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By locality of weak differentiability, it is enough to work near the exceptional point; translate it to zero. Let $g_i$ denote the classical derivative off zero, assigning any value at zero. For $\varphi \in C_c^{\infty }(\Omega )$ choose a smooth cutoff ${\chi }_{\epsilon }$ which is zero on $B_{\epsilon }$, one outside $B_{2\epsilon }$, satisfies $|D{\chi }_{\epsilon }|\le C/\epsilon$, and has its transition region contained in $\Omega$. The product $\varphi {\chi }_{\epsilon }$ avoids the singular point, so classical integration by parts givesThe middle term has absolute value at mostwhen $n\ge 2$. The other terms tend to $\int u{D}_i\varphi$ and $-\int g_i\varphi$, since $u$ is locally integrable and $g_i\in L_{\mathrm{loc}}^1$. Therefore $\boxed{D_{i}u=g_i\text{ weakly on all of }\Omega }$. Its value at the single point is irrelevant. This is bounded isolated-point removability for weak first derivatives.

For $n=1$, take $u(x)=0$ for $x<0$ and $u(x)=1$ for $x>0$ on $(-1,1)$, with any value at zero. It is bounded and $C^1$ off zero, with classical derivative zero there. Butso its distributional derivative is a Dirac delta. If a locally integrable weak derivative existed, it would be zero off zero and hence zero almost everywhere, contradicting this identity. The step function is the required counterexample; the cutoff error in dimension one need not vanish.

The reaction sign and the meaning of recoil translation matter here. In Lighthill elongated-body theory, put $\mathcal{D}={\partial }_t+U{\partial }_x$. The transverse momentum imparted to the fluid per unit length is $m\mathcal{D}h$, so the lateral force on the fish is $-\mathcal{D}(m\mathcal{D}h)$. Thus body mass and positive added mass enter effective inertia with the same sign. The periodic trailing-edge thrust, to quadratic order in slope, isThe reaction sign and this small-amplitude thrust formula can also be checked in the primary derivation of the elongated-body force, equations (2.6), (3.2) and the discussion after (3.5). For the long-body limit below, the recoiling leading edge contributes only at relative order ${\alpha }^{-2}$.

The requested recoil and thrust formulas cannot all hold for the literal fish described. In particular, the printed thrust would remain $-m{U}^2/4$ when $h_1=0$, although a straight, unperturbed fish has $h_t=h_x=0$ and zero reactive thrust. Its negative term is missing an amplitude-squared slope factor. The stated effective translational coefficient $M^{\prime }-1$ also has the wrong sign for positive added mass, and the literal centre of mass translation cannot be equated to the translation of an arbitrarily chosen undeformed body reference line. The PDF defines $I^{\prime }=I/[m(\alpha L{)}^3]$; the converted TeX incorrectly uses the square.

Here are the force balance and angular momentum calculations for a consistent interpretation of the stated uniform fish. They give a corrected recoil solution, as well as a direct test of the printed equations. Defineand let $R$ be actual centre of mass displacement. The prescribed deformation has mean $a(t)=h_{1}f(t)/[2(\alpha +1)]$. Its mass-weighted mean must be removed in the recoil correction in elongated-body theory:Indeed $\int hdx=\ell R$, since $\int \xi dx=0$ and $\int h_{0}dx=\ell a$. Introduce the two geometric integralsFor uniform mass per unit length, $I={\mu }_{b}J$ at leading small-amplitude order. Retaining the symbol $I$ makes the rotational momentum balance transparent.

The needed integrals of transverse acceleration and convective derivatives areThe second spatial derivative includes the hinge's slope jump, as a distributional derivative; discarding it would incorrectly remove the $U^2$ terms. Newton's lateral equation is $M\ddot{R}=-m\int {\mathcal{D}}^{2}hdx$. The angular momentum of the deforming fish is $I\dot{\theta }+{\mu }_{b}C\dot{f}$, so its time derivative equals $-m\int \xi {\mathcal{D}}^{2}hdx$. Consequently the consistent uniform-added-mass recoil equations areThese follow directly from the local Lighthill elongated-body theory model with the physical reaction sign. An alternative reference-line translation replaces $R$ by $R-a$, and changes the shape forcing, but cannot change $M+m\ell$ to a difference of two positive inertias. Leading-edge details of a uniform-depth idealization can require an end model; they do not alter this reaction-sign test or the zero-stroke counterexample.

