Finite element pde constrained optimization

By separability choose a dense subset $D\subseteq H$ that is at most countable. It is nonempty because $H$ contains $0$. If $D$ is finite, a finite listing from its bijection with a natural number, repeated periodically, gives a sequence; if $D$ is countably infinite, a bijection with $\mathbb{N}$ gives a sequence. Thus [F4] supplies a dense sequence $(v_j{)}_{j\ge 1}$ without any choice principle beyond the witnesses in the definition of countability.

For any bounded real sequence $(a_k)$ in $[-R,R]$, start with $I_0=[-R,R]$. At each stage split the current interval into its two closed halves and choose the left half if it contains infinitely many terms of the current sequence, and otherwise choose the right half. At least one half contains infinitely many terms. Choose the least index later than the preceding selected index whose term lies in the chosen half. The resulting nested intervals have lengths tending to zero, so [F1] gives a common point, and the selected subsequence converges to it. Every choice in this construction is the least natural index or the leftmost of two alternatives, so the extraction is determined by its input sequence.

For each $j$, the scalar sequence $⟨x_n,v_j{⟩}_H$ is bounded by $C\mathrm{\parallel }{v}_j{\mathrm{\parallel }}_H$ by [F3]. Apply the deterministic extraction in step 1.2 successively to these coordinate sequences, each time inside the previously obtained subsequence. This recursive rule produces nested infinite subsequences $N_1\supseteq N_2\supseteq \cdots$ on which the first, first two, and then first $j$ pairings converge, respectively. Let $n_k$ be the $k$th index of $N_k$. Nestedness makes $(n_k)$ strictly increasing, and for each fixed $j$ the sequence $⟨x_{n_k},v_j{⟩}_H$ converges because its tail is a subsequence of $N_j$.

For any $v\in H$, choose $v_j$ arbitrarily close to $v$. For indices $k,\mathrm{\ell }$, [F3] gives $\mathrm{\mid }⟨x_{n_k}-x_{n_{\mathrm{\ell }}},v{⟩}_H\mathrm{\mid }\le \mathrm{\mid }⟨x_{n_k}-x_{n_{\mathrm{\ell }}},v_j{⟩}_H\mathrm{\mid }+2C\mathrm{\parallel }v-v_j{\mathrm{\parallel }}_H\mathrm{.}$ The first term tends to zero as $k,\mathrm{\ell }\to \mathrm{\infty }$ for fixed $j$, and the second can be made arbitrarily small by density. Thus $⟨x_{n_k},v{⟩}_H$ converges for every $v$. Its limit $L(v)$ is linear and satisfies $\mathrm{\mid }L(v)\mathrm{\mid }\le C\mathrm{\parallel }v{\mathrm{\parallel }}_H$, so $L$ is a bounded linear functional. By [F2] there is a unique $x\in H$ with $L(v)=⟨x,v{⟩}_H$ for all $v$. Therefore $x_{n_k}\rightharpoonup x$.

The control $q_a$ belongs to $Q_{ad}$ because $q_a\le q_a\le q_b$ almost everywhere, so $Q_{ad}$ is nonempty. If $q,r\in Q_{ad}$ and $0\le t\le 1$, then $q_a\le tq+(1-t)r\le q_b$ almost everywhere; hence the convex combination is in $Q_{ad}$. The pointwise inequality also gives $\mathrm{\mid }q\mathrm{\mid }\le \mathrm{\mid }{q}_a\mathrm{\mid }+\mathrm{\mid }{q}_b\mathrm{\mid }$ almost everywhere for each $q\in Q_{ad}$, so $\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }{q}_a{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }{q}_b{\mathrm{\parallel }}_{L^2}$ and the box is bounded.

