[proofplan] We first construct the Euclidean projection $p$ of $v$ onto $K$ and set $q := v - p$. The variational inequality characterizing [projection onto a closed convex set](/theorems/647) implies that $q$ pairs nonpositively with every vector in $K$ and, using the conic structure, that $p \cdot q = 0$. This gives the required decomposition. We then show that the same orthogonality and polarity conditions force $q$ to be the projection of $v$ onto $K^\circ$, and finally prove uniqueness by comparing squared distances. [/proofplan] [step:Construct the projection of $v$ onto $K$] Fix $v \in \mathbb{R}^n$. Since $K$ is nonempty, choose $x_0 \in K$. Define the [continuous function](/page/Continuous%20Function) \begin{align*} \Phi: K \to [0,\infty), \qquad x \mapsto |v - x|^2. \end{align*} Let $\overline{B}(v, |v-x_0|)$ denote the closed Euclidean ball \begin{align*} \overline{B}(v, |v-x_0|) := \{x \in \mathbb{R}^n : |x-v| \leq |v-x_0|\}. \end{align*} Any minimizer of $\Phi$ over $K$ must lie in the compact set \begin{align*} K \cap \overline{B}(v, |v - x_0|). \end{align*} This set is nonempty because it contains $x_0$, closed because $K$ and $\overline{B}(v, |v - x_0|)$ are closed, and bounded by construction. Hence it is compact in $\mathbb{R}^n$, and $\Phi$ attains a minimum on it. Let $p \in K \cap \overline{B}(v, |v - x_0|)$ be such a minimizer. If $x \in K \setminus \overline{B}(v, |v-x_0|)$, then $|v-x|>|v-x_0|$, so $\Phi(x)>\Phi(x_0)\geq \Phi(p)$. Therefore this compact-set minimizer is a global minimizer of $\Phi$ over all of $K$. The minimizer is unique. If $p_1,p_2 \in K$ both minimize $\Phi$ and $p_1 \neq p_2$, then convexity of $K$ gives \begin{align*} m := \frac{p_1+p_2}{2} \in K. \end{align*} The parallelogram identity gives \begin{align*} |v-m|^2 = \frac{1}{2}|v-p_1|^2 + \frac{1}{2}|v-p_2|^2 - \frac{1}{4}|p_1-p_2|^2. \end{align*} Since $p_1 \neq p_2$, the final term is strictly negative, contradicting minimality. Thus $p$ is the unique Euclidean projection of $v$ onto $K$, so $p = P_K(v)$. Define \begin{align*} q := v - p \in \mathbb{R}^n. \end{align*} [guided] We need a well-defined vector $p$ before we can discuss the decomposition. The natural candidate is the nearest point in $K$ to $v$. Because $K$ may be unbounded, we first reduce the minimization to a compact set. Choose one point $x_0 \in K$, which is possible because $K$ is nonempty. Define \begin{align*} \Phi: K \to [0,\infty), \qquad x \mapsto |v - x|^2. \end{align*} If $x \in K$ satisfies $|v-x| > |v-x_0|$, then $\Phi(x) > \Phi(x_0)$, so such an $x$ cannot minimize $\Phi$. Let $\overline{B}(v, |v-x_0|)$ denote the closed Euclidean ball \begin{align*} \overline{B}(v, |v-x_0|) := \{x \in \mathbb{R}^n : |x-v| \leq |v-x_0|\}. \end{align*} Therefore every minimizer must lie in \begin{align*} K \cap \overline{B}(v, |v - x_0|). \end{align*} This set is nonempty, closed, and bounded in $\mathbb{R}^n$, hence compact. Since $\Phi$ is continuous, it attains its minimum on this compact set. Let $p \in K \cap \overline{B}(v, |v - x_0|)$ be a minimizer. For any $x \in K \setminus \overline{B}(v, |v-x_0|)$, we have $\Phi(x)>\Phi(x_0)\geq \Phi(p)$, so this $p$ is not merely a minimizer on the compact truncation; it is a global minimizer over all of $K$. Now we prove uniqueness, because later the theorem identifies $p$ with $P_K(v)$. Suppose $p_1,p_2 \in K$ are two minimizers. Since $K$ is convex, their midpoint \begin{align*} m := \frac{p_1+p_2}{2} \end{align*} belongs to $K$. The parallelogram identity gives \begin{align*} |v-m|^2 = \frac{1}{2}|v-p_1|^2 + \frac{1}{2}|v-p_2|^2 - \frac{1}{4}|p_1-p_2|^2. \end{align*} If $p_1 \neq p_2$, then $|p_1-p_2|^2 > 0$, so $|v-m|^2$ is strictly smaller than the common minimal value. This contradicts the definition of $p_1$ and $p_2$ as minimizers. Hence $p_1=p_2$, and the minimizer is unique. Thus $p=P_K(v)$. We now define the residual vector \begin{align*} q := v-p. \end{align*} The rest of the proof shows that this residual is exactly the polar projection. [/guided] [/step] [step:Derive the projection variational inequality] For every $x \in K$ and every $t \in [0,1]$, convexity of $K$ gives \begin{align*} p + t(x-p) = (1-t)p + tx \in K. \end{align*} Since $p$ minimizes $x \mapsto |v-x|^2$ over $K$, we have \begin{align*} |v-p|^2 \leq |v-p-t(x-p)|^2. \end{align*} Using $q=v-p$, this becomes \begin{align*} |q|^2 \leq |q-t(x-p)|^2. \end{align*} Expanding the square gives \begin{align*} 0 \leq -2t\,q\cdot(x-p) + t^2 |x-p|^2. \end{align*} For $t \in (0,1]$, divide by $t$: \begin{align*} 0 \leq -2\,q\cdot(x-p) + t |x-p|^2. \end{align*} Letting $t \to 0^+$ yields \begin{align*} q\cdot(x-p) \leq 0 \end{align*} for every $x \in K$. [/step] [step:Use the cone structure to prove $q \in K^\circ$ and $p\cdot q=0$] Because $K$ is a cone and $p \in K$, we have $tp \in K$ for every $t \geq 0$. Applying the variational inequality with $x=tp$ gives \begin{align*} q\cdot(tp-p) \leq 0. \end{align*} Equivalently, \begin{align*} (t-1)q\cdot p \leq 0 \end{align*} for every $t \geq 0$. Taking $t=0$ gives $q\cdot p \geq 0$, while taking $t=2$ gives $q\cdot p \leq 0$. Therefore \begin{align*} p\cdot q = 0. \end{align*} Now let $x \in K$. The variational inequality gives \begin{align*} q\cdot x \leq q\cdot p = 0. \end{align*} Since this holds for every $x \in K$, the definition of the polar cone gives $q \in K^\circ$. Thus \begin{align*} v=p+q, \qquad p \in K, \qquad q \in K^\circ, \qquad p\cdot q=0. \end{align*} [/step] [step:Identify $q$ as the projection of $v$ onto $K^\circ$] First note that $K^\circ$ is nonempty because $0 \in K^\circ$. It is convex because it is closed under convex combinations, and it is closed because \begin{align*} K^\circ = \bigcap_{x \in K}\{y \in \mathbb{R}^n : y\cdot x \leq 0\}, \end{align*} an intersection of closed half-spaces. Let $y \in K^\circ$. Since $p \in K$ and $y \in K^\circ$, we have $p\cdot y \leq 0$. Since $p\cdot q=0$, it follows that \begin{align*} p\cdot(y-q) = p\cdot y - p\cdot q \leq 0. \end{align*} Because $v-q=p$, this is exactly \begin{align*} (v-q)\cdot(y-q) \leq 0 \end{align*} for every $y \in K^\circ$. We now verify that this inequality makes $q$ the nearest point in $K^\circ$ to $v$. For every $y \in K^\circ$, \begin{align*} |v-y|^2 = |(v-q)+(q-y)|^2. \end{align*} Expanding