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Let $\Omega \subset \mathbb{R}^{n}$ be some open set. Let $f_{n},f\in C^{\infty}(\Omega)$. My question is: What does the following phrase mean? $f_{n}$ converges to $f$ in $C^{\infty}_{loc}(\Omega)$. What is the exact definition of such a convergence.

Does it mean the following? For each compact $K \subset \Omega$ and each integer $m \in \mathbb{N}_{0}$ there exists a subsequence of the sequence $(f_{n})$ which converges to $f$ in $C^{m}(K)$. If no, what is the right definition?

Ben

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    $\begingroup$ There shouldn't be any subsequences in your definition. $\endgroup$ Feb 23, 2016 at 17:42
  • $\begingroup$ ok. How does the right definition go? $\endgroup$
    – Ben
    Feb 23, 2016 at 17:52

1 Answer 1

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I would read it as:

For every compact set $K \subset \Omega$, every $m \ge 0$, and every indices $i_1, i_2, \dots, i_m$, we have $\partial_{i_1} \partial_{i_2} \dots \partial_{i_m} f_n \to \partial_{i_1} \partial_{i_2} \dots \partial_{i_m} f$ uniformly on $K$.

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  • $\begingroup$ Actually what I am trying to prove is the following: Let $f_{n} \in C^{\infty}(\mathbb{R})$ such that there exists a constant $C>0$ with $||f_{n}||_{C^{0}(\mathbb{R})} < C$ and $||f'_{n}||_{C^{0}(\mathbb{R})} < C$. Furthermore, we assume that for each $K \subset \mathbb{R}$ compact and each $k\in \mathbb{N}_{0}$ there exists a constant $C_{K,k}$ suhc that $||f^{(k)}_{n}||_{C^{0}(K)} < C_{K,k}$. Show: There exists a function $g \in C^{\infty}(\mathbb{R})$ such that $f_{n}\rightarrow g$ in $C^{\infty}_{loc}(\mathbb{R})$. $\endgroup$
    – Ben
    Feb 23, 2016 at 19:02
  • $\begingroup$ Is this statement correct? How can one show this? $\endgroup$
    – Ben
    Feb 23, 2016 at 19:03
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    $\begingroup$ I think that's a better question for math.stackexchange.com than here. It's certainly not true as stated (let $f_n = (-1)^n$) but it should be true for a subsequence. You should use Arzela-Ascoli infinitely many times. $\endgroup$ Feb 23, 2016 at 19:23
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    $\begingroup$ It looks more like homework than a research-level question so no, I won't. Sorry. $\endgroup$ Feb 23, 2016 at 19:27
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    $\begingroup$ This is not homework, its just a simplification of the objects I work with! Sorry for the missunderstanding! $\endgroup$
    – Ben
    Feb 23, 2016 at 19:32

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