[proofplan] We first record the basic analytic properties of support functions: compactness makes the supremum finite and attained, while boundedness of the convex body gives continuity on the sphere. Next we prove the reconstruction formula expressing a compact convex set as the intersection of its supporting half-spaces, which gives injectivity. The isometry follows by proving the Hausdorff-distance formula in both directions. Finally, we characterize the image by showing that support functions restrict to continuous sphere functions with sublinear homogeneous extensions, and conversely every such sublinear extension is the support function of the compact convex set cut out by its half-spaces. [/proofplan]

[step:Verify that support functions restrict to continuous functions on the sphere] Let $K \in \mathcal K^n$. For $a \in \mathbb R^n$ and $r \ge 0$, write $\overline{B}(a,r)=\{x \in \mathbb R^n: |x-a| \le r\}$ for the closed Euclidean ball. Since $K$ is compact, there exists $R_K \ge 0$ such that $K \subset \overline{B}(0,R_K)$. For $u,v \in \mathbb R^n$ and $y \in K$, \begin{align*} u \cdot y \le v \cdot y + |u-v|\,|y| \le v \cdot y + R_K |u-v|. \end{align*} Taking the supremum over $y \in K$ gives \begin{align*} h_K(u) \le h_K(v) + R_K |u-v|. \end{align*} Interchanging $u$ and $v$ yields \begin{align*} |h_K(u)-h_K(v)| \le R_K |u-v|. \end{align*} Thus $h_K$ is Lipschitz on $\mathbb R^n$, and in particular $h_K|_{S^{n-1}} \in C(S^{n-1})$. Hence $\Phi$ is well-defined. [/step]

[step:Recover a compact convex set from its support function] We prove that every $K \in \mathcal K^n$ satisfies \begin{align*} K = \{x \in \mathbb R^n : x \cdot u \le h_K(u) \text{ for all } u \in S^{n-1}\}. \end{align*} If $x \in K$, then $x \cdot u \le \sup_{y \in K} u \cdot y = h_K(u)$ for every $u \in S^{n-1}$. Conversely, let $x \in \mathbb R^n \setminus K$. Since $K$ is compact and the map \begin{align*} K &\longrightarrow \mathbb R \\ y &\longmapsto |x-y|^2 \end{align*} is continuous, there exists $y_0 \in K$ minimizing $|x-y|^2$ over $y \in K$. Because $x \notin K$, we have $x \ne y_0$. Define \begin{align*} u_0 = \frac{x-y_0}{|x-y_0|} \in S^{n-1}. \end{align*} For any $y \in K$ and any $t \in [0,1]$, convexity gives $(1-t)y_0+ty \in K$. Minimality of $y_0$ gives \begin{align*} |x-y_0|^2 \le |x-y_0-t(y-y_0)|^2. \end{align*} Expanding the right-hand side, \begin{align*} 0 \le -2t(x-y_0)\cdot (y-y_0)+t^2|y-y_0|^2. \end{align*} Dividing by $t>0$ and letting $t \downarrow 0$ gives \begin{align*} (x-y_0)\cdot (y-y_0) \le 0. \end{align*} Therefore $u_0 \cdot y \le u_0 \cdot y_0$ for every $y \in K$, so \begin{align*} h_K(u_0)=u_0 \cdot y_0 < u_0 \cdot x. \end{align*} Thus $x$ fails one of the half-space inequalities. The reconstruction formula follows, and consequently $\Phi$ is injective. [/step]

[step:Bound the support-function distance by the Hausdorff distance] Let $K,L \in \mathcal K^n$, and set \begin{align*} \delta = d_H(K,L). \end{align*} By the definition of the Hausdorff metric, for each $x \in K$ there exists $z \in L$ with $|x-z| \le \delta$, and for each $z \in L$ there exists $x \in K$ with $|x-z| \le \delta$. Fix $u \in S^{n-1}$. For every $x \in K$, choose $z \in L$ with $|x-z| \le \delta$. Then \begin{align*} u \cdot x = u \cdot z + u \cdot (x-z) \le h_L(u) + |u|\,|x-z| \le h_L(u)+\delta. \end{align*} Taking the supremum over $x \in K$ gives \begin{align*} h_K(u) \le h_L(u)+\delta. \end{align*} Interchanging $K$ and $L$ gives \begin{align*} h_L(u) \le h_K(u)+\delta. \end{align*} Hence \begin{align*} |h_K(u)-h_L(u)| \le \delta. \end{align*} Taking the supremum over $u \in S^{n-1}$ yields \begin{align*} \|\Phi(K)-\Phi(L)\|_\infty \le d_H(K,L). \end{align*} [/step]

