The proof establishes existence via the parallelogram law, uniqueness by a similar argument, and the variational characterisation by a calculus-of-variations argument on a quadratic [function](/page/Function) of a parameter. **Step 1 (Existence).** Let $\{w_k\} \subset K$ be a minimising [sequence](/page/Sequence): $\|v - w_k\|_V \to d := \inf_{w \in K}\|v - w\|_V$. By the parallelogram law, \begin{align*} \|w_k - w_j\|_V^2 = 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4\left\|v - \frac{w_k + w_j}{2}\right\|_V^2. \end{align*} Since $K$ is convex, $\frac{1}{2}(w_k + w_j) \in K$, so $\|v - \frac{1}{2}(w_k + w_j)\|_V^2 \ge d^2$. Therefore $\|w_k - w_j\|_V^2 \le 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4d^2 \to 0$ as $k, j \to \infty$. The sequence is Cauchy, and since $V$ is complete and $K$ is closed, it converges to some $u \in K$ with $\|v - u\|_V = d$. **Step 2 (Uniqueness).** Suppose $u_1, u_2 \in K$ both achieve the infimum. The parallelogram law gives $\|u_1 - u_2\|_V^2 = 2\|v - u_1\|_V^2 + 2\|v - u_2\|_V^2 - 4\|v - \frac{1}{2}(u_1 + u_2)\|_V^2 \le 4d^2 - 4d^2 = 0$, so $u_1 = u_2$. **Step 3 (Variational characterisation, forward direction).** Let $u = P_K(v)$. For any $w \in K$ and $t \in [0,1]$, the convexity of $K$ gives $u + t(w - u) \in K$. The function $\Psi(t) = \|v - u - t(w - u)\|_V^2$ attains its minimum at $t = 0$ (since $u$ is the closest point), so $\Psi'(0) \ge 0$. Computing: $\Psi'(0) = -2\langle v - u, w - u\rangle_V$, giving $\langle v - u, w - u\rangle_V \le 0$ for all $w \in K$. **Step 4 (Variational characterisation, reverse direction).** Suppose $u \in K$ satisfies $\langle v - u, w - u\rangle_V \le 0$ for all $w \in K$. Then for any $w \in K$: $\|v - w\|_V^2 = \|v - u\|_V^2 + 2\langle u - v, w - u\rangle_V + \|w - u\|_V^2 \ge \|v - u\|_V^2$ (since $\langle u - v, w - u\rangle_V = -\langle v - u, w - u\rangle_V \ge 0$). Hence $u = P_K(v)$. **Step 5 (Non-expansiveness).** Let $v_1, v_2 \in V$ and set $u_i = P_K(v_i)$. The characterisation gives $\langle v_1 - u_1, u_2 - u_1\rangle_V \le 0$ and $\langle v_2 - u_2, u_1 - u_2\rangle_V \le 0$. Adding: $\langle (v_1 - v_2) - (u_1 - u_2), u_2 - u_1\rangle_V \le 0$, which rearranges to $\|u_1 - u_2\|_V^2 \le \langle v_1 - v_2, u_1 - u_2\rangle_V \le \|v_1 - v_2\|_V \|u_1 - u_2\|_V$ by Cauchy–Schwarz. Dividing gives $\|P_K(v_1) - P_K(v_2)\|_V \le \|v_1 - v_2\|_V$.