[proofplan] We compare the two defining inclusions for Hausdorff distance with the corresponding inequalities for support functions. The main point is that for convex bodies $A$ and $B$, containment of $A$ in the closed $\varepsilon$-neighbourhood of $B$ is equivalent to $h_A(u) \leq h_B(u)+\varepsilon$ for every unit vector $u$. Applying this criterion twice, once with $(A,B)=(K,L)$ and once with $(A,B)=(L,K)$, identifies the admissible Hausdorff radii exactly with the uniform support-function error. [/proofplan]

[step:Define the parallel bodies used in the Hausdorff distance] For every nonempty compact set $A \subset \mathbb{R}^n$ and every $\varepsilon \geq 0$, define the closed $\varepsilon$-parallel body $A_\varepsilon \subset \mathbb{R}^n$ by \begin{align*} A_\varepsilon := \{x \in \mathbb{R}^n : \operatorname{dist}(x,A) \leq \varepsilon\}, \end{align*} where $\mathcal{P}(\mathbb{R}^n)$ denotes the power set of $\mathbb{R}^n$ and $\operatorname{dist}: \mathbb{R}^n \times \mathcal{P}(\mathbb{R}^n) \to [0,\infty)$ is given by \begin{align*} \operatorname{dist}(x,A) := \inf_{a \in A} |x-a|. \end{align*} Since $A$ is compact, the infimum in the definition of $\operatorname{dist}(x,A)$ is attained for each $x \in \mathbb{R}^n$. Equivalently, \begin{align*} A_\varepsilon = A + \overline{B}(0,\varepsilon), \end{align*} where \begin{align*} \overline{B}(0,\varepsilon) := \{z \in \mathbb{R}^n : |z| \leq \varepsilon\}. \end{align*} Indeed, if $x \in A_\varepsilon$, choose $a \in A$ with $|x-a|=\operatorname{dist}(x,A) \leq \varepsilon$; then $x=a+(x-a) \in A+\overline{B}(0,\varepsilon)$. Conversely, if $x=a+z$ with $a \in A$ and $z \in \overline{B}(0,\varepsilon)$, then $\operatorname{dist}(x,A) \leq |x-a|=|z| \leq \varepsilon$, so $x \in A_\varepsilon$. With this notation, the Hausdorff distance is \begin{align*} d_H(K,L) := \inf \{\varepsilon \geq 0 : K \subset L_\varepsilon \text{ and } L \subset K_\varepsilon\}. \end{align*} [/step]

