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The test-function space is $\mathcal{D}({\mathbb{R}}^n)=C_c^{\infty }({\mathbb{R}}^n)$. Its topology is the strict inductive limit topology of the spaces ${\mathcal{D}}_K$ of smooth functions supported in a fixed compact set $K$, each with the seminorms $p_{K,j}(\phi )={\max }_{|\alpha |\le j}{\sup }_K|{\partial }^{\alpha }\phi |$. In particular, a sequence converges in $\mathcal{D}$ precisely when its supports lie in one compact set and all its derivatives converge uniformly. The indices $\alpha$ use multi-index notation.

The distribution space ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ consists of continuous complex-linear forms on $\mathcal{D}$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\ge 0$ such thatWe use distributional convergence: $u_j\to u$ means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{D}$. The pairings are bilinear, with no complex conjugation. A test function $f$ is identified with its regular distribution $⟨f,\phi ⟩=\int f\phi$.

For a distribution $u$ and a test function $\psi$, their smoothing convolution with a test function isIt is a smooth function, with ${\partial }_x^{\alpha }(u\ast \psi )(x)=⟨u_y,{\partial }^{\alpha }\psi (x-y)⟩$. Indeed, for $x$ in a compact neighborhood all translated tests have support in one compact set, and their difference quotients converge in the space of test functions. Notice that $u\ast \psi$ need not be compactly supported.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }(B_1)$ with $\int \rho =1$, and write ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $\tilde{\rho }(x)=\rho (-x)$, thenThe right-hand test tends to $\phi$ in $\mathcal{D}$: its supports lie in $\mathrm{supp}\phi +{\overline{B}}_1$ for $\epsilon \le 1$, and every derivative converges uniformly by the approximate-identity argument. Thus the smooth regularizations converge to $u$ as distributions.

To obtain actual compactly supported approximants, choose a smooth cutoff function $\chi$ equal to one on $B_1$ and supported in $B_2$, and setFor each fixed test function $\phi$, the cutoff is identically one on its support once $j$ is large. Hence $⟨f_j,\phi ⟩=⟨u,{\tilde{\rho }}_{1/j}\ast \phi ⟩\to ⟨u,\phi ⟩$. This proves $\mathcal{D}$ is dense in ${\mathcal{D}}^{\prime }$, and in fact establishes the density of test functions in distributions and gives a convergent approximating sequence for each distribution. The expanding cutoff is essential when $u$ has noncompact support of a distribution.

For the radial limit, take a test function $\phi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $rdr=dt/2$ givesWhen $\phi$ is supported away from the origin, $A$ vanishes near $t=-1$ and extends to a compactly supported smooth function on the whole line. The folded sine approximation to a Dirac delta is exposed by setting $B(s)=A(s)+A(-s)$: integration by parts givesby the Riemann-Lebesgue lemma. ThereforeThis surface delta distribution is arclength measure on the unit circle. By the level-set normalization of a surface delta, it is $2\delta (|x{|}^2-1)$: the factor two cancels the gradient magnitude $|\nabla (|x{|}^2-1)|=2$ on the circle. It is not twice arclength measure.

For a test function that can meet the origin, the endpoint cannot be discarded. Now $A(-1)=\pi \phi (0)$. Its right derivative is integrable, with $A^{\prime }(-1+s)=(\pi /4)\Delta \phi (0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi \phi (0)+(\pi /4)(1+t)\Delta \phi (0)+O((1+t{)}^2)$ near $t=-1$. Applying integration by parts separately on the negative and positive intervals givesThe last two integrals tend to zero by the Riemann-Lebesgue lemma. Thus the radial quadratic oscillation defect in two dimensions gives the stronger oscillating point-mass defect formulaChoose $\phi$ supported in $B_{1/2}$ with $\phi (0)=1$. Its pairing is $-\pi \cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos (m+1)+\cos (m-1)=2\cos 1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos (2m)=2{\cos }^{2}m-1$ would then force $L=-1$. Consequently there is no limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

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.

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

Using multi-index notation, the Schwartz space isA sequence ${\phi }_m$ converges to $\phi$ in this Fréchet space when $p_{\alpha ,\beta }({\phi }_m-\phi )\to 0$ for every $\alpha ,\beta$. The space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ of tempered distributions is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$, and $u_m\to u$ there means weak convergence of distributions, namely $⟨u_m,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{S}$.

Continuity of a linear functional immediately implies that ${\phi }_m\to 0$ entails $⟨u,{\phi }_m⟩\to 0$. Conversely, enumerate the Schwartz seminorms as $p_1,p_2,\dots$. If $u$ were not continuous, then for each $m$ one could choose ${\phi }_m$ such thatEvery fixed seminorm tends to zero along this sequence, so ${\phi }_m\to 0$ in $\mathcal{S}$, contradicting the assumed sequential property. This is the sequential continuity criterion for a linear map on a metrizable topological vector space.