= Solution
To repeat <integration by parts> across whole periods, choose <periodic primitives with zero mean>. Set $h_0=f$, and recursively define
$$
h_{r+1}=Rh_r-\int_0^1Rh_r.
$$
Every $h_r$ has zero mean. Part (i) gives $h_{r+1}'=-h_r$ <almost everywhere>, and $h_{r+1}(0)=h_{r+1}(1)$ because $Rh_r$ vanishes at both endpoints. Its <periodic extension> is consequently locally an <absolutely continuous function>, with no jump at the integers. Part (ii) gives the uniform estimate
$$
\|h_{r+1}\|_{L^1}\leq2\|Rh_r\|_{L^1}\leq\|h_r\|_{L^1}\leq\|f\|_{L^1}.
$$
The <weak derivative> identity on the whole line is $D\overline h_{r+1}=-\overline h_r$. After scaling by the integer $j$ and repeating <integration by parts> $N$ times,
$$
\int_0^1\overline f(jx)\phi(x)\,dx
=j^{-N}\int_0^1\overline h_N(jx)\phi^{(N)}(x)\,dx.
$$
The <boundary terms> vanish because $\phi$ has <compact support> in $(0,1)$. The interval contains exactly $j$ periods, so
$$
\int_0^1|\overline h_N(jx)|\,dx=\int_0^1|h_N(t)|\,dt.
$$
Taking $N=m+1$ therefore yields
$$
\boxed{\left|j^m\int_0^1f_j\phi\right|
\leq j^{-1}\|f\|_{L^1}\|\phi^{(m+1)}\|_\infty\longrightarrow0.}
$$
This <rapid distributional decay of periodic oscillations> uses only <integrability> of $f$ and smoothness of the <test function>.
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