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I encountered this problem while solving a problem related to characterization of delta function upto constant multiple.

$\phi \in D(R)$, Space of compactly supported infinitely differentiable functions. If $\phi(0)=0$, Then there exist a $\psi \in D(R)$ such that $\phi = x\psi$

Original question is as follow:

If $xT=0$, where T is a distribution. Then, $T = c\delta$

If above proposition could be proved, then this characterization would follow.

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You are looking for Hadamard's lemma. http://en.wikipedia.org/wiki/Hadamard%27s_lemma

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  • $\begingroup$ What a manipulation! :D $\endgroup$ – Rahul Gupta Apr 27 '11 at 15:02
  • $\begingroup$ I know this question is very old, but I've run into the same problem recently. I see how Hadamard's Lemma gives you an infinitely differentiable $\psi$, but why is $\psi$ also compactly supported? $\endgroup$ – artificial_moonlet Oct 3 '14 at 1:45
  • $\begingroup$ if $\phi =x\psi$ then one of the two functions has compact support iff the other one has... $\endgroup$ – Dan Petersen Oct 3 '14 at 5:33
  • $\begingroup$ Ah, I misread the question prompt. I didn't realize we were also assuming $\phi(0) = 0$. $\endgroup$ – artificial_moonlet Oct 3 '14 at 12:28
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For the original question, you could use the fact that the support of $T$ can only be at the origin. Then a theorem (I forgot the name, it is in the distribution book of Kolk and Duistermaat) states that we can write

$T=\sum_{i=0}^{n}c_i\delta^{(i)}$

for some $n$. It easily follows that $c_i=0$ for $i>0$.

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