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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?

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  • $\begingroup$ 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? $\endgroup$ – Dieter Kadelka Feb 25 at 21:27
  • $\begingroup$ 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. $\endgroup$ – James Baxter Feb 26 at 1:04
  • $\begingroup$ Do you have this result for constant functions $g_k$? Of course uniform convergence then simply is convergence in $\mathbb{R}$. $\endgroup$ – Dieter Kadelka Feb 27 at 11:29
  • $\begingroup$ Yes, I have it haha. Also for any strictly positive or strictly negative g_k. $\endgroup$ – James Baxter Feb 27 at 13:06
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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 $s(0)$.

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  • $\begingroup$ Sorry, can you elaborate a bit on how (i) follows? $\endgroup$ – James Baxter Feb 28 at 1:25
  • $\begingroup$ @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. $\endgroup$ – Dieter Kadelka Feb 28 at 10:18

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