$(f\delta')(\phi)=-(f\phi)'(0)$? Or $-(f\phi')(0)$?

Question

This might be a stupid question but I can't find the answer elsewhere. Put $\delta(\phi)=\phi(0)$ for $\phi\in\mathcal{D}(\mathbb{R})$, the test function space on $\mathbb{R}$, and $f\in C^\infty(\mathbb{R})$. On one hand, since $\delta'(\phi)=-\delta(\phi')$, $(f\delta')(\phi)=-(f\delta)(\phi')=-\delta(f\phi')=-(f\phi')(0)$. On the other hand, however, $(f\delta')(\phi)=\delta'(f\phi)=-\delta((f\phi)')=-(f\phi)'(0)$. The two results clearly are not equal with general $f$ and $\phi$. Which one is correct? And why is the other not?

Answer 1

No, $(f\delta')(\phi)=-(f\delta)(\phi')$ isn't valid. The outermost operation operation if multiplication by $f$ so that must be taken care of first.

Had you had $(f\delta)'(\phi)$ then the derivative had been the outermost operation and you could rewrite the expression as $-(f\delta)(\phi').$

Answer 2

the step $f\delta'(\phi) = -f\delta(\phi')$ of the first answer is wrong, the correct answer is the second one.

Answer 3

The definition is $$ \langle fT,\phi\rangle=\langle T,f\phi\rangle. $$ Therefore, $$ \langle f\delta',\phi\rangle=\langle \delta',f\phi\rangle=-\langle\delta,(f\phi)'\rangle=-(f\phi)'(0). $$ Or if you prefer $$ f\delta'=f(0)\delta'-f'(0)\delta. $$