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The inverse-square radial vector field has zero ordinary divergence at every nonzero radius but flux $4\pi$ through every centred sphere. It is locally integrable in three dimensions. For a smooth compactly supported test function $\eta$, integrate by parts outside a ball of radius $ϵ$. The inward-oriented inner boundary gives $-\int \mathbf{F}\cdot \nabla \eta ={\int }_{S^2}\eta (ϵ\hat{\mathbf{r}})d\Omega$, which tends to $4\pi \eta (0)$. This proves the displayed distributional divergence, with the three-dimensional Dirac delta function representing the point source.

For $r=\sqrt{x^2+y^2+z^2}>0$, the radial vector field has components $F_i=x_{i}f(r)$. The product rule givesThus for $f(r)=r^{n-1}$,When $n=-2$, $\mathbf{F}=\hat{\mathbf{r}}/r^2$ and the ordinary divergence is zero at every nonzero radius. However, the field is singular at the origin. Its outward flux through a radius-$R$ sphere is $(1/R^2)(4\pi R^2)=4\pi$, so one cannot apply the smooth-field divergence theorem across the origin and conclude that this flux vanishes.

In the language of distributions, the complete distributional divergence identity is $\boxed{\nabla \cdot (\hat{\mathbf{r}}/r^2)=4\pi {\delta }^{(3)}(\mathbf{r})}$. Here ${\delta }^{(3)}$ is the three-dimensional Dirac delta function. Without that generalized interpretation, the correct statement is simply zero divergence away from the origin and undefined ordinary divergence at the singular point. Removing a small ball around the origin restores the ordinary theorem: the inner boundary has inward radial orientation and cancels the outer flux.

Away from the origin, the product rule and $\nabla r=\mathbf{r}/r$ giveThe field is undefined at $r=0$. It has outward flux $4\pi$ through any sphere centred there, so its global distributional divergence is $4\pi {\delta }^{(3)}(\mathbf{r})$, not the zero distribution. The pointwise result must not be extended through the singularity when applying the divergence theorem.

On an outward-oriented sphere of radius $R>0$, the electric field is normal with constant magnitude $A/R^2$. Thus its flux isIt is independent of $R$. There is no contradiction with solution: the classical divergence theorem cannot be applied to a full ball while pretending that $E$ is smooth at its center. On a punctured ball the outer flux and the inward-normal inner flux cancel, in agreement with zero ordinary divergence there.

The three-dimensional Dirac delta distribution ${\delta }^{(3)}(r)$ is supported at the origin and satisfies the sampling identityfor smooth compactly supported test functions $\phi$. It is not an ordinary function with a finite value at zero; its total integral is one, and translating it moves the point at which the test function is sampled.

The globally valid source equation isFor a direct weak verification, excise a ball of radius $\epsilon$ and integrate $-E\cdot \nabla \phi$ outside it. Since the ordinary divergence vanishes there, the result is $A{\int }_{S^2}\phi (\epsilon \hat{r})d\Omega \to 4\pi A\phi (0)$. This is precisely the displayed distributional divergence and restores the divergence theorem with the point charge included.

Away from the origin, the electric field is $E=Ar/r^3$, with $r$ here denoting the position vector in the numerator and its Euclidean norm in the denominator. The spherical-coordinate formula for divergence givesAt the origin the electric field is singular, so the ordinary divergence is undefined there. The equation above must not be extended to the whole space as an ordinary smooth-field identity. Its correct distributional divergence, derived in solution, includes the point source.

Write $\mathrm{TV}(u)=|Du|(\Omega )$ for the total variation seminorm on a domain. For a locally integrable real function its dual definition isThe bounded-variation space consists of functions $u\in L^1(\Omega )$ with finite $|Du|(\Omega )$, equipped with $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_1+|Du|(\Omega )$. In the equivalent distributional derivative description, $Du$ is a finite vector-valued Radon measure and $|Du|$ is its total variation measure.

To establish completeness of the bounded-variation space, let $(u_n)$ be Cauchy in this norm. Completeness of $L^1$ gives $u_n\to u$ in $L^1$. Given $\epsilon >0$, choose $N$ so that $\Vert u_n-u_m{\Vert }_{BV}<\epsilon$ whenever $m,n\ge N$. For fixed $n\ge N$, the given lower semicontinuity yieldsSince $\Vert u_n-u_m{\Vert }_1\to \Vert u_n-u{\Vert }_1$, we obtain $\Vert u_n-u{\Vert }_1+|D(u_n-u)|(\Omega )\le \epsilon$. In particular $u_n-u$ has finite variation; the triangle inequality then puts $u$ in the BV space. The same bound proves convergence in the full norm, so this is a Banach space.

