# Condition for differentiablility

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 $$g(x) = \psi(x)f(x)$$ 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 !

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.
• What do you mean by finite $f$ and $g$? – Davide Giraudo Feb 1 '12 at 12:29
• 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 $\mathbb R_-$. – Davide Giraudo Feb 1 '12 at 13:04