[step:Bound the Hausdorff distance by the support-function distance] Let \begin{align*} \varepsilon = \|\Phi(K)-\Phi(L)\|_\infty. \end{align*} For a non-empty set $A \subset \mathbb R^n$ and a point $w \in \mathbb R^n$, write \begin{align*} \operatorname{dist}(w,A)=\inf_{a \in A}|w-a|. \end{align*} We prove $K \subset L+\overline{B}(0,\varepsilon)$ and $L \subset K+\overline{B}(0,\varepsilon)$. Fix $x \in K$. Suppose, toward a contradiction, that $\operatorname{dist}(x,L)>\varepsilon$. Since $L$ is compact, there exists $z_0 \in L$ minimizing $|x-z|$ over $z \in L$. Define \begin{align*} u_0 = \frac{x-z_0}{|x-z_0|} \in S^{n-1}. \end{align*} For any $z \in L$ and any $t \in [0,1]$, convexity gives $(1-t)z_0+tz \in L$. Minimality of $z_0$ gives \begin{align*} |x-z_0|^2 \le |x-z_0-t(z-z_0)|^2. \end{align*} Expanding the right-hand side gives \begin{align*} 0 \le -2t(x-z_0)\cdot (z-z_0)+t^2|z-z_0|^2. \end{align*} Dividing by $t>0$ and letting $t \downarrow 0$ gives \begin{align*} (x-z_0)\cdot (z-z_0) \le 0. \end{align*} Dividing by $|x-z_0|>0$ gives $u_0 \cdot z \le u_0 \cdot z_0$ for every $z \in L$, and hence $h_L(u_0)=u_0 \cdot z_0$. Since $x \in K$, \begin{align*} h_K(u_0) \ge u_0 \cdot x. \end{align*} Therefore \begin{align*} h_K(u_0)-h_L(u_0) \ge u_0 \cdot x-u_0 \cdot z_0 = |x-z_0| = \operatorname{dist}(x,L)>\varepsilon, \end{align*} contradicting the definition of $\varepsilon$. Thus $\operatorname{dist}(x,L)\le \varepsilon$ for every $x \in K$. Now fix $z \in L$. If $\operatorname{dist}(z,K)>\varepsilon$, compactness of $K$ gives $x_0 \in K$ minimizing $|z-x|$ over $x \in K$. Define $v_0=(z-x_0)/|z-x_0| \in S^{n-1}$. For any $x \in K$ and $t \in [0,1]$, convexity gives $(1-t)x_0+tx \in K$, and minimality of $x_0$ gives \begin{align*} |z-x_0|^2 \le |z-x_0-t(x-x_0)|^2. \end{align*} Expanding, dividing by $t>0$, and letting $t \downarrow 0$ gives $(z-x_0)\cdot (x-x_0)\le 0$. Hence $v_0\cdot x\le v_0\cdot x_0$ for every $x\in K$, so $h_K(v_0)=v_0\cdot x_0$. Since $z\in L$, we have $h_L(v_0)\ge v_0\cdot z$, and therefore \begin{align*} h_L(v_0)-h_K(v_0) \ge v_0\cdot z-v_0\cdot x_0 = |z-x_0|=\operatorname{dist}(z,K)>\varepsilon, \end{align*} contradicting the definition of $\varepsilon$. Thus $\operatorname{dist}(z,K)\le \varepsilon$ for every $z \in L$. Hence \begin{align*} d_H(K,L)\le \varepsilon = \|\Phi(K)-\Phi(L)\|_\infty. \end{align*} Combining this with the previous step proves that $\Phi$ is an isometry. [/step]

[step:Show that support functions have sublinear homogeneous extensions] Let $K \in \mathcal K^n$. For $x \in \mathbb R^n$ and $t \ge 0$, \begin{align*} h_K(tx)=\sup_{y \in K} tx \cdot y = t \sup_{y \in K} x \cdot y = t h_K(x). \end{align*} For $x_1,x_2 \in \mathbb R^n$, \begin{align*} h_K(x_1+x_2) &= \sup_{y \in K} (x_1+x_2)\cdot y \\ &\le \sup_{y \in K} x_1\cdot y+\sup_{y \in K} x_2\cdot y \\ &= h_K(x_1)+h_K(x_2). \end{align*} Thus $h_K$ is sublinear. Since positive homogeneity gives \begin{align*} h_K(x)=|x|h_K\left(\frac{x}{|x|}\right) \end{align*} for $x \ne 0$, the positively homogeneous extension of $\Phi(K)=h_K|_{S^{n-1}}$ is exactly $h_K$. Therefore every function in the image of $\Phi$ has a sublinear positively homogeneous extension. [/step]

