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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.

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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 $\psi(x)=-\varphi(0)/x$.

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  • $\begingroup$ I've learned since posting this question that the hint is flawed. But thanks for the insight anyways. $\endgroup$ – artificial_moonlet Oct 3 '14 at 1:25

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