[step:Characterize one parallel-body inclusion by support functions]Let $A,B \subset \mathbb{R}^n$ be convex bodies and let $\varepsilon \geq 0$. We prove that \begin{align*} A \subset B_\varepsilon \quad\iff\quad h_A(u) \leq h_B(u)+\varepsilon \text{ for every } u \in S^{n-1}. \end{align*} Assume first that $A \subset B_\varepsilon$. Fix $u \in S^{n-1}$ and $a \in A$. Since $a \in B_\varepsilon$ and $B$ is compact, choose $b \in B$ such that $|a-b| \leq \varepsilon$. Then \begin{align*} a \cdot u = b \cdot u + (a-b)\cdot u \leq h_B(u) + |a-b|\,|u| \leq h_B(u)+\varepsilon, \end{align*} where the [first inequality](/theorems/2897) uses the [Cauchy-Schwarz Inequality](/theorems/432) in $\mathbb{R}^n$ and the fact that $|u|=1$. Taking the supremum over $a \in A$ gives \begin{align*} h_A(u) \leq h_B(u)+\varepsilon. \end{align*} Conversely, assume that \begin{align*} h_A(u) \leq h_B(u)+\varepsilon \end{align*} for every $u \in S^{n-1}$. Suppose, for contradiction, that $A \not\subset B_\varepsilon$. Then there exists $a_0 \in A$ with $\operatorname{dist}(a_0,B)>\varepsilon$. Since $B$ is compact, choose $b_0 \in B$ such that \begin{align*} |a_0-b_0|=\operatorname{dist}(a_0,B). \end{align*} Define the unit vector $u_0 \in S^{n-1}$ by \begin{align*} u_0 := \frac{a_0-b_0}{|a_0-b_0|}. \end{align*} We claim that \begin{align*} b \cdot u_0 \leq b_0 \cdot u_0 \end{align*} for every $b \in B$. Indeed, for any $b \in B$ and any $t \in [0,1]$, convexity gives \begin{align*} b_t := (1-t)b_0 + tb \in B. \end{align*} The minimality of $b_0$ implies \begin{align*} |a_0-b_0|^2 \leq |a_0-b_t|^2. \end{align*} Expanding the right-hand side gives \begin{align*} |a_0-b_t|^2 = |a_0-b_0 - t(b-b_0)|^2 = |a_0-b_0|^2 - 2t(a_0-b_0)\cdot(b-b_0) + t^2|b-b_0|^2. \end{align*} Subtracting $|a_0-b_0|^2$ and dividing by $t>0$ yields \begin{align*} 0 \leq -2(a_0-b_0)\cdot(b-b_0) + t|b-b_0|^2. \end{align*} Letting $t \downarrow 0$ gives \begin{align*} (a_0-b_0)\cdot(b-b_0) \leq 0. \end{align*} Dividing by $|a_0-b_0|>0$ gives \begin{align*} b \cdot u_0 \leq b_0 \cdot u_0. \end{align*} Therefore $h_B(u_0)=b_0\cdot u_0$, while \begin{align*} h_A(u_0) \geq a_0\cdot u_0. \end{align*} Hence \begin{align*} h_A(u_0)-h_B(u_0) \geq (a_0-b_0)\cdot u_0 = |a_0-b_0| = \operatorname{dist}(a_0,B) > \varepsilon, \end{align*} contradicting the assumed inequality. Thus $A \subset B_\varepsilon$.[/step]

[guided]We prove a reusable criterion: a convex body $A$ lies inside the $\varepsilon$-neighbourhood of a convex body $B$ exactly when every support value of $A$ is at most the corresponding support value of $B$ plus $\varepsilon$. First assume $A \subset B_\varepsilon$. Fix a unit vector $u \in S^{n-1}$ and a point $a \in A$. Because $a \in B_\varepsilon$, the distance from $a$ to $B$ is at most $\varepsilon$. Since $B$ is compact, there is a point $b \in B$ satisfying $|a-b| \leq \varepsilon$. Then \begin{align*} a \cdot u = b \cdot u + (a-b)\cdot u \leq h_B(u)+|a-b|\,|u| \leq h_B(u)+\varepsilon. \end{align*} The inequality uses two facts: $b \cdot u \leq h_B(u)$ by the definition of $h_B$, and $(a-b)\cdot u \leq |a-b|\,|u|$ by the [Cauchy-Schwarz Inequality](/theorems/432). Since $|u|=1$ and $|a-b|\leq \varepsilon$, the error is at most $\varepsilon$. Taking the supremum over all $a \in A$ gives \begin{align*} h_A(u) \leq h_B(u)+\varepsilon. \end{align*} Now assume instead that \begin{align*} h_A(u) \leq h_B(u)+\varepsilon \end{align*} for every $u \in S^{n-1}$. We prove $A \subset B_\varepsilon$ by contradiction. Suppose there is a point $a_0 \in A$ with $\operatorname{dist}(a_0,B)>\varepsilon$. Since $B$ is compact, choose $b_0 \in B$ minimizing the distance to $a_0$: \begin{align*} |a_0-b_0|=\operatorname{dist}(a_0,B). \end{align*} Because this distance is positive, the vector from $b_0$ to $a_0$ determines a unit direction. Define \begin{align*} u_0 := \frac{a_0-b_0}{|a_0-b_0|}. \end{align*} The key geometric point is that the nearest point $b_0$ exposes $B$ in the direction $u_0$. We verify this directly. Take any $b \in B$. For each $t \in [0,1]$, convexity of $B$ gives \begin{align*} b_t := (1-t)b_0 + tb \in B. \end{align*} Since $b_0$ is a nearest point of $B$ to $a_0$, we have \begin{align*} |a_0-b_0|^2 \leq |a_0-b_t|^2. \end{align*} Expanding, \begin{align*} |a_0-b_t|^2 = |a_0-b_0 - t(b-b_0)|^2 = |a_0-b_0|^2 - 2t(a_0-b_0)\cdot(b-b_0) + t^2|b-b_0|^2. \end{align*} After subtracting $|a_0-b_0|^2$ and dividing by $t>0$, \begin{align*} 0 \leq -2(a_0-b_0)\cdot(b-b_0) + t|b-b_0|^2. \end{align*} Letting $t \downarrow 0$ gives \begin{align*} (a_0-b_0)\cdot(b-b_0) \leq 0. \end{align*} Dividing by $|a_0-b_0|$ gives \begin{align*} b \cdot u_0 \leq b_0 \cdot u_0. \end{align*} Since this holds for every $b \in B$, the support value of $B$ in direction $u_0$ is \begin{align*} h_B(u_0)=b_0\cdot u_0. \end{align*} On the other hand, because $a_0 \in A$, \begin{align*} h_A(u_0) \geq a_0\cdot u_0. \end{align*} Therefore \begin{align*} h_A(u_0)-h_B(u_0) \geq (a_0-b_0)\cdot u_0 = |a_0-b_0| = \operatorname{dist}(a_0,B) > \varepsilon. \end{align*} This contradicts the assumed support-function inequality in the unit direction $u_0$. Hence no such $a_0$ exists, and $A \subset B_\varepsilon$.[/guided]

