[proofplan] We reformulate the variational inequality as a fixed-point problem by applying the Riesz Representation Theorem to encode both the bilinear form and the linear functional as inner products, then introducing a projection-based iteration map. We prove that the metric projection onto $K$ is non-expansive, use the coercivity and continuity bounds on the bilinear form to show the iteration map is a strict contraction for a suitable step-size parameter, and conclude existence and uniqueness via the Banach Fixed-Point Theorem. [/proofplan] [step:Represent the bilinear form and the functional via the Riesz map] By the [Riesz Representation Theorem](/theorems/221), every element of $V^*$ has a unique representative in $V$. Define $w \in V$ as the Riesz representative of $f$: \begin{align*} (w, x)_V &= f \circ x, \quad \forall x \in V. \end{align*} For each fixed $u \in V$, the map $x \mapsto a(u, x)$ is a continuous linear functional on $V$ with operator norm at most $M \|u\|_V$. By the Riesz Representation Theorem again, there exists a unique element $A(u) \in V$ satisfying \begin{align*} (A(u), x)_V &= a(u, x), \quad \forall x \in V. \end{align*} This defines a linear operator $A: V \to V$, $u \mapsto A(u)$. The variational inequality $a(u, v - u) \ge f \circ (v - u)$ for all $v \in K$ becomes \begin{align*} (A(u) - w, v - u)_V &\ge 0, \quad \forall v \in K. \end{align*} [guided] The strategy is to translate the variational inequality from the language of bilinear forms and dual pairings into the language of inner products and operators, where projection characterisations are available. By the [Riesz Representation Theorem](/theorems/221), every continuous linear functional on a Hilbert space is represented by an inner product with a unique element of the space. We apply this twice. First, $f \in V^*$ is a continuous linear functional, so there exists a unique $w \in V$ with \begin{align*} (w, x)_V &= f \circ x, \quad \forall x \in V. \end{align*} Second, for each fixed $u \in V$, the map $x \mapsto a(u, x)$ is linear in $x$. It is continuous because $|a(u, x)| \le M \|u\|_V \|x\|_V$ by the continuity hypothesis on $a$. The Riesz Representation Theorem provides a unique element $A(u) \in V$ with \begin{align*} (A(u), x)_V &= a(u, x), \quad \forall x \in V. \end{align*} This defines a linear operator $A: V \to V$, $u \mapsto A(u)$. Linearity of $A$ follows from the linearity of $a$ in its first argument: for $u_1, u_2 \in V$ and $\lambda \in \mathbb{R}$, the functional $x \mapsto a(u_1 + \lambda u_2, x) = a(u_1, x) + \lambda a(u_2, x)$ has Riesz representative $A(u_1) + \lambda A(u_2)$, so $A(u_1 + \lambda u_2) = A(u_1) + \lambda A(u_2)$. Substituting the Riesz representations into the variational inequality $a(u, v - u) \ge f \circ (v - u)$ yields \begin{align*} (A(u), v - u)_V &\ge (w, v - u)_V, \quad \forall v \in K, \end{align*} which rearranges to $(A(u) - w, v - u)_V \ge 0$ for all $v \in K$. [/guided] [/step] [step:Reformulate the variational inequality as a fixed-point problem for a projection map] Let $\rho \in \mathbb{R}_{>0}$. Multiplying $(A(u) - w, v - u)_V \ge 0$ by $-\rho$ reverses the inequality: \begin{align*} (\rho(w - A(u)), v - u)_V &\le 0. \end{align*} Adding and subtracting $u$ in the first argument gives \begin{align*} ((u + \rho(w - A(u))) - u, \; v - u)_V &\le 0, \quad \forall v \in K. \end{align*} By the characterisation of the metric projection onto a closed convex set, this holds for all $v \in K$ if and only if $u = P_K(u + \rho(w - A(u)))$. Equivalently, $u$ is a fixed point of the map \begin{align*} T: K &\to K, \quad x \mapsto