= Solution
Using $(Rf)'=\int_0^1f-f$ and <integration by parts>, with $\phi$ a <test function>, gives
$$
\int_0^1Rf\,\phi'=\int_0^1f\phi-\left(\int_0^1f\right)\left(\int_0^1\phi\right).
$$
For each $k\geq1$, apply the same identity to $R^kf$ and the <test function> $\phi^{(k)}$. Since $\int_0^1\phi^{(k)}=0$, it gives
$$
\int_0^1R^{k+1}f\,\phi^{(k+1)}=\int_0^1R^kf\,\phi^{(k)}.
$$
Starting from $k=1$ and iterating proves
$$
\boxed{\int_0^1f\phi-\left(\int_0^1f\right)\left(\int_0^1\phi\right)
=\int_0^1R^kf\,\phi^{(k)}\quad(k\geq1).}
$$
The intermediate means disappear because derivatives of a <test function> have zero integral; no assumption that $R^kf$ has zero mean is needed.
For the bound, write
$$
Rf(x)=\int_0^1\bigl(x-\mathbf1_{\{t<x\}}\bigr)f(t)\,dt.
$$
The <triangle inequality> and <Fubini's theorem> imply
$$
\begin{aligned}
\|Rf\|_{L^1}
&\leq\int_0^1|f(t)|\left[\int_0^t x\,dx+\int_t^1(1-x)\,dx\right]dt\\
&=\frac12\int_0^1\bigl[t^2+(1-t)^2\bigr]|f(t)|\,dt
\leq\frac12\|f\|_{L^1}.
\end{aligned}
$$
Thus $\boxed{c=\tfrac12\text{ is valid}}$, and every iterate $R^kf$ is in the <L1 space>.
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