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Use the orthonormal right-handed basisDirect multiplication gives $B(r)u=u$, $B(r)v=rv+w/\sqrt{2}$ and $B(r)w=-v/\sqrt{2}+rw$. Thus the matrix in this basis isThe last two columns are orthogonal and both have squared norm $r^2+1/2$. Therefore $B(r)$ is an orthogonal matrix precisely when $r^2+1/2=1$, giving the unique positive valueIts determinant is $r_0^2+1/2=1$, so it is a proper rotation matrix, not a reflection. The transverse block has cosine and sine both $1/\sqrt{2}$. ConsequentlyWith the oriented axis $u$, the angle is positive by the right-hand rule, since $u\times v=w$. The axis also follows from the unit-eigenvalue eigenspace. As a check, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

For fixed finite matrices, the matrix exponential and a Taylor expansion showThis agrees through degree two with $e^{h(A+B)}$, even when $A$ and $B$ do not commute. On a fixed time interval, $\Vert S(h)\Vert \le e^{h(\Vert A\Vert +\Vert B\Vert )}$ and the telescoping identityturns the $O(h^3)$ one-step defect into $O(N{h}^3)=O(h^2)$ global error for $Nh\le T$. Hence Strang splitting is second order in general; for commuting matrices it is exact.

For the spatial discretization, use $d=2/(M+1)$, $x_m=-1+md$ and the interior vector $U=(u_1,\dots ,u_M{)}^T$, with $u_0=u_{M+1}=0$. The central finite differences giveHere $\mathrm{tridiag}$ lists lower diagonal, main diagonal and upper diagonal, in that order. Thus $A^T=A$ and $B^T=-B$. Discrete summation by parts givesThe matrix exponential $e^{hA/2}$ is consequently a contraction in the Euclidean norm, while $e^{hB}$ is an orthogonal matrix for real $\alpha$. ThereforeRepeated split steps are contractive in the mesh-weighted norm $d\sum |u_m{|}^2$, uniformly in the spatial mesh and without a Courant–Friedrichs–Lewy condition. The split semidiscretization is unconditionally stable. No commutativity or shared eigenvectors are needed for this Strang splitting contraction for symmetric diffusion and skew advection. Its time-order proof is for fixed spatial matrices; as $d\to 0$, commutators can grow, so this stability result alone is not a uniform-in-mesh second-order error estimate.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.

For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.

Writing ${\mathbf{A}}_{\mu }=(A_{\mu 1},A_{\mu 2},A_{\mu 3})$, the cross product form of the gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$. In components this isCorrespondingly, the gauge field strength has components

A covariantly constant vector Higgs field has constant norm, becauseThe last equality uses antisymmetry of the gauge potential. On a connected region write this norm as $\varphi \ge 0$. If $\varphi =0$, the vector Higgs field is identically zero. If $\varphi >0$, the SO(3) group acts transitively on unit vectors, so choose a smooth local non-Abelian gauge transformation rotating $\Phi /\varphi$ to ${\mathbf{e}}_3$. In this unitary gauge, $\Phi =(0,0,\varphi {)}^T$ with $\varphi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.

For nonzero $\varphi$, substituting the constant aligned vector Higgs field into $D_{\mu }\Phi =0$ gives $g\varphi (A_{\mu 2},-A_{\mu 1},0{)}^T=0$. For nonzero gauge coupling, the most general compatible gauge potential and gauge field strength arewhere $a_{\mu }$ is an arbitrary real one-form. Thus the surviving gauge group is the $SO(2)$ stabilizer subgroup of ${\mathbf{e}}_3$. A covariantly constant vector Higgs field does not force the gauge field strength to vanish: it only requires the curvature to annihilate that field.

For the Higgs mechanism, choose a vacuum of the SO(3) invariant quartic scalar potential, so the constant norm is $v$. The canonically normalized kinetic term then containsThe two broken-direction gauge bosons have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ gauge boson remains massless. The two angular Goldstone bosons supply the massive vectors' longitudinal gauge-boson polarizations. The radial fluctuation remains a physical scalar field, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_{\mu }\Phi =0$ selects backgrounds without broken-direction gauge potentials; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the scalar potential.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

A Givens rotation is the identity outside rows and columns $p,q$, with the $2\times 2$ blockThe block's transpose times itself is the identity, so the whole matrix satisfies $\boxed{{\Omega }^T\Omega =I}$ and is an orthogonal matrix. In particular it preserves the Euclidean norm. To annihilate the lower entry of a column pair $(a,b{)}^T$, take $c=a/r$, $s=b/r$, $r=\sqrt{a^2+b^2}$; the rotated pair is $(r,0{)}^T$. If both entries are zero, take the identity rotation.

