[proofplan] We approximate the infinite-dimensional variational inequality by finite-dimensional problems on compact convex subsets $K_{F,R} = K \cap F \cap \overline{B}(0, R)$, where $F$ ranges over finite-dimensional subspaces of $V$. Brouwer's theorem provides solutions on each $K_{F,R}$. Coercivity yields a uniform bound $\|u_F\|_V \le M$ independent of $F$, which allows extending the inequality from $K_{F,R}$ to $K_F$. We convert to the Minty form via [Minty's Lemma](/theorems/108), define weakly compact sets $S_v$ for each $v \in K$, verify the finite intersection property using the Galerkin solutions, and extract a global solution from the nonempty intersection. [/proofplan] [step:Construct Galerkin approximations on finite-dimensional compact convex subsets] Let $\Lambda$ denote the directed set of all finite-dimensional subspaces of $V$. For each $F \in \Lambda$, define $K_F = K \cap F$. For $R > 0$, define \begin{align*} K_{F,R} &= \{v \in K_F \mid \|v\|_V \le R\}. \end{align*} Since $F$ is finite-dimensional, the Heine--Borel theorem implies that $K_{F,R}$ is compact. The set $K_{F,R}$ is also nonempty (it contains $0 \in K \cap F$ for $R > 0$) and convex. Define the inclusion $j_F: F \hookrightarrow V$ and its dual $j_F^*: V^* \to F^*$ by $j_F^*(f) = f \circ j_F$. The restricted operator \begin{align*} A_F: K_{F,R} &\to F^*, \quad x \mapsto j_F^*(A(j_F(x))) \end{align*} is continuous because $A$ is hemicontinuous. By the finite-dimensional existence theorem (Brouwer), there exists $u_{F,R} \in K_{F,R}$ satisfying \begin{align*} A(u_{F,R}) \circ (v - u_{F,R}) &\ge 0, \quad \forall v \in K_{F,R}. \end{align*} [guided] The idea of the Galerkin method is to solve the variational inequality on finite-dimensional slices of $K$ where compactness is available, then pass to the limit. In infinite dimensions, $K$ is generally not compact, so Brouwer's theorem does not apply directly. For each finite-dimensional subspace $F \le V$, the intersection $K_F = K \cap F$ is a closed convex subset of $F$. Intersecting further with the closed ball $\overline{B}(0, R) = \{v \in V : \|v\|_V \le R\}$ produces \begin{align*} K_{F,R} &= \{v \in K_F \mid \|v\|_V \le R\}. \end{align*} This set is closed and bounded in the finite-dimensional space $F$, hence compact by Heine--Borel. It is nonempty because $0 \in K$ (by hypothesis) and $0 \in F$, so $0 \in K_{F,R}$ for any $R > 0$. Convexity follows from the convexity of $K$, $F$, and the closed ball. To apply the finite-dimensional existence theorem, we need a continuous operator on $K_{F,R}$. Define the inclusion $j_F: F \hookrightarrow V$ (the identity on $F$) and its dual $j_F^*: V^* \to F^*$ by $j_F^*(f) = f \circ j_F$, which restricts a functional on $V$ to $F$. The operator $A_F = j_F^* \circ A \circ j_F: K_{F,R} \to F^*$ is continuous because $A$ is hemicontinuous on $K$. By the finite-dimensional existence result (which uses Brouwer's Fixed-Point Theorem), there exists $u_{F,R} \in K_{F,R}$ with \begin{align*} A_F(u_{F,R}) \circ (v - u_{F,R}) &\ge 0, \quad \forall v \in K_{F,R}. \end{align*} Unwinding the definition of $A_F$ and $j_F^*$, this is $A(u_{F,R}) \circ (v - u_{F,R}) \ge 0$ for all $v \in K_{F,R}$. [/guided] [/step] [step:Obtain a uniform bound on $\|u_{F,R}\|_V$ via coercivity] Since $0 \in K_{F,R}$, setting $v = 0$ in the finite-dimensional variational inequality gives \begin{align*} A(u_{F,R}) \circ (0 - u_{F,R}) &\ge 0, \end{align*} hence $A(u_{F,R}) \circ u_{F,R} \le 0$. By coercivity, $A(x) \circ x / \|x\|_V \to +\infty$ as $\|x\|_V \to \infty$. In particular, there exists $M > 0$ (independent of $F$) such that $A(x) \circ x > 0$ whenever $\|x\|_V > M$. Therefore $\|u_{F,R}\|_V \le M$. Fix $R > M$. Then $u_{F,R}$ lies in the interior of $\overline{B}(0, R)$ relative to $F$. [guided] The coercivity hypothesis provides the crucial a priori bound that prevents solutions from escaping to infinity. Testing the variational inequality at $v = 0 \in K_{F,R}$: \begin{align*} A(u_{F,R}) \circ (0 - u_{F,R}) &\ge 0, \quad \text{i.e.,} \quad A(u_{F,R}) \circ u_{F,R} \le 0. \end{align*} The coercivity condition $\lim_{\|x\|_V \to \infty} A(x) \circ x / \|x\|_V = +\infty$ means: for any $C > 0$, there exists $R_C > 0$ such that $A(x) \circ x > C\|x\|_V$ whenever $\|x\|_V > R_C$. Taking $C = 0$: there exists $M > 0$ such that $\|x\|_V > M$ implies $A(x) \circ x > 0$. Since $A(u_{F,R}) \circ u_{F,R} \le 0$, we must have $\|u_{F,R}\|_V \le M$. This bound $M$ depends only on $A$ and $K$, not on the subspace $F$ or the radius $R$. Fixing $R > M$ ensures $\|u_{F,R}\|_V \le M < R$, so $u_{F,R}$ lies strictly inside $\overline{B}(0, R)$ within $F$. [/guided] [/step] [step:Extend the finite-dimensional inequality from $K_{F,R}$ to all of $K_F$] Let $v \in K_F$ be arbitrary. Define $w_t = u_{F,R} + t(v - u_{F,R})$ for $t \in (0, 1)$. Since $\|u_{F,R}\|_V < R$, for $t > 0$ sufficiently small, $\|w_t\|_V \le R$, so $w_t \in K_{F,R}$. Substituting $v = w_t$ into the inequality from Step 1: \begin{align*} A(u_{F,R}) \circ (w_t - u_{F,R}) &\ge 0. \end{align*} Since $w_t - u_{F,R} = t(v - u_{F,R})$ and $t > 0$, dividing by $t$: \begin{align*} A(u_{F,R}) \circ (v - u_{F,R}) &\ge 0, \quad \forall v \in K_F. \end{align*} Write $u_F := u_{F,R}$ for this solution on $K_F$. [/step] [step:Convert to the Minty form and define weakly compact sets $S_v$] By [Minty's Lemma](/theorems/108) applied on the convex set $K_F$ with the monotone hemicontinuous operator $A$, the Stampacchia inequality $A(u_F) \circ (v - u_F) \ge 0$ for all $v \in K_F$ is equivalent to the Minty inequality: \begin{align*} A(v) \circ (v - u_F) &\ge 0, \quad \forall v \in K_F. \end{align*} For each $v \in K$, define \begin{align*} S_v &= \{x \in K \mid \|x\|_V \le M \text{ and } A(v) \circ (v - x) \ge 0\}. \end{align*} The set $\{x \in K \mid \|x\|_V \le M\}$ is closed, bounded, and convex in $V$, hence weakly compact (since bounded closed convex subsets of a Hilbert space are weakly compact). The set $\{x \in V \mid A(v) \circ (v - x) \ge 0\} = \{x \in V \mid A(v) \circ x \le A(v) \circ v\}$ is weakly closed because $A(v) \in V^*$ is a continuous linear functional (hence weakly continuous). The intersection $S_v$ is therefore weakly compact. [guided] The Minty reformulation is the key step that makes the weak compactness argument work. The Stampacchia inequality $A(u_F) \circ (v - u_F) \ge 0$ involves $A$ evaluated at the unknown $u_F$, which varies with $F$. The Minty form $A(v) \circ (v - u_F) \ge 0$ fixes $A$ at the known test point $v$, turning each inequality into a linear constraint on $u_F$. Applying [Minty's Lemma](/theorems/108) requires monotonicity and hemicontinuity of $A$, both given by hypothesis. The lemma yields the equivalence: $A(u_F) \circ (v - u_F) \ge 