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Suppose I have two functions that are Schwartz class, say $f,g \in S(\mathbb{R})$, and suppose I have another function $\psi(x)$ such that \begin{equation} g(x) = \psi(x)f(x) \end{equation} I would like to find a way to understand why this means that $\psi$ must be differentiable but I struggle to find a start as I am very new to Schwartz class functions.

I thought maybe smoothness is enough to begin with. In any case, away from points where $f(x) \neq 0$ I have no problems, but somehow I need to find a way to say something about the differentiability fo $\psi$ in general, provided $f,g$ are Schwartz (and not both identically zero).

Any help would be great !

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Even if $f$ and $g$ are not identically $0$, $\psi$ doesn't need to be differentiable. Indeed, take $f(x)=g(x)=$ a test function, whose support is contained in $[0,1]$ for example. Theses functions are in the Schwartz class, and the equality $g=\psi f$ is satisfied for each $\psi$ such that $\psi(x)=1$ for $x\in [0,1]$. But of course, out of this interval, $\psi$ is allowed to be anything, in particular not even continuous.

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  • $\begingroup$ This is true for finite f and g. But what about other non-finite functions from Schwartz class. ψ has to be differentiable everywhere on R in this case. $\endgroup$ – superM Feb 1 '12 at 12:26
  • $\begingroup$ What do you mean by finite $f$ and $g$? $\endgroup$ – Davide Giraudo Feb 1 '12 at 12:29
  • $\begingroup$ f and g have compact supports. $\endgroup$ – superM Feb 1 '12 at 12:40
  • $\begingroup$ We can take a function $\chi$ which is equal to $0$ on $\mathbb R_-$, $1$ on $[1,+\infty[$ and $0\leq \chi \leq 1$. Then put for example $f(x)=g(x)=e^{-x^2}\chi(x)$. $\psi$ has to be equal to $1$ on $\mathbb R_+$, but can be anything on $\mathbb R_-$. $\endgroup$ – Davide Giraudo Feb 1 '12 at 13:04
  • $\begingroup$ what if f and g aren't equal to 0 anywhere? $\endgroup$ – superM Feb 1 '12 at 13:11

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