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Let $a\in BV(\Omega )$, $b\in BV(\Omega )\cap L^{\infty }(\Omega )$ and ${\delta }_{t}c=c\circ {\Phi }_t-c$, where ${\Phi }_t$ is the local flow of $\phi e_j$ with nonnegative $\phi \in C_c^{\infty }(\Omega )$. The formula uses common oriented BV traces on a hypersurface; the trace difference of $a$ is zero almost everywhere on $J_b\setminus J_a$. It also holds for both negative-time increments with denominator positive $t$.

The BV slicing theorem reduces the calculation to one dimension. With a right-continuous representative of $a$ and $\mu =Da$, Fubini's theorem yieldsThe inner factor tends to $\phi (s)[b](s)$ and is bounded in absolute value by $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }$. Dominated convergence against $|\mu |$ leaves only its measure atoms at the countable slice jumps of $b$. That same bound times $|D{a}_y|$ permits integration over the transverse coordinates. No essential bound on $a$ is used, even if its slice values grow without bound. The formula is useful for opposite-flow fidelity identity for quadratic data with unbounded data and a bounded comparison function.

A smooth vector field is complete if its maximal integral curves of a vector field exist for every real time. Its local flow is then a global one-parameter group of diffeomorphisms; uniqueness of ordinary differential equations proves the group law and the inverse-time identity. On a compact manifold without boundary, every smooth vector field is complete: compactness gives a uniform positive local existence interval through every point, so a curve can be successively extended past any proposed finite endpoint. On $\mathbb{R}$, $x^2{\partial }_x$ is not complete because its solution $x(t)=x_0/(1-t{x}_0)$ has finite-time blowup for $x_0>0$.

On ${\mathbb{R}}^2$, the vector fields $X=y{\partial }_x$ and $Y=(x^2/2){\partial }_y$ are complete, since their local flows are $(x+ty,y)$ and $(x,y+t{x}^2/2)$. Their Lie bracket of vector fields is $-(x^2/2){\partial }_x+xy{\partial }_y$. Its integral curve of a vector field starting at $(2,0)$ is $(2/(1+t),0)$, which cannot continue through $t=-1$. The sum has an integral curve of a vector fieldthat cannot continue through $t=2\sqrt{3}$. Thus neither operation preserves complete vector fields on a general noncompact smooth manifold. This contrasts with a compact manifold without boundary, where every smooth vector field is complete.

The chain rule shows that $\alpha {\varphi }_t{\alpha }^{-1}$ is the local flow of the pushforward of a vector field ${\alpha }_{\ast }X$. If this equals $X$, uniqueness of integral curves of a vector field gives commutation. Conversely, differentiate the commuting identity at $t=0$ to obtain $d{\alpha }_p{X}_p=X_{\alpha (p)}$. All identities hold on their common domains; the vector field need not be complete.

For the local flow ${\varphi }_t$ of a smooth vector field $X$, pull a tensor field $T$ back to the fixed base point before differentiating. Pullback uses $d{\varphi }_t^{-1}$ on vector factors and the dual of $d{\varphi }_t$ on covector factors. The chain rule gives ${\mathcal{L}}_{X}f=Xf$. In coordinates, $d{\varphi }_t^{-1}Y({\varphi }_t(x))=Y+t(dYX-dXY)+O(t^2)$, so ${\mathcal{L}}_{X}Y=[X,Y]$. Differentiation of pullback preserves tensor contractions and obeys the Leibniz rule for tensor products; hence this is the tensor derivation defining the Lie derivative of a tensor field.

An integral curve follows a vector field by solving the displayed ordinary differential equation. For a smooth vector field, integral curves through initial points assemble into its local flow.

Assume $w\in BV\cap L^{\infty }$ minimizes scalar total variation denoising for $g\in BV\cap L^2$. The data $g$ may be unbounded. For the local flow of $\phi e_j$, use $(1-\theta )w+\theta w\circ {\Phi }_{\pm t}$ as two competitors. The total variation under opposite smooth flows and convexity of the total variation seminorm bound the sum of regularizer changes by $O(t^2)$. Minimality and the opposite-flow fidelity identity for quadratic data, evaluated using the BV jump-product limit with one bounded factor, implyLet $\theta \downarrow 0$. The integrands define finite signed Radon measures, since $w$ is bounded and the jump variation of $g$ is finite. Arbitrary nonnegative smooth $\phi$ make their densities nonnegative. Testing every coordinate direction removes the factor $|{\nu }_w\cdot e_j|$ and proves the stated inequality.

For $\xi$ in the Lie algebra of $G$, the left-invariant vector field is $l_{\xi }(g)=(d{L}_g{)}_e\xi$. Its global flow is right translation ${\Phi }^t=R_{\exp (t\xi )}$.

Let $X$ be a smooth vector field with local flow ${\varphi }_t$. The Lie derivative of a differential form isPulling back before taking the derivative compares forms in the same cotangent space. The Lie derivative of a differential form preserves degree and obeys the Leibniz rule for the wedge product of differential forms.

Cartan's magic formula expresses it using two intrinsic operations:where $d$ is the exterior derivative and ${\iota }_X$ is the interior product. In particular, it commutes with the exterior derivative.

On the real line, $X=x{\partial }_x$ has local flow ${\varphi }_t(x)=e^{t}x$. Hence ${\varphi }_t^{\ast }dx=e^{t}dx$ and ${\mathcal{L}}_{X}dx=dx$. The vector field vanishes at zero, but its differential still stretches the one-form there. For a volume form $\omega$, the equation ${\mathcal{L}}_X\omega =0$ defines a volume-preserving vector field.

