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

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.

• What's your definition of $||f||_{BV}$? – David C. Ullrich Dec 15 '15 at 22:12
• The definition I have is: $\| f \|_{BV} = | f(a) | + V_{a}^{b} f$. In a previous post I showed that $\| f \|_{BV}$ defines a norm in the space $BV[a, b]$. – Von Kar Dec 15 '15 at 22:17

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.
• I understand the first part of the demonstration where you made use of the triangle inequality to prove that $\sup_{x \in [a,b]} | f(x) | \le || f ||_{BV}$ but I don't understand why this inequality implies uniform convergence. Can you help me with this? – Von Kar Dec 16 '15 at 1:23
• Hmm. A true fact that you should be able to prove, that should explain this: $f_n\to f$ uniformly if and only if $\sup_{x\in[a,b]}|f_n(x)-f(x)|\to0$. – David C. Ullrich Dec 16 '15 at 1:51