For a fixed $u \in BV(\mathbb{R}^N)$, consider the function 
$h:(0,+\infty) \to BV(\mathbb{R}^N)$, given by


$h(t) = u (tx)$.


Is $h$ continuous?



About the continuity of a function on BV
Not in the strong topology: for instance take $u:=1\_{(-1,1)}$, i.e. the characteristic function of the interval $(-1,1)$ on $\mathbb R$ (in this example $N=1$). Then $\|h(t)-h(s)\|\_{BV}\ge 2$ whenever $s\neq t$, since the distributional derivative of the element $h(s)\in BV(\mathbb R)$ is $\delta\_{-1/s}-\delta\_{1/s}$.


**However**, continuity holds in an intermediate topology, namely the one given by the distance
$$d(u,v):=\|u-v\|\_{L^1}+\big|\|Du\|(\mathbb R^N)-\|Dv\|(\mathbb R^N)\big|,$$
which is good enough for many applications. The proof of continuity can be done by approximating your function $u$ with smooth, compactly supported functions (which can be done in this topology but not in the strong one).


