= Solution
Let $u^\dagger=u_J^\dagger$, $p^\dagger=A^*\mu^\dagger\in\partial J(u^\dagger)$, and $r_\alpha=Au_\alpha-f$. The <subgradient inequality>, the <dual pairing> bound, and $Au^\dagger=f$ give
$$
J(u_\alpha)-J(u^\dagger)\geq\langle p^\dagger,u_\alpha-u^\dagger\rangle
=\langle\mu^\dagger,r_\alpha\rangle
\geq-\|\mu^\dagger\|\|r_\alpha\|.
$$
Minimality of $u_\alpha$ in the <exact penalty method> gives
$$
\|r_\alpha\|+\alpha[J(u_\alpha)-J(u^\dagger)]\leq0.
$$
Combining them yields
$$
(1-\alpha\|\mu^\dagger\|)\|r_\alpha\|\leq0.
$$
The coefficient is strictly positive, so
$$
\boxed{Au_\alpha=f.}
$$
Now minimality gives $J(u_\alpha)\leq J(u^\dagger)$, and exact feasibility plus the defining minimality of $u^\dagger$ gives the reverse inequality. Thus $u_\alpha$ is a <J-minimizing solution>. Finally,
$$
\boxed{D_J^{p^\dagger}(u_\alpha,u^\dagger)
=J(u_\alpha)-J(u^\dagger)-\langle\mu^\dagger,A(u_\alpha-u^\dagger)\rangle=0.}
$$
This <exact-penalty threshold from a source condition> does not require one-homogeneity; convexity and the source subgradient suffice. If $\mu^\dagger=0$, the same argument works for every $\alpha>0$, interpreting the upper threshold as infinite.
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