= Solution
For each $n\geq1$, choose the continuous <probability density functions>
$$
f(u)=1,\qquad g_n(u)=1+\frac{u-1/2}{\sqrt n}.
$$
The second is at least $1/2$ and integrates to one. Their <expected values> differ by
$$
\psi(g_n)-\psi(f)=\frac1{\sqrt n}\int_0^1u(u-1/2)\,du=\frac1{12\sqrt n}.
$$
Writing $g_n/f=1+\Delta_n$ gives
$$
P_f\Delta_n^2=\frac1n\int_0^1(u-1/2)^2\,du=\frac1{12n}.
$$
The supplied product bound, also obtained from the <chi-squared divergence of product measures>, implies
$$
\|P_f^{(n)}-P_{g_n}^{(n)}\|_1^2
\leq\left(1+\frac1{12n}\right)^n-1
\leq e^{1/12}-1\leq\frac1{11}<1.
$$
For the final elementary estimate, compare the exponential series with the geometric series: $e^x\leq(1-x)^{-1}$ for $0\leq x<1$. Thus the overlap of the product laws is at least $1/2$. Part (b) now gives the explicit <mean-estimation minimax lower bound for continuous densities>
$$
\boxed{R_n^*\geq\frac14\cdot\frac1{144n}\cdot\frac12=\frac1{1152n}.}
$$
So \b[one may take $C=1/1152$, uniformly for all $n\geq1$]. The alternatives are allowed to depend on $n$, because the <minimax risk> takes a supremum over all continuous <probability density functions> separately at each sample size.
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