Let $H$ be a real Hilbert space and $K \in H$ non-empty, convex and closed. Let $A:H\times H\to \mathbb{R}$ be a bounded, coercive, symmetric bilinear form: $|A(u,v)|\leq\mu_1||u||||v||$ for all $u,v \in H$, $A(u,u)\geq \mu_2||u||^2$ for all $u \in H$ and $A(u,v)=A(v,u)$ for all $u,v \in H$. Let $f\in H^{\star}$ and consider the variational inequality \begin{equation} \label{eq:var} \begin{cases} u\in K \\ A(u,v-u)\geq f(v-u) \text{ for all } v\in K. \end{cases} \end{equation} Prove the existence of some $u\in K$ solving the above inequality as a follows:
a) Apply Riesz representation on $H,A$ to find an element $w\in H$ such that $A(w,v)=f(v)$ holds for all $v\in H$.
b) Consider the projection $u=P_K(w)$ of $w$ onto $K$. Show that $u$ solves the variational inequality.
c) Give a functional $J:K \to \mathbb{R}$ such that if $u\in K$ is a minimiser for $J$, then $u$ solves the variational inequality.
For a) we know by Riesz Representation theorem that for each $f\in H^{\star}$ there exists $w \in H$ such that $$\langle v,w\rangle_H=f(v)=(\langle w,v\rangle_H)$$ for all $v\in H$ ($H$ is real). I need to relate this with the bilinear form. I found this useful post Riesz representation theorem for bilinear forms which claims that for $u\in H$ there exists $w\in H$ such that $$A(u,v)=\langle v,w\rangle_H $$ for all $v\in H$. Then for $v\in H$ I see $$A(u,v)=f(v)$$ which looks like what I want but with $v$ instead of $w$. I guess I have to be careful here with the quantifiers and their order as well…
For b) I am trying to collect what I know and how that can show me the inequality. know by a) that $A\langle w,v-u\rangle=f(v,u)$ which looks quite similar to wwhat we want. Certainly $u\in K$ and I also know $\langle w,v-u\rangle=0$. To get the inequality in the desired direction I guess I need to use coercivity somehow and by Pythagoras I can tie it orthogonality? I can write something like $A(w-u,w-u)+A(u,u)\geq\mu_2^2|w|^2$ but that isn't very useful…
Point b) is not true as stated. You have to project $w$ w.r.t. the norm induced by $A$, i.e., you equip $H$ with the equivalent norm $$\| x \|_A := \sqrt{ A(x,x) }.$$