[step:Construct a convex body from a sublinear homogeneous extension] Let $f \in C(S^{n-1})$, and suppose its positively homogeneous extension $p_f:\mathbb R^n \to \mathbb R$ is sublinear. Define \begin{align*} K_f = \{x \in \mathbb R^n : x \cdot u \le f(u) \text{ for all } u \in S^{n-1}\}. \end{align*} Each set $\{x \in \mathbb R^n : x \cdot u \le f(u)\}$ is a closed half-space, so $K_f$ is closed and convex as an intersection of closed convex sets. We show $K_f$ is non-empty and compact. Apply the Hahn-Banach [extension theorem](/theorems/59) to the zero linear functional on the subspace $\{0\}\subset \mathbb R^n$, dominated by the sublinear function $p_f$. The domination hypothesis holds because the only point of the subspace is $0$ and positive homogeneity gives $p_f(0)=0$. Hence there exists a linear functional $\ell_0:\mathbb R^n\to\mathbb R$ such that \begin{align*} \ell_0(z)\le p_f(z) \end{align*} for every $z\in\mathbb R^n$. Since every linear functional on $\mathbb R^n$ is represented by Euclidean inner product with a unique vector, choose $y_0\in\mathbb R^n$ such that \begin{align*} \ell_0(z)=z\cdot y_0 \end{align*} for every $z\in\mathbb R^n$. For every $u\in S^{n-1}$, we have \begin{align*} u\cdot y_0=\ell_0(u)\le p_f(u)=f(u), \end{align*} so $y_0\in K_f$. Thus $K_f\ne\varnothing$. Since $f$ is continuous on the compact set $S^{n-1}$, define \begin{align*} M=\sup_{u \in S^{n-1}} |f(u)| < \infty. \end{align*} If $x \in K_f$ and $x \ne 0$, then with $u=x/|x|$ we have \begin{align*} |x| = x\cdot \frac{x}{|x|} \le f\left(\frac{x}{|x|}\right)\le M. \end{align*} Thus $K_f \subset \overline{B}(0,M)$. Since $K_f$ is closed and bounded in $\mathbb R^n$, the [Heine-Borel theorem](/theorems/315) implies that $K_f$ is compact. Hence $K_f \in \mathcal K^n$. [/step]

[step:Identify the constructed body's support function with the sublinear extension] We prove $h_{K_f}=p_f$ on $\mathbb R^n$. First let $x \in \mathbb R^n$. If $x=0$, then $h_{K_f}(0)=0=p_f(0)$. If $x \ne 0$, set $u=x/|x| \in S^{n-1}$. For every $y \in K_f$, \begin{align*} y\cdot u \le f(u). \end{align*} Multiplying by $|x|$ gives \begin{align*} x\cdot y \le |x|f(u)=p_f(x). \end{align*} Taking the supremum over $y \in K_f$ yields \begin{align*} h_{K_f}(x)\le p_f(x). \end{align*} For the reverse inequality, fix $x \in \mathbb R^n$. The sublinearity of $p_f$ gives \begin{align*} p_f(z) \le p_f(x)+p_f(z-x) \end{align*} for every $z \in \mathbb R^n$. For $x\ne 0$, write $\operatorname{span}\{x\}=\{tx:t\in\mathbb R\}$ for the one-dimensional linear subspace generated by $x$. Define the linear functional \begin{align*} \ell_x: \operatorname{span}\{x\} &\longrightarrow \mathbb R \end{align*} by $\ell_x(tx)=t p_f(x)$ for $t \in \mathbb R$. If $t\ge 0$, then positive homogeneity gives $\ell_x(tx)=p_f(tx)$. If $t<0$, the displayed sublinearity inequality applied with $z=0$ gives $p_f(x)+p_f(-x)\ge p_f(0)=0$, and therefore \begin{align*} \ell_x(tx)=t p_f(x)=|t|[-p_f(x)]\le |t|p_f(-x)=p_f(tx). \end{align*} Thus $\ell_x$ is dominated by $p_f$ on $\operatorname{span}\{x\}$. When $x=0$, take $\ell_0$ to be the zero functional on $\{0\}$, which is dominated by $p_f$ because $p_f(0)=0$. By the Hahn-Banach extension theorem, this dominated linear functional extends to a linear functional $\ell:\mathbb R^n\to\mathbb R$ satisfying $\ell(z)\le p_f(z)$ for all $z \in \mathbb R^n$. Since every linear functional on $\mathbb R^n$ is represented by Euclidean inner product with a unique vector, there exists a vector $y_x \in \mathbb R^n$ such that \begin{align*} \ell(z)=z\cdot y_x \end{align*} for all $z \in \mathbb R^n$. For every $u \in S^{n-1}$, \begin{align*} u\cdot y_x=\ell(u)\le p_f(u)=f(u), \end{align*} so $y_x \in K_f$. Hence \begin{align*} h_{K_f}(x)\ge x\cdot y_x=\ell(x)=p_f(x). \end{align*} Thus $h_{K_f}=p_f$ on $\mathbb R^n$, and in particular \begin{align*} \Phi(K_f)=h_{K_f}|_{S^{n-1}}=f. \end{align*} This proves the stated image characterization and completes the proof. [/step]