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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
    Commented Feb 25, 2019 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
    Commented Feb 26, 2019 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
    Commented Feb 27, 2019 at 11:29
  • $\begingroup$ Yes, I have it haha. Also for any strictly positive or strictly negative g_k. $\endgroup$
    – James Baxter
    Commented Feb 27, 2019 at 13:06
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    $\begingroup$ You can look at $s(\varepsilon)=\sum\limits_{k=0}^\infty g_k \varepsilon^k$ as a power series with coefficents in the Banach space of continuous function. As in the scalar case, as soon as the radius of convergence is positive, this function is continuously differentiable on the open disc where it converges. This answers i) and ii) if $s(\varepsilon)'$ means the derivative with respect to $\varepsilon$. Or do you mean the power series with coefficients $g'$ in ii)? $\endgroup$
    – Jochen Wengenroth
    Commented Mar 31, 2020 at 12:12

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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
    Commented Feb 28, 2019 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
    Commented Feb 28, 2019 at 10:18

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