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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.

Let ${\hat{u}}_{\alpha }$ minimize the nonsquared-residual objective. Comparison with $u^{†}$ givesThe source subgradient inequality givesConsequently,Thus forevery $0<\alpha <{\alpha }_0$ forces $A{\hat{u}}_{\alpha }=f$. The original comparison then gives $J({\hat{u}}_{\alpha })\le J(u^{†})$, so ${\hat{u}}_{\alpha }$ is itself $J$-minimizing. If $J$ is strictly convex, its restriction to the affine solution set has at most one minimizer, hence ${\hat{u}}_{\alpha }=u^{†}$. This is an exact penalty method.