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Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$, $\hslash =c=1$, and $\square ={\partial }_a{\partial }^a=-{\partial }_t^2+{\nabla }^2$. The Proca field has nonzero mass $m$. Vary its action, using the fact that the gauge field strength is an antisymmetric second-rank tensor:The integration by parts discards a boundary term, with the variation fixed at the boundary. Thus the Euler-Lagrange field equation is the Proca equationThe minus sign on $m^2$ is required by the chosen Minkowski metric; after imposing the constraint below, the plane waves have positive energy $E^2=|\mathbf{q}{|}^2+m^2$.

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.

It is useful to regulate the number of modes first, so the functional integral identities reduce to ordinary integration by parts. Let $C$ be the Gaussian covariance, $F=C^{-1}$, $W=e^{-S_1}$ and put a dot for $\Lambda {\partial }_{\Lambda }$. The matrix identity $\dot{F}=-F\dot{C}F$ givesThe second field derivative of the Gaussian isHence, after two integrations by parts,The imposed flow makes the bracket vanish. The remaining trace is independent of the fields and only changes the Gaussian normalization. Since the free Gaussian normalization is $\mathcal{N}=(\det (2\pi C){)}^{1/2}$, $\dot{\log \mathcal{N}}=\mathrm{Tr}(F\dot{C})/2$. ThereforeEquivalently, $Z$ is cutoff independent after discarding the stated overall rescaling. This is the Gaussian covariance differentiation identity behind the Polchinski equation.

With the Fourier convention above and functional derivatives satisfying $\delta \tilde{\varphi }(p)/\delta \tilde{\varphi }(q)={\delta }^{(4)}(p-q)$, contraction with $\dot{C}$ becomes $\int d^{4}p(2\pi {)}^4{\dot{C}}_{\Lambda }(p){\delta }^2/[\delta \tilde{\varphi }(p)\delta \tilde{\varphi }(-p)]$. Thus the numerator $(2\pi {)}^4$ in the printed flow is consistent with this derivative convention; it must not be changed independently of the convention.

Use unit speed and write the Cauchy data as $u(0)=f$, $u_t(0)=g$. For finite-energy data define the wave energy estimate quantityMultiply $u_{tt}-{\Delta }_{x}u=0$ by $u_t$ and use integration by parts. With compact support or sufficient decay the boundary flux is zero, soFor general finite-energy solutions, cutoff or approximation arguments justify this identity; equivalently the local estimate below, applied in both time directions and with radii tending to infinity, gives the same equality. Arbitrary smooth data need not have finite global energy; then the global bound with an infinite right side is uninformative, while the local estimate remains useful.

If $f\in H^1$ and $g\in L^2$, the fundamental theorem of calculus and the wave energy estimate further giveTogether these yield an a priori bound for $\Vert u(t){\Vert }_{H^1}+\Vert u_t(t){\Vert }_2$ on each bounded time interval. No existence assumption is proved by the estimate itself; it controls any sufficiently regular solution.

There is a normalization error in the PDF. With the printed unnormalized integral, testing a constant function equal to one would give a left side $|U|(1-|U|{)}^2$ and a right side zero. Thus that formulation fails whenever $|U|\ne 1$.

Use instead the average $v_U=|U{|}^{-1}{\int }_{U}v$. The Poincare-Wirtinger inequality, also called the Neumann-Poincare inequality, isIf no $C_P$ exists, subtract the average and normalize a violating sequence to obtain $w_j\in H^1(U)$ with ${\int }_U{w}_j=0$, $\Vert w_j{\Vert }_2=1$, and $\Vert \nabla w_j{\Vert }_2\to 0$. This sequence is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrachov compactness theorem supplies a subsequence converging strongly in $L^2(U)$ to $w$.

For every compactly supported test function $\phi$, integration by parts and these convergences give $\int w{\partial }_i\phi =0$. Thus $w$ has zero weak gradient. The Sobolev function with zero weak gradient result and connectedness make $w$ constant. Strong $L^2$ convergence preserves its zero integral, so $w=0$. It also preserves its norm one, a contradiction. This proves the correctly normalized Neumann-Poincare inequality.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincare inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

Multiply the viscous scalar conservation law by $-u_{xx}$, rather than estimating $F^{\prime \prime }$ after differentiating. Integration by parts yieldsThus $d\Vert u_x{\Vert }_2^2/dt\le (M^2/\epsilon )\Vert u_x{\Vert }_2^2$. Applying the Gronwall inequality givesOnly the assumed bound on $F^{\prime }$ is used; a global bound on $F^{\prime \prime }$ is not needed for this energy estimate.

