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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.

For this Neumann Poisson problem, interpret the forcing in $L^2(U)$, as is automatic if it is smooth up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior smooth but even $\int f\cdot 1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.

Suppose the weak solution is smooth on $\overline{U}$. Testing against compactly supported test functions gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and Green's first identity implyfor every smooth $v$ on $\overline{U}$. Every smooth boundary function has such an extension, so ${\partial }_{n}u=0$ on $\partial U$. This proves both the interior equation and the boundary condition.

Conversely, for $u\in C^2(\overline{U})$ satisfying the equation and zero normal derivative, Green's first identity gives the weak identity for all smooth $v$ on $\overline{U}$. The density of smooth functions in a Sobolev space and the Cauchy-Schwarz inequality extend it continuously to every $v\in H^1(U)$. Thus the classical solution is a weak solution, and the smooth weak solution is classical.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincaré inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.

The clamped second-order Sobolev space is $H_0^2(U)={\overline{C_c^{\infty }(U)}}^{H^2}$. On a smooth bounded domain the Sobolev trace theorem characterizes it by zero value and zero normal derivative on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.

For a smooth weak solution, compactly supported test functions and two integrations by parts giveso ${\Delta }^{2}u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u={\partial }_{n}u=0$ on $\partial U$. Thus it is a classical solution of the clamped biharmonic problem.

Conversely, a $C^4(\overline{U})$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported test function, two integrations by parts give ${\int }_U\Delta u\Delta v={\int }_{U}fv$. Both sides are continuous for the $H^2$ norm, so the defining density of $C_c^{\infty }(U)$ in $H_0^2(U)$ extends this equality to every required test. The two notions agree under the stated smoothness.

Use complex-linear distribution pairings, without conjugation. Write $\mathcal{D}=C_c^{\infty }({\mathbb{R}}^n)$ for the space of test functions. The smoothing convolution with a test function isOn a compact set of $x$ values, all the translated test functions have support in one compact set. Continuity of the distribution therefore permits differentiation in $x$, giving ${\partial }_x^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$ for every multi-index. In particular this convolution is a smooth function, even if $u$ is not tempered.

For the first associativity identity, integration against $\psi$ and the distribution pairing can be interchanged: the integrand has a common compact support in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. ConsequentlyThe common-support argument matters: a general distribution cannot be paired with arbitrary noncompact functions.

For convolution of distributions with a compactly supported factor, first take $v$ with compact support $K$ and defineThe inner pairing is interpreted using a cutoff function equal to one near $K$. It is smooth in $x$ and has support in $\mathrm{supp}\chi -K$. To see continuity, restrict $\chi$ to a fixed compact support $A$. The order of a distribution estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ givesThus $w$ is a distribution, not merely a formal iterated pairing.

Choose an additional cutoff function in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the tensor product of distributions permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives $u\ast v=v\ast u$. If $u$ rather than $v$ has compact support, use the same construction with the roles reversed; pairing $u$ against a smooth function is then legitimate.

Evaluating the resulting smoothing convolution with a test function givesIf $v$ has compact support, $v\ast \phi$ is itself a test function; if $u$ has compact support, its action on the smooth inner convolution uses a cutoff. This explains the meaning of the formula in either case. It also proves uniqueness: $(w\ast \phi )(0)=⟨w,\phi (-\cdot )⟩$, and reflection runs through all test functions. When both factors have compact support, the same definition gives $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$, their Minkowski sum.

The Schwartz space consists of smooth functions for which every seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }\varphi (x)|$ is finite. A tempered distribution is a continuous linear functional on this space. Fix the angular-frequency Fourier transform conventionThe Fourier transform isomorphism of the Schwartz space makes the dual definition continuous. For a compactly supported distribution, a fixed cutoff function $\chi =1$ near its support extends the action to smooth functions by $⟨u,g⟩=⟨u,\chi g⟩$. A finite-order estimate controls this by finitely many Schwartz space seminorms, so both compactly supported factors are tempered.

