Theorem 2

Let $f$ be a vector function defined and continuous on a closed bounded interval $I = [a,\ b]$ of $\mathbf{R}$ (where $a < b$) and taking values in a normed space E over $\mathbf{R}$; let $g$ be a continuous increasing real function on I. Suppose that $f$ and $g$ have right derivatives at all points of the relative complement in $[a,\ b[$ of a countable subset A of this interval (allowing $g'_r(x)$ to be infinite at some of the points $x \notin A$), and suppose that at each of these points we have

Under these hypotheses one has

The proof proceeds similarly to that of prop. 2. Let $\varepsilon > 0$ be arbitrary, and $(a_n)$ the sequence obtained by enumerating A in some order. Let J be the set of points $y \in I$ such that, for all $x$ such that $a \leq x \leq y$ one has

we shall show that $J = I$. One sees immediately, as in prop. 2, that J is an interval with left-hand endpoint $a$; if c is its right-hand endpoint then $c \in J$; indeed, for all $x < c$ one has (8), and a fortiori

from which, letting $x$ tend to $c$ in this inequality, it follows from the continuity of $f$ that $c$ satisfies (8).

Let us show that we must have $c = b$. So suppose that $c < b$ and that moreover $c \notin A$: then $f'_d(c)$ and $g'_r(c)$ exist and satisfy (6); suppose in the first place that $g'_r(c)$ (which is necessarily $\geq 0$) is finite; then one can always write $f'_d(c) = u g'_r(c)$, with $\| u \| \leq 1$; since the function $f(x) - u g(x)$ has zero right derivative at the point $c$ there must exist a $y$ such that $c < y \leq b$ and such that for $c \leq x \leq y$ one has

from which

and, taking account of (8), in which $x$ is replaced by $c$,

Thus one has $y \in J$, which is a contradiction. Suppose next that $c \notin A$ and that $g'_r(c) = +\infty$; then there is a $y$ such that $c < y \leq b$ and such that for $c \leq x \leq y$ one has on the one hand

while on the other hand

from which

and one concludes as above. Finally, if one has $c = a_k$, then there is a $y$ such that $c < y \leq b$, and such that for $c < x \leq y$ one has

from which, taking account of (8), with $x$ replaced by $c$,

which again entails a contradiction. The proof finishes as that of prop. 2.

Q.E.D.