Suppose $q_n\in Q_{ad}$ and $q_n\to q$ in $L^2(E)$. For almost every $x$, $q_n(x)\ge q_a(x)$, so $(q_a-q{)}^{+}\le \mathrm{\mid }{q}_n-q\mathrm{\mid }$ almost everywhere, where $s^{+}=\max \{s,0\}$. Therefore $\mathrm{\parallel }(q_a-q{)}^{+}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{q}_n-q{\mathrm{\parallel }}_2\to 0$ and $q\ge q_a$ almost everywhere. Similarly $(q-q_b{)}^{+}\le \mathrm{\mid }q-q_n\mathrm{\mid }$ gives $q\le q_b$ almost everywhere. Thus $q\in Q_{ad}$ and the box is norm closed.

Define $Z:L^2(E)\to L^2({\mathbb{R}}^d)$ by zero extension and $R:L^2({\mathbb{R}}^d)\to L^2(E)$ by restriction. The maps are well-defined on almost-everywhere classes, $RZ$ is the identity on $L^2(E)$, and $\mathrm{\parallel }Rf{\mathrm{\parallel }}_{L^2(E)}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2({\mathbb{R}}^d)}$.

Let $\mathcal{S}$ be the countable dense rational box-step family from [F1]. The set $R(\mathcal{S})$ is countable. For $q\in L^2(E)$, choose any $\epsilon >0$ and extend it by $Zq$. By density there is $s\in \mathcal{S}$ with $\mathrm{\parallel }Zq-s{\mathrm{\parallel }}_{L^2({\mathbb{R}}^d)}<\epsilon$; then $\mathrm{\parallel }q-Rs{\mathrm{\parallel }}_{L^2(E)}=\mathrm{\parallel }R(Zq-s){\mathrm{\parallel }}_2\le \mathrm{\parallel }Zq-s{\mathrm{\parallel }}_2<\epsilon$. Thus $R(\mathcal{S})$ is dense in $L^2(E)$, proving separability.

The integral pairing makes $L^2(E)$ a real Hilbert space by [F3], so it is both separable and Hilbert. Countable Choice is the only stated choice principle in [F1] and [F3].

Put $q(x)=\min \{q_b(x),\max \{q_a(x),r(x)\}\}$. This function is measurable and lies between $q_a$ and $q_b$; therefore $q\in Q_{ad}$ and $q\in L^2(E)$. For real numbers $a\le b$, the clipped value $c=\min \{b,\max \{a,s\}\}$ satisfies $(c-s)(t-c)\ge 0$ for every $t\in [a,b]$: if $s<a$, then $c=a$ and both factors are nonnegative; if $a\le s\le b$, then $c=s$ and the product is zero; if $s>b$, then $c=b$ and both factors are nonpositive. Applying this pointwise with $s=r(x)$ and $t=v(x)$ and integrating gives $(q-r,v-q{)}_{L^2}\ge 0$ for every $v\in Q_{ad}$.

For any $v\in Q_{ad}$, $\mathrm{\parallel }v-r{\mathrm{\parallel }}_2^2-\mathrm{\parallel }q-r{\mathrm{\parallel }}_2^2=\mathrm{\parallel }v-q{\mathrm{\parallel }}_2^2+2(q-r,v-q{)}_2\ge \mathrm{\parallel }v-q{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus $q$ minimizes the distance and equality can hold only when $v=q$ in $L^2$. Conversely, if $q$ is a distance minimizer, then for any $v\in Q_{ad}$ and $0<t\le 1$, $q+t(v-q)\in Q_{ad}$. Expanding $\mathrm{\parallel }q+t(v-q)-r{\mathrm{\parallel }}_2^2\ge \mathrm{\parallel }q-r{\mathrm{\parallel }}_2^2$, dividing by $2t$, and letting $t\downarrow 0$ gives $(q-r,v-q{)}_2\ge 0$.

