[proofplan] The proof first computes the convex conjugate of the gauge directly from the definitions, using absorption to ensure that $\gamma_C(x)$ is finite for every $x\in\mathbb{R}^n$. Membership in the polar set gives the upper bound $x \cdot y \leq \gamma_C(x)$ for every $x$, while failure of membership in the polar set gives a direction along which the conjugate supremum diverges. The support-function identity is then obtained from the [finite-dimensional bipolar theorem](/theorems/4108), whose separation-theorem content identifies the polar of $C^\circ$ with $\overline{\operatorname{conv}}(C\cup\{0\})$. [/proofplan] [step:Compute the conjugate on the polar set] Fix $y \in C^\circ$. We prove that $\gamma_C^*(y)=0$. Since $C$ is absorbing, for every $x\in\mathbb{R}^n$ there exists $r>0$ such that $x\in rC$; hence $\gamma_C(x)<+\infty$. Let $x \in \mathbb{R}^n$ and let $\lambda > \gamma_C(x)$. By the definition of $\gamma_C(x)$ as an infimum over a nonempty set of positive scalars, there exists a number $\mu > 0$ such that $\mu < \lambda$ and $x \in \mu C$. Hence there exists $c \in C$ with $x=\mu c$. Since $y \in C^\circ$, we have $c \cdot y \leq 1$, and therefore \begin{align*} x \cdot y = \mu(c \cdot y) \leq \mu < \lambda. \end{align*} Since this holds for every $\lambda > \gamma_C(x)$, it follows that \begin{align*} x \cdot y \leq \gamma_C(x). \end{align*} Thus \begin{align*} x \cdot y - \gamma_C(x) \leq 0 \end{align*} for every $x \in \mathbb{R}^n$. Taking the supremum over $x$ gives $\gamma_C^*(y) \leq 0$. Because $0 \in C$, we have $0 \in \lambda C$ for every $\lambda > 0$, so $\gamma_C(0)=0$. Evaluating the defining supremum at $x=0$ gives \begin{align*} \gamma_C^*(y) \geq 0 \cdot y - \gamma_C(0)=0. \end{align*} Therefore $\gamma_C^*(y)=0$. [/step] [step:Force the conjugate to be infinite outside the polar set] Fix $y \in \mathbb{R}^n \setminus C^\circ$. By the definition of $C^\circ$, there exists $x_0 \in C$ such that \begin{align*} x_0 \cdot y > 1. \end{align*} For each $t > 0$, since $x_0 \in C$, we have $t x_0 \in tC$, and hence \begin{align*} \gamma_C(t x_0) \leq t. \end{align*} Using $t x_0$ as a test point in the supremum defining $\gamma_C^*(y)$ gives \begin{align*} \gamma_C^*(y) \geq (t x_0) \cdot y - \gamma_C(t x_0) \geq t(x_0 \cdot y - 1). \end{align*} The coefficient $x_0 \cdot y - 1$ is strictly positive, so the right-hand side tends to $+\infty$ as $t \to \infty$. Therefore \begin{align*} \gamma_C^*(y)=+\infty. \end{align*} Combining this with the previous step proves \begin{align*} \gamma_C^*=\delta_{C^\circ}. \end{align*} [guided] We now prove that the conjugate cannot be finite when $y$ violates the polar inequalities. The condition $y \notin C^\circ$ means exactly that the defining inequality for the polar set fails for at least one point of $C$. Thus there is a point $x_0 \in C$ such that \begin{align*} x_0 \cdot y > 1. \end{align*} The point of using this particular $x_0$ is that the gauge is controlled on the ray generated by $x_0$. For every scalar $t > 0$, the point $t x_0$ lies in $tC$, because $x_0 \in C$. By the definition of the gauge as the infimum of all admissible scaling factors, this gives \begin{align*} \gamma_C(t x_0) \leq t. \end{align*} Now we test the conjugate supremum along the same ray. Since \begin{align*} \gamma_C^*(y)=\sup_{x \in \mathbb{R}^n}\bigl(x \cdot y-\gamma_C(x)\bigr), \end{align*} we may choose $x=t x_0$ and obtain \begin{align*} \gamma_C^*(y) \geq (t x_0)\cdot y-\gamma_C(t x_0). \end{align*} Using the estimate $\gamma_C(t x_0)\leq t$ gives \begin{align*} \gamma_C^*(y) \geq t(x_0 \cdot y)-t=t(x_0 \cdot y-1). \end{align*} Because $x_0 \cdot y-1>0$, this lower bound tends to $+\infty$ as $t \to \infty$. Hence the supremum defining $\gamma_C^*(y)$ is infinite: \begin{align*} \gamma_C^*(y)=+\infty. \end{align*} Together with the polar case, this proves precisely that $\gamma_C^*$ is $0$ on $C^\circ$ and $+\infty$ outside $C^\circ$, which is the identity \begin{align*} \gamma_C^*=\delta_{C^\circ}. \end{align*} [/guided] [/step] [step:Identify the support function as the gauge of the closed set] Let $S: \mathbb{R}^n \to (-\infty,\infty]$ be the function defined by \begin{align*} S(x)=\sigma_{C^\circ}(x)=\sup_{y \in C^\circ} x \cdot y. \end{align*} We prove that $S=\gamma_{\overline{C}}$. First fix $x \in \mathbb{R}^n$ and $\lambda > \gamma_{\overline{C}}(x)$. By the definition of the gauge as an infimum, there exists $\mu > 0$ such that $\mu < \lambda$ and $x \in \mu \overline{C}$. Choose $z \in \overline{C}$ such that $x=\mu z$. Since every $y \in C^\circ$ satisfies $c \cdot y \leq 1$ for all $c \in C$, and since the map $L_y: \mathbb{R}^n \to \mathbb{R}$ defined by $L_y(z)=z \cdot y$ is continuous, the same inequality holds for all $z \in \overline{C}$. Therefore \begin{align*} x \cdot y=\mu(z \cdot y)\leq \mu < \lambda \end{align*} for every $y \in C^\circ$. Taking the supremum over $y \in C^\circ$ and then using that this bound holds for every $\lambda > \gamma_{\overline{C}}(x)$ gives \begin{align*} S(x)\leq \gamma_{\overline{C}}(x). \end{align*} Conversely, if $S(x)=+\infty$, then $\gamma_{\overline{C}}(x)\leq S(x)$ holds immediately. It remains to treat the case $S(x)<+\infty$. Suppose $\lambda > S(x)$. Since $0 \in C^\circ$, we have $S(x)\geq 0$, so $\lambda>0$. The inequality $\lambda>S(x)$ implies \begin{align*} x \cdot y \leq \lambda \end{align*} for every $y \in C^\circ$, and therefore \begin{align*} \left(\frac{x}{\lambda}\right)\cdot y \leq 1 \end{align*} for every $y \in C^\circ$. By the finite-dimensional bipolar theorem, which is obtained from the strong separation theorem for closed convex sets, for every set $A \subset \mathbb{R}^n$ one has \begin{align*} (A^\circ)^\circ=\overline{\operatorname{conv}}(A\cup\{0\}). \end{align*} We apply this with $A=C$. Its hypotheses are satisfied because $C\subset\mathbb{R}^n$, and the theorem itself supplies the closure and convexification. Since $C$ is convex and $0\in C$, we have $\operatorname{conv}(C\cup\{0\})=C$, hence \begin{align*} (C^\circ)^\circ=\overline{C}. \end{align*} Thus $x/\lambda \in \overline{C}$, and hence $x \in \lambda \overline{C}$. Since this holds for every $\lambda > S(x)$, the definition of the gauge gives \begin{align*} \gamma_{\overline{C}}(x)\leq S(x). \end{align*} Therefore \begin{align*} \gamma_{\overline{C}}=\sigma_{C^\circ}. \end{align*} If $C$ is closed, then $\overline{C}=C$, so \begin{align*} \gamma_C=\sigma_{C^\circ}. \end{align*} [guided] Define $S: \mathbb{R}^n \to (-\infty,\infty]$ by \begin{align*} S(x)=\sigma_{C^\circ}(x)=\sup_{y \in C^\circ} x \cdot y. \end{align*} The goal is