# Convergence of a certain sum

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?

• Is the formulation of the problem correct? The answer seems trivial yes since $|g_k|$ are bounded and $\epsilon^k \to 0$ for $\epsilon \to 0$. Am I missing something? – Dieter Kadelka Feb 25 at 21:27
• The g_k aren’t bounded uniformly is the problem I think. Also a bunch of other stuff like the rate of convergence not being uniform in i, and the g_k being positive/negative. – James Baxter Feb 26 at 1:04
• Do you have this result for constant functions $g_k$? Of course uniform convergence then simply is convergence in $\mathbb{R}$. – Dieter Kadelka Feb 27 at 11:29
• Yes, I have it haha. Also for any strictly positive or strictly negative g_k. – James Baxter Feb 27 at 13:06

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 $$s(0)$$.
• @James Baxter: I've made the proof more explicit. ii) seems to be much more interesting, since $C^1$ is a dense subspace of $C^0$ and we still have convergence in $C^0$, It would be nice if you could present a proof of ii) now. – Dieter Kadelka Feb 28 at 10:18