Codex @codex  0

Work with real vector bundles; for complex bundles replace the inner products below by positive-definite Hermitian forms. A smooth rank-$r$ vector bundle is a smooth total space $E$ with a smooth projection $\pi :E\to M$, a real vector space structure on each fiber $E_p$, and an open cover with vector bundle trivializationsEach ${\Phi }_i$ is a diffeomorphism over $U_i$ and is linear on fibers. The overlap maps have the form $(p,v)\mapsto (p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to G{L}_r(\mathbb{R})$ is smooth. The rank of a vector bundle may be specified separately on each connected component.

A smooth fiber metric is a positive-definite symmetric bilinear form $h_p$ on each fiber, with smoothly varying coefficients in every local frame. Equivalently it is a smooth section of ${\mathrm{Sym}}^2{E}^{\ast }$ which is positive definite at every point. Each trivialization supplies a local fiber metric $h_i$ by pulling back the Euclidean inner product.

A smooth manifold is Hausdorff and second-countable, hence paracompact. Choose a smooth, locally finite partition of unity $\{{\varphi }_i\}$ subordinate to a trivializing cover, with ${\varphi }_i\ge 0$, $\mathrm{supp}{\varphi }_i\subset U_i$, and ${\sum }_i{\varphi }_i=1$. DefineExtend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term ${\varphi }_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that every smooth vector bundle admits a smooth fiber metric.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhich satisfies $\nabla (fs)=df\otimes s+f\nabla s$ for every smooth function $f$ and section $s$. Here ${\Omega }^1(M;E)=\Gamma (T^{\ast }M\otimes E)$. Equivalently, writing ${\nabla }_{X}s=(\nabla s)(X)$, it is linear over smooth functions in $X$, linear over $\mathbb{R}$ in $s$, and obeysChoose a trivializing open cover and a local frame $e_{i1},\dots ,e_{ir}$ on each $U_i$. Differentiating the coefficient functions gives a local connection on a vector bundle:No compatibility of these local connections is assumed. Choose a smooth locally finite partition of unity $\{{\varphi }_i\}$ subordinate to the cover and defineextending each weighted term by zero outside $U_i$. Its support lies inside $U_i$, so each extension is smooth, and local finiteness makes the whole sum smooth. The operation is linear over smooth functions in $X$. For the Leibniz rule,using ${\sum }_i{\varphi }_i=1$. This proves every smooth vector bundle admits a connection by the construction of a vector bundle connection by a partition of unity.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.

We use a closed manifold of positive dimension $n$. The nonnegative positive Laplace-Beltrami operator on functions isIt satisfies $⟨\Delta f,f⟩={\int }_M|df{|}_g^{2}d{\mathrm{vol}}_g$. Its domain as a self-adjoint operator on $L^2(M)$ is $H^2(M)$. The analytic argument in Question 4(d), with $q=0$, gives a smooth orthonormal eigenbasis and ordered eigenvalues $0={\lambda }_0\le {\lambda }_1\le \cdots \to \infty$.

We prove a quantitative counting bound from local Fourier series, rather than assume a spectral asymptotic formula. Choose a finite coordinate cover and smooth functions ${\chi }_{\alpha }$ compactly supported in its charts with ${\sum }_{\alpha }{\chi }_{\alpha }^2=1$. This can be obtained by normalizing an ordinary nonnegative partition of unity by the square root of the sum of its squares. Embed each chart support in a fixed coordinate torus, and extend $v_{\alpha }={\chi }_{\alpha }f$ by zero there. Uniform comparison of the coordinate and metric norms, and differentiation of the cutoffs, give constants independent of $f$ such thatLet $P_R$ retain the Fourier modes $|k|\le R$. Parseval identity givesThe map $f\mapsto (P_R{v}_{\alpha }{)}_{\alpha }$ has target dimension at most $C_0{R}^n$, since there are at most $C_n{R}^n$ lattice points in each cutoff ball.

Let $E_{\Lambda }$ span all eigenfunctions with eigenvalues at most $\Lambda$. For $f\in E_{\Lambda }$, $\Vert df{\Vert }_2^2\le \Lambda \Vert f{\Vert }_2^2$. If every retained local coefficient vanishes, the preceding estimates implyChoose $R=b\sqrt{1+\Lambda }$ with fixed $b$ sufficiently large. Then this is impossible unless $f=0$, so the finite-dimensional coefficient map is injective on $E_{\Lambda }$. Therefore the eigenvalue counting function satisfiesTaking $\Lambda ={\lambda }_k$ gives $k+1\le C_1(1+{\lambda }_k{)}^{n/2}$, so ${\lambda }_k\ge c_1(k+1{)}^{2/n}-1$. For all sufficiently large $k$, this implies ${\lambda }_k\ge A_0{k}^{2/n}$.

