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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 space of test functions isA sequence ${\phi }_m$ converges to $\phi$ in $\mathcal{D}(\mathbb{R})$ when all supports lie in one compact set $K$ andfor every nonnegative integer $j$. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is the continuous dual space of $\mathcal{D}(\mathbb{R})$, and $u_m\to u$ in ${\mathcal{D}}^{\prime }$ meansfor every test function $\phi$.

If a linear form $u$ is continuous, it clearly maps every null sequence to a scalar sequence tending to zero. Conversely, suppose it has this sequential property. For each compact $K$, its restriction to the Fréchet space ${\mathcal{D}}_K$ must be continuous: otherwise, for every $m$ one could choose ${\phi }_m\in {\mathcal{D}}_K$ such thatThen ${\phi }_m\to 0$ in $\mathcal{D}$ but its images do not tend to zero, a contradiction. Continuity on every ${\mathcal{D}}_K$ is precisely continuity for the strict inductive limit topology of $\mathcal{D}$, so $u\in {\mathcal{D}}^{\prime }$.

Use the convention ${\tau }_h\phi (x)=\phi (x-h)$. Translation and the distributional derivative are defined byTranslation, differentiation, and multiplication by $-1$ are continuous maps on $\mathcal{D}$, so these formulas define continuous linear functionals and hence distributions.

For fixed $\phi$, the difference quotient satisfiesThereforewhich proves $u^{\prime }={\lim }_{h\to 0}({\tau }_{-h}u-u)/h$ in ${\mathcal{D}}^{\prime }$.

For $-1<\lambda <0$, integration by parts after subtracting the value at zero givesThe subtraction makes the integrand locally integrable at zero, and $x^{\lambda }\to 0$ handles the other boundary.

The analogous Hadamard finite-part integral isIndeed, integrating from $\epsilon$ and combining the boundary term $\phi (\epsilon )\log \epsilon$ with the divergent constant part of the integral gives this limit.

Because $x_{+}^{\lambda }$ is a locally integrable function, its first distributional derivative has order at most one. It is not of order zero. Choose $\psi \in \mathcal{D}((0,1))$ with ${\int }_0^1\lambda t^{\lambda -1}\psi (t)dt\ne 0$ and put ${\psi }_{\epsilon }(x)=\psi (x/\epsilon )$. The sup norms stay bounded whileis unbounded as $\epsilon \downarrow 0$. This contradicts the local sup-norm estimate required of an order-zero distribution. Hence $(x_{+}^{\lambda }{)}^{\prime }$ has order exactly $\boxed{1}$.

Sonja Žunar presented joint work on topological aspects of smooth automorphic forms. Her 2022 slides explain the use of locally convex spaces and the strict inductive limit topology in organizing spaces of such functions. This supplies a topology compatible with an increasing sequence of function spaces, rather than treating automorphic forms only as an unstructured collection of functions.^[ref]

In the standard closed-stage setting, let $E_1\subseteq E_2\subseteq \cdots$ be Hausdorff locally convex spaces, with each $E_n$ a closed topological linear subspace of $E_{n+1}$. The strict inductive limit topology on $E={\bigcup }_n{E}_n$ is the finest topology that makes $E$ a locally convex space and makes every inclusion $E_n\to E$ continuous. Each stage retains its original topology as a closed subspace of the limit.^[ref]

For a locally convex space $F$, a linear map $T:E\to F$ is continuous exactly when every restriction $T{|}_{E_n}$ is continuous. This universal property concerns locally convex targets; it does not identify the construction with an unrestricted final topology in the category of all topological spaces.

For example, the union of finite-dimensional coordinate spaces ${\mathbb{R}}^n$, included by appending zeros, is the space of finitely supported sequences. The functional $\ell (x)={\sum }_j{x}_j$ is continuous for the strict inductive limit topology, since each restriction is a finite sum. Hence the coordinate unit vectors do not converge to zero, although they do in the coordinatewise product topology on ${\mathbb{R}}^{\mathbb{N}}$. The choice of topology changes convergence on the same algebraic vectors.