[proofplan] We first prove that the support function of a non-empty compact convex set is sublinear and has linear growth, using only the supremum definition and boundedness of the set. Conversely, starting from a sublinear function $p$ with linear growth, we define $K$ as the intersection of the half-spaces determined by $p$ and prove that this set is closed, convex, bounded, and non-empty. The key point is the reverse inequality $p(u) \leq h_K(u)$: for each non-zero $u$, we construct a linear functional dominated by $p$ and agreeing with $p$ on $u$, using a finite-dimensional Hahn-Banach extension argument proved inside the proof. Finally, the half-space description shows both $h_K=p$ and uniqueness of the representing compact convex set. [/proofplan]

[step:Derive sublinearity and linear growth from a compact convex set] Assume that $K \subset \mathbb{R}^n$ is non-empty, compact, and convex, and that $p = h_K$. Since $K$ is compact, the number \begin{align*} R_K := \sup_{x \in K} |x| \end{align*} is finite. Let $u,v \in \mathbb{R}^n$ and let $\lambda \geq 0$. The positive homogeneity of $h_K$ follows from \begin{align*} h_K(\lambda u) = \sup_{x \in K} \langle x,\lambda u\rangle = \lambda \sup_{x \in K} \langle x,u\rangle = \lambda h_K(u). \end{align*} For subadditivity, for each $x \in K$, \begin{align*} \langle x,u+v\rangle = \langle x,u\rangle + \langle x,v\rangle \leq h_K(u) + h_K(v). \end{align*} Taking the supremum over $x \in K$ gives \begin{align*} h_K(u+v) \leq h_K(u) + h_K(v). \end{align*} Thus $p$ is sublinear. It remains to prove the linear growth bound. For every $x \in K$, the [Cauchy-Schwarz inequality](/theorems/432) in the Euclidean inner product on $\mathbb{R}^n$ gives \begin{align*} \langle x,u\rangle \leq |\langle x,u\rangle| \leq |x|\,|u| \leq R_K |u|. \end{align*} Taking the supremum over $x \in K$ yields \begin{align*} p(u)=h_K(u)\leq R_K |u|. \end{align*} Since $K$ is non-empty, fix $x_0 \in K$. Then \begin{align*} p(u)=h_K(u)\geq \langle x_0,u\rangle \geq -|x_0|\,|u| \geq -R_K |u|. \end{align*} Combining the two inequalities gives \begin{align*} |p(u)| \leq R_K |u| \end{align*} for every $u \in \mathbb{R}^n$. [/step]

[step:Construct the candidate set from the half-space inequalities] Assume now that $p: \mathbb{R}^n \to \mathbb{R}$ is sublinear and that there exists $R \geq 0$ such that \begin{align*} |p(u)| \leq R|u| \qquad \text{for every } u \in \mathbb{R}^n. \end{align*} Define \begin{align*} K := \{x \in \mathbb{R}^n : \langle x,v\rangle \leq p(v) \text{ for every } v \in \mathbb{R}^n\}. \end{align*} For each $v \in \mathbb{R}^n$, define the closed half-space \begin{align*} H_v := \{x \in \mathbb{R}^n : \langle x,v\rangle \leq p(v)\}. \end{align*} The map \begin{align*} \varphi_v: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \langle x,v\rangle \end{align*} is continuous, so $H_v = \varphi_v^{-1}((-\infty,p(v)])$ is closed. Hence \begin{align*} K = \bigcap_{v \in \mathbb{R}^n} H_v \end{align*} is closed. Each $H_v$ is convex because, if $x_1,x_2 \in H_v$ and $t \in [0,1]$, then \begin{align*} \langle tx_1+(1-t)x_2,v\rangle = t\langle x_1,v\rangle + (1-t)\langle x_2,v\rangle \leq tp(v)+(1-t)p(v) = p(v). \end{align*} Therefore $K$, as an intersection of convex sets, is convex. Finally, define the closed Euclidean ball of radius $R$ by \begin{align*} \overline{B}(0,R) := \{y \in \mathbb{R}^n : |y| \leq R\}. \end{align*} If $x \in K$, then the defining inequality applied with $v=x$ gives \begin{align*} |x|^2 = \langle x,x\rangle \leq p(x) \leq |p(x)| \leq R|x|. \end{align*} If $x=0$, then $|x| \leq R$. If $x \neq 0$, division by $|x|$ gives $|x| \leq R$. Thus \begin{align*} K \subset \overline{B}(0,R). \end{align*} Consequently $K$ is bounded. Since $K$ is closed and bounded in finite-dimensional Euclidean space, it is compact by the finite-dimensional [Heine-Borel theorem](/theorems/315) (citing a result not yet in the wiki: [Heine-Borel theorem](/theorems/309)). [/step]

