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For each $0<r<1$, the spherical mean value property for harmonic functions givesInsert this into part 1(iii). The normalization of the radial mollifier is $d{\omega }_d{\int }_0^1\varphi (r)r^{d-1}dr=1$, henceA convolution with a smooth function of compact support is smooth wherever it is defined. Every point admits such an $\epsilon$, so $u\in C^{\infty }(\Omega )$.

The exterior derivative is the real-linear degree-one map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ characterized by $df(X)=Xf$ on smooth functions, the graded Leibniz ruleand $d^2=0$. In local coordinates $(x^1,\dots ,x^m)$, write $\alpha ={\sum }_I{\alpha }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$ using multi-index notation. Since $d{x}^i=d(x^i)$ and hence $d(d{x}^i)=0$, the defining rules forceThis proves local uniqueness, and the coordinate formulas agree on overlaps because the same rules are preserved by the pullback of a differential form. They also directly define an operator satisfying all three rules, proving existence.

An exact differential form is a form $\omega =d\eta$. Let $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ be the antipodal double covering map and let $a(x)=-x$. For a $2n$-form $\omega$ on real projective space, $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$, while the mapping degree of the antipodal map is $-1$. ThereforeBy the stated criterion, ${\pi }^{\ast }\omega =d\eta$. The invariant primitive under a finite group actionstill satisfies $d\bar{\eta }={\pi }^{\ast }\omega$ and descends to a form $\beta$ on ${\mathbb{RP}}^{2n}$. Since pullback through a covering is injective on differential forms, ${\pi }^{\ast }(d\beta -\omega )=0$ implies $d\beta =\omega$. Thus every $2n$-form on ${\mathbb{RP}}^{2n}$ is exact; this is the top-degree differential forms on even-dimensional real projective space are exact result.

The $k$th de Rham cohomology isthe closed differential forms modulo the exact ones. For the product, let $p:M\times S^1\to M$ be projection and choose a closed one-form $\nu$ on the circle with ${\int }_{S^1}\nu =1$. Averaging differential forms over the circle is cochain-homotopic to the identity, so every class has a rotation-invariant representative; if such a representative is exact, averaging a primitive gives an invariant primitive. Every invariant $k$-form has a unique decomposition $p^{\ast }\alpha +p^{\ast }\beta \wedge \nu$, andClosedness and exactness are therefore componentwise. Hence the mapis a well-defined bijection $H_{\mathrm{dR}}^k(M)\oplus H_{\mathrm{dR}}^{k-1}(M)\to H_{\mathrm{dR}}^k(M\times S^1)$, proving the de Rham cohomology of a product with a circle formula.

Setting $I=0$ in cylindrical magnetostatic pressure balance gives $p+B_z^2/(8\pi )=\mathrm{constant}$. The axial boundary value sets this constant to $p_0$. Thus the magnetic pressure support with vanishing axial field solution isThe azimuthal component of the magnetostatic Ampère-Maxwell equation givesSince $1-e^{-R^2/a^2}=R^2/a^2+O(R^4)$, the radial limit isThere is a regularity subtlety: $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ is not a smooth function of Cartesian position at the axis, and the nonzero limiting $j_{\varphi }$ multiplies an azimuthal unit vector with no unique direction there. These formulas solve the radial problem for $R>0$ and give the requested scalar limit, but the data in this part do not admit the fully smooth vector-field regularity assumed in part (a).

The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

For fixed $(x_1,x_2)$, apply the given one-variable identity to the smooth functionThe chain rule givesThereforewhereEvery derivative of the integrand is continuous. Repeated differentiation under the integral sign therefore shows that $h$ is smooth. This is the Second-order Hadamard lemma.

A phase function is a real smooth functionthat is positively homogeneous of degree one in $\theta$ and has no critical point in all variables:The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of all $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and multi-indices $\alpha ,\beta$,

For a cutoff $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero, define the oscillatory integral byRepeated integration by parts makes the limit meaningful and independent of the cutoff.

The singular support is the complement of the largest open set on which the distribution is represented by a smooth function. Suppose $x_0$ does not belong toOn a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|{\nabla }_{\theta }\Phi |$. The differential operatorsatisfies $L{e}^{i\Phi }=e^{i\Phi }$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_{\Phi }(a)$ is smooth near $x_0$, proving

For the stated distribution on ${\mathbb{R}}^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho =|x|$. The angular integral isTherefore, for $\rho >0$,The original amplitude is not Lebesgue integrable in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional Yukawa potential, satisfying $(-\Delta +1)u=4{\pi }^2{\delta }_0$ with the normalization used in the question.

