How to show that the followig function is distribution in $\mathfrak{D} =C_0^{\infty}(\Bbb R)$
$$\langle f,\varphi\rangle = \sum_{k=1}^{\infty}{\varphi^{(k)}(0)}$$
Distributional theory
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distribution-theory
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2I don't think it is; for example, if you have a function which is equal to $\exp$ on $(-1,1)$ and zero outside $(-2,2)$, it is in $C_0^\infty$ but the sum is $\infty$. For any particular finite truncation you have a distribution, of course... – 2017-02-05
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0Ian is right. You can have a stronger result - take any numeric sequence $\{a_n\}_{n\in\Bbb N}$. Then there exists a function $\phi\in C^\infty_0(\Bbb R)$ such that $\forall k\ge 0 \phi^{(k)}(0)=a_k$. Therefore, your series can diverge as fast as you want. – 2017-02-06
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0As TZakrevskiy said $T (\varphi) = \sum_{k=0}^\infty a_k \varphi^{(k)}(0)$ isn't continuous for the topology of $D(\mathbb{R})$ whenever $a_k$ doesn't end being $0$, see https://en.wikipedia.org/wiki/Non-analytic_smooth_function#Application_to_Taylor_series – 2017-02-06