= Exact-penalty threshold from a source condition
Suppose $Au^\dagger=f$ and $p^\dagger=A^*\mu^\dagger\in\partial J(u^\dagger)$. For $F_\alpha(u)=\|Au-f\|+\alpha J(u)$, the <subgradient inequality> and <dual pairing> bound give
$$
F_\alpha(u)-F_\alpha(u^\dagger)
\geq(1-\alpha\|\mu^\dagger\|)\|Au-f\|.
$$
Thus $0<\alpha\|\mu^\dagger\|<1$ forces every minimizer to solve $Au=f$ exactly. Comparison of penalties then shows it is a <J-minimizing solution>, and $D_J^{p^\dagger}(u,u^\dagger)=0$. The conclusion also holds for all $\alpha>0$ when $\mu^\dagger=0$. One-homogeneity is not needed for this argument.
Back to article page