For a smooth function $f$ and a vector field $X$, the Lie derivative is ${\mathcal{L}}_{X}f=X(f)=df(X)$. It is the derivative of $f$ along the local flow of $X$.

For vector fields $X,Y$, the Lie derivative is the Lie bracket of vector fields ${\mathcal{L}}_{X}Y=[X,Y]$. Pulling $Y$ back by the local flow of $X$ and differentiating gives $dY(X)-dX(Y)$. No affine connection is needed.

Let $w\in BV\cap L^{\infty }$ and $g\in BV\cap L^2$ on a bounded domain, $F_g(v)=\frac{1}{2}\Vert v-g{\Vert }_2^2$, and $w_{\pm t}=(1-\theta )w+\theta w\circ {\Phi }_{\pm t}$ for a smooth compactly supported local flow. The displayed formula is exact to an $o(t)$ remainder, and $g$ need not be bounded. Writing $J_t=\det D{\Phi }_t$, its cross-term identity isThe remainder is $o(t)$ because $1-J_{-t}=O(t)$ and weighted $L^1$ convergence follows fromFirst let $t\to 0$, then $K\to \infty$. The mass change of $w^2$ under opposite flows is $O(t^2)\Vert w{\Vert }_2^2$. Combined with the BV jump-product limit with one bounded factor, this evaluates the jump contribution without a bounded fidelity derivative.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

An integral curve of a vector field $X$ through $p$ is a smooth curve $c:I\to M$, on an interval containing zero, satisfying $c(0)=p$ and $\dot{c}(t)=X_{c(t)}$. The local flow is the smoothly varying family of these integral curves of a vector field:Here $D$ is an open neighborhood of $\{0\}\times M$; for the maximal local flow, its fibre over $p$ is the maximal existence interval of $c_p$. Local existence, uniqueness and smooth dependence for a smooth ordinary differential equation give this family. Uniqueness also gives ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever the relevant curves and compositions are defined, and ${\varphi }_{-t}={\varphi }_t^{-1}$ on their common domains.

Reparametrizing the integral curves of a vector field givesThus the flow of $2X$ is $\boxed{{\tilde{\varphi }}_t={\varphi }_{2t}}$, with domain $\{(t,p):(2t,p)\in D\}$.

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

Using the coordinate formula for the Lie bracket of vector fields, the two components areThus $\boxed{[X,Y]=-(x^2/2){\partial }_x+xy{\partial }_y}$, which is not the zero vector field.

The integral curves of a vector field $X$ solve $\dot{x}=y$, $\dot{y}=0$, whereas those of $Y$ solve $\dot{x}=0$, $\dot{y}=x^2/2$. Their local flows areBoth formulas exist for all real parameters, so both are complete vector fields. Direct composition givesAt $(1,0)$ their first coordinates differ by $ts/2$. Hence these local flows do not commute when $s,t$ are nonzero.

For the Lie bracket of vector fields, start at $(2,0)$. An integral curve of a vector field $[X,Y]$ isIt solves $\dot{x}=-x^2/2$, $\dot{y}=xy$, but $x(t)$ diverges as $t\downarrow -1$. It cannot extend through that finite time as a curve in ${\mathbb{R}}^2$. Complete vector fields are therefore not closed under the Lie bracket.

For the sum $X+Y$, its integral curves of a vector field obey $\dot{x}=y$, $\dot{y}=x^2/2$. Substitution of $x=(1+\alpha t{)}^{-2}$ and $y=\beta (1+\alpha t{)}^{-3}$ requiresChoosing $\alpha =-1/(2\sqrt{3})$ gives the integral curve of a vector fieldIt starts at $(1,1/\sqrt{3})$ and escapes to infinity at the finite time $2\sqrt{3}$. Complete vector fields are not closed under addition either. Both failures use the same pair of complete vector fields.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

Let ${\varphi }_s$ be the local flow of the smooth vector field $X$, with ${\varphi }_0=\mathrm{id}$ and $\frac{d}{ds}{\varphi }_s(p)=X_{{\varphi }_s(p)}$. The flow definition of the Lie derivative of a tensor field isFor a tensor field of type $(r,s)$, the mixed tensor pullback transports the tensor at ${\varphi }_s(p)$ back to $p$: it applies $(d{\varphi }_s{)}_p^{-1}$ to every contravariant factor and $(d{\varphi }_s{)}_p^{\ast }$ to every covariant factor. Thus every difference quotient belongs to the same tensor space at $p$. This definition gives another tensor field of the same type and uses no choice of affine connection. The tensor Lie derivative is defined wherever the local flow exists, including points where $X$ vanishes.

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.

Choose local coordinates $y^1,\dots ,y^{D-1}$ on the hypersurface $\Sigma$, and let ${\Phi }_s$ be the local flow of the vector field $X$. The map $F(s,y)={\Phi }_s(p(y))$ follows the integral curves of a vector field starting at $p(y)\in \Sigma$. At $s=0$, its differential maps the $s$ direction to $X$ and the remaining directions to a basis of $T_p\Sigma$. Transversality makes these vectors linearly independent, so the inverse function theorem makes $F$ a local manifold chart.

Set $x^0=s$ and transport the $y^i$ unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.