= Solution
Use the interior-domain definition of <total variation seminorm on a domain>, so there is no boundary-extension term at $\pm1$. For any admissible scalar <test function> $\varphi$,
$$
\int_{-1}^1u_R(x)\varphi'(x)\,dx
=\int_{-R}^R\varphi'(x)\,dx
=\varphi(R)-\varphi(-R)\leq2.
$$
Thus $\operatorname{TV}(u_R)\leq2$.
Choose $0<\varepsilon<\min\{R,1-R\}$. From the supplied <smooth bump function>, define
$$
\varphi(x)=e\,\omega_\varepsilon(x-R)-e\,\omega_\varepsilon(x+R).
$$
The two supports are disjoint and inside $(-1,1)$. Each bump has maximum $e^{-1}$, so $\|\varphi\|_\infty\leq1$, while $\varphi(R)=1$ and $\varphi(-R)=-1$. This attains the upper bound, proving
$$
\boxed{\operatorname{TV}(u_R)=2.}
$$
Equivalently, its <distributional derivative> is $Du_R=\delta_{-R}-\delta_R$, with total variation two. This is <total variation of an interval indicator>.
Back to article page
