Does distribution derivative generated by $C^{\infty}$ function forces the distribution to be compactly supported?

Question

Let $\phi \in C_C^{\infty}(\Bbb R)$. Then there exists a $\psi \in C_C^{\infty}(\Bbb R)$ such that $\psi' = \phi \iff \int_\Bbb R \phi = 0$. This is quite easy to be proved. From this it follows that if distribution derivative is zero, then the distribution generated by a constant function. Does it also follow that if distribution derivative is generated by $C^{\infty}$,then the distribution is compactly supported.

Answer 1

A distribution derivative can be $C^\infty$ without being compactly supported. It can be, for example, $1$. Remember, a function corresponds to a distribution if the function is locally integrable (it has an integral $<\infty$ on compact sets). Look at Rudin Functional Analysis 6.11-6.14.

Caveat: I'm not sure what you mean by "if distribution derivative is generated by $C^\infty$." I took you to mean that it corresponds to a function in $C^\infty$. Maybe you mean $C_c^\infty$? In that case, the answer is yes; look at Rudin 6.22, and consider that the integral of the product of two $C_c^\infty$ functions with nonoverlapping supports is zero.