For $f=\cos \omega t$, write $R=\mathrm{Re}(r{e}^{i\omega t})$ and $\theta =\mathrm{Re}(\tau e^{i\omega t})$, choosing the periodic response and omitting free rigid-motion constants. The two balances give the explicit recoil amplitudesAt the tail the displacement and slope amplitudes are $H_L=h_1+r+\ell \tau /2-h_1/[2(\alpha +1)]$ and $S_L=h_1/L+\tau$. Thus a corrected, directly calculable mean thrust iswithin the trailing-edge approximation, with any retained leading-edge contribution treated consistently. This expression has the correct dimensions and vanishes quadratically with stroke amplitude.

A particularly simple independent check is the limit $U=0$. For the uniform fish, $I={\mu }_{b}J$, the balances give $\ddot{R}=0$ and $\theta =-Cf/J$ for the periodic response. The tail displacement amplitude is then exactlyHence, to the requested relative order,The leading-edge contribution is only $O({\alpha }^{-2})$ here. This recoil reduction contradicts the printed positive correction, for example for $M^{\prime }=2$, $I^{\prime }=1/6$. It also shows explicitly why treating the prescribed shape's moving mass as stationary would give a different model.

For completeness, the positive correction quoted in the question has a simple conditional origin if its two displayed recoil equations are accepted as formal equations rather than derived physical balances. Put $J^{\prime }=I^{\prime }+1/6$ and keep $U/(\omega L)$ fixed as $\alpha \to \infty$, away from $M^{\prime }=1$. Adding those equations and then using the first gives their leading periodic responseUnder the additional reference-line interpretation $h=h_0+R+\xi \theta$, the in-phase tail amplitude is $h_1[1+1/(2\alpha (M^{\prime }-1))+1/(8\alpha J^{\prime })]$. Squaring it in the trailing-edge thrust gives the printed frequency correction. Even this formal route, however, gives the amplitude-consistent resultnot a term $-m{U}^2/4$ independent of $h_1$. The quadrature part of $R$ first changes the squared displacement at relative order ${\alpha }^{-2}$, while recoil changes the tail slope at that order as well. This conditional calculation explains the intended algebra without presenting the flawed starting equations as a physical theorem.

Reactive thrust is positive when the tail's lateral-motion term exceeds its streamwise slope penalty; for an unrecoiling hinged tail this requires $\omega L>U$. A real steady swimming speed additionally balances the thrust against resistive drag force, absent from this inertial idealization. Recoil can alter either term and is not universally thrust-enhancing. A pole at $M^{\prime }=1$ is not a physical negative-added-mass resonance: the genuine combined inertias remain positive.

The appropriate Poincare-Wirtinger inequality, with no zero-boundary assumption, isAssume $\Omega$ is nonempty, bounded, connected and has the stated cone property. If this failed, after subtracting the mean and normalizing there would be $u_j\in W^{1,p}(\Omega )$ withThe sequence is bounded in $W^{1,p}$. By the permitted Rellich-Kondrachov compactness theorem, a subsequence converges strongly in $L^p$ to $u$. For every compactly supported test function, integration by parts and these two convergences show that each distributional derivative of $u$ is zero. A function with zero weak gradient is constant on a connected open set: mollification makes it locally constant, and connectedness identifies the local constants.

Convergence in $L^p$ on a finite-measure set also preserves the integral, so $u_{\Omega }=0$ and the constant is zero. But strong convergence preserves $\Vert u{\Vert }_p=1$, a contradiction. This proves the inequality; connectedness prevents different constants on separate components.

Weak convergence of distributions means $⟨T_n,\phi ⟩\to ⟨T,\phi ⟩$ for every test function $\phi \in C_c^{\infty }(\Omega )$. The distributional derivative is defined byIt is linear and continuous: differentiation preserves compact support and bounds each test-function seminorm by one of higher derivative order, so composing it with the continuous functional $T$ again gives a distribution. Finally, for every fixed test function,Thus distributional differentiation preserves the stated convergence.

