[proofplan] Fix a direction $u \in \mathbb{R}^n$. The [support function of a Minkowski sum](/theorems/4098) is the supremum of $u \cdot (x+y)$ over pairs $(x,y) \in K \times L$, and the dot product separates this expression into a term depending only on $x$ and a term depending only on $y$. We prove that the supremum of this separated sum is the sum of the two separate suprema by matching upper and lower bounds. The scalar identity follows from pulling the non-negative scalar $\lambda$ through the supremum, with the case $\lambda=0$ handled separately. [/proofplan] [step:Show the support function values are finite] Fix $u \in \mathbb{R}^n$. Since $K$ and $L$ are compact subsets of $\mathbb{R}^n$, they are bounded, so there exist constants $R_K,R_L \geq 0$ such that $|x| \leq R_K$ for every $x \in K$ and $|y| \leq R_L$ for every $y \in L$. By the [Cauchy-Schwarz inequality](/theorems/432) for the Euclidean inner product, \begin{align*} |u \cdot x| &\leq |u|\,|x| \leq |u|R_K, \\ |u \cdot y| &\leq |u|\,|y| \leq |u|R_L. \end{align*} Thus $h_K(u)$ and $h_L(u)$ are finite [real numbers](/page/Real%20Numbers). [guided] Fix a direction $u \in \mathbb{R}^n$. Before manipulating suprema, we verify that the relevant quantities are finite real numbers. Compact subsets of $\mathbb{R}^n$ are bounded, so there are constants $R_K,R_L \geq 0$ such that $|x| \leq R_K$ for all $x \in K$ and $|y| \leq R_L$ for all $y \in L$. The Euclidean Cauchy-Schwarz inequality gives, for every $x \in K$ and $y \in L$, \begin{align*} |u \cdot x| &\leq |u|\,|x| \leq |u|R_K, \\ |u \cdot y| &\leq |u|\,|y| \leq |u|R_L. \end{align*} Therefore the sets $\{u \cdot x : x \in K\}$ and $\{u \cdot y : y \in L\}$ are bounded above and non-empty. Hence $h_K(u)$ and $h_L(u)$ are finite real numbers. [/guided] [/step] [step:Separate the supremum over the Minkowski sum] Define \begin{align*} \alpha &:= h_K(u)=\sup_{x \in K} u \cdot x, & \beta &:= h_L(u)=\sup_{y \in L} u \cdot y. \end{align*} For every $z \in K+L$, there exist $x \in K$ and $y \in L$ such that $z=x+y$. Hence \begin{align*} u \cdot z = u \cdot (x+y)=u \cdot x+u \cdot y \leq \alpha+\beta. \end{align*} Taking the supremum over $z \in K+L$ gives \begin{align*} h_{K+L}(u) \leq \alpha+\beta. \end{align*} Conversely, let $\varepsilon>0$. By the definition of supremum, there exist $x_\varepsilon \in K$ and $y_\varepsilon \in L$ such that \begin{align*} u \cdot x_\varepsilon &> \alpha-\frac{\varepsilon}{2}, & u \cdot y_\varepsilon &> \beta-\frac{\varepsilon}{2}. \end{align*} Since $x_\varepsilon+y_\varepsilon \in K+L$, \begin{align*} h_{K+L}(u) &\geq u \cdot (x_\varepsilon+y_\varepsilon) \\ &=u \cdot x_\varepsilon+u \cdot y_\varepsilon \\ &> \alpha+\beta-\varepsilon. \end{align*} Because $\varepsilon>0$ was arbitrary, $h_{K+L}(u) \geq \alpha+\beta$. Therefore \begin{align*} h_{K+L}(u)=h_K(u)+h_L(u). \end{align*} [guided] Set \begin{align*} \alpha &:= h_K(u)=\sup_{x \in K} u \cdot x, & \beta &:= h_L(u)=\sup_{y \in L} u \cdot y. \end{align*} The goal is to prove that the supremum over all sums $x+y$ equals $\alpha+\beta$. First prove the upper bound. Take any $z \in K+L$. By the definition of Minkowski sum, there are points $x \in K$ and $y \in L$ such that $z=x+y$. The Euclidean inner product is linear in its second argument, so \begin{align*} u \cdot z = u \cdot (x+y)=u \cdot x+u \cdot y. \end{align*} Since $\alpha$ is an upper bound for $\{u \cdot x : x \in K\}$ and $\beta$ is an upper bound for $\{u \cdot y : y \in L\}$, we get \begin{align*} u \cdot z \leq \alpha+\beta. \end{align*} This holds for every $z \in K+L$, so taking the supremum gives \begin{align*} h_{K+L}(u) \leq \alpha+\beta. \end{align*} For the reverse inequality, we use the approximation property of the supremum. Let $\varepsilon>0$. Since $\alpha=\sup_{x \in K} u \cdot x$, there exists $x_\varepsilon \in K$ such that \begin{align*} u \cdot x_\varepsilon > \alpha-\frac{\varepsilon}{2}. \end{align*} Similarly, since $\beta=\sup_{y \in L} u \cdot y$, there exists $y_\varepsilon \in L$ such that \begin{align*} u \cdot y_\varepsilon > \beta-\frac{\varepsilon}{2}. \end{align*} The point $x_\varepsilon+y_\varepsilon$ belongs to $K+L$, so \begin{align*} h_{K+L}(u) &\geq u \cdot (x_\varepsilon+y_\varepsilon) \\ &=u \cdot x_\varepsilon+u \cdot y_\varepsilon \\ &> \alpha+\beta-\varepsilon. \end{align*} Since this lower bound holds for every $\varepsilon>0$, it follows that \begin{align*} h_{K+L}(u) \geq \alpha+\beta. \end{align*} Combining the two inequalities yields \begin{align*} h_{K+L}(u)=h_K(u)+h_L(u). \end{align*} [/guided] [/step] [step:Pull a non-negative scalar through the support supremum] Let $\lambda \geq 0$. If $\lambda=0$, then $0K=\{0\}$, so \begin{align*} h_{0K}(u)=\sup_{z \in \{0\}} u \cdot z = 0 = 0\,h_K(u). \end{align*} Now suppose $\lambda>0$. For every $z \in \lambda K$, there exists $x \in K$ such that $z=\lambda x$, and therefore \begin{align*} u \cdot z = u \cdot (\lambda x)=\lambda(u \cdot x)\leq \lambda h_K(u). \end{align*} Taking the supremum over $z \in \lambda K$ gives \begin{align*} h_{\lambda K}(u)\leq \lambda h_K(u). \end{align*} Conversely, for every $\varepsilon>0$, choose $x_\varepsilon \in K$ such that \begin{align*} u \cdot x_\varepsilon > h_K(u)-\frac{\varepsilon}{\lambda}. \end{align*} Then $\lambda x_\varepsilon \in \lambda K$, and \begin{align*} h_{\lambda K}(u) &\geq u \cdot (\lambda x_\varepsilon) \\ &=\lambda(u \cdot x_\varepsilon) \\ &> \lambda h_K(u)-\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, \begin{align*} h_{\lambda K}(u)\geq \lambda h_K(u). \end{align*} Thus \begin{align*} h_{\lambda K}(u)=\lambda h_K(u). \end{align*} Since $u \in \mathbb{R}^n$ and $\lambda \geq 0$ were arbitrary, both asserted identities hold. [/step]