If $\| f_n - f \|_{BV} \rightarrow 0$, then $f_n$ converges uniformly to $f$ on $[a, b]$

Question

I'm learning about functions of bounded variation and need help with this theoretical problem:

Let $f_n : [a, b] \rightarrow \mathbb{R}$ a sequence of functions in $BV[a, b]$. Show that if $\| f_n - f \|_{BV} \rightarrow 0$, then $f_n$ converges uniformly to $f$ on $[a, b]$.

My thoughts:

Since $\| f_n - f \|_{BV} \rightarrow 0$ this means that the sequence of functions $f_n$ satisfies the Cauchy's criterion (am I right?). So for every $\varepsilon > 0$ there exists an $N$ so that $m, n > N$ implies that

$$\| f_n(x) - f_m(x) \|_{BV} < \varepsilon.$$

This is as far as I got.

Answer 1

If $||f||_{BV}=|f(a)|+V_a^bf$ then the definition of $V_a^b$ shows that $$|f(x)|\le|f(a)|+|f(x)-f(a)|\le||f||_{BV}\quad(x\in[a,b]).$$

Hence $$\sup_{x\in[a,b]}|f(x)|\le||f||_{BV},$$which says precisely that convergence in $BV$ implies uniform convergence.