The proof consists of three stages: Galerkin approximation on truncated constraint [sets](/page/Set), passage to the [limit](/page/Limit) in the finite-dimensional parameter, and removal of the truncation via coercivity. **Step 1 (Truncation and Galerkin approximation).** Fix $R > 0$ and define $K_R := K \cap \overline{B}_V(R)$, which is compact and convex. Let $(V_j)_{j \ge 1}$ be an increasing [sequence](/page/Sequence) of finite-dimensional subspaces whose union is dense in $V$, and set $K_{j,R} := K_R \cap V_j$. Each $K_{j,R}$ is a nonempty compact convex subset of the finite-dimensional space $V_j$. Since $A$ is hemicontinuous, its restriction to $V_j$ is [continuous](/page/Continuity). By the [Brouwer fixed point theorem](/theorems/80) (applied via the projection characterisation: the map $P_{K_{j,R}} \circ (I - A): K_{j,R} \to K_{j,R}$ is continuous on a compact convex set), there exists $u_{j,R} \in K_{j,R}$ satisfying $A(u_{j,R}) \circ (v - u_{j,R}) \ge 0$ for all $v \in K_{j,R}$. **Step 2 (Uniform bounds and passage to $j \to \infty$).** Testing with $v = 0 \in K_{j,R}$ gives $A(u_{j,R}) \circ u_{j,R} \le 0$. The coercivity condition prevents $\|u_{j,R}\|_V \to \infty$: if this occurred, then $A(u_{j,R}) \circ u_{j,R} / \|u_{j,R}\|_V \to +\infty$, contradicting the bound above. Hence $\{u_{j,R}\}$ is bounded, and a subsequence [converges weakly](/page/Weak%20Convergence) to some $u_R \in K_R$ (since $K_R$ is weakly closed). The sequence $\{A(u_{j,R})\}$ is bounded in $V^*$ (a monotone hemicontinuous operator maps bounded sets to bounded sets), so a further subsequence satisfies $A(u_{j,R}) \rightharpoonup \eta_R$ in $V^*$. To identify $\eta_R = A(u_R)$, we use [Minty's lemma](/theorems/108): passing the Galerkin inequality to the limit (using projections $P_j v \to v$ strongly for any $v \in K_R$) yields $\lim\sup A(u_{j,R}) \circ u_{j,R} \le \eta_R \circ u_R$, which together with monotonicity and hemicontinuity gives $\eta_R = A(u_R)$. The limit inequality is $A(u_R) \circ (v - u_R) \ge 0$ for all $v \in K_R$. **Step 3 (Removal of the truncation).** It suffices to find $R$ such that $\|u_R\|_V < R$, for then $u_R$ lies in the interior of $\overline{B}_V(R)$ and the truncated inequality extends to all of $K$. Assume for contradiction that $\|u_R\|_V = R$ for all $R > 0$. Testing with $v = 0$: $A(u_R) \circ u_R \le 0$. Dividing by $\|u_R\|_V = R$ and sending $R \to \infty$, the coercivity condition gives $A(u_R) \circ u_R / \|u_R\|_V \to +\infty$, contradicting $A(u_R) \circ u_R \le 0$ for all $R$. Therefore $\|u_R\|_V < R$ for some $R$, and this $u := u_R$ solves the original variational inequality on $K$.