Codex @codex  0

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 $k$.

For any nonnegative integer $\ell$, choose $k>\ell +n/2$. The Sobolev embedding theorem on compact interior subsets gives $u\in C_{\mathrm{loc}}^{\ell }$. Taking all $\ell$ proves $u\in C^{\infty }(\Omega )$, for its smooth representative. The first gain from $L^2$ to $H^1$ is the step supplied by the mollified equation; assuming $H^1$ in the initial definition would miss that step.

For an integer $s\ge 0$, the Sobolev space iswhere $D^{\beta }$ is a weak derivative. For $p<\infty$ one may use the Sobolev norm $({\sum }_{|\beta |\le s}\Vert D^{\beta }u{\Vert }_p^p{)}^{1/p}$; for $p=\infty$ use the maximum of the finitely many essential-supremum norms.

For $1\le p<d$, the Sobolev inequality is $\Vert u{\Vert }_{p^{\ast }}\le C_{d,p}\Vert \nabla u{\Vert }_p$, with Sobolev conjugate exponent $p^{\ast }=dp/(d-p)$. For $d<p<\infty$, Morrey's inequality supplies a continuous representative satisfyingAt $p=\infty$ the representative is Lipschitz continuous. At the critical exponent $p=d>1$, first-order Sobolev regularity gives every finite $L^q$ embedding for $q\ge d$, with an inhomogeneous norm, but generally no $L^{\infty }$ embedding. The one-dimensional endpoint $W^{1,1}(\mathbb{R})\hookrightarrow L^{\infty }(\mathbb{R})$ is an exception.

For the proof of Morrey's inequality, start with a smooth $u$ and write $u_{B(x,r)}$ for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables givesThe last step is the Holder inequality; integrability of the kernel to power $p^{\prime }$ is exactly $(d-1)p^{\prime }<d$. For $r=|x-y|$, translate the averaging ball along the segment from $x$ to $y$. The fundamental theorem of calculus along a line segment and the Holder inequality giveCombining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with $r=1$, together with $|u_{B(x,1)}|\le |B_1{|}^{-1/p}\Vert u{\Vert }_p$, gives the supremum estimate. Density of smooth functions in a Sobolev space then gives a uniformly convergent sequence of smooth representatives, preserving both bounds. For $p=\infty$, mollification gives the Lipschitz version.

For the decay conclusion assume $d<p<\infty$. The representative is uniformly continuous. If $|u(x_j)|\ge 2\theta >0$ along points escaping to infinity, the Hölder bound gives a radius $\rho >0$, independent of $j$, on which $|u|\ge \theta$. A subsequence has disjoint radius-$\rho$ balls, each contributing at least ${\theta }^p|B_{\rho }|$ to $\Vert u{\Vert }_p^p$, a contradiction. This is uniformly continuous integrable functions vanish at infinity.

The finite-$p$ restriction is necessary. If the printed range includes $p=\infty$, its decay assertion is false: $u\equiv 1$ belongs to $W^{1,\infty }$ but does not tend to zero.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

The general-order Cauchy-Kovalevskaya theorem applies to analytic data on a non-characteristic hypersurface. In local coordinates $(t,z)$, a real analytic system of order $m$ in an unknown vector $U$ is in normal form whenwith $F$ real analytic near the initial jet. Analytic data ${\partial }_t^{j}U(0,z)=h_j(z)$ for $0\le j<m$ determine a unique real analytic solution near each initial point. More general systems, including different orders for the components, have the same conclusion after solving for the highest normal derivatives: the corresponding coefficient matrix, or highest-derivative Jacobian for a nonlinear system, must be invertible. This is the system's non-characteristic hypersurface condition. The uniqueness asserted by the Cauchy-Kovalevskaya theorem is initially uniqueness in the real analytic class.

A real analytic coordinate change flattens a real analytic initial hypersurface. Non-characteristicity permits solving for its highest normal derivatives by the real analytic implicit function theorem. Introduce all derivatives through order $m-1$ as additional unknowns. Their normal derivatives are either another jet variable, a tangential first derivative of a jet variable, or the right-hand side of the original equation. This gives a first-order real analytic system. The derivative-compatibility identities have zero initial data and are preserved by the system; equivalently, its real analytic coefficient recurrence reproduces the derivatives of $U$. This explains both reductions, rather than assuming arbitrary real analytic hypersurfaces are already flat.

