Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 326 / 3

For $u\in L^1(\Omega )$, the total variation seminorm on a domain isUsing $C_c^1$ instead gives the same definition. The zero test field shows nonnegativity, and $\mathrm{TV}(0)=0$, so this is a proper extended-real function. Each test field defines a linear functional of $u$; taking their supremum proves convexity, equivalentlyIt is not strictly convex: two distinct constant functions both have zero total variation, as does every convex combination of them.

The function of bounded variation on a domain space isTotal variation is not coercive on $BV(\Omega )$: for $u_k\equiv k$ on a domain of positive measure, $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$, whereas $\mathrm{TV}(u_k)=0$. This is noncoercivity of total variation on constants; a mean-zero constraint can remove the constant obstruction.

An absolutely one-homogeneous functional satisfieswith $J(0)=0$. The printed strictly positive codomain cannot hold at zero; the standard interpretation is a nonnegative, possibly extended-real functional.

For $p\in \partial J(u)$, apply the subgradient inequality at $v=0$:so $⟨p,u⟩\ge J(u)$. Apply it at $v=2u$ and use one-homogeneity:which gives the reverse inequality. ThereforeThis is the Euler identity for a convex one-homogeneous functional; differentiability is unnecessary.

Use the interior-domain definition of total variation seminorm on a domain, so there is no boundary-extension term at $\pm 1$. For any admissible scalar test function $\phi$,Thus $\mathrm{TV}(u_R)\le 2$.

Choose $0<\epsilon <\min \{R,1-R\}$. From the supplied smooth bump function, defineThe two supports are disjoint and inside $(-1,1)$. Each bump has maximum $e^{-1}$, so $\Vert \phi {\Vert }_{\infty }\le 1$, while $\phi (R)=1$ and $\phi (-R)=-1$. This attains the upper bound, provingEquivalently, its distributional derivative is $D{u}_R={\delta }_{-R}-{\delta }_R$, with total variation two. This is total variation of an interval indicator.