it preserve the involution. The algebra character description of the spectrum and the spectral-radius equality giveThus it is injective and isometric, and its range is closed because $A$ is complete. The range contains constants, is closed under complex conjugation, and separates points of ${\Phi }_A$: two distinct algebra characters differ on some element of $A$.

The complex Stone-Weierstrass theorem says that a self-adjoint unital subalgebra of $C(K)$ separating points is uniformly dense. For completeness, its familiar lattice argument explains the last step. The real part of its uniform closure is closed under absolute value, by polynomial approximation of $|t|$ on bounded real intervals, and hence under finite maxima and minima. Constants and point separation permit a real function matching any given real continuous $g$ at any chosen pair of points. Fixing the first point, take a maximum of finitely many such functions to obtain a function above $g-\epsilon$ everywhere and equal to $g$ at that first point. It is below $g+\epsilon$ in a neighborhood of that point. A finite cover by these neighborhoods and the minimum of their associated functions then lies between $g-\epsilon$ and $g+\epsilon$ everywhere. Real and imaginary parts give density for complex functions.

Apply this to $\mathcal{G}(A)$. Its range is both dense and closed, so it is all of $C({\Phi }_A)$. ThereforeThis proves the commutative theorem in the setting of the question. For the zero algebra the corresponding compact space is empty and the representation is $C(\varnothing )=\{0\}$.

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.

Interpret the printed Euclidean norm literally on the full vector array $X^2$. It is a single global norm, not the sum of pixelwise gradient lengths in discrete isotropic total variation. Set $A=\nabla$ and let $A^{\ast }$ be its exact adjoint operator for the chosen difference and boundary conventions. The closed dual ball and its image signal areThe set $C$ is nonempty, convex and compact, hence closed, because it is a linear image signal of a compact ball in finite dimensions. Euclidean norm duality givesthe support function of $C$.

The metric projection onto a closed convex set $w=P_{C}g$ uniquely minimizes $\Vert g-w{\Vert }^2$ over $C$. Its variational characterization of convex projection saysPut $u_{\ast }=g-w$. This inequality says precisely that ${\sigma }_C(u_{\ast })=⟨u_{\ast },w⟩$. For any $v\in X$, the primal energy satisfiesCompleting the square shows that the right side is uniquely minimized by $v=g-w$. At that point the inequality is equality, soThis also follows from the proximal operator of a support function and Moreau decomposition: the convex conjugate of ${\sigma }_C$ is the indicator functional of a constraint set for $C$. The squared fidelity is strictly convex and coercive, so the primal minimizer exists and is unique.

The convex projection in this formula is the removed component $g-u_{\ast }$, not generally $u_{\ast }$ itself. A metric convex projection onto one fixed closed convex set is idempotent. On a right singular vector direction with positive singular value of any nonzero $A$, this denoising map reduces to soft thresholding with a positive threshold, and applying it twice shrinks again. It therefore cannot be such a convex projection for all data. This qualifies the printed convex projection wording while giving the required global gradient-norm projection residual.

Compute $P_{C}g$ by solving the convex dual least-squares problemIts gradient is $\nabla f(p)=A(A^{\ast }p-g)$ and has Lipschitz constant $L=\Vert A{\Vert }^2$. Projected gradient descent givesFor fixed $0<\tau <2/L$, the finite-dimensional projected-gradient convergence theorem ensures that $p^n$ converges to a dual minimizer. Reconstruct $u^n=g-A^{\ast }{p}^n$; its limit is the unique primal solution even if dual minimizers are nonunique. If $A=0$, the primal solution is simply $g$ and no iteration is needed. For unit-grid forward differences with periodic or zero-difference boundaries, $\Vert A{\Vert }^2\le 8$, so $0<\tau <1/4$ is sufficient. Grid-spacing factors or other boundary stencils change this bound.

The normalization here is global. If a pixelwise sum of gradient lengths had instead been intended, the dual feasible set would be a product of pixelwise balls and convex projection would normalize each block separately; it is a different regularizer and should not be silently substituted.