The box is compact and $\varphi$ is continuous, so [F2] supplies a minimizer $x_{\ast }$. If ${\mathrm{\ell }}_i<u_i$ and $x_{\ast ,i}={\mathrm{\ell }}_i$, the feasible right derivative is $g_i\ge 0$; if $x_{\ast ,i}=u_i$, the feasible left derivative gives $g_i\le 0$; and if ${\mathrm{\ell }}_i<x_{\ast ,i}<u_i$, both signs of perturbation are feasible, so $g_i=0$. For a fixed coordinate with ${\mathrm{\ell }}_i=u_i$, assign lower status if $g_i\ge 0$ and upper status if $g_i\le 0$ (either if $g_i=0$).

Assign each nonfixed coordinate of $x_{\ast }$ its lower, upper, or free status according as it is at the lower bound, at the upper bound, or strictly between them; assign fixed coordinates as in step 1.1. Every nonempty principal block $H_{FF}$ is positive definite: for $0\mathrm{\ne }y\in {\mathbb{R}}^F$, extend $y$ by zeros to $x\in {\mathbb{R}}^n$ to get $y^T{H}_{FF}y=x^{T}Hx>0$. Hence $H_{FF}$ is nonsingular; if $F=\mathrm{\varnothing }$, its system is vacuous. The free subsystem has the unique solution $x_{\ast ,F}$ by its zero free gradient. Thus the algorithm recovers $x_{\ast }$, whose free coordinates are strictly interior by their assignment and whose gradient signs hold by step 1.1. It is retained by [F1], so the algorithm terminates within $3^n$ checks.

Let $x$ be a retained candidate and $g=Hx+c$. For any feasible $y$, the checked coordinate signs give $g_i(y_i-x_i)\ge 0$ on lower-active coordinates, the same inequality on upper-active coordinates, and zero on free coordinates; hence $g^T(y-x)\ge 0$. The exact expansion $\varphi (y)-\varphi (x)=g^T(y-x)+\frac{1}{2}(y-x{)}^{T}H(y-x)$ is nonnegative, and is strictly positive when $y\mathrm{\ne }x$ because $H$ is positive definite. Thus every retained candidate is the unique minimizer, so the first one returned by the algorithm is.

For a cell $K$, [F4] makes ${\bar{r}}_K=\mathrm{\mid }K{\mathrm{\mid }}^{-1}{\int }_{K}r$ finite. For any constant value $z_K$, ${\int }_K\mathrm{\mid }{z}_K-r{\mathrm{\mid }}^2=\mathrm{\mid }K\mathrm{\mid }\mathrm{\mid }{z}_K-{\bar{r}}_K{\mathrm{\mid }}^2+{\int }_K\mathrm{\mid }r-{\bar{r}}_K{\mathrm{\mid }}^2,$ because the cross term vanishes by ${\int }_K(r-{\bar{r}}_K)=0$. The second term is independent of $z_K$.

Among real $z_K\in [q_a{\mathrm{\mid }}_K,q_b{\mathrm{\mid }}_K]$, the square $\mathrm{\mid }{z}_K-{\bar{r}}_K{\mathrm{\mid }}^2$ is uniquely minimized by the clipped value: if ${\bar{r}}_K$ is below the interval the nearest endpoint is the lower bound; if it is inside, it is itself; if it is above, the nearest endpoint is the upper bound. Thus the displayed cell value uniquely minimizes the cell integral, and it lies in the required interval by [F1].

The cell choices are independent, so summing the cellwise minima from step 2.1 over the partition up to its null facets [F3] proves that $P_{Q_h}r$ is the unique global minimizer. For $v\in Q_h$, convexity makes $p+t(v-p)$ feasible, where $p=P_{Q_h}r$ and $0<t\le 1$. Minimality gives $\mathrm{\parallel }p+t(v-p)-r{\mathrm{\parallel }}_2^2-\mathrm{\parallel }p-r{\mathrm{\parallel }}_2^2=2t(p-r,v-p{)}_2+t^2\mathrm{\parallel }v-p{\mathrm{\parallel }}_2^2\ge 0.$ Divide by $2t$ and let $t\downarrow 0$ to obtain the projection variational inequality $(p-r,v-p{)}_2\ge 0$.

