Spatial BV, finite perimeter, and TV conventions — Examples and Counterexamples

Extend ${\mathrm{\partial }}_1{\varphi }_1$ by zero outside $B$. It is integrable because it is bounded and supported in a finite-measure disk [F5]. Fubini [F4] expresses its integral as the integral over $y\in (-1,1)$ of the slice integrals over $[-a(y),a(y)]$, where $a(y)=\sqrt{1-y^2}$. Each slice integrand is continuous, so [F6] identifies its Lebesgue integral with its Riemann integral, and [F7] evaluates that integral as ${\varphi }_1(a(y),y)-{\varphi }_1(-a(y),y)$. In the right graph $x=a(y)$, [F8] gives $d{\mathcal{H}}^1=dy\mathrm{/}a(y)$ and ${\nu }_1=a(y)$; in the left graph $x=-a(y)$, it gives $d{\mathcal{H}}^1=dy\mathrm{/}a(y)$ and ${\nu }_1=-a(y)$. Thus the slice difference equals ${\int }_{\mathrm{\partial }B}{\varphi }_1{\nu }_{1}d{\mathcal{H}}^1$. The same argument with the coordinates exchanged gives the second component, hence ${\int }_B\mathrm{div}\varphi dx={\int }_{\mathrm{\partial }B}\varphi \cdot \nu d{\mathcal{H}}^1$.

By the distributional derivative convention [F1], step 1.1 yields $D{\chi }_B=-\nu {\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\partial }B}$. The circle has finite arclength, so this is a finite vector Radon measure and ${\chi }_B\in BV({\mathbb{R}}^2)$.

The parametrization $\theta \mapsto (\cos \theta ,\sin \theta )$ has speed one on $[0,2\pi ]$, so ${\mathcal{H}}^1(\mathrm{\partial }B)=2\pi$. For every admissible test field, step 1.1 gives ${\int }_B\mathrm{div}\varphi dx={\int }_{\mathrm{\partial }B}\varphi \cdot \nu d{\mathcal{H}}^1\le 2\pi \mathrm{.}$ Choose a smooth radial cutoff $q$ supported in $(1\mathrm{/}2,3\mathrm{/}2)$, with $0\le q\le 1$ and $q(1)=1$, and set $\varphi (x)=q(\mathrm{\mid }x\mathrm{\mid })x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on that annulus and zero outside. By [F3] this is smooth, compactly supported and has norm at most one; on $\mathrm{\partial }B$ it equals $\nu$. Its flux is $2\pi$. The supremum in [F2] is therefore exactly $2\pi$, proving the asserted total variation and perimeter.

The intervals are disjoint and have Lebesgue measure $2^{-(k+1)}$ by [F4], so $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_1={\sum }_{k=1}^N{2}^{-(k+1)}<1\mathrm{/}2$. Integration by parts on each interval, justified by [F5] and [F6], shows that $D{u}_N$ is an atomic signed measure: its jump at $1\mathrm{/}2$ has magnitude $1$, its $N-1$ interior jumps at $2^{-2},\dots ,2^{-N}$ have magnitude $2$, and its jump at $2^{-(N+1)}$ has magnitude $1$. These are all jumps in $(0,1)$, so $u_N\in BV((0,1))$ and $TV(u_N;(0,1))=1+2(N-1)+1=2N$.

The sequence converges in $L^1$ to $u=(-1{)}^k$ on each $I_k$ and $u=0$ on $(1\mathrm{/}2,1)$, since $\mathrm{\mid }{u}_N-u\mathrm{\mid }\le 1$ and its support lies in $(0,2^{-(N+1)})$, whose measure tends to zero by [F4]. For each $M$, choose disjoint small intervals around $x_j=2^{-j}$, $1\le j\le M$, avoiding every other dyadic jump, and smooth bumps equal to one at these points, with signs chosen opposite to the jumps of $u$. Their sum is a compactly supported smooth test function ${\varphi }_M$ with $\mathrm{\mid }{\varphi }_M\mathrm{\mid }\le 1$. Its support is bounded away from zero, so only finitely many intervals $I_k$ meet it; integration by parts on those intervals, using [F5] and [F6], gives ${\int }_0^{1}u{\varphi }_M^{\mathrm{\prime }}dx=1+2(M-1)=2M-1.$ Thus the test-field supremum for $u$ is infinite. If $u$ were in $BV((0,1))$, [F2] would make that supremum equal its finite total variation, a contradiction. This proves the refuted assertion fails despite the uniform $L^1$ bound.

For every $\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, extend $\varphi$ by zero to a smooth compactly supported function on ${\mathbb{R}}^n$. It and each derivative are integrable by [F4]. Fubini [F5] reduces ${\int }_{{\mathbb{R}}^n}{\mathrm{\partial }}_i\varphi$ to one-dimensional integrals along coordinate lines; on each line the derivative is continuous with compact support, so [F6] identifies its Lebesgue integral with its Riemann integral and [F7] makes that integral zero. Thus $⟨D_{i}u,\varphi ⟩=-c{\int }_{\mathrm{\Omega }}{\mathrm{\partial }}_i\varphi dx=0$ for every coordinate, so $Du=0$ and $T{V}_N(u;\mathrm{\Omega })=0$ by [F1].

On the disk, the zero extension of $c$ is $c{\chi }_B$. Its distributional derivative is $cD{\chi }_B$, so homogeneity of the partition variation and [F2] give $T{V}_D(c;B)=\mathrm{\mid }c\mathrm{\mid }TV({\chi }_B;{\mathbb{R}}^2)=2\pi \mathrm{\mid }c\mathrm{\mid }$.