Is the space of continuous functions with bounded variation separable?

Question

Let $$BVC([0,1]) = \{f:\mathbb [0,1] \to \mathbb R,\, f\in BV([0,1])\cap C([0,1]),\, f(0)=0\},$$ and $$\|f\|_{BVC}=\mathrm{Var}(f).$$

Is $BVC([0,1])$ separable?

Answer 1

We will build an uncountable discrete set of functions belonging to $BVC([0,1])$.

Let $\alpha=\alpha _1\alpha _2...$ is infinite binary sequence. We will build the analogue of Cantor function for every such sequence.

Let us consider the next construct (or see informal description below).

Let $F_\alpha (0)=0, F_\alpha (1)=1$;

Step 1. For $x\in [\frac13;\frac23]\quad $ let $\quad F_\alpha(x)=F_\alpha(0)+\dfrac{1+2\alpha_1}{4}(F_\alpha(1)-F_\alpha(0))$;

Step 2.1. For $x\in [\frac19;\frac29]\quad $ let $\quad F_\alpha(x)=F_\alpha(0)+\dfrac{1+2\alpha_2}{4}(F_\alpha(\frac13)-F_\alpha(0))$;

Step 2.2. For $x\in [\frac79;\frac89]\quad $ let $\quad F_\alpha(x)=F_\alpha(\frac23)+\dfrac{1+2\alpha_2}{4}(F_\alpha(1)-F_\alpha(\frac23))$;

and so on.

If we replace all the coefficients $\dfrac{1+2\alpha_i}4$ by $\dfrac12$, we obtain the Cantor function. In other words, in the classic construction, the vertical intervals is always divided in the ratio 1:1, but here we divide it in the ratio 1:3 or 3:1 (depending on the $\alpha _i$).

In the same way we can close this construction if we extend the function by continuity. Monotonicity is obvious. So $F_\alpha \in BVC[0,1]$ for any binary sequence $\alpha $.

It remains to prove only that $||F_\alpha-F_\beta||_{BVC}\ge 1$ for $\alpha \ne \beta$.

Let $\alpha_i = \beta _i, i=1,...,n-1$, and $\alpha_n \ne \beta _n$. Then $F_\alpha=F_\beta$ for Step 1,...,$n-1$. Let $\{p_k, k=0,...,2^n\}$ be partition that corresponds to Step $n$ (with $p_0=0, p_{2^n}=1$).

Then $\mathrm{Var}(F_\alpha-F_\beta)\ge 2\sum\limits_k (|F_\alpha(p_k)-F_\beta(p_k)|)=2\cdot (\frac34-\frac14)(F_\alpha(1)-F_\beta(0))=1$.