P_K(x - \rho(A(x) - w)). \end{align*} [guided] Why reformulate as a fixed point? Because we want to apply the [Banach Fixed-Point Theorem](/theorems/289), and for that we need a contraction on a complete metric space. The projection map $P_K$ is the key ingredient that keeps iterates inside $K$. Starting from $(A(u) - w, v - u)_V \ge 0$ for all $v \in K$, we multiply by $-\rho$ (with $\rho > 0$) to reverse the inequality: \begin{align*} (\rho(w - A(u)), v - u)_V &\le 0, \quad \forall v \in K. \end{align*} We rewrite $\rho(w - A(u)) = (u - \rho(A(u) - w)) - u$ by adding and subtracting $u$, so the inequality becomes \begin{align*} ((u - \rho(A(u) - w)) - u, \; v - u)_V &\le 0, \quad \forall v \in K. \end{align*} The characterisation of the metric projection onto a closed convex set $K$ states: $u = P_K(z)$ if and only if $u \in K$ and $(z - u, v - u)_V \le 0$ for all $v \in K$. Setting $z = u - \rho(A(u) - w)$, the inequality above is exactly the projection characterisation, so $u = P_K(u - \rho(A(u) - w))$. Define the map $T: K \to K$ by $T(x) = P_K(x - \rho(A(x) - w))$. Then $u$ solves the variational inequality if and only if $u = T(u)$. [/guided] [/step] [step:Establish that the metric projection $P_K$ is non-expansive] [claim:Non-expansiveness of $P_K$] For all $x, y \in V$, the metric projection satisfies $\|P_K(x) - P_K(y)\|_V \le \|x - y\|_V$. [/claim] [proof] Let $x, y \in V$ and set $p = P_K(x)$, $q = P_K(y)$. By the projection characterisation, for all $z \in K$: \begin{align*} (x - p, z - p)_V &\le 0, \\ (y - q, z - q)_V &\le 0. \end{align*} Since $p, q \in K$, we substitute $z = q$ into the first inequality and $z = p$ into the second: \begin{align*} (x - p, q - p)_V &\le 0, \\ (y - q, p - q)_V &\le 0. \end{align*} Writing $p - q = -(q - p)$ in the second inequality and adding the two: \begin{align*} (x - p, q - p)_V + (y - q, -(q - p))_V &\le 0, \\ ((x - p) - (y - q), q - p)_V &\le 0, \\ ((x - y) - (p - q), -(p - q))_V &\le 0. \end{align*} Expanding: \begin{align*} -(x - y, p - q)_V + \|p - q\|_V^2 &\le 0. \end{align*} Hence $\|p - q\|_V^2 \le (x - y, p - q)_V$. By the Cauchy--Schwarz inequality applied to the inner product $(\cdot, \cdot)_V$ with the pair $(x - y, p - q)$: \begin{align*} \|p - q\|_V^2 &\le \|x - y\|_V \|p - q\|_V. \end{align*} If $\|p - q\|_V = 0$ the claim holds. Otherwise, dividing by $\|p - q\|_V > 0$ yields $\|p - q\|_V \le \|x - y\|_V$. [/proof] [/step] [step:Bound the contraction factor of $T$ using coercivity and continuity of $a$] For $x, y \in K$, non-expansiveness of $P_K$ gives \begin{align*} \|T(x) - T(y)\|_V^2 &\le \|(x - y) - \rho(A(x) - A(y))\|_V^2. \end{align*} Expanding the squared norm via the inner product: \begin{align*} \|T(x) - T(y)\|_V^2 &\le \|x - y\|_V^2 - 2\rho\,(A(x) - A(y), x - y)_V + \rho^2 \|A(x) - A(y)\|_V^2. \end{align*} Since $A$ is linear, $A(x) - A(y) = A(x - y)$. By coercivity of $a$: \begin{align*} (A(x - y), x - y)_V &= a(x - y, x - y) \ge \alpha \|x - y\|_V^2. \end{align*} By continuity of $a$, we bound $\|A(x - y)\|_V$. For any $z \in V$ with $z \neq 0$: \begin{align*} \|A(z)\|_V^2 &= (A(z), A(z))_V = a(z, A(z)) \le M \|z\|_V \|A(z)\|_V, \end{align*} so $\|A(z)\|_V \le M \|z\|_V$. Substituting both bounds: \begin{align*} \|T(x) - T(y)\|_V^2 &\le (1 - 2\rho\alpha + \rho^2 M^2)\,\|x - y\|_V^2. \end{align*} [guided] We want to show that $T$ shrinks distances. The key is to translate the abstract properties of $a$ (coercivity and continuity) into quantitative estimates on the operator $A$. For $x, y \in K$, non-expansiveness of $P_K$ established above yields \begin{align*} \|T(x) - T(y)\|_V &= \|P_K(x - \rho(A(x) - w)) - P_K(y - \rho(A(y) - w))\|_V \\ &\le \|(x - \rho(A(x) - w)) - (y - \rho(A(y) - w))\|_V \\ &= \|(x - y) - \rho(A(x) - A(y))\|_V. \end{align*} The $w$ terms cancel because they appear with opposite signs. Squaring both sides: \begin{align*} \|T(x) - T(y)\|_V^2 &\le \|(x - y) - \rho(A(x-y))\|_V^2 \\ &= \|x - y\|_V^2 - 2\rho\,(A(x-y), x-y)_V + \rho^2 \|A(x-y)\|_V^2. \end{align*} Now we bound the two terms involving $A$. For the coercivity term, $a(x-y, x-y) \ge \alpha \|x-y\|_V^2$ gives \begin{align*} (A(x-y), x-y)_V &= a(x-y, x-y) \ge \alpha \|x-y\|_V^2. \end{align*} For the operator norm term, we compute $\|A(z)\|_V$ for any $z \in V$. Using the definition $(A(z), \cdot)_V = a(z, \cdot)$ evaluated at $A(z)$: \begin{align*} \|A(z)\|_V^2 &= (A(z), A(z))_V = a(z, A(z)) \le M \|z\|_V \|A(z)\|_V. \end{align*} Dividing by $\|A(z)\|_V > 0$ (the case $A(z) = 0$ is immediate) gives $\|A(z)\|_V \le M\|z\|_V$, and squaring yields $\|A(z)\|_V^2 \le M^2 \|z\|_V^2$. Substituting into the expansion: \begin{align*} \|T(x) - T(y)\|_V^2 &\le \|x-y\|_V^2 - 2\rho\alpha\|x-y\|_V^2 + \rho^2 M^2 \|x-y\|_V^2 \\ &= (1 - 2\rho\alpha + \rho^2 M^2)\,\|x-y\|_V^2. \end{align*} [/guided] [/step] [step:Choose $\rho$ to make $T$ a strict contraction and apply the Banach Fixed-Point Theorem] Define $k(\rho) = \sqrt{1 - 2\rho\alpha + \rho^2 M^2}$. The condition $k(\rho) < 1$ requires $1 - 2\rho\alpha + \rho^2 M^2 < 1$, equivalently $\rho(\rho M^2 - 2\alpha) < 0$. Since $\rho > 0$, this holds when $\rho < 2\alpha / M^2$. Fixing any $\rho \in (0, 2\alpha / M^2)$, we obtain \begin{align*} \|T(x) - T(y)\|_V &\le k(\rho)\,\|x - y\|_V, \quad k(\rho) < 1. \end{align*} The set $K$ is a closed subset of the complete Hilbert space $V$, hence a complete [metric space](/page/Metric%20Space). By the [Banach Fixed-Point Theorem](/theorems/289), $T$ has a unique fixed point $u \in K$. This $u$ is the unique solution to the variational inequality $a(u, v - u) \ge f \circ (v - u)$ for all $v \in K$. [guided] We need the contraction factor $k(\rho) = \sqrt{1 - 2\rho\alpha + \rho^2 M^2}$ to satisfy $k(\rho) < 1$. Squaring, this requires \begin{align*} 1 - 2\rho\alpha + \rho^2 M^2 &< 1. \end{align*} Subtracting $1$ from both sides gives $\rho^2 M^2 - 2\rho\alpha < 0$, which factors as $\rho(\rho M^2 - 2\alpha) < 0$. Since $\rho > 0$, the first factor is positive, so we need the second factor to be negative: $\rho M^2 - 2\alpha < 0$, i.e., $\rho < 2\alpha / M^2$. Why does such a $\rho$ always exist? The coercivity constant $\alpha > 0$ and the continuity constant $M > 0$ are given hypotheses, so $2\alpha / M^2 > 0$ and the interval $(0, 2\alpha / M^2)$ is nonempty. In fact, the optimal choice $\rho^* = \alpha / M^2$ minimises $k(\rho)^2 = 1 - \alpha^2 / M^2$, giving the tightest contraction. Fixing any $\rho \in (0, 2\alpha / M^2)$, the map $T: K \to K$ satisfies \begin{align*} \|T(x) - T(y)\|_V &\le k(\rho)\,\|x - y\|_V, \quad k(\rho) \in [0, 1). \end{align*} This makes $T$ a strict contraction. The [Banach Fixed-Point Theorem](/theorems/289) requires two ingredients: a complete metric space and a strict contraction mapping that space into itself. We verify both: $K$ is a closed subset of the complete Hilbert space $V$, and a closed subset of a complete [metric space](/page/Metric%20Space) is itself complete. The map $T$ sends $K$ into $K$ by construction, since $P_K$ always lands in $K$. Therefore $T$ has exactly one fixed point $u \in K$, and $u = T(u) = P_K(u - \rho(A(u) - w))$ is equivalent to the variational inequality $a(u, v - u) \ge f \circ (v - u)$ for all $v \in K$. [/guided] [/step]