Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 64 (Measure and Image)

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.

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.

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.

For a scalar $u\in BV(\Omega )$, put $E_t=\{x\in \Omega :u(x)>t\}$. The coarea formula for BV functions is the measure identityfor every Borel $A\subseteq \Omega$. For almost every real $t$, $E_t$ is a finite-perimeter set, with $\mathrm{Per}(E_t;A)=|D{\chi }_{E_t}|(A)$. In particular $\mathrm{TV}(u)={\int }_{\mathbb{R}}\mathrm{Per}(E_t;\Omega )dt$. This is relative perimeter: it counts interfaces inside $\Omega$, and does not add the boundary jump of a zero extension.

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 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.

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.

First apply Lebesgue decomposition to the derivative measure relative to ${\mathcal{L}}^n$:Here $\nabla u$ is the almost-everywhere approximate gradient, rather than an assertion that $u$ belongs to $W^{1,1}$. Split the singular part into its jump part of a BV derivative and Cantor part of a BV derivative:An approximate jump point has a unit normal ${\nu }_u$ and distinct finite BV traces on a hypersurface $u^{+},u^{-}$, obtained as mean limits on the corresponding two half-balls. Their set $J_u$ is the jump set of a BV function, countably $(n-1)$-rectifiable. Reversing the normal swaps the BV traces on a hypersurface and leaves the displayed measure unchanged. The approximate discontinuity set $S_u$ differs from $J_u$ only by an ${\mathcal{H}}^{n-1}$-null set.

The remaining $D^{c}u=D^{s}u-D^{j}u$ is singular to Lebesgue measure and gives zero mass to every set with sigma-finite ${\mathcal{H}}^{n-1}$ measure. It is diffuse rather than a second jump contribution. In one dimension the three parts are illustrated by an affine function, a step function and the Cantor function, respectively. Countably many jumps therefore do not imply that the singular derivative has no Cantor part of a BV derivative.

Put $\mu =Du$, a finite signed Radon measure, and defineFor $\phi \in C_c^1((a,b))$, Fubini's theorem for the signed measure givesThus $D{F}_l=\mu =Du$. A distribution on a connected interval with zero derivative is constant: any compactly supported test of integral zero is the derivative of a compactly supported test, so it pairs to zero with $u-F_l$. Choose the resulting constant $c$. Thenare one-sided representatives of a one-dimensional BV function. The first equals $u$ almost everywhere. Their difference is $\mu (\{t\})$, nonzero at at most countably many points, so the second also equals $u$ almost everywhere.

Finite-measure continuity applied to $|\mu |$ proves $F_l(s)\to F_l(t)$ as $s\uparrow t$, and $F_r(s)\to F_r(t)$ as $s\downarrow t$. Moreover the opposite one-sided limits are $F_l(t+)=F_r(t)$ and $F_r(t-)=F_l(t)$. Consequently both representatives are continuous exactly where $\mu (\{t\})=0$. For each positive integer $m$, there are only finitely many measure atoms of magnitude at least $1/m$, since their total magnitudes are bounded by $|\mu |((a,b))$. Their union is countable, proving the requested discontinuity bound. This does not require the jump points to be isolated; they can be dense.

The hint's one-dimensional statement is consistent with this construction: zero ${\mathcal{H}}^0$ measure means an empty set, so $S_u\subseteq J_u$. This concerns approximate discontinuities of the BV space class; arbitrary changes of representative at points could create artificial pointwise discontinuities. The interval-mass representatives above remove that ambiguity and give the required one-sided continuity directly.