Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 326 / 2 / v / Solution

Let $w=K^{\ast }Kv\in \partial J(u^{†})$ from the normal-operator source condition, and set $e=u-u^{†}$. Since $u^{†}$ is a least-squares solution, the normal equation for a linear inverse problem gives $K^{\ast }(K{u}^{†}-f)=0$. Expand the objective difference and use the Bregman distance definition:The cross term with the residual of the least-squares solution vanishes by the normal equation; exact data are not required. Compare the minimizing $u_{\alpha }$ with $z=u^{†}-\alpha v$. At $z$, the completed square is zero. Dividing the resulting inequality by $\alpha$ proves the stronger shifted-comparator Bregman boundDropping the nonnegative squared term gives the requested estimate. If the comparator lies outside the effective domain of $J$, its distance is infinite and the inequality is valid but gives no finite error bound.