 Solution

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.