to show that the sublinear function obtained by taking support over the polar set is exactly the gauge of the [closed set](/page/Closed%20Set) $\overline{C}$. We first prove $S(x)\leq \gamma_{\overline{C}}(x)$. Fix $x \in \mathbb{R}^n$ and choose $\lambda > \gamma_{\overline{C}}(x)$. By the definition of the gauge, $\gamma_{\overline{C}}(x)$ is the infimum of all positive scalars $r$ such that $x\in r\overline{C}$. Hence there is a scalar $\mu>0$ with $\mu<\lambda$ and $x\in \mu\overline{C}$. Choose $z\in\overline{C}$ such that $x=\mu z$. Now fix $y\in C^\circ$. By the definition of the polar set, $c\cdot y\leq 1$ for every $c\in C$. The map $L_y: \mathbb{R}^n\to\mathbb{R}$ given by $L_y(z)=z\cdot y$ is continuous, so the inequality extends from $C$ to its closure: $z\cdot y\leq 1$ for every $z\in\overline{C}$. Applying this to the chosen $z$ gives \begin{align*} x\cdot y=\mu(z\cdot y)\leq \mu<\lambda. \end{align*} Because this holds for every $y\in C^\circ$, taking the supremum over $C^\circ$ gives $S(x)\leq \lambda$. Since the same argument works for every $\lambda>\gamma_{\overline{C}}(x)$, we obtain \begin{align*} S(x)\leq \gamma_{\overline{C}}(x). \end{align*} We now prove the reverse inequality. Fix $x\in\mathbb{R}^n$. If $S(x)=+\infty$, then $\gamma_{\overline{C}}(x)\leq S(x)$ is automatic, so there is nothing to prove for this $x$. We therefore assume $S(x)<+\infty$ and choose $\lambda>S(x)$. Since $0\in C^\circ$, the supremum defining $S(x)$ is at least $x\cdot 0=0$, so $\lambda>0$. The inequality $\lambda>S(x)$ implies $x\cdot y\leq \lambda$ for every $y\in C^\circ$, hence \begin{align*} \left(\frac{x}{\lambda}\right)\cdot y\leq 1 \end{align*} for every $y\in C^\circ$. This says exactly that $x/\lambda\in (C^\circ)^\circ$. We identify this bipolar set using the finite-dimensional bipolar theorem. The theorem states that for every subset $A\subset\mathbb{R}^n$, \begin{align*} (A^\circ)^\circ=\overline{\operatorname{conv}}(A\cup\{0\}), \end{align*} with the closure taken in the Euclidean topology; the separation-theorem input is what rules out any point outside the closed convex hull. We apply it with $A=C$. Since $C$ is convex and $0\in C$, adjoining $0$ and taking the convex hull do not change the set: $\operatorname{conv}(C\cup\{0\})=C$. Therefore \begin{align*} (C^\circ)^\circ=\overline{C}. \end{align*} Thus $x/\lambda\in\overline{C}$, so $x\in\lambda\overline{C}$. By the definition of the gauge, this gives $\gamma_{\overline{C}}(x)\leq\lambda$. Since the argument holds for every $\lambda>S(x)$, we conclude \begin{align*} \gamma_{\overline{C}}(x)\leq S(x). \end{align*} Combining the two inequalities gives $S=\gamma_{\overline{C}}$, that is, \begin{align*} \gamma_{\overline{C}}=\sigma_{C^\circ}. \end{align*} If $C$ is closed, then $\overline{C}=C$, and therefore \begin{align*} \gamma_C=\sigma_{C^\circ}. \end{align*} [/guided] [/step] [step:State the nonclosed conclusion using the closed convex hull] When $C$ is not assumed closed, the argument above proves \begin{align*} \gamma_{\overline{C}}=\sigma_{C^\circ}. \end{align*} Because $C$ is convex and $0\in C$, the set $\overline{C}$ is exactly the closed convex hull of $C\cup\{0\}$. Hence the support-function representation holds after replacing $C$ by its closed convex hull containing $0$, as claimed. [/step]