Let $M=\Vert F^{\prime }{\Vert }_{\infty }$. The sharper energy estimate uses the divergence structure of the viscous scalar conservation law. Multiply by $u$ and integrate over the line. With $G(s)={\int }_0^{s}r{F}^{\prime }(r)dr$,because the decay makes $G(u)$ tend to zero at both ends. Integration by parts in the diffusion term therefore givesOne may take $C_0=0$, uniformly in $\epsilon$. This does not require $F(0)=0$.

If a bound explicitly involving $M$ and $\epsilon$ is desired, retaining the transport term and using the Cauchy-Schwarz inequality and the elementary inequality $ab\le (a^2+b^2)/2$ givesThe Gronwall inequality then gives the valid but weaker choice $C_0=M^2/\epsilon$. The exact cancellation explains why its divergence is unnecessary.

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.

For a finite-radius centrally regular stellar polytrope with $0<n<5$ and $n\ne 1$, eliminate ${\rho }_c$ from $R=a{\xi }_1$ and $M=4\pi a^3{\rho }_c{\omega }_n$. The resulting polytropic mass-radius relation isHere ${\xi }_1,{\omega }_n$ are dimensionless functions of $n$. At $n=1$ no such single-valued mass-as-a-power-of-radius relation exists at fixed $K$: the radius is fixed instead. At $n=3$ the exponent is zero and $M=4\pi {\omega }_3(K/(\pi G){)}^{3/2}$ is independent of central mass density. For the incompressible case, $M=(4\pi {\rho }_c/3)R^3$ at fixed mass density. Regular $n\ge 5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.

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 Arzela-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 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.

Take real $\kappa$ and use the L2 norm on the spatial interval. Existence, uniqueness and continuous dependence are the three requirements of Hadamard well-posedness. An energy method supplies the decisive estimate. For a smooth solution with homogeneous Dirichlet boundary conditions, integration by parts givesThe drift contributes only a boundary term, which vanishes. The Poincare inequality $\Vert u_x{\Vert }_2^2\ge {\pi }^2\Vert u{\Vert }_2^2$ further givesApply the same argument to the difference of two solutions to obtain uniqueness and continuous dependence on the initial data.

For existence, use the Dirichlet gauge transform for constant drift: $w=e^{\kappa x/2}u$ satisfies $w_t=w_{xx}-{\kappa }^{2}w/4$ with zero boundary values. Expanding $w_0$ in its Fourier sine series givesFor $u_0\in L^2(0,1)$ the series defines a solution continuous in $L^2$ down to $t=0$ and smooth for positive time; multiplication by the fixed bounded exponentials preserves this interpretation. Its energy estimate follows by approximation with smooth initial data. For a classical solution at the initial corners, require the usual smoothness and boundary compatibility instead. The problem is well posed in $L^2$, with a contraction estimate independent of the initial data.

The finite element method replaces an infinite-dimensional variational problem by a finite-dimensional one, usually using functions that are polynomial on small mesh elements. Ritz method and Galerkin method describe how the discrete equations are selected; neither intrinsically requires piecewise polynomials, though those spaces make assembly local and sparse.

For a concrete realization of the two-point problem, choose homogeneous Dirichlet boundary conditions at both endpoints of $(0,1)$. The differential expression alone is not a complete boundary value problem, so this boundary choice is an explicit assumption. Let $p\in W^{1,\infty }$ with $p\ge p_0>0$, $q\in L^{\infty }$ with $q\ge 0$, and $f\in L^2$. Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_2$. By the Poincare inequality this is equivalent to its full first-order Sobolev norm. Multiplying by a test function and applying integration by parts givesThe endpoint terms vanish by the essential boundary condition. The form is a symmetric bilinear form, bounded byand it is a coercive bilinear form:The load is a bounded linear functional. The Lax-Milgram theorem gives a unique weak solution and a bound $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\ast }}/p_0$. Under the stated coefficient regularity it solves the differential equation in the usual weak sense.