Their Fourier transform of a compactly supported distribution is the smooth function $\hat{u}(\lambda )=⟨u,e^{-i\lambda \cdot x}⟩$. Applying the compact-support convolution definition to the exponential givesThe convolution theorem has no extra factor with this normalization.

For the spherical surface measure convolution, put $k=|\lambda |$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area givesAt $k=0$ the removable value is $4\pi a^2$, the total sphere area. Hence $\widehat{u_a\ast u_b}(\lambda )=16{\pi }^{2}ab\sin (ak)\sin (bk)/k^2$.

For $r=|x|>0$, angular integration in Fourier inversion now givesThis conditional integral can be made rigorous by first inserting $e^{-\epsilon k}$ and then taking $\epsilon \downarrow 0$ in tempered distributions. The supplied sine identity gives an integral of $\pi /4$ when $|a-b|<r<a+b$, and zero off that interval. Thusas a regular distribution. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use ${\int }_0^{\infty }{e}^{-\epsilon k}\sin (sk)dk/k=\arctan (s/\epsilon )$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a locally integrable function in three dimensions. Dominated convergence theorem therefore identifies the distributional limit with the displayed density.

Changing the two endpoint sphere values does not change the regular distribution; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,exactly the product of the original sphere areas.

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.

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 $\chi \in C_c^{\infty }(X)$.

A compactly supported distribution of finite order of a distribution $m$ has a smooth Fourier transform satisfying $|\hat{u}(\lambda )|\le C(1+|\lambda |{)}^m$, by applying the finite-order estimate to a fixed cutoff times the exponential. Thus its weighted squared transform is bounded by $C⟨\lambda {⟩}^{2(s+m)}$. This is integrable precisely in the sufficient range $2(s+m)<-n$, and provesThis is the negative Sobolev regularity of a compactly supported distribution; the strict inequality is important, and is not a claim that this sufficient threshold is optimal for each distribution.

To establish elliptic regularity without assuming the answer as an a priori smoothness hypothesis, first obtain a global constant-coefficient gain. For a compactly supported distribution $w$ with $P(D)w\in H^q$, the polynomial lower bound at high frequency givesOn $|\lambda |\le R$, the transform of $w$ is smooth and bounded. Therefore $P(D)w\in H^q$ implies $w\in H^{q+N}$. Possible low-frequency zeros of $P$ are harmless; we never divide by them.

Two elementary mapping facts supply the variable-coefficient argument. Distributional derivatives of order $j$ map $H^r$ into $H^{r-j}$. Sobolev multiplication by a smooth cutoff is bounded on $H^r$ for every real $r$, including negative ones. Indeed, for a compactly supported smooth $a$, the Peetre weight inequalityand $\widehat{aw}=(2\pi {)}^{-n}\hat{a}\ast \hat{w}$ reduce the bound to Young's convolution inequality with the integrable kernel $⟨\zeta {⟩}^{|r|}|\hat{a}(\zeta )|$. Smooth coefficients only need this property on compact subsets, where they can be multiplied by another cutoff.

Write the lower-order part as $B={\sum }_{|\alpha |<N}{f}_{\alpha }{D}^{\alpha }$, with $N\ge 1$, and suppose provisionally that $u\in H_{\mathrm{loc}}^r$ on a relatively compact neighborhood. For a cutoff function $\chi$ supported there,The bracket is the operator commutator. Its order is at most $N-1$: in the product rule, every surviving term has at least one derivative falling on $\chi$. Choose a second cutoff function equal to one near $\mathrm{supp}\chi$ when estimating the products. The derivative and smooth-multiplication bounds then giveThe global gain just proved applies to the compactly supported distribution $\chi u$, and yieldsThis is the cutoff bootstrap for local elliptic regularity; it works for real indices, not just nonnegative integers.

