[proofplan] We reformulate the variational inequality as a fixed-point problem: $u \in K$ solves $a(u, v-u) \ge f(v-u)$ for all $v \in K$ if and only if $u = P_K(u - \rho \mathcal{J}(\mathcal{A}u - f))$ for any $\rho > 0$, where $\mathcal{A}: V \to V^*$ is the operator induced by $a$, $\mathcal{J}: V^* \to V$ is the Riesz isomorphism, and $P_K$ is the metric projection. We then show that the resulting map $T_\rho(v) = P_K(v - \rho\mathcal{J}(\mathcal{A}v - f))$ is a strict contraction for $\rho \in (0, 2\alpha/M^2)$, where $\alpha$ is the coercivity constant and $M$ is the continuity constant, and apply the [Banach Fixed Point Theorem](/theorems/289). [/proofplan] [step:Introduce the operator $\mathcal{A}$ and the Riesz map $\mathcal{J}$] Associate with the bilinear form $a$ a bounded [linear operator](/page/Bounded%20Linear%20Operator) $\mathcal{A}: V \to V^*$ defined by \begin{align*} (\mathcal{A}u)(v) = a(u, v) \quad \text{for all } u, v \in V. \end{align*} By continuity of $a$, $\|\mathcal{A}u\|_{V^*} \le M\|u\|_V$, so $\mathcal{A}$ is bounded with $\|\mathcal{A}\|_{\mathcal{L}(V, V^*)} \le M$. Let $\mathcal{J}: V^* \to V$ denote the inverse of the Riesz isomorphism, the isometric isomorphism satisfying \begin{align*} (\mathcal{J}\phi, v)_V = \phi(v) \quad \text{for all } \phi \in V^*, \, v \in V. \end{align*} In particular, $\|\mathcal{J}\phi\|_V = \|\phi\|_{V^*}$ for all $\phi \in V^*$. [/step] [step:Reformulate the variational inequality as a fixed-point equation for $P_K$] The variational inequality $a(u, v - u) \ge f(v - u)$ for all $v \in K$ can be written as \begin{align*} (\mathcal{A}u - f)(v - u) \ge 0 \quad \text{for all } v \in K. \end{align*} Applying $\mathcal{J}$ and using $(\mathcal{J}\phi, w)_V = \phi(w)$: \begin{align*} (\mathcal{J}(\mathcal{A}u - f), v - u)_V \ge 0 \quad \text{for all } v \in K. \end{align*} For any $\rho > 0$, multiplying by $\rho$ preserves the inequality: \begin{align*} (u - [u - \rho\mathcal{J}(\mathcal{A}u - f)], v - u)_V \ge 0 \quad \text{for all } v \in K. \end{align*} By the [variational characterisation of the metric projection](/theorems/87), this holds if and only if \begin{align*} u = P_K(u - \rho\mathcal{J}(\mathcal{A}u - f)). \end{align*} Define the map \begin{align*} T_\rho: V &\to K, \\ v &\mapsto P_K(v - \rho\mathcal{J}(\mathcal{A}v - f)). \end{align*} The variational inequality is equivalent to $u = T_\rho(u)$. [guided] The goal is to convert the variational inequality into a form where a contraction mapping argument applies. The projection characterisation from [Theorem 87](/theorems/87) says: $u = P_K(z)$ if and only if $(u - z, v - u)_V \ge 0$ for all $v \in K$. Starting from the variational inequality $a(u, v-u) \ge f(v-u)$, rewrite it as: \begin{align*} (\mathcal{A}u - f)(v - u) \ge 0 \quad \text{for all } v \in K. \end{align*} Apply the Riesz map: since $(\mathcal{J}\phi, w)_V = \phi(w)$, this becomes \begin{align*} (\mathcal{J}(\mathcal{A}u - f), v - u)_V \ge 0 \quad \text{for all } v \in K. \end{align*} Multiplying by $\rho > 0$ and rewriting: \begin{align*} (u - [u - \rho\mathcal{J}(\mathcal{A}u - f)], v - u)_V \ge 0 \quad \text{for all } v \in K. \end{align*} Setting $z = u - \rho\mathcal{J}(\mathcal{A}u - f)$, the projection characterisation gives $u = P_K(z)$. This is precisely the fixed-point equation $u = T_\rho(u)$. Why introduce the parameter $\rho$? The bilinear form $a$ is not necessarily symmetric, so $\mathcal{A}$ is not self-adjoint, and we cannot directly apply spectral methods. The parameter $\rho$ gives a free scaling that we will optimise to ensure the contraction constant is less than $1$. [/guided] [/step] [step:Prove $T_\rho$ is a contraction for $\rho \in (0, 2\alpha/M^2)$] Let $u, w \in V$. By the [non-expansiveness of $P_K$](/theorems/91): \begin{align*} \|T_\rho(u) - T_\rho(w)\|_V^2 &\le \|(u - w) - \rho\mathcal{J}(\mathcal{A}(u - w))\|_V^2. \end{align*} Set $\xi = u - w$. Expand the squared norm using the Riesz property $(\mathcal{J}\phi, z)_V = \phi(z)$: \begin{align*} \|\xi - \rho\mathcal{J}(\mathcal{A}\xi)\|_V^2 &= \|\xi\|_V^2 - 2\rho(\mathcal{J}(\mathcal{A}\xi), \xi)_V + \rho^2\|\mathcal{J}(\mathcal{A}\xi)\|_V^2 \\ &= \|\xi\|_V^2 - 2\rho\, a(\xi, \xi) + \rho^2\|\mathcal{A}\xi\|_{V^*}^2. \end{align*} By coercivity, $a(\xi, \xi) \ge \alpha\|\xi\|_V^2$. By continuity, $\|\mathcal{A}\xi\|_{V^*} \le M\|\xi\|_V$. Substituting: \begin{align*} \|T_\rho(u) - T_\rho(w)\|_V^2 &\le \|\xi\|_V^2 - 2\rho\alpha\|\xi\|_V^2 + \rho^2 M^2\|\xi\|_V^2 \\ &= (1 - 2\rho\alpha + \rho^2 M^2)\|\xi\|_V^2. \end{align*} Define $k_\rho^2 = 1 - 2\rho\alpha + \rho^2 M^2$. The condition $k_\rho < 1$ requires $1 - 2\rho\alpha + \rho^2 M^2 < 1$, i.e., $\rho(M^2\rho - 2\alpha) < 0$, which holds for $\rho \in (0, 2\alpha/M^2)$. [guided] The key estimate combines three ingredients: non-expansiveness of $P_K$, coercivity of $a$, and continuity of $a$. Starting from the non-expansiveness of the projection: \begin{align*} \|T_\rho(u) - T_\rho(w)\|_V \le \|(u - w) - \rho\mathcal{J}(\mathcal{A}(u-w))\|_V. \end{align*} Set $\xi = u - w$ and square both sides. Expanding $\|\xi - \rho\mathcal{J}(\mathcal{A}\xi)\|_V^2$: \begin{align*} &= (\xi - \rho\mathcal{J}(\mathcal{A}\xi), \xi - \rho\mathcal{J}(\mathcal{A}\xi))_V \\ &= \|\xi\|_V^2 - 2\rho(\mathcal{J}(\mathcal{A}\xi), \xi)_V + \rho^2\|\mathcal{J}(\mathcal{A}\xi)\|_V^2. \end{align*} For the cross term: $(\mathcal{J}(\mathcal{A}\xi), \xi)_V = (\mathcal{A}\xi)(\xi) = a(\xi, \xi) \ge \alpha\|\xi\|_V^2$ by coercivity. This contributes $-2\rho\alpha\|\xi\|_V^2$. For the last term: $\|\mathcal{J}(\mathcal{A}\xi)\|_V = \|\mathcal{A}\xi\|_{V^*} \le M\|\xi\|_V$ by continuity. This contributes $+\rho^2 M^2\|\xi\|_V^2$. Combining: \begin{align*} \|T_\rho(u) - T_\rho(w)\|_V^2 \le (1 - 2\rho\alpha + \rho^2 M^2)\|u - w\|_V^2. \end{align*} The quadratic $q(\rho) = \rho^2 M^2 - 2\rho\alpha$ is negative when $0 < \rho < 2\alpha/M^2$, so $k_\rho = \sqrt{1 - 2\rho\alpha + \rho^2 M^2} < 1$ in this range. The optimal choice $\rho = \alpha/M^2$ minimises $k_\rho$, giving $k_\rho = \sqrt{1 - \alpha^2/M^2}$. [/guided] [/step] [step:Apply the Banach Fixed Point Theorem to obtain the unique solution] Choose $\rho = \alpha/M^2 \in (0, 2\alpha/M^2)$. Then $T_\rho: V \to V$ is a strict contraction on the [complete metric space](/page/Complete%20Metric%20Space) $V$. By the [Banach Fixed Point Theorem](/theorems/289), there exists a unique $u \in V$ with $T_\rho(u) = u$. Since $T_\rho(V) \subseteq P_K(V) = K$, we have $u \in K$. By the equivalence established above, $u$ is the unique element of $K$ satisfying $a(u, v - u) \ge f(v - u)$ for all $v \in K$. [/step]