The Ritz method minimizes the energyover $V$, or over a conforming finite-dimensional subspace $V_h\subset V$. Its first variation is $a(v,w)-\ell (w)$; symmetry and coercivity make the functional strictly convex. In fact, if $u$ is the weak solution,Thus the energy minimizer is exactly the weak solution. The Galerkin method instead directly asks for $u_h\in V_h$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. For this symmetric coercive problem, Ritz-Galerkin equivalence for a symmetric coercive form shows that the discrete minimizer and Galerkin solution coincide. For a nonsymmetric problem, a Galerkin formulation can still apply but the same quadratic minimization generally cannot represent its full bilinear form; appropriate coercivity or inf-sup conditions must then justify the chosen spaces.

Subtract the continuous and discrete weak equations to obtain Galerkin orthogonality $a(u-u_h,v_h)=0$. Since $a$ defines the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity givesIn a general reference norm, the Céa lemma yieldsThe proof uses $a(e,e)=a(e,u-v_h)$ together with coercivity and continuity. Dense approximation spaces therefore imply convergence; the estimate reduces a numerical error problem to an approximation problem.

For implementation, partition the interval by nodes $0=x_0<\cdots <x_N=1$. Choose continuous piecewise-linear functions and the interior piecewise-linear hat functions ${\varphi }_j$ as a basis, with the two boundary coefficients prescribed. Write $u_h={\sum }_j{U}_j{\varphi }_j$. The coefficients satisfyThe stiffness matrix is symmetric positive definite: $U^{T}KU=a(u_h,u_h)>0$ for nonzero interior coefficients. The local supports make it tridiagonal in this one-dimensional linear-element case.

On an element of length $h_e$, for constant element coefficients $p_e,q_e$, the two-node matrix and constant-load vector areFor variable coefficients, integrate $p,q,f$ against the same shape functions rather than silently replacing them by constants. Element contributions are added at shared nodes, and boundary values are eliminated or lifted into the right side. A sparse direct solve or a suitable iterative method gives the nodal coefficients. Numerical quadrature should preserve the required accuracy and, with positive coefficients and weights, the positive energy structure.

A finite element interpolation estimate gives ${\inf }_{v_h}\Vert u-v_h{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ for these elements, where $h$ is the largest element size and the solution has the stated regularity. The Céa lemma then gives first-order energy error. If the dual elliptic problem has $H^2$ regularity, the Aubin–Nitsche duality argument gives second-order $L^2$ error: for $e=u-u_h$, solve $a(v,z)=(e,v{)}_{L^2}$, then use $a(e,z)=a(e,z-I_{h}z)$ to gain one further factor of $h$. Higher-order elements improve rates when the solution is sufficiently smooth.

Nonzero Dirichlet data are handled by a boundary lifting and an affine trial space. Neumann boundary conditions enter naturally through the integrated boundary term; Robin terms modify both the form and the load. With pure Neumann conditions and $q=0$, constants lie in the kernel: existence needs the load/flux compatibility condition, uniqueness needs a mean constraint, and the preceding Dirichlet coercivity argument cannot simply be reused. Conforming approximation, boundary conditions and coercivity are the ingredients connecting the finite-element construction to a justified Ritz or Galerkin solution.

For $u\in {\mathcal{E}}^{\prime }$, choose a cutoff function $\chi =1$ on a neighborhood of its support. Its extension to smooth functions makesindependent of the choice of cutoff. For real frequency, the definition of the distributional Fourier transform gives $F(\lambda )=\hat{u}(\lambda )$: interchange the pairing with the integral of a Schwartz function, using the finite-order estimate on the cutoff's compact support. On each compact set of complex frequencies the exponential and all its $x$ derivatives have uniformly convergent power series. Continuity of the distribution allows termwise differentiation, withConsequently $F$ is entire on ${\mathbb{C}}^n$.

For the precise exponential type, a fixed enlarged support would give an unnecessarily enlarged radius. Instead use the shrinking-cutoff exponential-type estimate. On one fixed compact neighborhood of the closed radius-$\delta$ ball, $u$ has finite order of a distribution $m$. Choose ${\chi }_{\epsilon }=1$ near that ball, supported in the radius-$(\delta +\epsilon )$ ball, with $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1/2$. The finite-order estimate givesTake $\epsilon =[2(1+|z|){]}^{-1}$. Its extra exponential is bounded by $e^{1/2}$, while the derivative cost is polynomial. ThusThe entire function was defined with a fixed cutoff; only its bound uses a frequency-dependent cutoff, so no holomorphic dependence is lost. This proves the required estimate with some $N\ge 0$ without enlarging $\delta$.

