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Put $h=H^2(P_0,P_1)$ with the paper's normalized Hellinger distance. The Hellinger affinity isThe last bound follows from the Cauchy-Schwarz inequality and the unit integrals of the two probability density functions. Using a common dominating probability measure, for example $(P_0+P_1)/2$, makes the product calculation immediate. The probability density functions of the product measures factor, and Tonelli's theorem givesThe Squared Hellinger distance is independent of the choice of common dominating measure, so this calculation applies to the original distributions. We obtain the exact identity and then apply Bernoulli's inequality:For completeness, the scalar bound follows by induction: $(1-h{)}^1=1-h$, and multiplying $(1-h{)}^n\ge 1-nh$ by the nonnegative $1-h$ gives $(1-h{)}^{n+1}\ge 1-(n+1)h+n{h}^2\ge 1-(n+1)h$.

Use two translated uniform distributions, with parameters $0$ and $a=1/(2n)$. Their distance in the usual metric is $a$, so take the separation radius $\delta =a/2=1/(4n)$ and the loss function $\Phi (r)=r^2$. The supplied testing reduction, equivalently the Le Cam two-point lemma, and part (a)(i) implyTo see the reduction directly, turn any estimator into a binary hypothesis test by choosing the closer of $0$ and $a$. If the test is wrong, the absolute estimation error is at least $a/2$. The worst quadratic risk is at least its average over these two parameters, giving precisely this bound.

Under Lebesgue measure on ${\mathbb{R}}^n$, the two sample probability density functions are the indicators of the cubes $[0,1{]}^n$ and $[a,a+1{]}^n$. Their intersection is $[a,1{]}^n$, of volume $(1-a{)}^n$. Thereforewhere the inequality is Bernoulli's inequality. ConsequentlyThis also fits the Hellinger distance approach: the two one-sample affinities give $H^2(U_0,U_a)=a$, while equal-height indicator densities imply $V(U_0^n,U_a^n)=H^2(U_0^n,U_a^n)$. Part (a)(ii) thus yields the same bound $V\le na=1/2$. The change of support produces Squared Hellinger distance proportional to $a$, explaining the parameter scale $a\asymp n^{-1}$ and quadratic risk scale $n^{-2}$.

With exact data $f=K{u}^{†}$, the iteration error obeysThe source condition for quadratic regularization $u^{†}=K^{\ast }v$ excludes any null space component. Using its singular system of a compact operator coefficients,By Bernoulli's inequality, $(1+\tau {\sigma }^2{)}^k\ge 1+k\tau {\sigma }^2$. Also $1+k\tau {\sigma }^2\ge 2\sqrt{k\tau }\sigma$. ThereforeSumming the squared coefficients and using Bessel inequality yieldsThis is an a priori error rate from the source condition for quadratic regularization, with constant independent of $k$.

It is sufficient thatand that the stopping rule obeyFor example, $k(\delta )=\lfloor {\delta }^{-1}\rfloor$ works.

For exact data $f\in \mathcal{D}(A^{†})$, each factor $(1-\tau {\sigma }_i^2{)}^k$ tends to zero. The Picard criterion makes $⟨f,y_i⟩/{\sigma }_i$ square summable, while $|1-(1-\tau {\sigma }_i^2{)}^k|$ is uniformly bounded. The dominated convergence theorem on the resulting series yields $u_k(f)\to A^{†}f$.

For noisy data with $\Vert f^{\delta }-f\Vert \le \delta$, the filter representation givesIndeed, when $\tau {\sigma }^2\le 1$, Bernoulli's inequality gives $1-(1-\tau {\sigma }^2{)}^k\le \min (k\tau {\sigma }^2,1)$; when $\tau {\sigma }^2>1$, the quotient is uniformly bounded because $\tau \Vert A{\Vert }^2<2$. Hence the triangle inequality giveswhich proves that early-stopped Landweber iteration is a convergent regularization of an inverse problem.

Replace the same last $k$ vectors by $(100,\dots ,100)$. Applying part a to each independent Gaussian coordinate shows that the probability its contaminated coordinate median exceeds $ϵ$ is at least $1-e^{-c_{ϵ}n}$. Independence across coordinates and Bernoulli's inequality giveFor $n$ sufficiently large as a function of $d$ and $ϵ$, the last expression is at least $1/2$. On this event,which proves the stated lower bound for the supremum over adversarial perturbations.

Put $u=c^{-1}-1>0$, so $c=(1+u{)}^{-1}$. By Bernoulli's inequality, $(1+u{)}^n\ge 1+nu$, and henceMore explicitly, for $\epsilon >0$ choose an integer $N>1/(u\epsilon )$. For $n\ge N$, the right-hand side is less than $\epsilon$. Thus the geometric sequence has limit of a sequence zero, either directly by the definition or by the squeeze theorem.

True. Since $a_n^n\to 2$, there is $N_0$ such that $1<a_n^n<3$ for $n\ge N_0$. Positivity then gives $a_n>1$ on this tail.

Fix $\epsilon >0$. If $a_n\ge 1+\epsilon$, Bernoulli's inequality givesFor $n>2/\epsilon$, this contradicts $a_n^n<3$. Therefore eventually $1<a_n<1+\epsilon$, proving

False. Choose the positive sequenceIt has limit of a sequence one, but Bernoulli's inequality givesThus $(a_n^n)$ is not a bounded sequence.