[proofplan] We construct a continuous self-map $T: K \to K$ by composing the perturbation $x \mapsto x - A(x)$ with the metric projection $P_K$ onto the compact convex set $K$. Brouwer's Fixed-Point Theorem provides a fixed point, and the projection characterisation converts this fixed point into a solution of the variational inequality. Uniqueness under strict monotonicity follows by testing the two variational inequalities against each other and deriving a contradiction. [/proofplan] [step:Construct a continuous self-map on $K$ via the metric projection] Define the metric projection onto the closed convex set $K$: \begin{align*} P_K: \mathbb{R}^n &\to K, \quad x \mapsto \operatorname{argmin}_{z \in K} \|x - z\|_{\mathbb{R}^n}. \end{align*} Define the composite map \begin{align*} T: K &\to K, \quad x \mapsto P_K(x - A(x)). \end{align*} The map $x \mapsto x - A(x)$ is continuous because $A$ is continuous by hypothesis. The projection $P_K$ is non-expansive (hence continuous). The composition of continuous maps is continuous, so $T: K \to K$ is continuous. [guided] The goal is to produce a self-map of $K$ whose fixed points solve the variational inequality. The natural candidate is $T(x) = P_K(x - A(x))$: we shift $x$ in the direction $-A(x)$ and project back onto $K$. The intuition is that if $u$ solves $(A(u), v - u)_{\mathbb{R}^n} \ge 0$ for all $v \in K$, then $-A(u)$ points "inward" at $u$ relative to $K$, so $u - A(u)$ projects back to $u$ itself. We verify that $T$ is well-defined. The metric projection $P_K: \mathbb{R}^n \to K$ assigns to each $z \in \mathbb{R}^n$ the unique minimiser of $\|z - y\|_{\mathbb{R}^n}$ over $y \in K$. Existence and uniqueness follow from the fact that $K$ is nonempty, closed, and convex: the distance function $y \mapsto \|z - y\|_{\mathbb{R}^n}^2$ is strictly convex and coercive on the closed convex set $K$, so it attains a unique minimum. Next we verify continuity. The projection $P_K$ is non-expansive: for all $z_1, z_2 \in \mathbb{R}^n$, \begin{align*} \|P_K(z_1) - P_K(z_2)\|_{\mathbb{R}^n} &\le \|z_1 - z_2\|_{\mathbb{R}^n}. \end{align*} In particular, $P_K$ is Lipschitz with constant $1$, hence continuous. The map $x \mapsto x - A(x)$ sends $K \subset \mathbb{R}^n$ into $\mathbb{R}^n$ and is continuous because $A: K \to \mathbb{R}^n$ is continuous by hypothesis and the identity map is continuous. The composition $T = P_K \circ (\operatorname{id} - A)$ is therefore continuous as a composition of continuous maps, and it maps $K$ into $K$ because $P_K$ has range $K$. [/guided] [/step] [step:Apply Brouwer's Fixed-Point Theorem to obtain a fixed point] The set $K \subset \mathbb{R}^n$ is nonempty, compact, and convex. The map $T: K \to K$ is continuous. By Brouwer's Fixed-Point Theorem, there exists $u \in K$ with $T(u) = u$, i.e., \begin{align*} P_K(u - A(u)) &= u. \end{align*} [/step] [step:Convert the fixed-point equation into the variational inequality via the projection characterisation] By the projection characterisation, $u = P_K(z)$ for $z = u - A(u)$ if and only if $u \in K$ and \begin{align*} (z - u, v - u)_{\mathbb{R}^n} &\le 0, \quad \forall v \in K. \end{align*} Substituting $z = u - A(u)$: \begin{align*} ((u - A(u)) - u, \; v - u)_{\mathbb{R}^n} &\le 0, \quad \forall v \in K, \\ (-A(u), \; v - u)_{\mathbb{R}^n} &\le 