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Write the residual of an inverse problem as $r_k=K{u}_{\delta }^{(k)}-f^{\delta }$. The iteration can be rearranged asApplying $K$ gives $(I+\tau K{K}^{\ast })r_{k+1}=r_k$. Since $K{K}^{\ast }$ is a positive operator, its resolvent operator $(I+\tau K{K}^{\ast }{)}^{-1}$ has operator norm at most one. HenceMore explicitly, taking the inner product of the residual equation with $r_{k+1}$ givesThis directly proves the same monotonicity by the Cauchy-Schwarz inequality, including the case $r_{k+1}=0$. An unresolvable data component in $\mathcal{R}(K{)}^{\perp }$ remains unchanged.