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The L1 norm properties of absolute homogeneity and the triangle inequality follow from the corresponding properties of the absolute value and the integral. If $\Vert f{\Vert }_1=0$, the nonnegative continuous function $|f|$ has zero integral. Were $|f(x_0)|>0$, continuity would make it bounded below by a positive number on a nontrivial interval, contradicting the zero integral. Thus $f=0$, proving positive definiteness.

Put $M=\Vert f{\Vert }_{\infty }$ and choose the continuous cutoff functionThen $g_n(0)=g_n(1)=0$, $\Vert g_n{\Vert }_{\infty }\le M$, and $f-g_n$ is supported in the two endpoint intervals of total length $2/n$. Therefore

Now suppose ${\int }_0^{1}fg=0$ for every $g\in \mathcal{S}$. Apply this to the above $g_n$. By the uniform norm bound,Hence ${\int }_0^1{f}^2=0$, and continuity again gives $f=0$.

Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ and $D_j=-i{\partial }_{x_j}$. Thus $D_j$ has Fourier symbol ${\xi }_j$. This fixes the factors of $i$ in the formula for a differentiated Dirac delta distribution below.

With the Japanese bracket $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$ consists of smooth functions $a$ such that, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the symbol class order, whereas differentiation in $\theta$ lowers it.

A phase function is real-valued and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is a positively homogeneous function of degree one in $\theta$, and has nonzero total differential:The nonvanishing condition concerns both sets of variables, not just ${\nabla }_{\theta }\Phi$. The homogeneous convention is imposed away from $\theta =0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a smooth function of $x$.

Choose a cutoff function $\rho (\theta )$ equal to one near zero. The low-frequency part with oscillatory integral amplitude $\rho a$ is an ordinary convergent integral and defines a smooth function: all $x$ derivatives of $\Phi$ are $O_K(|\theta |)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta \ne 0$,Then $L{e}^{i\Phi }=e^{i\Phi }$. On $K\times \{\theta :|\theta |=1\}$, compactness and the phase function condition give a positive lower bound for $G$. Homogeneity consequently gives $G\ge c_K|\theta {|}^2$. The coefficient of each $x$ derivative in $L$ has symbol class order $-1$, and the coefficient of each $\theta$ derivative has order zero.

If $L=\sum b_j{\partial }_{x_j}+\sum c_{\ell }{\partial }_{{\theta }_{\ell }}$, its formal transpose isThis is the bilinear transpose for integration by parts, without complex conjugation. It lowers the symbol class order by one. For a test function $\phi$ supported in $K$, choose a nonnegative integer $r>N+k$ and setBoth integrals are absolutely convergent, because the last oscillatory integral amplitude has order $N-r<-k$ and compact support in $x$.

To verify that this defines the intended oscillatory integral, take any $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and insert $\chi (\epsilon \theta )$ in the original integral. Repeated integration by parts gives the preceding expression with $(L^t{)}^r$ applied also to this cutoff function. Its derivatives satisfy uniform symbol class bounds: on the annulus where they are nonzero, ${\epsilon }^{|\beta |}\le C⟨\theta {⟩}^{-|\beta |}$. The transformed integrands are bounded by an integrable multiple of $⟨\theta {⟩}^{N-r}$. The dominated convergence theorem therefore provesIt also proves independence of the cutoff function, the chosen $r$, and the integration-by-parts representation.

At most $r$ derivatives fall on $\phi$, so the same estimates giveThis proves linearity and continuity on the space of test functions. It also proves the finite order of an oscillatory integral distribution: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.

For $x_0\in X$, take $k=n$, $\Phi =(x-x_0)\cdot \theta$, and $a=(2\pi {)}^{-n}{\theta }^{\alpha }$. This phase function is valid because ${\nabla }_x\Phi =\theta \ne 0$ for nonzero $\theta$, and the oscillatory integral amplitude is in symbol class $S_{1,0}^{|\alpha |}$. The required identity isIndeed, extending a test function by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expressionHere Fourier inversion applies because $\phi$ is a Schwartz function. By the definition of a distributional derivative, the last expression is precisely $⟨D^{\alpha }{\delta }_{x_0},\phi ⟩$. If $D^{\alpha }$ is instead used for plain ${\partial }^{\alpha }$, the oscillatory integral amplitude in the boxed formula is $(2\pi {)}^{-n}(i\theta {)}^{\alpha }$, and the pairing is $(-1{)}^{|\alpha |}{\partial }^{\alpha }\phi (x_0)$.

