1
$\begingroup$

For every $\epsilon \in [0, 1)$ and $n$ in $\mathbb N$, let $g(n, \epsilon)$ be a function $\mathbb R \to \mathbb R$.

Suppose for all $\epsilon \in [0, 1)$, $g(n, \epsilon)$ is $C^n$ smooth and that for each $n$, $g(n, \epsilon)$ converges uniformly in $C^n$ norm to $0$ as $e \to 1-$. Suppose further that for every $\epsilon$, $\lim_{N \to \infty}$ $ \sum_{n = 1}^N g(n, \epsilon) \epsilon^n$ converges uniformly.

If $F := \lim_{\epsilon \to 1-} \sum_{n = 1}^\infty g(n, \epsilon) \epsilon^n$ exists in the uniform sense, is $F$ necessarily $C^\infty$ smooth?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ In the middle paragraph, shouldn't it be $\lim_{N\to\infty}$? Also, in the last paragraph, shouldn't the sum be up to infinity? $\endgroup$
    – Wojowu
    Commented Mar 17, 2019 at 14:55
  • 1
    $\begingroup$ Unless I've misunderstood, this has easy counterexamples. Take a sequence of smooth functions $f_n$ converging uniformly to a non-smooth function. Set $g(n,\epsilon)=\epsilon^{-n}f_n$ if $\lceil 1/(1-\epsilon)\rceil=n,$ and set $g(n,\epsilon)=0$ otherwise. This gives $\sum_{n=1}^\infty g(n,\epsilon)=f_{\lceil 1/(1-\epsilon)\rceil}.$ $\endgroup$
    – Dap
    Commented Mar 17, 2019 at 15:08
  • $\begingroup$ Ah, you’re right. $\endgroup$
    – James Baxter
    Commented Mar 17, 2019 at 15:48

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .