I am currently looking at these online notes on PDEs, page 59.
How does it follow that if $f^R = \phi(x/R) f(x)$
$ \phi(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 1 & \text{if $|x|\leq 1,$}\\ 0 & \text{if $|x|\geq 2.$}\\ \end{array}\right. $
then
$\partial^\alpha f^R=\phi^R\partial^\alpha f+\frac{1}{R}h^R$
How have we worked out the derivative here? Is it a weak derivative? The function is not even defined in between $|x| \in (1,2)$. We the Leibnitz rule has been employed, but how is $h^R$ known to be bounded?
As far as I can see we have $\partial^\alpha f^R = \phi^R \partial^\alpha f+ \frac{1}{R}(\partial^\alpha \phi )(x/R) $