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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 $\boxed{1}$.