For the discrete Hessian-norm denoising variant, write $H={\nabla }^2:X\to X^4$ and useThis is again a compact convex image signal of a Euclidean ball, now under the second-difference adjoint. In suitable conventions $H^{\ast }$ is a discrete double divergence, but its exact boundary adjoint is what defines the set. It lies in $(\ker H{)}^{\perp }$, so components in $\ker H$ are preserved by denoising. Interior second derivatives annihilate affine image signals; whether all such image signals remain in the kernel depends on the boundary convention. With a pixelwise Hessian norm, the corresponding balls would instead be four-component blocks.

A rigorous Mumford–Shah functional permits nonsmooth image signals and free discontinuities. One classical admissible class consists of relatively closed countably rectifiable sets $K\subset \Omega$ with finite Hausdorff measure ${\mathcal{H}}^1(K)$, and $u\in W^{1,2}(\Omega \setminus K)\cap L^{\infty }(\Omega )$ with finite energy. No exterior boundary values are prescribed. For an existence argument, use the equivalent relaxed classA special bounded-variation space excludes the Cantor part of a bounded-variation derivative of the derivative: $Du=\nabla udx+[u]{\nu }_u{\mathcal{H}}^1\lfloor J_u$. The jump set of a bounded-variation function $J_u$ is the relaxed image edge set. Clipping to $[-M,M]$ decreases squared fidelity, does not increase the gradient term, and does not create jumps, so this bound loses no minimizers.

Take a minimizing sequence and compare with a constant image signal. Its $L^2$ gradient norms and jump lengths are bounded. Alsoso the sequence is bounded in $BV$. The SBV compactness theorem for bounded values, superlinear gradient growth and bounded jump measure yields an $L^1$ limit in $SBV$, weak convergence of gradients in $L^2$, and lower semicontinuity of both the Dirichlet term and the jump measure. The uniform value bound upgrades convergence to $L^2$, so fidelity converges. This proves existence of a relaxed minimizer. Essential closedness of Mumford–Shah jump sets then supplies a relatively closed representative $K=\overline{J_u}\cap \Omega$, without added length, and $u\in W^{1,2}(\Omega \setminus K)$. This completes the outline for the classical pair problem. Arbitrary Hausdorff convergence of image edge sets alone is not an adequate substitute for these compactness and regularity results. No uniqueness is claimed for segmentation.

As $\alpha \to \infty$ with $\beta$ fixed, bounded energy forces $\nabla u\to 0$ in $L^2$. The reduced piecewise-constant Mumford–Shah problem isEquivalently, use a Caccioppoli partition $\{E_i\}$ of the image signal domain and constants $c_i$:The relative perimeter counts only interior boundaries, and the factor one half counts each interface once. Adjacent equal-valued regions can be merged, removing unnecessary boundaries.

For fixed $K$, let $E_i$ be its positive-area regions. Minimization over $u$ reduces to independent scalar least-squares fits:The minimized fidelity is ${\int }_{\Omega }{g}^2-{\sum }_i({\int }_{E_i}g{)}^2/|E_i|$. Thus region means in piecewise-constant segmentation give the optimal grey values for a fixed segmentation.

For a fixed full spatial function $u$, the image edge set must contain its jumps; any extra curve only adds length. The optimal choice is its essential jump set, with a relatively closed representative when appropriate. There is no independent relocation of boundaries while that full function is held fixed. A different common alternating step fixes the values $c_i$ but allows the labels $E_i$ to move. It minimizes the fidelity-plus-perimeter partition functional above. Without the perimeter term each point takes its nearest grey value; with it, interface length is penalized. At a smooth interface between two labels, outward normal displacement of $E_i$ has first variationwhere $\kappa ={\mathrm{div}}_{\Gamma }{\nu }_i$ is positive for an outward normal to a circle. The stationary segmentation interface curvature balance isThis is the geometric interpretation of optimizing boundaries with fixed grey levels, and distinguishes it from fixing the whole spatial image signal.

