 Solution

Choose a fixed positive Landweber relaxation parameter $\tau$ and rearrange the normal equation for a linear inverse problem as the fixed point equationSuccessive substitution gives Landweber iterationSet $B=A^{\ast }A$ and $P=I-\tau B$. Induction yields the closed polynomial-operator expressionFor $n=0$ the sum is empty, giving zero. The expression is well defined without an inverse of $A^{\ast }A$, which can have a kernel of a linear map and an unbounded inverse on its range. Each iterate lies in the closure of $\mathrm{Ran}{A}^{\ast }$, so initialization at zero eliminates the arbitrary null-space component of a solution. The unrelaxed choice $\tau =1$ gives $x_{n+1}=(I-A^{\ast }A)x_n+A^{\ast }y$ and the corresponding sum with $\tau =1$; it needs the convergence restriction in part (d).