Is this proposition correct? Will you please give a contour example if it is wrong?
If $J \in L^2(\mathbb{R}) \cap C^1 (\mathbb{R})$, $f \in C^{\infty}(\mathbb{R})$, $|f'|\leq K$, where $K > 0$ is a constant, then
(1) $J' \in L^2(\mathbb{R})$;
(2) $f \cdot J' \in L^2 (\mathbb{R})$.
I was considering as follows when I asked this question:
(1) If $J \in L^2(\mathbb{R})$, then $\exists N>0$ sufficiently large and $\exists \epsilon>0$ sufficiently small, s.t. $J=o(x^{-(1/2+ϵ)})$ for $|x|> N$.
Because $J \in C^1(\mathbb{R})$, $J'$ is continuous for $|x| \leq N$.
As $J=o(x^{-(1/2+ϵ)})$, $J' = o(x^{-(3/2+ϵ)})$ for $|x| > N$. $J' \in L^2(\mathbb{R})$.
(2) Because $f \in C^{\infty}(\mathbb{R})$ and $|f'|\leq K$, $|f| \leq K|x|+1 $, for all $|x|> N$.
$J' = o(x^{-(3/2+ϵ)})$, $|f| \leq K|x|+1 $, so $f \cdot J' = o(x^{-(1/2+ϵ)})$, for $|x| > N$. Because $f \cdot J'$ is continuous for $|x| \leq N $, $f \cdot J' \in L^2(\mathbb{R})$.
I tried this example $J = \frac{\sin (x)}{ x }$, $f = x $.
But I found my thoughts were wrong which are marked in bold.
By the comment from @PhoemueX, a contour example for this proposition was constructed as $J = \frac{\sin (x) \sin (e^x) }{ x }$.
This counterexample is in terms of continuous but non-differentiable functions because it's easy to describe. At the points of non-differentiability one can smooth out the function to the degree required, e.g. by convolution with mollifiers etc.
Take $f\equiv 1$. $J$ is a function whose graph is an infinite sequence of hats (by "hat" I mean like the graph of $1-|x|$) of height 1 on the $x$-axis, consecutive hats being separated by unit length and $J(x)=0$ in between hats. The base width of the $n^{\rm th}$ hat is $1/n$. Thus $J$ is $L^2$ and $|J'(x)| = n$ a.e. on the hats (and $0$ in-between). So, $J'$ is not $L^2$.