For the first example, the principal symbol of the Laplace equation at the conormal $d{x}_1$ is $1$, so the plane is a non-characteristic hypersurface. The theorem therefore applies to real analytic $u{|}_{x_1=0}$ and ${\partial }_{1}u{|}_{x_1=0}$:This does not give Hadamard well-posedness in smooth or Sobolev space norms. For $n\ge 2$, the harmonic functionshave zero value data and normal derivative data $e^{-\sqrt{N}}\cos (N{x}_2)$ tending to zero with every fixed tangential derivative. At any fixed $x_1=\delta >0$, however, $u_N(\delta ,0)$ diverges. Thus the elliptic Cauchy problem for a partial differential equation is unstable despite real analytic solvability. The exceptional case $n=1$ is the elementary affine ordinary differential equation and has no such tangential high-frequency instability.

For the heat equation, viewed as a second-order equation in all $(t,x)$ variables, the principal symbol is $|\xi {|}^2$. It vanishes on $dt$, so $t=0$ is a characteristic hypersurface andWriting $u_t={\Delta }_{x}u$ does not repair its hypotheses: the right side has second spatial derivatives, exceeding the first-order normal-form allowance. Indeed real analytic initial data need not produce a jointly real analytic solution through time zero. In one dimension, $u_0(x)=(1+x^2{)}^{-1}$ would forceThe time Taylor coefficients $(-1{)}^k(2k)!/k!$ have zero radius of convergence.

Nevertheless the forward heat equation has Hadamard well-posedness in standard $L^2$ or Sobolev space classes for $t\ge 0$; its Fourier multiplier $e^{-t|\xi {|}^2}$ gives stable smoothing. Backward evolution has the growing multiplier $e^{|t||\xi {|}^2}$ and is unstable. Neither forward stability nor backward instability is decided simply by whether the real analytic theorem applies.

Use the interior $\Omega =\{1/2<|x|<2\}$ for the Sobolev space notation; the closed shell in the PDF specifies its boundary. Put $A(x)=\mathrm{diag}(1,1,g(|x|))$. A weak solution has zero boundary trace in the sense of the Sobolev trace theorem, equivalently $u\in H_0^1(\Omega )$, and satisfiesThe minus sign is necessary because the operator in the paper is positive divergence, not its negative. One can first test with $C_c^{\infty }$ functions and then use density in the zero-boundary Sobolev space. Smoothness of $g$ makes its restriction to the compact radial interval bounded, so both sides define continuous functionals on this space. Merely belonging to $H^1$ without the zero trace would omit the boundary condition.

Adopt the sign convention that a weak supersolution for $Lu=D_i(a^{ij}{D}_{j}u)$ satisfies $Lu\le 0$, or $\int a^{ij}{D}_{j}u{D}_i\phi \ge 0$ for every nonnegative compactly supported test function. For a nonnegative supersolution, the Weak Harnack inequality gives some $q>0$ and $C_H\ge 1$, depending only on $n,\lambda ,\Lambda$, such thatHere $|E{|}^{-1}{\int }_E$ denotes the average over $E$; equivalently the right side gives a lower bound for the essential infimum. One may take a sufficiently small positive exponent, so no assumption $q\ge 1$ is needed. The inequality is unchanged under translations and dilations of the Euclidean balls.

Since a Sobolev space element is defined up to changes on a null set, the infimum is initially essential. After choosing a continuous representative it is the ordinary infimum. Without choosing such a representative, an arbitrary pointwise infimum cannot be bounded by an integral.

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzela-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.

For any $R>0$, set $U_R(y)=u(Ry)$ and $a_R^{ij}(y)=a^{ij}(Ry)$. These coefficients retain the same bounds and ellipticity, and a change of variables in the weak formulation shows that $U_R$ is a nonnegative weak subsolution on $B_1$. Part (a) gives the scale-correct estimateThe constant is independent of $R$. Fix a Euclidean ball $B_s$ and let $R\to \infty$ with $R>2s$. The right side tends to zero, so $u=0$ almost everywhere on $B_s$. Since $s$ is arbitrary, the Liouville theorem for integrable nonnegative elliptic subsolutions provesThis is $u\equiv 0$ as a Sobolev space element. Any continuous representative is identically zero pointwise; an arbitrary measurable representative can be modified at isolated points without changing the weak equation. Integrability and the explicit scaling force the conclusion.