Let $p=P_{Q_h}r$ and $q=P_{Q_h}s$. Apply the variational inequality from step 3.1 with $v=q$ and $v=p$, respectively. Adding gives $\mathrm{\parallel }p-q{\mathrm{\parallel }}_2^2\le (r-s,p-q{)}_2$. Hölder with exponents $2,2$ then yields $\mathrm{\parallel }p-q{\mathrm{\parallel }}_2\le \mathrm{\parallel }r-s{\mathrm{\parallel }}_2$ when $p\mathrm{\ne }q$; if $p=q$ the inequality is immediate.

Let $A=\{x\in E:q(x)<q_a(x)\}$. Since $E$ is bounded, choose $M\ge 1$ with $E\subset (-M,M{)}^d$. The single-box cover in the definition of Lebesgue outer measure gives $\mu (E)\le (2M{)}^d<\mathrm{\infty }$, so ${\mathbf{1}}_A\in L^2(E)$. If $A$ had positive measure, $f=(q_a-q){\mathbf{1}}_A$ would be nonnegative, integrable by [F2], and positive on a set of positive measure. Thus [F3] gives ${\int }_{E}f>0$. Weak convergence tested against ${\mathbf{1}}_A$ gives ${\int }_A(q-q_a)={\lim }_{n\to \mathrm{\infty }}{\int }_A(q_n-q_a)\ge 0$, contradicting ${\int }_A(q-q_a)=-{\int }_{E}f<0$. Therefore $q\ge q_a$ almost everywhere.

Let $B=\{x\in E:q(x)>q_b(x)\}$. As in step 1.1, ${\mathbf{1}}_B\in L^2(E)$. If $B$ had positive measure, $f=(q-q_b){\mathbf{1}}_B$ would be nonnegative, integrable by [F2], and positive on a set of positive measure. Thus [F3] gives ${\int }_{E}f>0$. Weak convergence tested against ${\mathbf{1}}_B$ gives ${\int }_B(q-q_b)={\lim }_n{\int }_B(q_n-q_b)\le 0$, contradicting ${\int }_B(q-q_b)={\int }_{E}f>0$. Therefore $q\le q_b$ almost everywhere, and [F1] gives $q\in Q_{ad}$.

The weak-derivative test identities are linear, and [F4] makes each derivative coordinate intrinsic, so $W^{1,2}(\mathrm{\Omega };\mathbb{R})$ is a vector space with the inner product induced from $L^2(\mathrm{\Omega }{)}^{d+1}$; it induces the displayed full Sobolev norm and is definite because its zeroth coordinate is $v$. Let $(v_n)$ be Cauchy in this norm. Its function coordinates and each of its finitely many weak-derivative coordinates are Cauchy in $L^2(\mathrm{\Omega })$ because each coordinate norm is bounded by the full $W^{1,2}$ norm. By [F3], there are $v,g_1,\dots ,g_d\in L^2(\mathrm{\Omega })$ with $v_n\to v$ and $D_i{v}_n\to g_i$ in $L^2$ for each $i$.

For every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $i$, the weak-derivative identity for $v_n$ is ${\int }_{\mathrm{\Omega }}{v}_n{D}_i\phi =-{\int }_{\mathrm{\Omega }}(D_i{v}_n)\phi$. By [F5], both test functions are in $L^2$; Cauchy–Schwarz lets both integrals pass to the limits in [F3], giving ${\int }_{\mathrm{\Omega }}v{D}_i\phi =-{\int }_{\mathrm{\Omega }}{g}_i\phi$. Thus each $g_i$ is a weak derivative of $v$. By uniqueness in [F4], $D_{i}v=g_i$ as an $L^2$ class, and the component convergences from step 1.1 give $v_n\to v$ in the full $W^{1,2}$ norm. Hence $W^{1,2}(\mathrm{\Omega };\mathbb{R})$ is complete.

