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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.