[proofplan] Fix a direction $u \in \mathbb{R}^n$. The defining observation is that the linear functional $z \mapsto \langle z,u\rangle$ splits additively on a Minkowski sum: $\langle x+y,u\rangle=\langle x,u\rangle+\langle y,u\rangle$. This gives the upper bound by taking suprema, and compactness gives maximizers in $K$ and $L$ that realize the reverse inequality. The parallel-body formula follows by applying the same identity to the closed Euclidean ball and computing its support function directly. [/proofplan] [step:Compare every value on $K+L$ with the two separate support functions] Fix $u \in \mathbb{R}^n$. Define the restricted linear functional \begin{align*} \varphi_{u,K+L}: K+L &\to \mathbb{R} \\ z &\mapsto \langle z,u\rangle . \end{align*} Let $z \in K+L$. By the definition of $K+L$, there exist $x \in K$ and $y \in L$ such that $z=x+y$. Therefore, \begin{align*} \langle z,u\rangle &=\langle x+y,u\rangle \\ &=\langle x,u\rangle+\langle y,u\rangle \\ &\leq h_K(u)+h_L(u). \end{align*} Taking the supremum over all $z \in K+L$ gives \begin{align*} h_{K+L}(u) \leq h_K(u)+h_L(u). \end{align*} [guided] Fix a direction $u \in \mathbb{R}^n$. The support function measures the largest value of the linear functional in this direction, so we name that functional on the Minkowski sum: \begin{align*} \varphi_{u,K+L}: K+L &\to \mathbb{R} \\ z &\mapsto \langle z,u\rangle . \end{align*} Now take an arbitrary point $z \in K+L$. By the definition of the Minkowski sum, this means that $z$ has a decomposition $z=x+y$ for some $x \in K$ and $y \in L$. The Euclidean inner product is linear in its first argument, hence \begin{align*} \langle z,u\rangle &=\langle x+y,u\rangle \\ &=\langle x,u\rangle+\langle y,u\rangle . \end{align*} The first term is bounded above by the supremum of the same functional over $K$, and the second term is bounded above by the supremum over $L$: \begin{align*} \langle x,u\rangle+\langle y,u\rangle \leq h_K(u)+h_L(u). \end{align*} Since this inequality holds for every $z \in K+L$, taking the supremum over $K+L$ yields \begin{align*} h_{K+L}(u) \leq h_K(u)+h_L(u). \end{align*} [/guided] [/step] [step:Use compactness to choose maximizers that attain the reverse inequality] Define the restricted linear functionals \begin{align*} \varphi_{u,K}: K &\to \mathbb{R} \\ x &\mapsto \langle x,u\rangle \end{align*} and \begin{align*} \varphi_{u,L}: L &\to \mathbb{R} \\ y &\mapsto \langle y,u\rangle . \end{align*} These functions are continuous, and $K$ and $L$ are compact and non-empty. Hence there exist points $x_u \in K$ and $y_u \in L$ such that \begin{align*} h_K(u)=\langle x_u,u\rangle, \qquad h_L(u)=\langle y_u,u\rangle . \end{align*} Since $x_u+y_u \in K+L$, we have \begin{align*} h_{K+L}(u) &\geq \langle x_u+y_u,u\rangle \\ &=\langle x_u,u\rangle+\langle y_u,u\rangle \\ &=h_K(u)+h_L(u). \end{align*} Together with the previous inequality, this proves \begin{align*} h_{K+L}(u)=h_K(u)+h_L(u). \end{align*} [guided] To prove the opposite inequality, we need actual points in $K$ and $L$ where the two suprema are attained. Define \begin{align*} \varphi_{u,K}: K &\to \mathbb{R} \\ x &\mapsto \langle x,u\rangle \end{align*} and \begin{align*} \varphi_{u,L}: L &\to \mathbb{R} \\ y &\mapsto \langle y,u\rangle . \end{align*} Both maps are continuous because they are restrictions of the linear functional $v \mapsto \langle v,u\rangle$ on $\mathbb{R}^n$. The sets $K$ and $L$ are compact and non-empty, so each continuous real-valued function attains its supremum on the corresponding set. Therefore there exist points $x_u \in K$ and $y_u \in L$ satisfying \begin{align*} h_K(u)=\langle x_u,u\rangle, \qquad h_L(u)=\langle y_u,u\rangle . \end{align*} The point $x_u+y_u$ belongs to $K+L$ by definition of Minkowski sum. Evaluating the support functional at this one admissible point gives a lower bound for the supremum: \begin{align*} h_{K+L}(u) &\geq \langle x_u+y_u,u\rangle \\ &=\langle x_u,u\rangle+\langle y_u,u\rangle \\ &=h_K(u)+h_L(u). \end{align*} The first step gave the reverse inequality, so the two sides are equal: \begin{align*} h_{K+L}(u)=h_K(u)+h_L(u). \end{align*} [/guided] [/step] [step:Compute the support function of the closed Euclidean ball] Let $\varepsilon \geq 0$, and define \begin{align*} B_\varepsilon := \overline{B}(0,\varepsilon) = \{b \in \mathbb{R}^n : |b| \leq \varepsilon\}. \end{align*} For $u=0$, we have $h_{B_\varepsilon}(0)=0=\varepsilon |0|$. Suppose $u \neq 0$. For every $b \in B_\varepsilon$, the Euclidean [Cauchy-Schwarz inequality](/theorems/432) gives \begin{align*} \langle b,u\rangle \leq |b|\,|u| \leq \varepsilon |u|. \end{align*} Define \begin{align*} b_u := \varepsilon \frac{u}{|u|} \in \mathbb{R}^n . \end{align*} Then $|b_u|=\varepsilon$, so $b_u \in B_\varepsilon$, and \begin{align*} \langle b_u,u\rangle = \left\langle \varepsilon \frac{u}{|u|},u\right\rangle = \varepsilon |u|. \end{align*} Thus \begin{align*} h_{B_\varepsilon}(u)=\varepsilon |u| \end{align*} for every $u \in \mathbb{R}^n$. [/step] [step:Apply the Minkowski sum identity to the parallel body] By definition, \begin{align*} K_\varepsilon = K+B_\varepsilon. \end{align*} The set $B_\varepsilon$ is non-empty, compact, and convex. Applying the identity already proved with $L=B_\varepsilon$ gives, for every $u \in \mathbb{R}^n$, \begin{align*} h_{K_\varepsilon}(u) &=h_{K+B_\varepsilon}(u) \\ &=h_K(u)+h_{B_\varepsilon}(u) \\ &=h_K(u)+\varepsilon |u|. \end{align*} This proves the stated parallel-body formula. [/step]