Theorem 1

Let $A$ be a set filtered by a filter $\mathfrak{F}$, and $(f_\alpha)_{\alpha \in A}$ a family of vector functions with values in a complete normed space $E$ over $\mathbf{R}$, defined on an interval $I \subset \mathbf{R}$: for each $\alpha \in A$ let $g_\alpha$ be a primitive of $f_\alpha$. We suppose that: 1. with respect to the filter $\mathfrak{F}$ the functions $f_\alpha$ converge uniformly on every compact subset of $I$ to a function $f$; 2. there is a point $a \in I$ such that, with respect to the filter $\mathfrak{F}$, the family $(g_\alpha(a))$ has a limit in $E$. Under these hypotheses the functions $g_\alpha$ converge uniformly (with respect to $\mathfrak{F}$) on every compact subset of $I$ to a primitive $g$ of $f$.

By the remark at the beginning of this subsection we can restrict ourselves to the case where $I$ is a compact interval.

First let us show that the $g_\alpha$ converge uniformly on $I$ to a continuous function $g$. By hypothesis, for every $\varepsilon > 0$ there is a set $M \in \mathfrak{F}$ such that, for any two indices $\alpha, \beta$ belonging to $M$, one has $\|f_\alpha(x) - f_\beta(x)\| \leq \varepsilon$ for each $x \in I$; in consequence one has (I, p. 15, th. 2)

where $l$ denotes the length of $I$; since by hypothesis $g_\alpha(a)$ approaches a limit with respect to $\mathfrak{F}$, it follows from the Cauchy criterion that the $g_\alpha$ converge uniformly on $I$. It remains to show that the limit $g$ of the $g_\alpha$ is a primitive of $f$.

For each integer $n > 0$ let $\alpha_n$ be an index such that $\|f(x) - f_{\alpha_n}(x)\| \leq 1/n$ on $I$; it is clear that the sequence $(f_{\alpha_n})$ converges uniformly to $f$ and that the sequence $(g_{\alpha_n})$ converges uniformly to $g$ on $I$. Let $H_n$ be the countable subset of $I$ where $f_{\alpha_n}$ is not the derivative of $g_{\alpha_n}$, and let $H$ be the union of the $H_n$, which is thus a countable subset of I; we shall see that at every point $x \in I$ not belonging to H the function g has a derivative equal to $f(x)$. Indeed, one sees as above that for every $m \geq n$ and every $y \in I$ one has

Letting $m$ increase indefinitely one also has

for every $y \in I$; now, there exists an $h > 0$ such that, for $|y - x| \leq h$ and $y \in I$, one has $\| g_\alpha(y) - g_{\alpha_n}(x) - f_{\alpha_n}(x)(y - x) \| \leq |y - x| / n$; since, on the other hand, we have $\| f(x) - f_{\alpha_n}(x) \| \leq 1 / n$, we finally obtain

for $y \in I$ and $|y - x| \leq h$, which completes the proof.