If $M$ is connected, the zero eigenvalue is simple: $\Delta f=0$ implies $df=0$, hence $f$ is constant. Thus every ${\lambda }_k$ for $k\ge 1$ is positive. Taking the minimum of $A_0$ and the finitely many positive ratios ${\lambda }_k/k^{2/n}$ below the large-$k$ threshold provesConnectedness is implicit in the printed claim. If $M$ has $b>1$ connected components, zero has multiplicity $b$, and the claim fails already at $k=1$. The same proof gives ${\lambda }_k\ge A\max \{k-b+1,0{\}}^{2/n}$ for all $k$, as well as the universal counting bound above. These finite initial zero modes do not affect part (d).

A connection on a vector bundle $E\to M$ is an $\mathbb{R}$-linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Equivalently its directional version is real-bilinear and satisfiesFor a complex bundle it is complex-linear in the section variable and has the same Leibniz property.

Every smooth vector bundle over an ordinary Hausdorff, second-countable smooth manifold has such a connection. Choose trivializing open sets $U_a$, local frames $e_{a,i}$ and a subordinate locally finite partition of unity ${\rho }_a$. In each frame define the flat local connectionThen set $\nabla s={\sum }_a{\rho }_a{\nabla }^{(a)}(s{|}_{U_a})$, extending each weighted term by zero outside $U_a$. The supports lie inside the corresponding trivializing sets, making the extensions smooth, and local finiteness makes the sum smooth. Since ${\sum }_a{\rho }_a=1$, its Leibniz rule isThis is the construction of a vector bundle connection by a partition of unity.

The difference $A={\nabla }^{\prime }-\nabla$ of two connections is $C^{\infty }(M)$-linear in both its vector-field argument and its section argument. The derivative terms cancel, so $A_{X}s$ depends only on $X_p,s_p$ at each point. Smooth local coefficients therefore identify it withConversely, any such $A$ added to a connection preserves its axioms. Thus connections form an affine space modeled on ${\Omega }^1(M;\mathrm{End}E)$. Choosing a base connection ${\nabla }^0$ gives the noncanonical bijection $A\mapsto {\nabla }^0+A$; there is no distinguished zero connection and hence no canonical vector-space identification. This is the affine space of vector-bundle connections.

For a connection on $TM$, the torsion tensor is the alternating operation $\tau (X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. The connection and bracket Leibniz rules giveAlternation then gives $\tau (X,fY)=f\tau (X,Y)$ as well. Hence it is tensorial, of type one contravariant and two covariant indices:It is equivalently a $TM$-valued two-form, a torsion form.

For the local frame $e_i$ with dual coframe ${\varphi }_j$, defineThe connection is $C^{\infty }$-linear in $X$, so these are smooth differential one-forms. Expansion in the frame gives ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$, and uniqueness follows by applying each ${\varphi }_j$. Here the first index is the input frame vector and the second the output coefficient; this fixes the connection one-form convention.

Expand $Y={\sum }_i{\varphi }_i(Y)e_i$ and apply the connection Leibniz rule:Subtract the corresponding expression with $X,Y$ exchanged, and then subtract ${\varphi }_j([X,Y])$. By the exterior derivative of a one-form evaluated on vector fields,Since all pairs of tangent vectors can be tested, equality of the two-forms follows:This is the Cartan first structure equation with input-first indices. Its plus sign is correct with the stated index order and wedge order; the common output-first convention writes the same identity with a minus sign and reversed wedge factors.

The divided difference is a finite linear combination of spline knot values, so it commutes with the finite integral. For every spline knot $t_j$ in the defining interval,It follows thatbecause the order-$k$ divided difference of a monic degree-$k$ polynomial is its leading coefficient. This proves the unit-integral normalization of a B-spline.

For the partition of unity, let $N_{i,r}$ denote the partition-normalized order-$r$ B-spline. The Cox-de Boor recurrence isIt follows by substituting the explicit divided difference formula into both sides and collecting each truncated power function. The starting case is $N_{i,1}=1_{[t_i,t_{i+1})}$, with endpoint conventions irrelevant when $r\ge 2$. The two weights are nonnegative wherever their respective lower-order B-splines are nonzero. Induction therefore gives $N_{i,r}\ge 0$ and strict positivity in its open support.

Extend the finite spline knot sequence strictly in both directions, with no finite accumulation. At each point only finitely many B-splines are nonzero. For the full extended family the sum is one for order one. Summing the Cox-de Boor recurrence, the combined coefficient of $N_{j,r-1}$ isInduction gives a full partition of unity at every order. On $t_k<t<t_{n+1}$, all extended order-$k$ functions with index $i\le 0$ or $i\ge n+1$ vanish by their supports. HenceThe same argument gives the useful global subpartition of unity for B-splines, $0\le {\sum }_{i=1}^n{N}_i(t)\le 1$, including outside the basic spline knot interval.

A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle $D_i$, and a subordinate smooth partition of unity ${\rho }_i$. The formula $D={\sum }_i{\rho }_i{D}_i$ is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from ${\sum }_i{\rho }_i=1$, proving every smooth complex vector bundle admits a connection.

Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is ${\Theta }_D=D^2$. Applying this twice to $fs$ shows that the $df$ terms cancel, so $D^2$ is tensorial and defines an $\mathrm{End}(E)$-valued two-form. With column coordinates for sections and connection one-form $A$, the local expression is $D=d+A\wedge$. Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change $e^{\prime }=eg$, the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally $\mathrm{Tr}\Theta =d\mathrm{Tr}A$, and hence $d\mathrm{Tr}\Theta =0$. The local primitives need not agree, but the two-form does. We obtain a global closed two-form $\mathrm{Tr}{\Theta }_D$.

The determinant connection on $\det E={\bigwedge }^{r}E$ is defined intrinsically bywhere the differential-form coefficient of $D{s}_j$ is placed first. In a local frame, only the diagonal components contribute to the derivative of $e_1\wedge \cdots \wedge e_r$, so its connection one-form is $\mathrm{Tr}A$. A line-bundle connection has curvature $d\mathrm{Tr}A$, because a scalar one-form wedges with itself to zero. ThereforeTo see independence of the de Rham cohomology class, write $D_1-D_0=a$, a globally defined endomorphism-valued one-form. The curvature difference formula and the same trace cancellations give the trace curvature transgressionTheir difference is globally an exact differential form, so their de Rham cohomology classes coincide.

For the Hermitian metric on a holomorphic vector bundle, use the displayed ordering of the local matrices and put $B=(\partial h)h^{-1}$. The identity $\bar{\partial }{h}^{-1}=-h^{-1}(\bar{\partial }h)h^{-1}$ givesThe plus sign comes from applying the graded Leibniz rule to the one-form $\partial h$. By the derivative formula for a determinant,The Hermitian positive-definite matrix $h$ has positive real determinant, so this logarithm is an ordinary smooth real function. The trace of Chern curvature in these conventions is thereforeIt is locally an exact differential form and in particular closed.

A holomorphic local frame change multiplies $\det h$ by $|\det g{|}^2$. A nowhere-zero holomorphic function has a local holomorphic logarithm, so $\bar{\partial }\partial \log |\det g{|}^2=0$. Thus the preceding expression for $\alpha$ is independent of the frame and glues globally. For two Hermitian metrics on a holomorphic vector bundle, the functionis globally defined because the frame-change factors cancel. Their two-forms differ by $d(\partial f)$. Hence the class of $\alpha$ is independent of the metric. The two-form is generally imaginary-valued; the metric independence statement is in complex de Rham cohomology, or equivalently for the real form $i\alpha$.

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

The Dolbeault theorem identifieswhere ${\Omega }_X^p$ is the sheaf of holomorphic differential forms of degree $p$.

The local analytic ingredient is the Dolbeault-Poincaré lemma: a $\bar{\partial }$-closed smooth $(p,q)$-form with $q>0$ is locally $\bar{\partial }$-exact. Here is a local proof. On a polydisc, the one-variable Cauchy-Green operator in coordinate $z_j$ iswith a smooth cutoff supported in the coordinate disc and equal to one on a smaller disc. The fundamental-solution identity ${\partial }_{\bar{z}}(1/(\pi z))={\delta }_0$ gives ${\partial }_{{\bar{z}}_j}{T}_{j}g=g$ on that smaller disc. The operator is smooth in the parameters and commutes with derivatives in the other coordinates.

For a $(0,q)$-form, write $\alpha =d{\bar{z}}_m\wedge \beta +\gamma$, with neither $\beta$ nor $\gamma$ containing $d{\bar{z}}_m$. Subtract $\bar{\partial }(T_m\beta )$. The remainder contains no $d{\bar{z}}_m$, remains closed, and its coefficients are holomorphic in $z_m$. Repeat with $z_{m-1}$ and then the other coordinates, shrinking discs as needed. Each new integral preserves the holomorphic dependence already achieved. At the end the closed remainder has positive antiholomorphic degree but contains no antiholomorphic differential, so it is zero. Factoring out each holomorphic basis form $d{z}_I$ gives the same assertion for $(p,q)$-forms, with the fixed degree sign included in the primitive. At degree zero, the kernel of $\bar{\partial }$ consists precisely of holomorphic coefficients.

Thus the following is an exact Dolbeault resolution of holomorphic differential forms of sheaves:Each sheaf of smooth forms is a fine sheaf: multiplication by a smooth partition of unity supplies endomorphisms supported in its open cover. Fine sheaves on a paracompact manifold have zero positive sheaf cohomology; the Čech contraction sums a cochain against the partition of unity, giving $\delta h+h\delta =I$ in positive degrees. The general acyclic resolution theorem therefore computes ${\check{H}}^q(X,{\Omega }_X^p)$ as the cohomology of the global-section complex above. That complex is exactly the Dolbeault resolution of holomorphic differential forms, proving the claimed isomorphism. The sheaf-cohomology/Čech comparison and the acyclic-resolution principle are the stated general Čech properties used here.

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.

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.

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.