= Solution
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> $p_{K,j}(\varphi)=\max_{|\alpha|\leq j}\sup_K|\partial^\alpha\varphi|$. In particular, \b[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 $\mathcal D'(\mathbb R^n)$ consists of continuous complex-linear forms on $\mathcal D$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\geq0$ such that
$$
|\langle u,\varphi\rangle|\leq C_K\max_{|\alpha|\leq q_K}\sup_K|\partial^\alpha\varphi|,\qquad \operatorname{supp}\varphi\subset K.
$$
We use <distributional convergence>: \b[$u_j\to u$ means $\langle u_j,\varphi\rangle\to\langle u,\varphi\rangle$ for every $\varphi\in\mathcal D$]. The pairings are bilinear, with no complex conjugation. A <test function> $f$ is identified with its <regular distribution> $\langle f,\varphi\rangle=\int f\varphi$.
For a <distribution> $u$ and a <test function> $\psi$, their <smoothing convolution with a test function> is
$$
\boxed{(u*\psi)(x)=\langle u_y,\psi(x-y)\rangle.}
$$
It is a <smooth function>, with $\partial_x^\alpha(u*\psi)(x)=\langle u_y,\partial^\alpha\psi(x-y)\rangle$. 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 $u*\psi$ need not be compactly supported.
Choose a nonnegative <mollifier> $\rho\in C_c^\infty(B_1)$ with $\int\rho=1$, and write $\rho_\varepsilon(x)=\varepsilon^{-n}\rho(x/\varepsilon)$. If $\widetilde\rho(x)=\rho(-x)$, then
$$
\langle u*\rho_\varepsilon,\varphi\rangle=\langle u,\widetilde\rho_\varepsilon*\varphi\rangle.
$$
The right-hand test tends to $\varphi$ in $\mathcal D$: its supports lie in $\operatorname{supp}\varphi+\overline B_1$ for $\varepsilon\leq1$, 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 set
$$
\boxed{f_j(x)=\chi(x/j)(u*\rho_{1/j})(x)\in\mathcal D(\mathbb R^n).}
$$
For each fixed <test function> $\varphi$, the cutoff is identically one on its support once $j$ is large. Hence $\langle f_j,\varphi\rangle=\langle u,\widetilde\rho_{1/j}*\varphi\rangle\to\langle u,\varphi\rangle$. This proves \b[$\mathcal D$ is dense in $\mathcal D'$], 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> $\varphi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $r\,dr=dt/2$ gives
$$
\langle u_m,\varphi\rangle=\int_{-1}^{\infty}m\sin(m|t|)A(t)\,dt,
\qquad A(t)=\frac12\int_0^{2\pi}\varphi\big(\sqrt{1+t}(\cos\omega,\sin\omega)\big)\,d\omega.
$$
When $\varphi$ 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> gives
$$
\int_0^{\infty}m\sin(ms)B(s)\,ds=B(0)+\int_0^{\infty}\cos(ms)B'(s)\,ds\longrightarrow2A(0),
$$
by the <Riemann-Lebesgue lemma>. Therefore
$$
\boxed{u_m\longrightarrow\delta_{S^1}\quad\text{in }\mathcal D'(\mathbb R^2\setminus\{0\}),\qquad
\langle\delta_{S^1},\varphi\rangle=\int_0^{2\pi}\varphi(\cos\omega,\sin\omega)\,d\omega.}
$$
This <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 $A(-1)=\pi\varphi(0)$. Its right derivative is integrable, with $A'(-1+s)=(\pi/4)\Delta\varphi(0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi\varphi(0)+(\pi/4)(1+t)\Delta\varphi(0)+O((1+t)^2)$ near $t=-1$. Applying <integration by parts> separately on the negative and positive intervals gives
$$
\langle u_m,\varphi\rangle=2A(0)-A(-1)\cos m-\int_{-1}^{0}\cos(mt)A'(t)\,dt+\int_0^{\infty}\cos(mt)A'(t)\,dt.
$$
The 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> formula
$$
\boxed{u_m=\delta_{S^1}-\pi\cos m\,\delta_0+o_{\mathcal D'}(1).}
$$
Choose $\varphi$ supported in $B_{1/2}$ with $\varphi(0)=1$. Its pairing is $-\pi\cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos(m+1)+\cos(m-1)=2\cos1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos(2m)=2\cos^2m-1$ would then force $L=-1$. Consequently \b[there is no limit in $\mathcal D'(\mathbb R^2)$].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-327-circle-and-point-defect.png]
{title=The stable unit-circle arclength contribution and the oscillating point-mass coefficient at the origin}
{height=360}
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