[step:Translate the two Hausdorff inclusions into one uniform inequality] Apply the criterion from the previous step with $(A,B)=(K,L)$. For every $\varepsilon \geq 0$, \begin{align*} K \subset L_\varepsilon \quad\iff\quad h_K(u) \leq h_L(u)+\varepsilon \text{ for every } u \in S^{n-1}. \end{align*} Apply the same criterion with $(A,B)=(L,K)$. Then \begin{align*} L \subset K_\varepsilon \quad\iff\quad h_L(u) \leq h_K(u)+\varepsilon \text{ for every } u \in S^{n-1}. \end{align*} Thus both inclusions hold if and only if, for every $u \in S^{n-1}$, \begin{align*} h_K(u)-h_L(u) \leq \varepsilon \quad\text{and}\quad h_L(u)-h_K(u) \leq \varepsilon. \end{align*} Equivalently, \begin{align*} |h_K(u)-h_L(u)| \leq \varepsilon \end{align*} for every $u \in S^{n-1}$. [/step]

[step:Take the infimum over admissible radii] Since $K$ and $L$ are compact, define finite constants \begin{align*} R_K := \sup_{x \in K} |x| \quad\text{and}\quad R_L := \sup_{y \in L} |y|. \end{align*} For every $u \in S^{n-1}$, the [Cauchy-Schwarz Inequality](/theorems/432) gives $|x \cdot u| \leq |x|$ for $x \in K$ and $|y \cdot u| \leq |y|$ for $y \in L$, so $|h_K(u)| \leq R_K$ and $|h_L(u)| \leq R_L$. Hence $|h_K(u)-h_L(u)| \leq R_K+R_L$ for every $u \in S^{n-1}$. Define the finite number $\Delta \in [0,\infty)$ by \begin{align*} \Delta := \sup_{u \in S^{n-1}} |h_K(u)-h_L(u)|. \end{align*} The previous step shows that, for every $\varepsilon \geq 0$, \begin{align*} K \subset L_\varepsilon \text{ and } L \subset K_\varepsilon \quad\iff\quad \Delta \leq \varepsilon. \end{align*} Therefore the set of admissible radii in the definition of $d_H(K,L)$ is exactly \begin{align*} \{\varepsilon \geq 0 : \Delta \leq \varepsilon\}. \end{align*} Its infimum is $\Delta$. Hence \begin{align*} d_H(K,L) = \Delta = \sup_{u \in S^{n-1}} |h_K(u)-h_L(u)|. \end{align*} This proves the formula. [/step]