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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 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

For distributions $u$ on ${\mathbb{R}}^m$ and $v$ on ${\mathbb{R}}^n$, their tensor product acts on a joint test function by $⟨u\otimes v,\Phi ⟩=⟨u_x,⟨v_y,\Phi (x,y)⟩⟩$. The inner function is smooth with compact support in the projection of $\mathrm{supp}\Phi$, and finite order of a distribution estimates prove continuity. Reversing the pairings gives the same value: finite sums of product kernels are dense in the required smooth seminorms, and the two orders agree on product kernels. Its support is contained in the Cartesian product of the two supports. This provides a rigorous replacement for formal integration in separate distribution variables.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.