There is always a legitimate starting index. Around any fixed point choose ${\chi }_0=1$ on a smaller neighborhood. The compactly supported distribution ${\chi }_{0}u$ has the negative Sobolev regularity proved above, so $u\in H_{\mathrm{loc}}^{r_0}$ on that neighborhood for some finite $r_0$. The index need not be uniform over all of $X$. Repeating the one-step gain finitely many times reaches $s+N$, or the target is already reached if $r_0\ge s+N$. Since the point was arbitrary,For $N=0$ the result follows directly by dividing by the nonzero constant.

Finally, if $Lu=0$, its forcing belongs to every Local Sobolev space. The gain consequently gives every local Sobolev order for $u$. For each integer $j\ge 0$, choose an order larger than $j+n/2$ and apply the Sobolev embedding theorem to a localized solution. It has a $C^j$ representative; the representatives agree because they represent the same distribution. Thus every distributional solution of $Lu=0$ is smooth on $X$.

A weak solution is an equivalence class $u\in L^{\infty }((0,\infty )\times \mathbb{R})$ satisfying, for every test function $\phi \in C_c^{\infty }([0,\infty )\times \mathbb{R})$,This is integration by parts for the nonconservative transport equation. In particular, the weak formulation contains $c_x\phi$ as well as $c{\phi }_x$. The last integral encodes the initial condition; a pointwise boundary value of an arbitrary Lebesgue space representative is not the definition. Compact support of the test function makes the identity meaningful even if $c$ is unbounded.

Use the same formula for a measurable representative of $u_0$. Dilation preserves null sets and gives $\Vert u{\Vert }_{L^{\infty }}=\Vert u_0{\Vert }_{L^{\infty }}$. To check the weak formulation, change variables $x=e^{t}y$ and set $\psi (t,y)=e^t\phi (t,e^{t}y)$. ThenThe spacetime integral becomes $\int u_0(y){\psi }_t(t,y)dydt=-\int u_0(y)\phi (0,y)dy$, as required.

For uniqueness, take any bounded weak solution and write $v(t,y)=u(t,e^{t}y)$. Choosing $\phi (t,x)=e^{-t}\psi (t,e^{-t}x)$ in the weak formulation givesFor tensor test functions $\psi (t,y)=\eta (t)\theta (y)$, this says that $\int v(t,y)\theta (y)dy$ is distributionally constant and has value $\int u_0\theta$. A countable dense family of spatial test functions identifies $v=u_0$ almost everywhere. Thus the displayed solution is the unique bounded weak solution.

For the inflow transport boundary condition, the weak formulation isHere $\phi$ is a smooth compactly supported test function on the closed quadrant. Put $z=x-at$ and $\psi (t,z)=\phi (t,z+at)$. The transformed region is $t>\tau (z):=\max (0,-z/a)$. The formula in the preceding solution is $v(t,z)=b(z)$, with $b(z)=u_0(z)$ for $z\ge 0$ and $b(z)=f(-z/a)$ for $z<0$. Consequently the first integral equals $-\int b(z)\psi (\tau (z),z)dz$. Its positive-$z$ portion cancels the initial integral; the substitution $z=-at$ in its negative-$z$ portion cancels the inflow integral, including the factor $a$. This proves weak existence for arbitrary bounded data without corner compatibility.

For uniqueness, subtract two weak solutions. Interior test functions in these characteristic coordinates give ${\partial }_{t}v=0$ distributionally, so $v(t,z)=b(z)$ on each vertical ray. This follows first on rectangles compactly contained in $t>\tau (z)$ using tensor test functions, and then on the whole region by overlapping rectangles. In the full weak formulation the remaining lower-boundary term is $-\int b(z)\psi (\tau (z),z)dz=0$. Arbitrary smooth traces supported separately on $z>0$ and $z<0$ force $b=0$ there. The single point $z=0$ has zero measure. Thus the two-branch formula is the unique bounded weak solution.

Use complex-linear distributions, so the pairing contains no complex conjugation. The space of smooth functions is $\mathcal{E}(X)=C^{\infty }(X)$, with seminormsThus $f_j\to f$ means uniform convergence of every derivative on every compact subset of $X$. A cofinal compact exhaustion with $K_j\subset \mathrm{int}{K}_{j+1}$ makes $p_j=p_{K_j,j}$ an increasing defining family. Such an exhaustion exists for every open set; for example, use bounded sets staying a positive distance from its complement and enlarge them slightly. This is a Fréchet space.

