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An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.

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.

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 $X$.

Reflect $u$ across the two endpoints of $I$ into a slightly larger interval, giving an $H^1$ extension, and multiply by a smooth cutoff function equal to one near $\overline{I}$. Convolve the resulting function on $\mathbb{R}$ with a mollifier. The convolutions $u_{\epsilon }$ are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in $L^2$ proves both $u_{\epsilon }\to u$ and $u_{\epsilon }^{\prime }\to u^{\prime }$ in $L^2(I)$, hence convergence in the Sobolev norm.

To obtain convergence almost everywhere, select a sequence with $\Vert u_n-u{\Vert }_2\le 2^{-n}$. Then $\int {\sum }_n|u_n-u{|}^2<\infty$, so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in $H^1$ and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.

Choose a fixed smooth cutoff function with $\Vert {\zeta }^{\prime }{\Vert }_{\infty }=C_0$ and set $J=(-1/2,1/2)$. The Caccioppoli inequality gives $\Vert u^{\prime }{\Vert }_{L^2(J)}\le 2R{C}_0\Vert u{\Vert }_{L^2(-1,1)}$. Since the length of $J$ is one, the corrected interval Sobolev supremum estimate yields $\Vert u{\Vert }_{L^{\infty }(J)}\le \Vert u{\Vert }_{L^2(J)}+\Vert u^{\prime }{\Vert }_{L^2(J)}$, and the Hölder seminorm is at most $\Vert u^{\prime }{\Vert }_{L^2(J)}$. ThereforeFor locally $H^1$ weak solutions with $u\in L^2(-1,1)$, density of smooth functions in a Sobolev space justifies the test ${\zeta }^{2}u$, and the previous representative estimates already apply to $H^1$. Mollifying a solution need not preserve the equation with the same measurable coefficient, so approximation is used for admissible tests and Sobolev estimates, not to assert that the mollified function solves the original equation. The required constant depends only on $R$.

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.

Use a Lipschitz continuous cutoff function $\eta$ which equals one on $B_R(x_0)$, vanishes outside $B_{2R}(x_0)$, takes values in $[0,1]$, and has $|\nabla \eta |\le 2/R$. Its gradient is supported in the annulus $A_R=B_{2R}(x_0)\setminus B_R(x_0)$. The function ${\eta }^2(u-c)$ is an admissible test function in the zero-boundary Sobolev space for the weak solution, by approximation with smooth compactly supported test functions. If ${\Lambda }_0$ bounds the operator norm of $(a^{ij})$, we can take ${\Lambda }_0=n\Lambda$. The weak equation and uniform ellipticity giveYoung inequality bounds the right side byAfter absorption, this annular Caccioppoli inequality isThe dimension is fixed in the notation $C(\lambda ,\Lambda )$ of the question; with the entrywise coefficient bound its dependence on $n$ is absorbed there. If the larger ball merely lies in $B$ without its closure being compactly contained, approximation from smaller cutoff functions gives the same admissible test function and estimate. Crucially, the right side uses only the annulus, where the cutoff function varies. The constant $c$ is arbitrary because constants have zero gradient.

Choose nested interior open sets $U\subset \subset V\subset \subset W\subset \subset \Omega$. For small $\sigma$, part (b) gives a smooth equation on $V$ of the formThe divergence-forcing interior H1 estimate isOne can obtain this estimate directly by testing the smooth equation against ${\eta }^2{u}_{\sigma }$ and applying Young inequality, with a cutoff function equal to one on $U$ and supported in $V$. The approximate identity and the $L^2$ convolution bound give, uniformly in $\sigma$,with analogous bounds for $(cu{)}_{\sigma }$ and $f_{\sigma }$. The coefficients need only be bounded on $W$. Therefore $u_{\sigma }$ is bounded in $H^1(U)$. It converges to $u$ in $L^2(U)$, and weak sequential compactness in a Hilbert space in this Sobolev space gives $u\in H^1(U)$. Since $U$ was arbitrary, $u\in H_{\mathrm{loc}}^1(\Omega )$.

Now expand the distributional derivative:Its right side belongs to $L_{\mathrm{loc}}^2$, so the interior $H^2$ elliptic regularity estimate gives $u\in H_{\mathrm{loc}}^2$. More generally, if $u\in H_{\mathrm{loc}}^m$ for an integer $m\ge 1$, multiplication by the smooth coefficients puts the right side in $H_{\mathrm{loc}}^{m-1}$. Interior elliptic regularity then gives $u\in H_{\mathrm{loc}}^{m+1}$. This elliptic regularity bootstrap proves $u\in H_{\mathrm{loc}}^k$ for every integer $k$.