# Shifting a smooth function of compact support

Let $\varphi : \mathbb{R} \rightarrow \mathbb{R}$ be an infinitely differentiable function of compact support. Define $$\psi (x) := \begin{cases} \frac{\varphi (x) - \varphi (0)}{x}, & \text{x \neq 0} \\ \varphi'(0), & \text{x = 0} \end{cases}.$$

Is $\psi$ be infinitely differentiable and compactly supported?

Some context: $\psi$ is given as a hint to prove that for a distribution $f$ with the property $xf = 0$ (in the sense of distributions), $f = C \delta_0$, for some constant $C$. If $\psi$ has the appropriate properties, then the proof for the main question follows quickly. After some tinkering, however, I am not convinced that $\psi$ has either necessary property.

Yes, $\psi$ is $C^\infty$-smooth because Quotient of two smooth functions is smooth.
If $\varphi(0)=0$, then $\psi$ has compact support, because in this case $\psi(x)=0$ whenever $\varphi(x)=0$.
But if $\varphi(0)\ne 0$, then $\psi$ does not have compact support. Indeed, for large enough $x$ we have $\varphi(x)=0$ and therefore $\psi(x)=-\varphi(0)/x$.