Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 326 / 4 / c

Let $u^{†}=u_J^{†}$, $p^{†}=A^{\ast }{\mu }^{†}\in \partial J(u^{†})$, and $r_{\alpha }=A{u}_{\alpha }-f$. The subgradient inequality, the dual pairing bound, and $A{u}^{†}=f$ giveMinimality of $u_{\alpha }$ in the exact penalty method givesCombining them yieldsThe coefficient is strictly positive, soNow minimality gives $J(u_{\alpha })\le J(u^{†})$, and exact feasibility plus the defining minimality of $u^{†}$ gives the reverse inequality. Thus $u_{\alpha }$ is a J-minimizing solution. Finally,This exact-penalty threshold from a source condition does not require one-homogeneity; convexity and the source subgradient suffice. If ${\mu }^{†}=0$, the same argument works for every $\alpha >0$, interpreting the upper threshold as infinite.