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in another thread an offtopic question came to my mind. Consider a test/bump function $\phi$, then consider all its derivatives $\phi^{(k)}$, of course they are bounded by constants $M_k$, next form the set $M_0, M_1, M_2,\ldots$ off all those bounds. As was pointed out in another thread this set itself may not be bounded, so heres my question. Is this enough to conclude that the series $$ \sum_{k=1}^{\infty} a_k \phi^{(k)}(0) \qquad a_k \in \mathbb{R} $$ is not in $\mathcal{D}'$ for $\phi \in \mathcal{D}$?

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Could you please say what the $a_k$'s are? – Mercy King Jun 20 '12 at 23:05
they are just arbitrary real numbers, i added it to the equation. – Stefan Jun 20 '12 at 23:17
up vote 1 down vote accepted

Unless your series converges you cannot say that it belongs to $\mathcal{D}'$. Suppose for instance that only finitely many $\phi^{(k)}(0)$'s are $0$. Then consider the sequence $a_k$ defined by $a_k=1/\phi^{(k)}(0)$ if $\phi^{(k)}(0) \ne 0$ and $a_k=0$ otherwise. We see that the series $\sum_{k=1}^\infty a_k\phi^{(k)}(0)$ diverges!

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The series is not properly defined - do you mean $$\sum^{\infty}_{k=1}a_{k}M_{k}$$ instead? If you want to reason with $\mathcal{D}'$, then I guess you mean distributions. Yes, select $\phi=\sin[x]$, then $|\phi^{k}(x)|\le 1$ and we have have $\sum 1=\infty$. I guess you are suggesting that since $$M_{k}$$ is not bounded as $k\rightarrow \infty$, $a_{k}M_{k}$ could well blow up. This make sense since even for $\frac{1}{x}$ on an interval $[\epsilon,1]$, its derivatives's absolute value can be arbitrarily large.

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no, i mean $\sum_{k=1}^{\infty} a_k \phi^{(k)}(0)$, i forgot to write $(0)$ after $\phi^{(k)}$, i changed it now! – Stefan Jun 20 '12 at 23:18
Well, $0$ is just an arbitrarily point, so I guess my example still carries through. – Bombyx mori Jun 20 '12 at 23:19
yes, my intuitiv reasoning goes like this, with your example $\phi = \sin(x)$, the value one at zero is taken infinitely often by the derivates, so the series should diverge, am i right? – Stefan Jun 20 '12 at 23:21
yes, you convinced me :) – Stefan Jun 20 '12 at 23:22
oh, just to point it out, your second example $\frac{1}{x}$ is not a distribution! – Stefan Jun 20 '12 at 23:23

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