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Is every tempered distribution of finite order?

It seems that yes with the definition.

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2 Answers 2

Yes. Every tempered distribution is bounded with respect to one of the seminorms $\|\phi\|_{k,m} = \sum_{|\alpha|\le k} \sum_{|\beta|\le m} \sup_x |x^\alpha D^\beta \phi(x)|$, and therefore is of finite order. In fact we can write any tempered distribution as $D^\beta g$ for some polynomially bounded continuous function $g$ and some multi-index $\beta$.

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ok, thanks for answering. –  Nestor Jan 15 '12 at 19:30

Is there really no typo in the question? Well, of course there are tempered distribution of finite order. Regular distributions are of order zero, the point mass is of order one, its derivative of order two and so on... If you ask for order infinity, the answer is still yes...

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Excuse me, I meant : is every tempered distribution of finite order? –  Nestor Jan 15 '12 at 18:45
    
No, not of order infinity. –  Robert Israel Jan 15 '12 at 19:05
    
Sorry for misreading the question. And of course, Robert is right: Distributions of infinite order are not tempered. –  Dirk Jan 16 '12 at 7:31

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