0$ for all $v \in K_F$ if and only if $A(v) \circ (v - u_F) \ge 0$ for all $v \in K_F$. For each $v \in K$, the set $S_v = \{x \in K : \|x\|_V \le M, \; A(v) \circ (v - x) \ge 0\}$ captures all points in the bounded part of $K$ that satisfy the Minty inequality for the specific test vector $v$. Why is $S_v$ weakly compact? In a Hilbert space, the closed ball $\overline{B}(0, M)$ is weakly compact (by the Banach--Alaoglu theorem applied to the reflexive space $V$). The set $K$ is convex and (norm-)closed, hence weakly closed. Their intersection $\{x \in K : \|x\|_V \le M\}$ is weakly closed inside the weakly compact ball, hence weakly compact. The half-space $\{x \in V : A(v) \circ x \le A(v) \circ v\}$ is weakly closed because $x \mapsto A(v) \circ x$ is a continuous linear functional on $V$, and continuous linear functionals are weakly continuous. The intersection $S_v$ of a weakly compact set and a weakly closed set is weakly compact. [/guided] [/step] [step:Verify the finite intersection property and extract a global solution] Let $v_1, \ldots, v_m \in K$ and let $F = \operatorname{span}\{v_1, \ldots, v_m\}$. Each $v_i \in K_F$, and the Galerkin solution $u_F$ satisfies $\|u_F\|_V \le M$ and $A(v_i) \circ (v_i - u_F) \ge 0$ for each $i$. Therefore $u_F \in S_{v_i}$ for all $i$, so \begin{align*} \bigcap_{i=1}^m S_{v_i} &\neq \varnothing. \end{align*} The family $\{S_v\}_{v \in K}$ of weakly compact sets satisfies the finite intersection property. By the general topological principle that a family of compact sets with the finite intersection property has nonempty total intersection: \begin{align*} \bigcap_{v \in K} S_v &\neq \varnothing. \end{align*} Let $u \in \bigcap_{v \in K} S_v$. Then $u \in K$, $\|u\|_V \le M$, and $A(v) \circ (v - u) \ge 0$ for all $v \in K$. This is the Minty variational inequality on $K$. By the reverse implication of [Minty's Lemma](/theorems/108): \begin{align*} A(u) \circ (v - u) &\ge 0, \quad \forall v \in K. \end{align*} [guided] The finite intersection property argument connects the finite-dimensional solutions to a global solution. For any finite collection $v_1, \ldots, v_m \in K$, let $F = \operatorname{span}\{v_1, \ldots, v_m\}$. This is a finite-dimensional subspace, so we have a Galerkin solution $u_F \in K_F$ with $\|u_F\|_V \le M$. Since each $v_i \in F$, we have $v_i \in K_F$, and the Minty inequality on $K_F$ gives $A(v_i) \circ (v_i - u_F) \ge 0$. Combined with $u_F \in K$ and $\|u_F\|_V \le M$, this means $u_F \in S_{v_i}$ for every $i \in \{1, \ldots, m\}$, so $u_F \in \bigcap_{i=1}^m S_{v_i}$. Since every finite subcollection of $\{S_v\}_{v \in K}$ has nonempty intersection, and each $S_v$ is weakly compact, the finite intersection property for compact sets (in the weak topology) guarantees \begin{align*} \bigcap_{v \in K} S_v &\neq \varnothing. \end{align*} Pick any $u$ in this intersection. By definition of $S_v$, for every $v \in K$ we have $u \in K$, $\|u\|_V \le M$, and $A(v) \circ (v - u) \ge 0$. This is the Minty variational inequality on the full set $K$. To recover the Stampacchia form, we apply the reverse implication of [Minty's Lemma](/theorems/108). The hypotheses of the lemma (monotonicity and hemicontinuity of $A$, convexity and closedness of $K$) are all in force. Therefore $A(u) \circ (v - u) \ge 0$ for all $v \in K$, completing the existence proof. [/guided] [/step]