Suppose $ g_i: [0, 1] \to \Bbb R$, $i\in\Bbb N$, are $C^1$ functions and that there is some $c > 0$ such that for every $0 < \epsilon < c$, the functions $$ s(\epsilon)_i := \sum_{k=0}^i {\epsilon}^k g_k $$ converge uniformly to a $C^1$ function $s(\epsilon)$.
As $\epsilon \to 0$ does
i) $s(\epsilon) \to s(0)$ uniformly?
ii) $s(\epsilon)’ \to s(0)’$ uniformly?
Proof of i): First let $(B,\|.\|) := (C^0,\|.\|_\infty)$. For the following reasoning the Banach space $(B,\|.\|)$ may be arbitrary. Let $(g_k)_{k \in N_0}$ be any sequence in $B$, such that for some $\epsilon > 0$ and $s(\epsilon)_n := \sum_{k=0}^n \epsilon^k g_k$ $s(\epsilon) := \lim_{n \to \infty} s(\epsilon)_n$ exists. Put $u_k := s(\epsilon)_k - s(\epsilon)_{k-1} = \epsilon^k g_k$. Since $s(\epsilon)_k$ is converging, necessarily $\lim_{k \to \infty} u_k = 0$, in particular $U := \sup_{k \in N_0} \|u_k\| < \infty$. Now for $0 < \delta < \epsilon$ $s(\delta)_n = \sum_{k=0}^n \left(\frac{\delta}{\epsilon}\right)^k u_k$, $\|s(\delta)_m - s(\delta)_n\| \leq \frac{(\delta/\epsilon)^{m+1}}{1-\delta/\epsilon}U$ for $0 \leq m < n$, thus $s(\delta) = \lim_{n \to \infty} s(\delta)_n$ exists with $\|s(\delta)_0 - s(\delta)\| \leq \frac{\delta/\epsilon}{1-\delta/\epsilon}U$. But $s(\delta)_0$ is nothing else but $s(0)$.