For the converse, set $V(z)=e^{iz\cdot y}U(z)$. Its restriction to real frequencies has polynomial growth. Define the inverse tempered distribution $v$ byRapid decay of the Schwartz function transform proves convergence and continuity, and Fourier inversion gives $\hat{v}=V$ on real frequencies.

We prove its support directly by contour shifting, rather than assuming the support conclusion of the Paley–Wiener–Schwartz theorem. Let $\omega$ be a real unit vector and take a test function $\varphi$ supported in $x\cdot \omega \ge \delta +ϵ$ for some $ϵ>0$. For every integer $M\ge 0$, integration by parts in the real-frequency Fourier integral givesIndeed, move $(1-{\Delta }_x{)}^M$ from the oscillatory exponential to $e^{-t\omega \cdot x}\varphi$; its derivatives supply at most $2M$ powers of $t$.

The product $V(z)\hat{\varphi }(-z)$ is entire. Rotate coordinates so that $\omega$ is the first coordinate vector and apply Cauchy integral theorem to a rectangle in that one complex coordinate, integrating the remaining real coordinates afterwards. For fixed $t$, choose $2M>N+n+2$; the displayed decay estimate, uniformly on the intervening imaginary segment, makes the vertical faces vanish as the real rectangle width tends to infinity. The horizontal integrals are absolutely convergent. HenceUsing $(1+|\xi +it\omega |{)}^N\le C(1+t{)}^N(1+|\xi |{)}^N$ and the growth hypothesis bounds this pairing byIt tends to zero as $t\to \infty$, so the pairing vanishes. Every point outside the closed radius-$\delta$ ball lies in such a separating half-space; a finite partition of unity for the support of a test function outside the ball proves $\mathrm{supp}v\subset \overline{B}(0,\delta )$.

Translate this compactly supported distribution by $y$: define $⟨u,\varphi ⟩=⟨v,\varphi (\cdot +y)⟩$. The Translation property of the Fourier transform givesThe compact-support entire extension agrees with $U$ everywhere by the identity theorem, applied successively in the complex coordinates. Fourier inversion also gives uniqueness. This completes both directions of the ball version of the Paley–Wiener–Schwartz theorem.

For the wave equation, take a real constant $c\ne 0$. The Fourier transform method for the wave equation givesThis inverse tempered distribution is twice differentiable in $t$: time derivatives introduce polynomial frequency factors, still integrable against every Schwartz function. It has the specified initial displacement and zero initial velocity and satisfies the equation distributionally.

To apply the support theorem, replace the real norm by the entire wave cosine multiplierThis series is entire and equals $\cos (ct\sqrt{z\cdot z})$ regardless of the square-root choice. Write $z=a+ib$ and $q=\sqrt{z\cdot z}$. Since $|z\cdot z|\le |a{|}^2+|b{|}^2$,Together with $|\cos w|\le e^{|\mathrm{Im}w|}$, this yields $|C_t(z)|\le e^{|c|t|\mathrm{Im}z|}$. The forward estimate for the translated initial support now givesThe converse therefore proves finite propagation with the stated speed:In particular, $\sqrt{z\cdot z}$ itself need not be entire; it is the even cosine series that provides the required entire extension.

For completeness this construction identifies the distributional Cauchy solution even without initially assuming spatial temperedness. The zero-displacement wave multiplier is $S_{\tau }(z)={\int }_0^{\tau }{C}_s(z)ds$, entire with bound $|S_{\tau }(z)|\le \tau e^{|c|\tau |\mathrm{Im}z|}$. Given a compactly supported smooth $\varphi$ and final time $T$, the backward solution $w(s)={\mathcal{F}}^{-1}[S_{T-s}\hat{\varphi }]$ is smooth, has support in one compact ball for $0\le s\le T$ by the same support theorem, and satisfies $w(T)=0$, $w_s(T)=-\varphi$. For the difference $h$ of two distributional solutions with zero initial data,All pairings use compactly supported test functions. The expression vanishes initially and equals $⟨h(T),\varphi ⟩$ finally. Hence $h(T)=0$, establishing uniqueness in the usual time-differentiable distributional solution class.

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.