For $\lambda >0$, minimize the smoothing spline objective over the Sobolev space of functions with square-integrable second derivative on the observed interval:The roughness penalty measures curvature; its null space consists of affine functions. To see why the minimizer is a natural cubic spline, let $g$ be the natural spline interpolating the values of any competitor $m$ at its distinct observation locations, and put $h=m-g$. Integration by parts twice on each knot interval gives $\int g^{\prime \prime }{h}^{\prime \prime }=0$: $g^{\prime \prime \prime \prime }=0$ on each interval, $h$ vanishes at the knots, $g^{\prime \prime }$ is continuous, and $g^{\prime \prime }$ vanishes at the outer endpoints. Hencewhile both functions have the same residual sum of squares. This proves the reduction to a finite-dimensional spline space.

For distinct $X_i$ and $n\ge 2$, use its $n$-dimensional basis, write $B_{ij}=b_j(X_i)$, and define the positive semidefinite matrixThe penalized least squares problem is $\Vert Y-B\beta {\Vert }^2+\lambda {\beta }^T\Omega \beta$, soUse linear continuation outside the observation interval. The fitted-value matrix is $S_{\lambda }=B(B^{T}B+\lambda \Omega {)}^{-1}{B}^T$; its trace gives the effective degrees of freedom. As $\lambda$ decreases to zero, the natural spline approaches interpolation, with trace $n$; as it grows without bound, the fit approaches ordinary least squares on $1,x$, with trace two. For $\lambda =0$ the criterion alone does not determine values between observations; the natural interpolant is the limiting convention. Repeated observation locations are combined by using distinct knots and their multiplicities as weights, rather than pretending the spline design has $n$ independent interpolation values.

The range of the Volterra integration operator is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a Sobolev space element that is an absolutely continuous function. Indeed $Ku$ has weak derivative $u$ and zero initial trace; conversely the fundamental theorem of calculus reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.

The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its distributional derivative is a Dirac delta distribution, not an $L^2$ function. ThusFor a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative norm is $\sqrt{n}$, illustrating unstable differentiation.

The supplied SVD gives $K({\sigma }_j{u}_j)={\sigma }_j^2{v}_j$, or more simply $K^{†}({\sigma }_j{v}_j)=u_j$. Since ${\sigma }_j{v}_j\to 0$ but $\Vert u_j\Vert =1$, the Moore–Penrose inverse is discontinuous. Its domain is the dense, nonclosed range above, and there it is the weak derivative.

Take the indicator function $u(x_1,x_2)={\chi }_{\{x_1>1/2\}}$ inside $(0,1{)}^2$. For a compactly supported test vector field, integration in $x_1$ and then $x_2$ givesThe vertical derivative term integrates to zero. The absolute pairing is at most one, and test vector fields with ${\xi }_1=-1$ along all but arbitrarily short endpoint pieces approach one. Thus $|Du|(\Omega )=1$. More explicitly, the distributional derivative is $Du=e_1{\mathcal{H}}^1\upharpoonright \{(1/2,x_2):0<x_2<1\}$, where ${\mathcal{H}}^1$ is Hausdorff measure along the segment. Also $\Vert u{\Vert }_1=1/2$.

A Sobolev space function in $W^{1,1}(\Omega )$ has each weak derivative represented by an $L^1$ density with respect to planar Lebesgue measure. The nonzero segment measure here is singular with respect to that measure, so this requirement fails. Equivalently, the Sobolev fundamental theorem of calculus on lines would require an absolutely continuous representative on almost every horizontal slice, whereas each slice has a jump. ThereforeThis is the bounded-variation step outside W11 example.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincare inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincare inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

Let $u_j\in C_c^{\infty }({\mathbb{R}}^n)$ tend to $u$ in the Sobolev space $W^{1,p}$, using density of smooth functions in a Sobolev space. Write $\alpha =1-n/p>0$. Applying the Morrey inequality on a cube to a unit cube centered at $x$, and bounding its average by the Holder inequality, givesfor every smooth $v$. Hence $u_j$ is a Cauchy sequence in the supremum norm and has a bounded continuous limit $u^{\ast }$ by uniform convergence. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^{\ast }$ almost everywhere.

For $x\ne y$, choose $r=4|x-y|$, so that $y\in Q_r(x)$. The second estimate from part (a), followed by passage to the limit, givesThus $u^{\ast }$ is a Hölder continuous function andIt is the unique continuous representative: two continuous functions agreeing almost everywhere agree everywhere.

For $1\le p<n$, take a smooth cutoff function $\chi$ equal to one near zero and setIts weak derivative has size $O(|x{|}^{-a-1})$ near zero, whose $p$th power is integrable because $p(a+1)<n$. Thus $u\in W^{1,p}$, but it is essentially unbounded in every neighborhood of zero and has no continuous representative. The value assigned at zero is irrelevant.

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 Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives