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If $g$ is smooth and $\nabla g\ne 0$ on $S=\{g=0\}$, transverse local coordinates give $⟨\delta (g),\phi ⟩={\int }_S\phi /|\nabla g|dS$. Thus ${\delta }_S=|\nabla g|\delta (g)$ as distributions. In particular ${\delta }_{S^1}=2\delta (|x{|}^2-1)$. Omitting this Jacobian confuses the delta of a defining function with a geometric surface delta distribution.

For a moving level surface, $n=\nabla S/|\nabla S|$, $v_n=-S_t/|\nabla S|$, and the surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$. Piecewise classical conservation laws acquire a distributional flux defect $[b]\cdot n-v_n[a]$. Vanishing of this defect is the Rankine-Hugoniot condition.

Choose the shock wave normal to point from $S<0$ to $S>0$:The latter identity follows by differentiating $S({\mathbf{x}}_s(t),t)=0$ along the moving surface. Write $[a]=a^{+}-a^{-}$ and $[\mathbf{b}]={\mathbf{b}}^{+}-{\mathbf{b}}^{-}$. The distributional derivative of the Heaviside step function is ${\partial }_{t}H(S)=S_t\delta (S)$ and $\nabla H(S)=\nabla S\delta (S)$. The regular terms cancel using the conservation equations on each side, leavingThus the explicitly requested vector isOnly its normal component matters; adding tangential velocity to the surface parametrization changes neither result. The invariant surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$, so the formula is the moving-interface conservation jump identity ${\partial }_{t}a+\nabla \cdot \mathbf{b}=([\mathbf{b}]\cdot \mathbf{n}-v_n[a]){\delta }_s$.

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 ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

For the geometric surface delta distribution on the radius-$a$ sphere in ${\mathbb{R}}^3$, $⟨u_a,\varphi ⟩=a^2{\int }_{S^2}\varphi (a\omega )d\sigma (\omega )$. Angular integration gives the Fourier transform ${\hat{u}}_a(\xi )=4\pi a\sin (a|\xi |)/|\xi |$, with removable value $4\pi a^2$ at zero. The convolution of distributions with a compactly supported factor gives the regular distributionThis density has total mass $16{\pi }^2{a}^2{b}^2$. Values on endpoint spheres do not affect the distribution. When $a=b$, the inverse-distance singularity is locally integrable in three dimensions and is not an additional point mass. The annulus expresses the triangle inequality for the sum of two vectors of fixed lengths.