Stampacchia's variational inequality

By [F1] fix $f\in H$ with $F(w)=⟨w,f⟩$ for all $w$; by [F2] fix $A\in \mathcal{B}(H)$ with $a(u,v)=⟨Au,v⟩$, $\mathrm{\parallel }A\mathrm{\parallel }\le M$ and $⟨Au,u⟩\ge \alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2$ for all $u$. The real bilinear form need not be symmetric by [F9].

If $H=\{0\}$, then $K=\{0\}$ and $u=0$ is the unique solution, since $a(0,0)=F(0)=0$. Otherwise choose $z_0\mathrm{\ne }0$; boundedness and coercivity give $\alpha \mathrm{\parallel }{z}_0{\mathrm{\parallel }}^2\le a(z_0,z_0)\le M\mathrm{\parallel }{z}_0{\mathrm{\parallel }}^2$, so $0<\alpha \le M$. Choose $\rho :=\alpha \mathrm{/}{M}^2$ and $q:=\sqrt{1-{\alpha }^2\mathrm{/}{M}^2}$, which satisfies $0\le q<1$ and $q^2=1-2\rho \alpha +{\rho }^2{M}^2$. For $z\in H$ the expansion of [F3] together with $⟨Az,z⟩\ge \alpha \mathrm{\parallel }z{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }Az\mathrm{\parallel }\le M\mathrm{\parallel }z\mathrm{\parallel }$ [F2, F8] gives $\mathrm{\parallel }(I-\rho A)z{\mathrm{\parallel }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2-2\rho ⟨Az,z⟩+{\rho }^2\mathrm{\parallel }Az{\mathrm{\parallel }}^2\le (1-2\rho \alpha +{\rho }^2{M}^2)\mathrm{\parallel }z{\mathrm{\parallel }}^2=q^2\mathrm{\parallel }z{\mathrm{\parallel }}^2$. Hence the map $T(w):=P_K(w-\rho (Aw-f))$ satisfies $\mathrm{\parallel }T(w)-T(w^{\mathrm{\prime }})\mathrm{\parallel }\le \mathrm{\parallel }(I-\rho A)(w-w^{\mathrm{\prime }})\mathrm{\parallel }\le q\mathrm{\parallel }w-w^{\mathrm{\prime }}\mathrm{\parallel }$ for all $w,w^{\mathrm{\prime }}\in K$ by the nonexpansiveness of $P_K$ [F4], that is, $T$ is a contraction of $K$ with constant $q<1$ [F6].

The set $K$ is a nonempty closed subset of the complete metric space $H$, hence complete for the subspace metric [F5]; the contraction $T:K\to K$ of step 2.1 therefore has exactly one fixed point $u\in K$ by [F6], that is, $u=P_K(u-\rho (Au-f))$.

For $u\in K$, the fixed point equation $u=P_K(u-\rho (Au-f))$ is equivalent, by [F7] applied with $x=u-\rho (Au-f)$, to $⟨u-(u-\rho (Au-f)),v-u⟩\ge 0$ for every $v\in K$, that is, to $\rho ⟨Au-f,v-u⟩\ge 0$ for every $v\in K$; since $\rho >0$ this is equivalent to $⟨Au-f,v-u⟩\ge 0$, hence to $a(u,v-u)\ge F(v-u)$ for every $v\in K$ by step 1.1.

(Uniqueness and conclusion) Let $u,u^{\mathrm{\prime }}\in K$ both satisfy the variational inequality. Testing the inequality for $u$ at $v=u^{\mathrm{\prime }}$ and the inequality for $u^{\mathrm{\prime }}$ at $v=u$ and adding gives $a(u,u^{\mathrm{\prime }}-u)+a(u^{\mathrm{\prime }},u-u^{\mathrm{\prime }})\ge F(u^{\mathrm{\prime }}-u)+F(u-u^{\mathrm{\prime }})=0$; bilinearity turns the left-hand side into $-a(u-u^{\mathrm{\prime }},u-u^{\mathrm{\prime }})$, so $a(u-u^{\mathrm{\prime }},u-u^{\mathrm{\prime }})\le 0$, and coercivity gives $\alpha \mathrm{\parallel }u-u^{\mathrm{\prime }}{\mathrm{\parallel }}^2\le a(u-u^{\mathrm{\prime }},u-u^{\mathrm{\prime }})\le 0$, hence $u=u^{\mathrm{\prime }}$. The zero-dimensional case was settled in step 2.1; in the remaining case steps 3.1 and 4.1 exhibit the unique fixed point $u\in K$ which solves the inequality, and the uniqueness argument just given makes it the only solution. This proves the statement, Countable Choice having entered only through [F1] and [F4] [A1].