The statement as printed is false without an integrability condition. On any nonempty open set, the constant functions $f_n\equiv n$ are harmonic and therefore subharmonic, but their supremum is $+\infty$, which is not locally integrable.

A valid version assumes $g={\sup }_n{f}_n\in L_{\mathrm{loc}}^1(\Omega )$. First, the maximum of finitely many subharmonic functions is subharmonic. For a distributional proof, mollify two functions on smaller interior domains and replace their maximum byThis is smooth, convex, and nondecreasing in each argument. The chain rule shows that its Laplacian applied to two smooth subharmonic functions is nonnegative: the first-derivative terms multiply their nonnegative Laplacians, and the Hessian term is nonnegative by convexity. Mollifier convergence in $L_{\mathrm{loc}}^1$ and the uniform bound $|M_{\delta }-\max |\le \delta /2$ allow both regularizations to be removed. Continuity of distributional derivatives under this convergence proves the finite-maximum assertion.

Now $g_N={\max }_{n\le N}{f}_n$ increases to $g$ and satisfies $|g_N|\le |f_1|+|g|$ almost everywhere. The dominated convergence theorem gives convergence in $L_{\mathrm{loc}}^1$. For every nonnegative test function,Thus $g$ is subharmonic in the distributional sense. This is the locally integrable supremum theorem for subharmonic functions. For the pointwise upper semicontinuous convention, take its canonical representative; the raw supremum need not already have that regularity.

The locally integrable logarithm has distributional derivative $(\log |x|{)}^{\prime }=\mathrm{PV}(1/x)$. By the Fourier transform of a derivative and the previous result,For $k\ne 0$, this gives $-\pi /|k|$. At zero the expression must be extended as a Hadamard finite-part integral, not treated as a locally integrable function. For example, define the finite-part inverse-absolute-value distribution byThen $kT=\mathrm{sgn}k$. Different reference cutoffs change $T$ only by a multiple of the Dirac delta distribution. The general solution of the transform equation is consequentlyIndeed multiplication by $k$ annihilates only a delta term among distributions supported at zero; delta derivatives would generate additional nonzero terms. This supplies the precise interpretation of the paper's formula, and the convention-dependent constant $C$ need not be found.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

Write $\mathrm{TV}(u)=|Du|(\Omega )$ for the total variation seminorm on a domain. For a locally integrable real function its dual definition isThe bounded-variation space consists of functions $u\in L^1(\Omega )$ with finite $|Du|(\Omega )$, equipped with $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_1+|Du|(\Omega )$. In the equivalent distributional derivative description, $Du$ is a finite vector-valued Radon measure and $|Du|$ is its total variation measure.

To establish completeness of the bounded-variation space, let $(u_n)$ be Cauchy in this norm. Completeness of $L^1$ gives $u_n\to u$ in $L^1$. Given $\epsilon >0$, choose $N$ so that $\Vert u_n-u_m{\Vert }_{BV}<\epsilon$ whenever $m,n\ge N$. For fixed $n\ge N$, the given lower semicontinuity yieldsSince $\Vert u_n-u_m{\Vert }_1\to \Vert u_n-u{\Vert }_1$, we obtain $\Vert u_n-u{\Vert }_1+|D(u_n-u)|(\Omega )\le \epsilon$. In particular $u_n-u$ has finite variation; the triangle inequality then puts $u$ in the BV space. The same bound proves convergence in the full norm, so this is a Banach space.

For the disk data, put $B=B(0,R)$ and assume $\alpha >0$. The exact total variation denoising of a disk isHere $(s{)}_{+}=\max (s,0)$ is the positive part. A total variation calibration certifies global optimality, including competitors that are not radial or piecewise constant. Define the bounded vector fieldIt has $|z|\le 1$. Its normal component is continuous across the circle, so the distributional divergence has no extra boundary measure. Direct differentiation gives $q=\mathrm{div}z=(2/R){\chi }_B$. The dual definition implies $\int vq\le \mathrm{TV}(v)$: for the standard $BV\cap L^2$ domain one can cut $z$ off at radius $L$, with the error bounded by $C{L}^{-1}{\int }_{L<|x|<2L}|v|\to 0$, and then smooth the test field. In the larger homogeneous bounded-variation space, the same error is bounded by $C\Vert v{\Vert }_{L^2(L<|x|<2L)}\to 0$. Thus both usual whole-plane formulations give the same certificate.

