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Suppose $J$ is nonnegative and sequentially lower semicontinuous, and its sublevel sets are strongly sequentially compact. For a solvable exact equation $Au=f$, every sequence of residual-method minimizers with $\delta \to 0$ has a strongly convergent subsequence, and every limit is a J-minimizing solution. Indeed, comparison with an exact penalty minimizer bounds all reconstructions in one compact sublevel set; the residual bound gives $\Vert A{u}_{\delta }-f\Vert \le 2\delta$. Continuity of $A$ and lower semicontinuity of $J$ identify each subsequential limit as an exact minimizer. Thus the distance to a set of exact penalty minimizers tends to zero. Uniqueness of the exact minimizer upgrades this to full strong convergence.

Uniqueness cannot be omitted: for $A=0$, $f_{\delta }=f=0$, and $J(u)=\max \{|u|-1,0\}$ on $\mathbb{R}$, the sublevel sets are compact intervals but the minimizers $u_{1/k}=(-1{)}^k$ do not converge.

Suppose $A{u}^{†}=f$ and $p^{†}=A^{\ast }{\mu }^{†}\in \partial J(u^{†})$. For $F_{\alpha }(u)=\Vert Au-f\Vert +\alpha J(u)$, the subgradient inequality and dual pairing bound giveThus $0<\alpha \Vert {\mu }^{†}\Vert <1$ forces every minimizer to solve $Au=f$ exactly. Comparison of penalties then shows it is a J-minimizing solution, and $D_J^{p^{†}}(u,u^{†})=0$. The conclusion also holds for all $\alpha >0$ when ${\mu }^{†}=0$. One-homogeneity is not needed for this argument.

The assumptions prove convergence to the set of J-minimizing solutions, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.

Let $u^{†}$ be any J-minimizing solution, whose existence is granted. It is feasible for every noisy-data constraint because $A{u}^{†}=f$ and $\Vert f-f_{\delta }\Vert \le \delta$. Minimality therefore gives $J(u_{\delta })\le J(u^{†})$. All reconstructions lie in this strongly sequentially compact space sublevel set. For any sequence ${\delta }_k\downarrow 0$, pass to a subsequence with $u_{{\delta }_k}\to u_{\ast }$. Boundedness of $A$ and the triangle inequality implyso $A{u}_{\ast }=f$. By sequential lower semicontinuity,Thus $u_{\ast }$ is a J-minimizing solution. The same compactness argument shows the distance to a set of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full strong convergence.

For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $A=0$, $f=f_{\delta }=0$, and $J(u)=\max \{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\ge 0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1{)}^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is compact-sublevel convergence of the residual method.

The source condition in variational regularization at an exact solution $u^{†}$ is the existence of a source element ${\mu }^{†}\in {\mathcal{V}}^{\ast }$ such thatFor a Hilbert data space one identifies ${\mathcal{V}}^{\ast }$ with $\mathcal{V}$. If $u$ satisfies $Au=f$ and $A^{\ast }\mu \in \partial J(u)$, then every other exact solution $v$ obeys the subgradient inequalityHence $u$ is a J-minimizing solution. This is source-condition certification of a penalty-minimizing solution.

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.

Given $\Vert f_{\delta }-f\Vert \le \delta$, the residual method minimizes a penalty $J(u)$ subject to $\Vert Au-f_{\delta }\Vert \le \delta$. Every exact solution is feasible, so a reconstruction $u_{\delta }$ has penalty no larger than that of any J-minimizing solution.

If $Au=f$ and $A^{\ast }\mu \in \partial J(u)$, then $u$ is a J-minimizing solution. For every other exact solution $v$, the subgradient inequality gives $J(v)\ge J(u)+⟨\mu ,Av-Au⟩=J(u)$. This is a certificate of global optimality supplied by the range of the adjoint operator.