For the disk data, put $B=B(0,R)$ and assume $\alpha >0$. The exact total variation denoising of a disk isHere $(s{)}_{+}=\max (s,0)$ is the positive part. A total variation calibration certifies global optimality, including competitors that are not radial or piecewise constant. Define the bounded vector fieldIt has $|z|\le 1$. Its normal component is continuous across the circle, so the distributional divergence has no extra boundary measure. Direct differentiation gives $q=\mathrm{div}z=(2/R){\chi }_B$. The dual definition implies $\int vq\le \mathrm{TV}(v)$: for the standard $BV\cap L^2$ domain one can cut $z$ off at radius $L$, with the error bounded by $C{L}^{-1}{\int }_{L<|x|<2L}|v|\to 0$, and then smooth the test field. In the larger homogeneous bounded-variation space, the same error is bounded by $C\Vert v{\Vert }_{L^2(L<|x|<2L)}\to 0$. Thus both usual whole-plane formulations give the same certificate.

Use the perimeter identity $\mathrm{TV}({\chi }_B)=2\pi R$ and $|B|=\pi R^2$. If $c=1-2\alpha /R>0$, then $\int c{\chi }_{B}q=2\pi Rc=\mathrm{TV}(c{\chi }_B)$ and $u_{\ast }-g+\alpha q=0$. Consequently every competitor $v$ satisfiesIf $\alpha \ge R/2$, replace $z$ by $z_{\alpha }=(R/(2\alpha ))z$. Its norm is still at most one, its divergence is $q_{\alpha }={\chi }_B/\alpha$, and equality in the calibration holds at $u_{\ast }=0$. The identical comparison proves optimality and uniqueness of zero, including the threshold $\alpha =R/2$. The only general results used are completeness of $L^1$, lower semicontinuity of variation, the indicator-perimeter identity, the distributional integration-by-parts/dual variation formula and the quadratic norm identity. For $\alpha =0$, the unique squared-error minimizer is simply $g$.

Let $B=B(0,R)$, with $R>0$. Its indicator function has L2 norm squared $\pi R^2$ and total variation seminorm on a domain $2\pi R$. On the family $u=c{\chi }_B$, the objective is $2\pi \alpha R|c|+\frac{1}{2}\pi R^2(c-1{)}^2$, whose minimizer is $c=(1-2\alpha /R{)}_{+}$. To prove global optimality, a total variation calibration is needed.

Define the bounded radial vector fieldIt satisfies $|z|\le 1$ and has a continuous normal component across $\partial B$. Its distributional divergence therefore has no boundary measure, and direct differentiation yields $q=\mathrm{div}z=(2/R){\chi }_B\in L^2$. Although $z$ is not compactly supported, multiply it by a smooth radial cutoff equal to one through radius $T$ and zero beyond $2T$. The extra divergence has magnitude $O(R/T^2)$ on an annulus of area $O(T^2)$, hence L2 norm $O(R/T)$. Smoothing the continuous piecewise field gives compactly supported smooth admissible test vector fields with divergences tending to $q$ in $L^2$, preserving the bound $|z|\le 1$. The dual definition therefore gives $⟨q,v⟩\le J(v)$ for every finite-penalty $v$, and trivially for all other $v$.

Moreover $⟨q,{\chi }_B⟩=(2/R)\pi R^2=2\pi R=J({\chi }_B)$. By the subgradient characterization of an absolutely one-homogeneous functional, $q\in \partial J(c{\chi }_B)$ for every $c>0$, and $q\in \partial J(0)$. If $0<\alpha <R/2$, choose $c=1-2\alpha /R$; then $(g-c{\chi }_B)/\alpha =q$. If $\alpha \ge R/2$, choose $u=0$; then $g/\alpha =(R/(2\alpha ))q$ belongs to $\partial J(0)$, since multiplying the dual bound by a number in $[0,1]$ preserves it. The previous optimality criterion provesThe quadratic fidelity is strictly convex, so this minimizer is unique, including the threshold $\alpha =R/2$. The disk retains its radius on the positive branch and disappears on the zero branch; this is total variation denoising of a disk.