Fix $I$ and write $I^{\prime }$ for the complementary column indices. Over $U_I$, decompose $M=(M_I,M_{I^{\prime }})$ and $v=(v_I,v_{I^{\prime }})$. The equation $Mv=0$ is equivalent toMatrix inversion is smooth on the invertible matrices, sois a smooth, fiberwise-linear trivialization $U_I\times {\mathbb{R}}^{n-m}\to {\pi }^{-1}(U_I)$. The sets $U_I$ cover the base, proving that $\pi :E_{m,n}\to X_{m,n}$ is a vector bundle of rank $n-m$. It is the kernel bundle of a constant-rank family of linear maps, and its rank also follows from the rank-nullity theorem.

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

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

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

Take a nonnegative radial mollifier ${\eta }_{\epsilon }$ supported in $B_{\epsilon }(0)$, with $\epsilon <\mathrm{dist}(x,\partial \Omega )$. Writing the convolution in polar coordinates and applying the spherical mean value property for harmonic functions at every radius shows directly thatThe convolution is smooth, so $u$ agrees locally with a smooth function. Since $x$ was arbitrary, $u\in C^{\infty }(\Omega )$.

On either open half-line, part (i) says $\hat{u}=-v^{\prime }$. Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. ThereforewhereandwhereBoth functions are smooth on their respective half-lines, proving the required assertion.

A phase function is a real smooth function $\Phi$ on $X\times ({\mathbb{R}}^k\setminus \{0\})$ that is positively homogeneous of degree one in $\theta$ and has nonzero total differential $d_{x,\theta }\Phi$. The symbol classconsists of smooth amplitudes for which, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,

Choose a smooth cutoff function $\chi$ equal to one near zero. The associated oscillatory integral is defined on a test function $f$ byOn the compact $x$-support of $f$, use an integration-by-parts operator $L$ satisfying $L{e}^{i\Phi }=e^{i\Phi }$. Repeated application of its formal adjoint lowers the effective symbol order until the integral is absolutely convergent. The resulting bounds involve only finitely many derivatives of $f$, prove that the limit is independent of $\chi$, and give the seminorm estimate required for

For a Riemannian metric on an ambient manifold and an embedded submanifold, let $\pi :TN{|}_M\to TM$ be orthogonal projection. Project the restriction of a connection to an embedded submanifold to obtain $D_{X}Y=\pi {\nabla }_{X}Y$. Since $\pi Y=Y$, the Leibniz rule and linearity over smooth functions hold, making $D$ an affine connection. If the ambient connection is the Levi-Civita connection, pairing with tangent vectors shows that $D$ is a metric connection; projecting the zero-torsion identity shows it is a torsion-free connection. By uniqueness, $D$ is the induced Levi-Civita connection.

For a smooth function $f$ and vector field $X$,on every tensor field, with $X\otimes df$ interpreted as the endomorphism $Y\mapsto Y(f)X$. Both sides vanish on functions; on vector fields this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$.

A Schwartz function is an element of the Schwartz space: it is a smooth function and every derivative decreases faster than every inverse polynomial. Equivalently, $x^{\alpha }{D}^{\beta }f(x)$ defines a bounded function for every pair of multi-indices $\alpha ,\beta$.

For a smooth function $f(x,y)$,where the smooth remainder is

For any distribution $E\in {\mathcal{D}}^{\prime }$ and any test function $f$, the convolution $(E\ast f)(x)=⟨E_y,f(x-y)⟩$ is a smooth function. On a compact set of $x$ values, the translated test functions and all their derivatives have supports in one fixed compact set, so distributional continuity permits every derivative:For constant-coefficient operators, $P(D)(E\ast f)=(P(D)E)\ast f$. Hence a fundamental solution of a linear differential operator supplies a smooth particular solution for every compactly supported smooth datum. No temperedness of $E$ is needed.

A map between smooth manifolds is smooth when its expression in every pair of coordinate charts is a smooth function.

If $u\in W_{\mathrm{loc}}^{1,2}$ and $\Phi$ is continuously differentiable with bounded derivative on the range in use, then $D(\Phi (u))={\Phi }^{\prime }(u)Du$ almost everywhere. One proves this first for smooth functions and then uses density of smooth functions in a Sobolev space and boundedness of ${\Phi }^{\prime }$. On compact sets one may truncate $\Phi$ away from the range of a bounded $u$.

A tensor derivation is a real-linear operation preserving every tensor type, obeying the tensor-product Leibniz rule, and commuting with every tensor contraction. A derivation $D$ of smooth functions and a real-linear operator on vector fields satisfying $D(fY)=fDY+(Df)Y$ extend uniquely to a tensor derivation. On a differential one-form it must satisfyThis expression is linear over smooth functions in $Y$. In a local frame $D{e}_a=A^b{}_a{e}_b$, the dual rule is $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Apply the product rule to every coefficient and frame factor to define the extension; these dual signs cancel under contraction. Frame changes agree by differentiating the inverse matrix. Cutoffs prove locality and hence uniqueness from local expansions.