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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.

A problem is well posed in the sense of Hadamard when a solution exists for every admissible datum, is unique, and depends continuously on the datum. It is ill posed if any one of these three properties fails.

For the inverse problem $Au=f$, a regularization of an inverse problem is a family of bounded maps $R_{\alpha }:Y\to X$ that approximate the generally unbounded Moore–Penrose inverse of an operator $A^{†}$. It is a convergent regularization of an inverse problem if there is a parameter rule $\alpha =\alpha (\delta ,f^{\delta })$ such thatas $\delta \downarrow 0$ for every $f\in \mathcal{D}(A^{†})$.