By [F4], $V=H_0^1(\mathrm{\Omega };\mathbb{R})$ is the closure of the linear space $C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$ in $W^{1,2}$. It is therefore a closed linear subspace of that complete space, and the inherited inner product is exactly the one in Linear coercive elliptic PDE control problem and reduced objective. Thus $V$ is a real Hilbert space under Countable Choice.

For fixed $q$, the functional $F_q(v)=⟨f+Bq,v⟩$ is bounded and linear, with $\mathrm{\parallel }{F}_q{\mathrm{\parallel }}_{V^{\ast }}\le \mathrm{\parallel }f+Bq{\mathrm{\parallel }}_{V^{\ast }}$. Applying [F2] to $a$ and $F_q$ now gives the unique state and the stated bound.

Define the adjoint form $a^{\ast }(p,v):=a(v,p)$. It is bounded with the same constant $M$, and $a^{\ast }(p,p)=a(p,p)\ge \beta \mathrm{\parallel }p{\mathrm{\parallel }}_V^2$, so it is coercive with the same $\beta$. The functional $G_q(v)=(Ju(q)-u_d,Jv{)}_H$ is bounded by $\mathrm{\parallel }J\mathrm{\parallel }\mathrm{\parallel }Ju(q)-u_d{\mathrm{\parallel }}_H\mathrm{\parallel }v{\mathrm{\parallel }}_V$. Apply [F2] to $a^{\ast }$ and $G_q$; the result is the unique $p(q)$ with $a(v,p(q))=G_q(v)$ and the stated bound. No symmetry of $a$ is used.

Steps 1.1, 2.1, and 3.1 establish the real Hilbert structure required by [F2] using Countable Choice through the real $L^2$ completeness result [F3]. Steps 4.1 and 5.1 give the state, adjoint, uniqueness, and both estimates.

For $h\in U$, let $w_h=u(q+h)-u(q)$. Subtracting the two state equations gives $a(w_h,v)=⟨Bh,v⟩$ for all $v$, so $w_h$ depends linearly and boundedly on $h$ by the state estimate in [F1]. Thus $Sh=J{w}_h$ is a bounded linear map from $U$ to $H$. The functional $h\mapsto ⟨Bh,p(q)⟩$ is bounded, so [F2] gives the unique $B^{\ast }p(q)\in U$ in the displayed definition.

Expanding the two squared norms in $\widehat{J}(q+h)$ yields $\widehat{J}(q+h)-\widehat{J}(q)=(Ju(q)-u_d,J{w}_h{)}_H+\alpha (q,h{)}_U+\frac{1}{2}\mathrm{\parallel }Sh{\mathrm{\parallel }}_H^2+\frac{\alpha }{2}\mathrm{\parallel }h{\mathrm{\parallel }}_U^2\mathrm{.}$ The adjoint equation with $v=w_h$ and the state variation equation with test $p(q)$ give $(Ju(q)-u_d,J{w}_h{)}_H=a(w_h,p(q))=⟨Bh,p(q)⟩=(B^{\ast }p(q),h{)}_U$. The last two terms are bounded by a constant times $\mathrm{\parallel }h{\mathrm{\parallel }}_U^2$, so the Fréchet derivative is $B^{\ast }p(q)+\alpha q$.

Since the state map is affine with linear part $S$, the same exact expansion at any $q$ shows the second derivative is $D^2\widehat{J}[h,k]=(Sh,Sk{)}_H+\alpha (h,k{)}_U$, represented by $S^{\ast }S+\alpha I$. Hence $((S^{\ast }S+\alpha I)h,h)=\mathrm{\parallel }Sh{\mathrm{\parallel }}_H^2+\alpha \mathrm{\parallel }h{\mathrm{\parallel }}_U^2\ge \alpha \mathrm{\parallel }h{\mathrm{\parallel }}_U^2$ and $\mathrm{\parallel }{S}^{\ast }S+\alpha I\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}^2+\alpha$. The constant Hessian gives the stated Lipschitz-gradient bound; the exact quadratic expansion gives strong convexity.