Is $AC[a,b]$ closed in $(BV[a,b],TV)$?

Question

Consider $BV[a,b]$ the space of all bounded variation functions on a real interval $[a,b]$, endowed with the total variation norm $TV$. $AC[a,b]$, the space of absolutely continuous functions, is a subspace of $BV$. Is it closed?

Answer 1

The quantity $\|f\| = |f(a)| + \|f\|_{TV}$ is a natural norm on $BV[a,b].$ If $f\in AC,$ then this norm equals $|f(a)| + \int_a^b|f'|.$ Suppose $f_n$ is a sequence in $AC$ that is Cauchy in this norm. Then $f_n(a)$ is a Cauchy sequence in $\mathbb R,$ and $f_n'$ is Cauchy in $L^1[a,b].$ Thus $f_n(a) \to c$ in $ \mathbb R$ and $f_n'\to g$ in $L^1[a,b].$ We find then that $f_n(x) \to c + \int_a^x g$ in $BV.$ The last function is in $AC,$ proving $AC$ is closed in $BV.$