the square gives \begin{align*} |v-y|^2 = |v-q|^2 + |q-y|^2 + 2(v-q)\cdot(q-y). \end{align*} Since $(v-q)\cdot(y-q) \leq 0$, we have $(v-q)\cdot(q-y) \geq 0$. Hence \begin{align*} |v-y|^2 \geq |v-q|^2. \end{align*} Thus $q$ minimizes the squared distance from $v$ to $K^\circ$. By the uniqueness argument for projections onto nonempty closed convex sets already proved above, $q=P_{K^\circ}(v)$. [guided] The residual $q=v-p$ is already known to lie in $K^\circ$. To prove that it is the projection of $v$ onto $K^\circ$, we must show that no other vector $y \in K^\circ$ is closer to $v$. First, $K^\circ$ is a nonempty closed convex set. It is nonempty because $0\cdot x=0$ for every $x \in K$, so $0 \in K^\circ$. It is convex because if $y_1,y_2 \in K^\circ$ and $s \in [0,1]$, then for every $x \in K$, \begin{align*} ((1-s)y_1+sy_2)\cdot x = (1-s)y_1\cdot x + sy_2\cdot x \leq 0. \end{align*} It is closed because it can be written as the intersection \begin{align*} K^\circ = \bigcap_{x \in K}\{y \in \mathbb{R}^n : y\cdot x \leq 0\}, \end{align*} and each set in this intersection is a closed half-space. Now fix $y \in K^\circ$. Since $p \in K$ and $y \in K^\circ$, the defining inequality of the polar cone gives \begin{align*} p\cdot y \leq 0. \end{align*} We also proved $p\cdot q=0$. Therefore \begin{align*} p\cdot(y-q) = p\cdot y - p\cdot q \leq 0. \end{align*} Because $v-q=p$, this becomes \begin{align*} (v-q)\cdot(y-q) \leq 0. \end{align*} This is the projection inequality for the point $q$ relative to the set $K^\circ$. To convert that inequality into a distance comparison, expand the squared distance from $v$ to an arbitrary $y \in K^\circ$: \begin{align*} |v-y|^2 = |(v-q)+(q-y)|^2. \end{align*} Using the Euclidean [inner product](/page/Inner%20Product) expansion, \begin{align*} |v-y|^2 = |v-q|^2 + |q-y|^2 + 2(v-q)\cdot(q-y). \end{align*} The inequality $(v-q)\cdot(y-q) \leq 0$ is equivalent to $(v-q)\cdot(q-y) \geq 0$, so the final two terms after $|v-q|^2$ are nonnegative. Hence \begin{align*} |v-y|^2 \geq |v-q|^2. \end{align*} Thus $q$ is a minimizer of the distance from $v$ to $K^\circ$. Since $K^\circ$ is nonempty, closed, and convex, the uniqueness argument for projections applies to $K^\circ$, so $q=P_{K^\circ}(v)$. [/guided] [/step] [step:Prove uniqueness of the orthogonal polar decomposition] Suppose $a \in K$ and $b \in K^\circ$ satisfy \begin{align*} v=a+b, \qquad a\cdot b=0. \end{align*} For any $x \in K$, since $b \in K^\circ$ we have $b\cdot x \leq 0$. Hence \begin{align*} b\cdot(a-x)=b\cdot a-b\cdot x \geq 0. \end{align*} Expanding the squared distance gives \begin{align*} |v-x|^2 = |a+b-x|^2. \end{align*} Thus \begin{align*} |v-x|^2 = |a-x|^2 + |b|^2 + 2b\cdot(a-x) \geq |b|^2. \end{align*} At $x=a$, equality holds: \begin{align*} |v-a|^2 = |b|^2. \end{align*} Therefore $a$ minimizes the squared distance from $v$ to $K$. By uniqueness of the projection onto $K$, $a=P_K(v)=p$. Then \begin{align*} b=v-a=v-p=q. \end{align*} So the pair $(p,q)$ is the unique pair in $K \times K^\circ$ satisfying \begin{align*} v=p+q, \qquad p\cdot q=0. \end{align*} This completes the proof. [/step]