[proofplan] We prove both directions of the equivalence. For necessity, we exploit convexity of $K$ to construct a one-parameter family of competitors $w_t = (1-t)u + tw$ inside $K$ and compare $\|v - u\|_V^2$ against $\|v - w_t\|_V^2$. Expanding the squared norm and sending $t \to 0^+$ extracts the variational inequality. For sufficiency, we expand $\|v - w\|_V^2$ by inserting $\pm\, u$ and use the variational inequality to bound the cross term from below, establishing that $u$ is the global minimizer. [/proofplan] [step:Necessity — derive the variational inequality from the minimization property] Assume that $u = P_K(v)$, so that $u$ minimizes $\|v - w\|_V$ over all $w \in K$. Let $w \in K$ be arbitrary. Because $K$ is convex, the convex combination \begin{align*} w_t &:= (1-t)u + tw = u + t(w - u) \end{align*} belongs to $K$ for every $t \in [0, 1]$. The minimality of $u$ therefore gives \begin{align*} \|v - u\|_V^2 &\leq \|v - w_t\|_V^2 \end{align*} We expand the right-hand side using bilinearity of the inner product: \begin{align*} \|v - w_t\|_V^2 &= \|(v - u) - t(w - u)\|_V^2 \\ &= ((v - u) - t(w - u),\, (v - u) - t(w - u))_V \\ &= \|v - u\|_V^2 - 2t(v - u, w - u)_V + t^2\|w - u\|_V^2 \end{align*} Substituting into the distance inequality and cancelling $\|v - u\|_V^2$ from both sides yields \begin{align*} 0 &\leq -2t(v - u, w - u)_V + t^2\|w - u\|_V^2 \end{align*} For any $t \in (0, 1]$ we may divide by $2t > 0$ to obtain \begin{align*} (v - u, w - u)_V &\leq \frac{t}{2}\|w - u\|_V^2 \end{align*} The left-hand side is independent of $t$. Sending $t \to 0^+$, the right-hand side vanishes, and we conclude \begin{align*} (v - u, w - u)_V &\leq 0 \end{align*} The algebraic equivalence follows by expanding linearly: $(v - u, w - u)_V = (v, w - u)_V - (u, w - u)_V \leq 0$ rearranges to $(u, w - u)_V \geq (v, w - u)_V$. [/step] [step:Sufficiency — recover minimality from the variational inequality] Assume that $u \in K$ satisfies $(v - u, w - u)_V \leq 0$ for all $w \in K$. Let $w \in K$ be arbitrary. We decompose $v - w$ by inserting $\pm\, u$: \begin{align*} \|v - w\|_V^2 &= \|(v - u) + (u - w)\|_V^2 \\ &= ((v - u) + (u - w),\, (v - u) + (u - w))_V \\ &= \|v - u\|_V^2 + 2(v - u, u - w)_V + \|u - w\|_V^2 \end{align*} The variational inequality states $(v - u, w - u)_V \leq 0$. Negating the second argument reverses the sign: \begin{align*} (v - u, u - w)_V &= -(v - u, w - u)_V \geq 0 \end{align*} Since $\|u - w\|_V^2 \geq 0$ as well, we obtain \begin{align*} \|v - w\|_V^2 &\geq \|v - u\|_V^2 + 0 + 0 = \|v - u\|_V^2 \end{align*} This holds for every $w \in K$, so $u$ minimizes the distance from $v$ to $K$, i.e., $u = P_K(v)$. [/step]