A QR decomposition of an $m\times n$ matrix, $m\ge n$, is $A=QR$, where $Q$ is an $m\times m$ orthogonal matrix and $R$ is $m\times n$ and upper trapezoidal: $R_{ij}=0$ for $i>j$. In particular its bottom $m-n$ rows vanish.

Successive left Givens rotations annihilate the entries below the diagonal, column by column. When working on column $j$, rotate rows $j$ and $i>j$; their earlier columns are already zero, so the earlier eliminations persist. If $G_k\cdots G_{1}A=R$, thenA product of orthogonal matrices is orthogonal. The construction remains valid for a rank-deficient matrix, although uniqueness is not asserted. This is QR decomposition by Givens rotations.

The orthogonal matrix $Q$ preserves the Euclidean norm, soThe first three entries of $Rx-Q^{T}b$ can be set to zero by backward substitution in a triangular system, while its fourth entry is always $3$. Solving the triangular system gives $x_3=16/5$, $x_2=(35-2x_3)/5=143/25$, and $x_1=(98/25-x_2-x_3)/3=-5/3$. Thus the unique least-squares solution and residual norm areUniqueness follows from the nonzero diagonal entries of the top triangular block, equivalently the full column rank of $A$.

Use the scaled proximal operator ${\mathrm{prox}}_{\gamma E}$; take $\gamma =1$ for each unscaled resolvent.

For the quadratic case, write $D^{T}D=\mathrm{diag}(d_1,\dots ,d_n)$ and $\xi =Qx$. Since $Q$ is an orthogonal matrix, the quadratic proximal term becomes $\Vert \xi -Qz{\Vert }^2/2$. The subgradient optimality condition givesAll $d_j\ge 0$, so each diagonal entry is positive, including when $d_j=0$. Multiplying by the inverse diagonal matrix givesThe proximal operator of Poisson data fidelity separates pointwise. At each $t$, the scalar optimality equation on $x>0$ isSince $y>0$, the two roots have opposite signs. Only the positive root is admissible in the logarithm:The negative root is outside the effective domain. The second derivative $1+\gamma y/x^2$ of the proximal objective is positive, establishing the unique pointwise minimizer. The integral formula is understood on the function spaces where these quantities are admissible.

Finally, for $E(x)=|x|$, positive and negative minimizers obey $x-z+\gamma =0$ and $x-z-\gamma =0$, respectively. At zero the subdifferential condition is $z\in [-\gamma ,\gamma ]$. Combining these cases yields soft thresholding:Thus $\gamma =1$ supplies all three requested closed forms.

Componentwise,soRepeated application gives $(A^k{)}^T=(A^T{)}^k$. Transposing the matrix exponential term by term therefore yieldsMultiplication of the two power series givesHenceIf $e^{tA}$ is an orthogonal matrix for every real $t$, thenIts coefficient of $t$ must vanish, sothat is, $A$ is a skew-symmetric matrix.

For unit $u$, put $A=I-2u{u}^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an orthogonal matrix. The affine reflection in a hyperplane is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the Euclidean norm and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an Euclidean isometry fixing the hyperplane pointwise.

For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:Changing both signs of $u,c$ leaves the same reflection. It belongs to the orthogonal group $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\Vert p{\Vert }^2-\Vert q{\Vert }^2)/(2\Vert p-q\Vert )$,To prove the finite reflection decomposition of a Euclidean isometry, first note that a distance-preserving map $g$ fixing zero preserves inner products by the polarization identity. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)={\sum }_i{x}_{i}g(e_i)$. Thus $g$ is orthogonal.

Every orthogonal transformation is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^{\perp }$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.

For an arbitrary isometry $f$, if $f(0)\ne 0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. ThereforeThe glide reflection $(x,y)\mapsto (x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.