I am looking for a proof and reference of the following statement.
Let $\mathcal{F} = \{f: [a,b] \to \mathbb{R} : f(a) = 0 \;\text{ and } \;V_{a}^{b}(f) \leq C\}$. The set $\mathcal{F}$ is compact.
Some necessary details.
The quantity$$V_{a}^{b}(f) = \sup_{\mathcal{P} \subset [a,b]} \sum_{i=1}^{n_{\mathcal{P}}-1} |f(x_{i})-f(x_{i-1})|$$ is the total variation of a function $f$.
The compactness is with respect to the typical topology in the space of functions of bounded variation. That is, with norm $$\|f\|_{BV} = \|f\|_{L_1} + V_{a}^{b}(f).$$
Here's a quick attempt. I've shown that $V_{a}^{b}(f) \leq C \Rightarrow \|f\|_{\infty} \leq C$, and I can then use the completeness of $L_{\infty}([a,b])$. The problem with this approach is that I can't recover that a limit of a sequence in $\mathcal{F}$ is of variation less than or equal to $C$.
UPDATE
After a night of thinking on it I think this result is false.
Let $q_{n}$ be an enumeration of $(\mathbb{Q}\setminus\{0\}) \cap [a,b]$, which gives a sequence of all the rationals in $[a,b]$ excluding zero. Define the sequence of functions $f_n: [a,b] \to \mathbb{R}$ by $$f_{n}(x) = \begin{cases} 1 & :x = q_n \\ 0 & :\text{ otherwise}. \end{cases}$$ We have $f_{n} \in \mathcal{F}$ with $C = 1$. But this sequence can have no cauchy subsequence because
$$V_{a}^{b}(f_{n}-f_{m}) = \sup_{\mathcal{P}} \sum_{i=1}^{n_{P}-1} |f_{n}(x_{i})-f_{m}(x_{i}) - (f_{n}(x_{i-1}) - f_{m}(x_{i-1}))| = 2$$ when $n \neq m$, which follows from the density of the irrationals. Therefore this sequence does not have a convergent subsequence, which shows that $\mathcal{F}$ is not compact.
However, I did find the following statement: The set $\mathcal{F}$ is compact in the $L_1$ topology. This follows from the compactness theorem here, which, sadly, is offered without proof. Any references are still appreciated.