The compactly supported distribution space is the continuous dual space ${\mathcal{E}}^{\prime }(X)$. We use its weak-star topology: $u_j\to u$ precisely when $⟨u_j,f⟩\to ⟨u,f⟩$ for every $f\in \mathcal{E}(X)$. A stronger standard choice is the strong dual topology, which requires uniform convergence on bounded subsets of $\mathcal{E}(X)$; specifying the weak convention avoids conflating these definitions.

A continuous linear functional necessarily takes null sequences to zero. Conversely, suppose a linear functional $u$ fails to be continuous. For each $j$, it is unbounded on $\{f:p_j(f)\le 1\}$; otherwise scaling would give a continuity estimate. We can therefore choose $f_j$ withFor each fixed $l$, $p_l(f_j)\le p_j(f_j)\to 0$ once $j\ge l$. This contradicts the assumed null-sequence property. Hence sequential continuity characterizes the continuous dual:This is the sequential continuity criterion in a metrizable vector space applied to a linear functional.

Continuity also yields one compact $K⋐X$, an integer $m$ and a constant $C$ such that $|u(f)|\le C{p}_{K,m}(f)$. Thus $u$ has compact support in $K$ and finite order of a distribution. Conversely, a compactly supported distribution extends to $\mathcal{E}(X)$ by $u(f)=u(\eta f)$, where a cutoff function $\eta$ equals one near its support of a distribution. The local finite-order estimate makes this extension continuous and independent of $\eta$. This identifies the dual definition with compactly supported distributions.

Here is an explicit compact continuous-derivative representation of a distribution. Extend $u$ by zero to ${\mathbb{R}}^n$ using a cutoff function inside $X$, retaining finite order $m$. Put $r=m+2$ andThis locally $C^m$ function satisfies ${\partial }^{\gamma }E={\delta }_0$. Indeed, each one-dimensional factor has $r$th distributional derivative equal to the Dirac delta distribution. The convolution $F=u\ast E$ is continuous: the finite-order estimate extends $u$ to $C^m$ functions near its compact support, and translated $E$ varies continuously in their $C^m$ norms. Equivalently, mollify $E$ and use the estimate to obtain local uniform convergence of the convolved functions. Distributional differentiation gives ${\partial }^{\gamma }F=u$.

Choose $\chi \in C_c^{\infty }(X)$ equal to one near $\mathrm{supp}u$. Since $\chi u=u$, repeated Leibniz rule givesEvery coefficient function on the right is continuous and compactly supported in $X$. Thus the required representation is finite:The equality first holds on test functions and then on all smooth functions after inserting a common cutoff, so it holds in ${\mathcal{E}}^{\prime }(X)$.

A general distribution need not admit one finite such sum, even if compact support is not required of the continuous functions. On $X=\mathbb{R}$, considerThe sum is locally finite, so it defines a distribution. If it were a finite sum ${\sum }_{l=0}^M{\partial }^l{g}_l$ with all $g_l$ continuous, then on every fixed compact set its action would be bounded by test derivatives through order $M$. Near an integer $j>M$, however, it is exactly ${\delta }_j^{(j)}$, whose order is $j$. To see the contradiction directly, choose a test function $\psi$ with ${\psi }^{(j)}(0)\ne 0$ and put ${\psi }_{\epsilon }(x)={\epsilon }^j\psi ((x-j)/\epsilon )$. All derivative norms through order $M$ tend to zero, while $⟨{\delta }_j^{(j)},{\psi }_{\epsilon }⟩=(-1{)}^j{\psi }^{(j)}(0)$. This is a distribution of unbounded order. In dimensions $n>1$, the same example uses point masses at $(j,0,\dots ,0)$ and derivatives in the first coordinate.