[step:Extend supporting linear functionals dominated by the sublinear function]We prove the finite-dimensional dominated extension fact needed for the reverse support inequality. [claim:Finite-dimensional dominated extension] Let $E \subset \mathbb{R}^n$ be a linear subspace, let \begin{align*} L: E \to \mathbb{R} \end{align*} be linear, and suppose that \begin{align*} L(y) \leq p(y) \qquad \text{for every } y \in E. \end{align*} Then there exists a linear functional \begin{align*} \ell: \mathbb{R}^n \to \mathbb{R} \end{align*} such that $\ell|_E=L$ and \begin{align*} \ell(v) \leq p(v) \qquad \text{for every } v \in \mathbb{R}^n. \end{align*} [/claim] [proof] It is enough to extend across one new dimension at a time. Suppose $E \neq \mathbb{R}^n$ and choose $a \in \mathbb{R}^n \setminus E$. Define the enlarged subspace \begin{align*} E_a := E + \operatorname{span}\{a\}. \end{align*} Every element of $E_a$ has a unique representation $y+ta$ with $y \in E$ and $t \in \mathbb{R}$. We first choose a real number $\alpha$ so that the formula \begin{align*} L_a(y+ta) := L(y) + t\alpha \end{align*} defines an extension dominated by $p$. For $y,z \in E$, subadditivity of $p$ and domination of $L$ on $E$ give \begin{align*} L(y)+L(z) = L(y+z) \leq p(y+z) = p((y-a)+(z+a)) \leq p(y-a)+p(z+a). \end{align*} Therefore \begin{align*} L(y)-p(y-a) \leq p(z+a)-L(z) \end{align*} for every $y,z \in E$. Hence \begin{align*} \sup_{y \in E}\bigl(L(y)-p(y-a)\bigr) \leq \inf_{z \in E}\bigl(p(z+a)-L(z)\bigr) \end{align*} in the extended order. The endpoint quantities are finite [real numbers](/page/Real%20Numbers). Indeed, fixing $z=0$ in the preceding inequality gives \begin{align*} L(y)-p(y-a) \leq p(a)-L(0)=p(a) \end{align*} for every $y \in E$, so the supremum is bounded above. Fixing $y=0$ gives \begin{align*} -p(-a)=L(0)-p(-a) \leq p(z+a)-L(z) \end{align*} for every $z \in E$, so the infimum is bounded below. Since $0 \in E$, both indexing sets are non-empty, and therefore the interval \begin{align*} \left[ \sup_{y \in E}\bigl(L(y)-p(y-a)\bigr), \inf_{z \in E}\bigl(p(z+a)-L(z)\bigr) \right] \end{align*} is a non-empty real interval. Choose \begin{align*} \alpha \in \mathbb{R} \end{align*} inside this interval. Define \begin{align*} L_a: E_a &\to \mathbb{R} \\ y+ta &\mapsto L(y)+t\alpha. \end{align*} The unique representation in $E_a$ makes $L_a$ well-defined, and the formula is linear. We verify domination by $p$. If $t=0$, then $L_a(y)=L(y)\leq p(y)$. If $t>0$, put $z=t^{-1}y \in E$. Since $\alpha \leq p(z+a)-L(z)$, positive homogeneity of $p$ gives \begin{align*} L_a(y+ta) = t(L(z)+\alpha) \leq t p(z+a) = p(y+ta). \end{align*} If $t<0$, put $s=-t>0$ and $z=s^{-1}y \in E$. Since $\alpha \geq L(z)-p(z-a)$, we obtain \begin{align*} L_a(y+ta) = L(y)-s\alpha = s(L(z)-\alpha) \leq s p(z-a) = p(y-sa) = p(y+ta). \end{align*} Thus $L_a \leq p$ on $E_a$. Repeating this one-dimensional extension finitely many times along a basis completion from $E$ to $\mathbb{R}^n$ produces the desired linear functional $\ell: \mathbb{R}^n \to \mathbb{R}$. [/proof] Now fix $u \in \mathbb{R}^n$ with $u \neq 0$. Define \begin{align*} E_u := \operatorname{span}\{u\} \end{align*} and define the linear functional \begin{align*} L_u: E_u &\to \mathbb{R} \\ tu &\mapsto t\,p(u). \end{align*} This is well-defined because $u \neq 0$. We check that $L_u$ is dominated by $p$ on $E_u$. If $t \geq 0$, then positive homogeneity gives \begin{align*} L_u(tu)=t\,p(u)=p(tu). \end{align*} If $t<0$, then $-t>0$, and subadditivity applied to $u+(-u)=0$ gives \begin{align*} 0=p(0)\leq p(u)+p(-u), \end{align*} so $-p(u)\leq p(-u)$. Hence \begin{align*} L_u(tu) = (-t)(-p(u)) \leq (-t)p(-u) = p(tu). \end{align*} By the finite-dimensional dominated extension claim, there exists a linear functional \begin{align*} \ell_u: \mathbb{R}^n \to \mathbb{R} \end{align*} such that $\ell_u|_{E_u}=L_u$ and \begin{align*} \ell_u(v)\leq p(v) \qquad \text{for every } v \in \mathbb{R}^n. \end{align*} Using the standard Euclidean basis $(e_1,\dots,e_n)$ of $\mathbb{R}^n$, define \begin{align*} x_u := \sum_{i=1}^n \ell_u(e_i)e_i \in \mathbb{R}^n. \end{align*} For every $v=\sum_{i=1}^n v_i e_i \in \mathbb{R}^n$, \begin{align*} \langle x_u,v\rangle = \sum_{i=1}^n \ell_u(e_i)v_i = \ell_u(v). \end{align*} Therefore \begin{align*} \langle x_u,v\rangle \leq p(v) \qquad \text{for every } v \in \mathbb{R}^n, \end{align*} so $x_u \in K$. Moreover, \begin{align*} \langle x_u,u\rangle = \ell_u(u)=L_u(u)=p(u). \end{align*}[/step]

[guided]The obstacle is that the definition of $K$ gives only upper bounds of the form $\langle x,u\rangle \leq p(u)$. To prove that the support function actually reaches the value $p(u)$, we need to manufacture a point $x_u \in K$ whose pairing with $u$ is exactly $p(u)$. We first prove the finite-dimensional extension principle behind this construction. Let $E \subset \mathbb{R}^n$ be a linear subspace and let $L:E\to\mathbb{R}$ be linear with $L(y)\leq p(y)$ for every $y\in E$. Suppose $a\notin E$ and set \begin{align*} E_a := E+\operatorname{span}\{a\}. \end{align*} Every element of $E_a$ has a unique form $y+ta$, with $y\in E$ and $t\in\mathbb{R}$, because $a\notin E$. We seek an extension of the form \begin{align*} L_a: E_a &\to \mathbb{R} \\ y+ta &\mapsto L(y)+t\alpha, \end{align*} where $\alpha\in\mathbb{R}$ is still to be chosen. The domination condition $L_a\leq p$ imposes upper and lower bounds on $\alpha$. For $y,z\in E$, subadditivity of $p$ gives \begin{align*} L(y)+L(z) = L(y+z) \leq p(y+z) = p((y-a)+(z+a)) \leq p(y-a)+p(z+a). \end{align*} Rearranging, \begin{align*} L(y)-p(y-a) \leq p(z+a)-L(z) \end{align*} for all $y,z\in E$. Thus every lower bound candidate is below every upper bound candidate: \begin{align*} \sup_{y \in E}\bigl(L(y)-p(y-a)\bigr) \leq \inf_{z \in E}\bigl(p(z+a)-L(z)\bigr). \end{align*} We must also check that these are real endpoint values rather than infinite extended values. Fixing $z=0$ in the preceding inequality gives \begin{align*} L(y)-p(y-a) \leq p(a)-L(0)=p(a) \end{align*} for every $y\in E$, so the supremum is bounded above. Fixing $y=0$ gives \begin{align*} -p(-a)=L(0)-p(-a) \leq p(z+a)-L(z) \end{align*} for every $z\in E$, so the infimum is bounded below. Since $0\in E$, both families are non-empty. Therefore the interval \begin{align*} \left[ \sup_{y \in E}\bigl(L(y)-p(y-a)\bigr), \inf_{z \in E}\bigl(p(z+a)-L(z)\bigr) \right] \end{align*} is a non-empty real interval, and we may choose $\alpha\in\mathbb{R}$ in it. Now define $L_a(y+ta)=L(y)+t\alpha$. If $t=0$, then $L_a(y)=L(y)\leq p(y)$. If $t>0$, put $z=t^{-1}y$. Since $\alpha\leq p(z+a)-L(z)$, we have \begin{align*} L_a(y+ta) = t(L(z)+\alpha) \leq t p(z+a) = p(y+ta). \end{align*} If $t<0$, write $s=-t>0$ and put $z=s^{-1}y$. Since $\alpha\geq L(z)-p(z-a)$, we get \begin{align*} L_a(y+ta) = s(L(z)-\alpha) \leq s p(z-a) = p(y+ta). \end{align*} Thus $L_a$ is still dominated by $p$. Repeating this construction along a basis completion extends $L$ to a linear functional on all of $\mathbb{R}^n$ dominated by $p$. We now apply the