As $\beta \to \infty$ with $\alpha$ fixed, a constant competitor bounds the minimum independently of $\beta$, forcing ${\mathcal{H}}^1(J_u)\to 0$. compactness in the relaxed formulation leaves no jump or Cantor part of a bounded-variation derivative, so the limit is in $W^{1,2}(\Omega )$ on the connected rectangle. The reduced edge-free Mumford–Shah limit isA set of zero length can be omitted; this does not impose a zero image signal or a Dirichlet boundary value. Comparison with any fixed $H^1$ competitor and lower semicontinuity justify the limit minimization.

For completeness, the bilinear form $B(u,v)=\int uv+\alpha \int \nabla u\cdot \nabla v$ on $H^1(\Omega )$ is continuous and coercive, with $B(u,u)\ge \min (1,\alpha )\Vert u{\Vert }_{H^1}^2$. The right-hand side $\int gv$ is bounded because $g\in L^2$ on the bounded rectangle. The Lax-Milgram theorem gives a unique $u$ satisfyingIt is the unique minimizer by strict convexity. Formally,with the Neumann condition understood through this weak formulation. Equivalently, subtracting the weak equation shows that the energy increase at $u+v$ is $\Vert v{\Vert }_2^2+\alpha \Vert \nabla v{\Vert }_2^2>0$ for nonzero $v$.

Taking the essential supremum in the characteristic solution gives the sharper estimateIndeed, the bijective characteristic flow map preserves each spatial-velocity essential supremum. If $H_t={\mathrm{esssup}}_{0\le s\le t,x,v\in \mathbb{R}}|h(s,x,v)|$, this proves $\Vert f_t{\Vert }_{\infty }\le \Vert f_0{\Vert }_{\infty }+t{H}_t$. A time-independent source has $H_t=\Vert h{\Vert }_{L^{\infty }({\mathbb{R}}^2)}$, which is the displayed form. For a time-dependent source, the same symbol must mean a bound uniform over the elapsed time interval; its norm at the final time alone need not bound the accumulated forcing.

The bound is sharp. Take $f_0=1$ and $h=1$, giving $f(t,x,v)=1+t$ and equality for every $t\ge 0$. Both functions are smooth and bounded, although their finite-$p$ integrals over the whole plane are infinite.

Choose $f_0(x,v)=e^{-(x^2+v^2)}$ and $h(t,x,v)=x^2+v^2$. The initial Gaussian function is smooth and belongs to every finite Lp space, and is also bounded. The source is smooth and nonzero. The accumulated source along the backward characteristic curve isIts quadratic-form eigenvalues are $(e^{2t}-1)/2$ and $(1-e^{-2t})/2$. Both are positive for $t>0$, soConsequently $\Vert f_t{\Vert }_p=\infty$ for every $t>0$ and every finite $p\ge 1$; it is unbounded, so its $L^{\infty }$ norm is infinite as well. This spatially nonintegrable forcing in transport example avoids any ambiguity about whether the last endpoint is included in “all $p$”.

For each $p\ge 1$ with $f_0\in L^p$, a sufficient condition is $h\in L^1([0,T];L^p({\mathbb{R}}^2))$ on every finite interval. The Minkowski integral inequality and measure preservation give the finite-time Lp bound for Hamiltonian transportFor all finite $p$ simultaneously, impose, for example, $h\in L_{\mathrm{loc}}^1([0,\infty );L^1\cap L^{\infty })$. The elementary bound $\Vert h_s{\Vert }_p\le \Vert h_s{\Vert }_1^{1/p}\Vert h_s{\Vert }_{\infty }^{1-1/p}$ and the Holder inequality in time show that its $L^p$ norms are locally integrable for every $1\le p<\infty$. If a bounded initial value is also required, the same assumption controls the $p=\infty$ endpoint by the previous part.