= Solution
There is a genuine qualification: \b[the first printed inequality is valid as a useful general assertion for $0<\eta<1$, but is false for arbitrary $\eta>0$.] Also the displayed <likelihood ratio> presupposes $P_{f_1}\ll P_{f_0}$. We first prove the intended bound and then give a counterexample to the unrestricted quantifier.
For any <estimator> $T$, set $A=\{d(T,f_0)<r_n\}$. By the <triangle inequality>, $d(T,f_1)\ge r_n$ on $A$, while $d(T,f_0)\ge r_n$ on $A^c$. Hence
$$
\max_{j=0,1}r_n^{-1}\mathbb E_{f_j}d(T,f_j)
\ge\frac12\{P_{f_0}(A^c)+P_{f_1}(A)\}.
$$
Let $C=\{|Z-1|\le\eta\}$ with $0<\eta<1$. On $C$, $Z\ge1-\eta$. Therefore
$$
P_{f_0}(A^c)+P_{f_1}(A)
\ge (1-\eta)\{P_{f_0}(A^c)+P_{f_0}(A\cap C)\}
\ge(1-\eta)P_{f_0}(C)
\ge(1-\eta)\left(1-\frac{\mathbb E_{f_0}|Z-1|}{\eta}\right).
$$
The last step is <Markov inequality>. Taking the supremum over the model and then the infimum over <estimators> proves the intended <metric two-point risk bound>. At $\eta=1$ the right side is zero. For $\eta>1$ a negative right side is also harmless, but a positive right side need not be a lower bound.
Here is an explicit counterexample with positive <probability density functions> and $n=3$. Let $A_0=[0,1/2)$, $\delta=1/100$, and take the model consisting of
$$
f_0=2(1-\delta)\mathbf1_{A_0}+2\delta\mathbf1_{A_0^c},\qquad
f_1=2\delta\mathbf1_{A_0}+2(1-\delta)\mathbf1_{A_0^c}.
$$
Use the <L2 norm> <metric>, for which $D=d(f_0,f_1)=2(1-2\delta)$, and take $r_n=D/2$ for every $n$. The majority-half decision selects $f_0$ if at least two observations lie in $A_0$. Each error probability is $p=3\delta^2-2\delta^3=0.000298$, so its maximum normalized <risk function> is $2p=0.000596$. This is also the <likelihood-ratio test>, giving $\mathbb E_{f_0}|Z-1|=2(1-2p)$. At $\eta=3/2$ the proposed right side is $1/12-2p/3>0.083$, contradicting the risk of this explicit <estimator>. Thus the intended range restriction cannot be omitted.
For the second request use the intended <L2 norm> <metric> $d(f,g)=(\int_0^1(f-g)^2)^{1/2}$. The PDF defines its square, not the square itself as a <metric>. Put
$$
g=\mathbf1_{[0,1/2)}-\mathbf1_{[1/2,1]},\qquad
f_0=1,\quad f_{1,n}=1+a_ng,\quad a_n=\frac1{8\sqrt n},\quad r_n=\frac1{16\sqrt n}.
$$
Both <probability density functions> belong to the model, and $d(f_0,f_{1,n})=a_n=2r_n$. The two-point choice may depend on $n$, because the lower bound is applied separately at each sample size. Under $f_0$, the product <likelihood ratio> $Z_n=\prod_i(1+a_ng(X_i))$ satisfies
$$
\mathbb E_0Z_n=1,\qquad
\mathbb E_0(Z_n-1)^2=(1+a_n^2)^n-1\le e^{1/64}-1<\frac1{16}.
$$
This is the product <chi-squared divergence>. By <Cauchy-Schwarz inequality>, $\mathbb E_0|Z_n-1|<1/4$. With $\eta=1/2$, the proved <metric two-point risk bound> has normalized right side greater than $1/8$. Consequently
$$
\boxed{\inf_T\sup_{f\in\mathcal P_\Psi}\sqrt n\,\mathbb E_fd(T,f)\ge\frac1{128}>0.}
$$
The proof works for every $n\ge1$ despite the introductory $n>2$. <Measurability> of the functions defining the <probability density functions> is understood. This establishes the requested lower bound; it does not assert a matching upper bound over this unrestricted class.
Back to article page
