Proposition 6

Supposing that the hypotheses of prop. 5 are satisfied, let $a(\alpha)$, $b(\alpha)$ be two functions defined on $V$, with values in $I$; if the derivatives $a'(\alpha_0),\ b'(\alpha_0)$ exist and are finite then the function $g(\alpha) = \int_{a(\alpha)}^{b(\alpha)} f(t, \alpha)\, dt$ admits at $\alpha_0$ a derivative given by the formula

Indeed, for all $\alpha \in V$ distinct from $\alpha_0$ one can write

By prop. 5 of II, p. 74, the first integral on the right-hand side tends to $\int_{a(\alpha_0)}^{b(\alpha_0)} f'_\alpha(t, \alpha_0)\, dt$ as $\alpha$ tends to $\alpha_0$. In the second integral we replace $f(t, \alpha)$ by $f(b(\alpha_0), \alpha_0)$ and show that the difference tends to 0. We put $M = \max (\|f(b(\alpha_0), \alpha_0)\|, |b'(\alpha_0)| + 1)$; the function $b(\alpha)$ being continuous at the point $\alpha_0$ and the function $f$ continuous at the point $(b(\alpha_0), \alpha_0)$, for every $\varepsilon$ such that $0 < \varepsilon < 1$ there exists an $r > 0$ such that the relation $|\alpha - \alpha_0| \leq r$ entails that $\|f(t, \alpha) - f(b(\alpha_0), \alpha_0)\| \leq \varepsilon$ for all $t$ belonging to the interval with endpoints $b(\alpha_0)$ and $b(\alpha)$; thus one may also suppose that the relation $|\alpha - \alpha_0| \leq r$ entails $\left| \frac{b(\alpha) - b(\alpha_0)}{\alpha - \alpha_0} - b'(\alpha_0) \right| \leq \varepsilon$.

By the mean value formula (II, p 62, formula (17)) one thus has

and consequently

which shows that $\frac{1}{\alpha - \alpha_0} \int_{b(\alpha_0)}^{b(\alpha)} f(t, \alpha)\, dt$ tends to $b'(\alpha_0) f(b(\alpha_0), \alpha_0)$. In the same way one shows that $\frac{1}{\alpha - \alpha_0} \int_{a(\alpha_0)}^{a(\alpha)} f(t, \alpha)\, dt$ tends to $a'(\alpha_0) f(a(\alpha_0), \alpha_0)$.