Not every distribution is one oscillatory integral of the stated class. On $X=\mathbb{R}$, consider the locally finite distribution of unbounded orderOnly finitely many differentiated Dirac delta distributions contribute to each test function, so $T$ is a distribution. Near $j$ it is exactly $D^j{\delta }_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta \in C_c^{\infty }((-1/4,1/4))$ with ${\eta }^{(j)}(0)\ne 0$ and useThe derivatives through order $r$ remain bounded for $0<\epsilon \le 1$, whereasAny single oscillatory integral with finite $k$ and a fixed finite symbol class order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty open set in positive dimension by choosing points that leave every compact set and taking locally finite differentiated Dirac delta distributions there. This obstruction concerns a single fixed-order oscillatory integral, rather than local representations or an infinite sum of them.

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

The continuous coefficients are uniformly continuous on a compact neighborhood of $\overline{W}$. For every $x_0\in \overline{W}$, choose a ball $B_r(x_0)⋐U$ small enough thatfor all $i,j$. Part 4(b), with the frozen symmetric matrix $A=(a^{ij}(x_0))$, then applies on this ball.

Choose a finite collection of these balls and a smooth partition of unity $({\eta }_k)$ that sums to one near $\overline{W}$, with each ${\eta }_k$ supported in its corresponding ball. Applying part 4(b) to ${\eta }_{k}u$ givesSince $(a^{ij})$ is symmetric, the Leibniz rule gives the commutator formulaThe coefficients and the finitely many derivatives of the cutoff functions are bounded, soFinally $u={\sum }_k{\eta }_{k}u$ near $\overline{W}$. Summing the finite set of local estimates provesThis is the coefficient-freezing interior second-derivative estimate.

The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by $x$ gives $x\mathrm{pv}(1/x)=1$.

If $v$ is any other solution, $w=v-\mathrm{pv}(1/x)$ obeys $xw=0$. To identify this kernel of multiplication by a coordinate, choose a cutoff function $\eta$ equal to one near zero. Every test function has the form $\phi =\phi (0)\eta +x\psi$ with $\psi \in \mathcal{D}$. Thus $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$, proving $w=c{\delta }_0$. ThereforeFor real distributions the constants are real.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

Choose a cutoff function $\eta$ supported in $B_{\rho }(x)$, equal to one on $B_{{\rho }^{\prime }}(x)$, and satisfying $|D\eta |\le (\rho -{\rho }^{\prime }{)}^{-1}$. Use ${\eta }^{2}u$ as a test function in the weak formulation:The Cauchy-Schwarz inequality followed by Young inequality givesand henceThis is the Caccioppoli inequality. The numerical constant is convention-dependent and is normally absorbed into the displayed estimate; replacing the radii by fixed intermediate radii gives the stated form with one universal constant.

Choose a radial mollifier $\rho \in C_c^{\infty }$ with integral one and support in the unit ball, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one there. For a radial tempered distribution $T$, the functionsare radial Schwartz functions. Smoothness comes from convolution of a tempered distribution with a Schwartz function, and compact support comes from the cutoff. Pairing with $\phi$ gives $⟨f_j,\phi ⟩=⟨T,{\check{\rho }}_{1/j}\ast (\chi (\cdot /j)\phi )⟩$. With $q_m(\phi )={\max }_{|\beta |\le m}{\sup }_x(1+|x|{)}^m|{\partial }^{\beta }\phi (x)|$, the cutoff tail and the mean value theorem giveThe continuity estimate for $T$ therefore proves $f_j\to T$ even in the strong dual topology. Inserting a cutoff is essential because convolution alone need not give rapid decay.

In one base and one phase variable, take $\Phi =x\theta$ and $a=x\eta (\theta )/|\theta |$, with an even smooth cutoff function equal to zero near zero and one for large $|\theta |$. This is a symbol of order $-1$. Its oscillatory integral is $-2x\log |x|$ plus a smooth function near zero. Thus it is singular there even though the restricted function $a(0,\cdot )$ has empty support. The joint closed conic support of an oscillatory amplitude retains the limiting base point and its stationary directions. A support-sensitive regularity theorem must use that conic neighborhood information, not merely vanishing of an amplitude at one base point.