extension result to the one-dimensional subspace generated by a fixed non-zero vector $u$. Define \begin{align*} E_u := \operatorname{span}\{u\} \end{align*} and \begin{align*} L_u:E_u &\to \mathbb{R} \\ tu &\mapsto t\,p(u). \end{align*} For $t\geq 0$, positive homogeneity gives $L_u(tu)=p(tu)$. For $t<0$, subadditivity gives \begin{align*} 0=p(0)\leq p(u)+p(-u), \end{align*} so $-p(u)\leq p(-u)$, and therefore \begin{align*} L_u(tu)=(-t)(-p(u))\leq (-t)p(-u)=p(tu). \end{align*} Thus $L_u\leq p$ on $E_u$. The extension result gives a linear functional $\ell_u:\mathbb{R}^n\to\mathbb{R}$ such that $\ell_u(v)\leq p(v)$ for every $v\in\mathbb{R}^n$ and $\ell_u(u)=p(u)$. Represent this functional by a vector using the Euclidean inner product: with the standard basis $(e_1,\dots,e_n)$, define \begin{align*} x_u := \sum_{i=1}^n \ell_u(e_i)e_i. \end{align*} Then for every $v=\sum_{i=1}^n v_i e_i$, \begin{align*} \langle x_u,v\rangle = \sum_{i=1}^n \ell_u(e_i)v_i = \ell_u(v) \leq p(v). \end{align*} Hence $x_u\in K$, and at the chosen direction $u$, \begin{align*} \langle x_u,u\rangle=\ell_u(u)=p(u). \end{align*} This is the point of the extension argument: it builds an actual supporting point of $K$ in the direction $u$.[/guided]

[step:Identify the support function of the constructed set] We prove that $h_K=p$. First, for every $x\in K$ and every $u\in\mathbb{R}^n$, the definition of $K$ gives \begin{align*} \langle x,u\rangle \leq p(u). \end{align*} Taking the supremum over $x\in K$ yields \begin{align*} h_K(u) \leq p(u) \end{align*} for every $u\in\mathbb{R}^n$. For the reverse inequality, let $u\neq 0$. The previous step gives $x_u\in K$ such that \begin{align*} \langle x_u,u\rangle=p(u). \end{align*} Since $h_K(u)$ is the supremum of $\langle x,u\rangle$ over all $x\in K$, we obtain \begin{align*} h_K(u)\geq \langle x_u,u\rangle=p(u). \end{align*} Together with $h_K(u)\leq p(u)$, this gives $h_K(u)=p(u)$ for every $u\neq 0$. For $u=0$, positive homogeneity of $p$ gives \begin{align*} p(0)=p(0\cdot 0)=0p(0)=0. \end{align*} Also, \begin{align*} h_K(0)=\sup_{x\in K}\langle x,0\rangle=0. \end{align*} Hence $h_K(0)=p(0)$. Therefore $h_K=p$ on all of $\mathbb{R}^n$. The same construction also proves that $K$ is non-empty when $n\geq 1$: choosing any non-zero $u\in\mathbb{R}^n$ produces $x_u\in K$. If $n=0$, then $\mathbb{R}^0=\{0\}$, and the definition of $K$ gives $K=\{0\}$ because the only possible inequality is $\langle 0,0\rangle\leq p(0)=0$. Thus $K$ is non-empty in every dimension. [/step]

[step:Recover the compact convex set from its support function] It remains to justify the asserted formula and uniqueness. We have already defined \begin{align*} K = \{x\in\mathbb{R}^n : \langle x,u\rangle \leq p(u) \text{ for every } u\in\mathbb{R}^n\} \end{align*} and proved that this set is non-empty, compact, convex, and satisfies $h_K=p$. Suppose $C\subset\mathbb{R}^n$ is another non-empty compact convex set with $h_C=p$. For every $x\in C$ and every $u\in\mathbb{R}^n$, \begin{align*} \langle x,u\rangle \leq h_C(u)=p(u), \end{align*} so $C\subset K$. Conversely, let $x\in K$. If $x\notin C$, then, since $C$ is compact and convex, the finite-dimensional [strict separation theorem](/theorems/1044) gives a vector $u\in\mathbb{R}^n$ such that \begin{align*} \langle x,u\rangle > \sup_{y\in C}\langle y,u\rangle = h_C(u)=p(u) \end{align*} (citing a result not yet in the wiki: strict separation theorem for a point and a compact convex set). This contradicts the defining inequality for $x\in K$. Hence $K\subset C$, and therefore $K=C$. Thus the representing compact convex set is exactly the half-space intersection stated in the theorem. [/step]