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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.

We prove the required Hahn-Banach theorem for a seminorm directly. First suppose the vector space is real. At an intermediate extension stage, let $g:E\to \mathbb{R}$ be a linear functional with $|g(y)|\le p(y)$, and let $z\notin E$. Extending to $E+\mathbb{R}z$ amounts to choosing $c=f(z)$. The upper domination $f\le p$ requiresThe interval is nonempty: for all $y,w\in E$,Its endpoints are finite since inserting zero gives bounds $-p(z)$ and $p(z)$, while every lower candidate is below every upper candidate. Choose $c$ in the interval and put $f(y+tz)=g(y)+tc$. For $t>0$, the upper bound with $w=y/t$ proves $f(y+tz)\le p(y+tz)$. For $t<0$, writing $s=-t>0$ and applying the lower bound with $y/s$ proves the same inequality. The case $t=0$ is already known. Applying this domination to $-x$ and using $p(-x)=p(x)$ gives $|f(x)|\le p(x)$.

Partially order all dominated linear functional extensions of the original $g$ by extension of their domains and values. They form a nonempty set. The union along any chain is a well-defined dominated linear functional on a vector subspace, so every chain has an upper bound. The Zorn lemma gives a maximal extension. If its domain were not $X$, the preceding one-dimensional construction would enlarge it, a contradiction. Thus the required extension exists on $X$.

For a complex vector space, apply the proved real result to $\mathrm{Re}g$ on the underlying real vector subspace. Let $U:X\to \mathbb{R}$ be the resulting real linear functional, with $|U(x)|\le p(x)$, and setIt is additive and real-linear, and $f(ix)=if(x)$, so it is complex-linear. On $Y$, the identity $\mathrm{Re}g(iy)=-\mathrm{Im}g(y)$ proves $f(y)=g(y)$. Choose $\alpha$ with $|\alpha |=1$ and $\alpha f(x)=|f(x)|$ when $f(x)\ne 0$. ThenThe bound is immediate if $f(x)=0$. This proves the real and complex seminorm forms of Hahn-Banach theorem without invoking any version of that theorem.

For the distance to a set assertion, let $d=d(z,Y)>0$. Define on $Y+\mathbb{K}z$ the linear functional $g(y+\alpha z)=\alpha d$. The decomposition is unique because $z\notin Y$, and for $\alpha \ne 0$,For $\alpha =0$ the bound is immediate. Apply the just-proved Hahn-Banach theorem with $p(x)=\Vert x\Vert$ to get $f\in X^{\ast }$, $f{|}_Y=0$, $f(z)=d$, and $\Vert f\Vert \le 1$. Choose $y_n\in Y$ with $\Vert z-y_n\Vert \to d$. Sinceand $d>0$, passage to the limit gives $\Vert f\Vert \ge 1$. Thus $\boxed{f{|}_Y=0,f(z)=d(z,Y),\Vert f\Vert =1}$. This version of the Hahn-Banach distance formula needs neither a closed $Y$ nor an attained distance to a set.

For the Banach limit construction below, take the real l-infinity sequence space, let $S(x_1,x_2,\dots )=(x_2,x_3,\dots )$ be its left shift on bounded sequences, put $Y=(I-S){\ell }^{\infty }$, and let $1=(1,1,\dots )$. This is a vector subspace and its distance to a set from $1$ determines the construction.

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.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives