a problem concerning continuous functions of bounded variation

Question

Here is a problem:

Suppose $f,g: [a,b]\rightarrow \mathbb{R}$ are both continuous and of bounded variation. Show that the set $\{(f(t),g(t))\in\mathbb{R}^2: t\in [a,b]\}$ CANNOT cover the entire unit square $[0,1]\times [0,1]$.

Thanks for all helps!

Answer 1

Say $\gamma(t)=(f(t),y(t))$, so $\gamma$ is an $\Bbb R^2$-valued function of bounded variation.

Divide $[0,1]^2$ into $n^2$ squares $Q_j$ of side length $1/n$, and let $z_j$ be the center of $Q_j$. Choose $t_j$ with $\gamma(t_j)=z_j$.

Now let $s_1<s_2,\dots<s_{n^2}$ be the sequence $(t_j)$ arranged in increasing order. Then $$\sum_{j=1}^{n^2-1}|\gamma(s_j)-\gamma(s_{j+1})|\ge\sum_{j=1}^{n^2-1}\frac1n\sim n,$$so $\gamma$ cannot have bounded variation. (Note: As @gerw points out we never used continuity; a space-filling "curve" cannot have bounded variation, continuous or not.)