0, \quad \forall v \in K. \end{align*} Multiplying by $-1$ reverses the inequality: \begin{align*} (A(u), v - u)_{\mathbb{R}^n} &\ge 0, \quad \forall v \in K. \end{align*} This is the variational inequality, confirming that $u$ is a solution. [guided] The projection characterisation theorem states: given a closed convex set $K$ and a point $z \in \mathbb{R}^n$, the element $p = P_K(z)$ is the unique element of $K$ satisfying $(z - p, v - p)_{\mathbb{R}^n} \le 0$ for all $v \in K$. Geometrically, this says $z - p$ makes an obtuse angle with every direction $v - p$ pointing into $K$ from $p$. Applying this with $z = u - A(u)$ and $p = u$ (since $P_K(u - A(u)) = u$): \begin{align*} ((u - A(u)) - u, \; v - u)_{\mathbb{R}^n} &\le 0, \quad \forall v \in K. \end{align*} The $u$ terms cancel in the first argument, leaving $(-A(u), v - u)_{\mathbb{R}^n} \le 0$. Extracting the factor of $-1$ by linearity and multiplying both sides by $-1$: \begin{align*} (A(u), v - u)_{\mathbb{R}^n} &\ge 0, \quad \forall v \in K. \end{align*} This is exactly the finite-dimensional variational inequality. [/guided] [/step] [step:Prove uniqueness under strict monotonicity by contradiction] Assume $A$ is strictly monotone and suppose $u_1, u_2 \in K$ are both solutions with $u_1 \neq u_2$. Then \begin{align*} (A(u_1), v - u_1)_{\mathbb{R}^n} &\ge 0 \quad \forall v \in K, \\ (A(u_2), v - u_2)_{\mathbb{R}^n} &\ge 0 \quad \forall v \in K. \end{align*} Setting $v = u_2$ in the first and $v = u_1$ in the second (both valid since $u_1, u_2 \in K$): \begin{align*} (A(u_1), u_2 - u_1)_{\mathbb{R}^n} &\ge 0, \\ (A(u_2), u_1 - u_2)_{\mathbb{R}^n} &\ge 0. \end{align*} Adding these two inequalities (using $u_1 - u_2 = -(u_2 - u_1)$ in the second): \begin{align*} (A(u_1) - A(u_2), u_2 - u_1)_{\mathbb{R}^n} &\ge 0. \end{align*} Multiplying by $-1$: \begin{align*} (A(u_1) - A(u_2), u_1 - u_2)_{\mathbb{R}^n} &\le 0. \end{align*} But strict monotonicity requires $(A(u_1) - A(u_2), u_1 - u_2)_{\mathbb{R}^n} > 0$ for $u_1 \neq u_2$. This is a contradiction, so $u_1 = u_2$. [guided] The uniqueness argument uses a standard "test against each other" technique for variational inequalities. Given two solutions $u_1, u_2 \in K$, each satisfies the variational inequality for all test vectors in $K$. The key observation is that $u_2 \in K$ is a valid test vector for $u_1$'s inequality, and vice versa. Setting $v = u_2$ in $u_1$'s inequality and $v = u_1$ in $u_2$'s: \begin{align*} (A(u_1), u_2 - u_1)_{\mathbb{R}^n} &\ge 0, \\ (A(u_2), u_1 - u_2)_{\mathbb{R}^n} &\ge 0. \end{align*} Writing $u_1 - u_2 = -(u_2 - u_1)$ in the second inequality and using linearity of the inner product in the second argument: \begin{align*} -(A(u_2), u_2 - u_1)_{\mathbb{R}^n} &\ge 0. \end{align*} Adding to the first inequality: \begin{align*} (A(u_1) - A(u_2), u_2 - u_1)_{\mathbb{R}^n} &\ge 0. \end{align*} Negating both sides (which reverses the inequality) and using $(u_2 - u_1) = -(u_1 - u_2)$: \begin{align*} (A(u_1) - A(u_2), u_1 - u_2)_{\mathbb{R}^n} &\le 0. \end{align*} Strict monotonicity demands $(A(x) - A(y), x - y)_{\mathbb{R}^n} > 0$ whenever $x \neq y$. Applied to $x = u_1$, $y = u_2$, this gives a strict positive lower bound, contradicting $\le 0$. Therefore $u_1 = u_2$, and the solution is unique. [/guided] [/step]