Use the perimeter identity $\mathrm{TV}({\chi }_B)=2\pi R$ and $|B|=\pi R^2$. If $c=1-2\alpha /R>0$, then $\int c{\chi }_{B}q=2\pi Rc=\mathrm{TV}(c{\chi }_B)$ and $u_{\ast }-g+\alpha q=0$. Consequently every competitor $v$ satisfiesIf $\alpha \ge R/2$, replace $z$ by $z_{\alpha }=(R/(2\alpha ))z$. Its norm is still at most one, its divergence is $q_{\alpha }={\chi }_B/\alpha$, and equality in the calibration holds at $u_{\ast }=0$. The identical comparison proves optimality and uniqueness of zero, including the threshold $\alpha =R/2$. The only general results used are completeness of $L^1$, lower semicontinuity of variation, the indicator-perimeter identity, the distributional integration-by-parts/dual variation formula and the quadratic norm identity. For $\alpha =0$, the unique squared-error minimizer is simply $g$.

The shear $(t,x,v)\mapsto (t,x+tv,v)$ is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus $F,G$ are locally integrable functions. On a compact time interval $I$ and compact label set $K\subset {\mathbb{R}}^6$, the integral ${\int }_K{\int }_I|G|dtdxdv$ is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that $G(\cdot ,x,v)\in L_{\mathrm{loc}}^1(\mathbb{R})$ for almost every $(x,v)$.

To transform the weak equation, choose $\varphi (t,x,v)=\rho (x-tv,v)\psi (t)$, where $\rho \in C_c^{\infty }({\mathbb{R}}^6)$ and $\psi \in C_c^{\infty }(\mathbb{R})$. Including a velocity cutoff in $\rho$ ensures the required compact support. The transport derivative of this test function is $\rho (x-tv,v){\psi }^{\prime }(t)$. Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which ${\partial }_{t}F=G$ as one-dimensional distributions.

For each remaining label, subtract the primitive $H(t)={\int }_0^{t}G(s)ds$. The distributional derivative of $F-H$ is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, $\tilde{F}(t)=C+H(t)$. The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.

The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes $3k\psi (0)$. The jump condition for an impulse is thereforeContinuity at zero gives $C_{+}=C_{-}=C$. The two one-sided derivatives are $-3kC/2$ and $+3kC/2$, whose difference is exactly $-3kC$. Hence the matching condition introduces no further restriction on $C$, and the globally bounded solution isThis continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.

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.

Use $D_j=-i{\partial }_{x_j}$, so the Fourier transform of $D^{\alpha }u$ is ${\lambda }^{\alpha }\hat{u}$. Distinguish the full degree-$N$ polynomial $P(\lambda )$ from its homogeneous top-degree part $p_N(\lambda )$, the principal symbol. The elliptic differential operator condition isIt does not require the full polynomial to be nonzero at small frequencies. Compactness of the unit sphere gives $c_0={\min }_{|\theta |=1}|p_N(\theta )|>0$. Homogeneity and the lower-degree remainder implyFor sufficiently large $|\lambda |$ the second term is at most half the first, so the high-frequency lower bound for an elliptic polynomial isHere $⟨\lambda ⟩$ is the Japanese bracket. The order-zero case just means a nonzero constant and is immediate.

For real $s$, the Sobolev space definition with the current Fourier normalization isChanging the harmless constant in the norm gives the same space. A distribution on $X$ belongs to the Local Sobolev space $H_{\mathrm{loc}}^s(X)$ when $\chi u$, extended by zero outside $X$, belongs to $H^s({\mathbb{R}}^n)$ for every test function $\chi \in C_c^{\infty }(X)$.