Exercise 5, § 2

Let $\mathbf{f}$ be the vector mapping of $I = [0,\ 1]$ into $\mathbf{R}^3$ defined as follows: for $0 \leq t \leq \frac{1}{4}$, $\mathbf{f}(t) = (-4t,\ 0,\ 0)$; for $\frac{1}{4} \leq t \leq \frac{1}{2}$ let $\mathbf{f}(t) = (-1,\ 4t - 1,\ 0)$; for $\frac{1}{2} \leq t \leq \frac{3}{4}$ let $\mathbf{f}(t) = (-1,\ 1,\ 4t - 2)$; finally, for $\frac{3}{4} \leq t \leq 1$ take $\mathbf{f}(t) = (4t - 4,\ 1,\ 1)$: Show that the convex set generated by the set $\mathbf{f}'_d(I)$ is not identical to the closure of the set of values of $\frac{\mathbf{f}(y) - \mathbf{f}(x)}{y - x}$ as $(x,\ y)$ runs through the set of pairs of distinct points of $I$ (cf. exerc. 4 a)).