For $u\in L^1(\Omega)$, the <total variation seminorm on a domain> is
$$
\operatorname{TV}(u)=\sup_{\substack{\varphi\in C_c^\infty(\Omega;\mathbb R^n)\\\|\varphi\|_\infty\leq1}}
\int_\Omega u\,\operatorname{div}\varphi\,dx.
$$
Using $C_c^1$ instead gives the same definition. The zero test field shows nonnegativity, and $\operatorname{TV}(0)=0$, so this is a <proper extended-real function>. Each test field defines a linear functional of $u$; taking their supremum proves <convexity>, equivalently
$$
\operatorname{TV}(tu+(1-t)v)\leq t\operatorname{TV}(u)+(1-t)\operatorname{TV}(v),\qquad0\leq t\leq1.
$$
It is \b[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 is
$$
BV(\Omega)=\{u\in L^1(\Omega):\operatorname{TV}(u)<\infty\},\qquad
\|u\|_{BV}=\|u\|_{L^1}+\operatorname{TV}(u).
$$
Total variation is \b[not coercive on $BV(\Omega)$]: for $u_k\equiv k$ on a domain of positive measure, $\|u_k\|_{BV}=k|\Omega|\to\infty$, whereas $\operatorname{TV}(u_k)=0$. This is <noncoercivity of total variation on constants>; a mean-zero constraint can remove the constant obstruction.

Solution

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

On a bounded domain of positive measure, the constant functions $u_k\equiv k$ satisfy $\operatorname{TV}(u_k)=0$ while $\|u_k\|_{BV}=k|\Omega|\to\infty$. Thus the <total variation seminorm on a domain> is neither coercive in the $BV$ norm nor strictly convex. Additional control of the mean can remove this obstruction; the <Poincaré inequality for total variation> supplies the needed bound on suitable connected domains.

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 Incoming links: Noncoercivity of total variation on constants