Bounds on Hausdorff distance via singular values For some $\delta>0$, let $X$ and $X_\delta$ be two bounded convex polytopes in $\mathbb{R}^n$, defined as $X = \{x \in \mathbb{R}^n : Ax \leq b \}$ and $X_\delta = \{x \in \mathbb{R}^n : Ax \leq b + \delta u\}$ respectively, where $A \in \mathbb{R}^{m\times n}$ (assume $m\geq n$ and $\mathrm{Rank} A = n$), $b \in \mathbb{R}^m$ and $u = (1,\ldots,1)^\mathrm{T} \in \mathbb{R}^m$.
I would like to have an estimate on their Hausdorff distance $d_H(X,X_\delta)$, that is, upper/lower bounds that are fairly easy to compute.
What do you suggest?

Here is an (incomplete) idea: Since $X\subseteq X_\delta$, $d_H(X,X_\delta) = \inf \{ \varepsilon \geq 0: X_\delta \subseteq X \oplus \varepsilon \mathcal{B}\}$, where $\mathcal{B}$ is the closed unit ball (we work with the Euclidean norm, $||\cdot||_2$). Now (*) take $x' \in \partial X_\delta := \{x \in \mathbb{R}^n : Ax' = b + \delta u \}$ and $x \in \partial X := \{x \in \mathbb{R}^n : Ax = b  \}$. It follows that $A(x'-x) = \varepsilon u$; denoting $x'-x = \varepsilon d$ for some $d \in \mathcal{B}$, we get $Ad = u\dfrac{\delta}{\varepsilon}$. 
From the SVD of $A$ we get that $\sigma_{min}(A) \leq ||u||_2 \dfrac{\delta}{\varepsilon} \leq \sigma_{max}(A)$, and noting that $||u||_2 = \sqrt{m}$, the upper bound $\varepsilon \leq \dfrac{\delta \sqrt{m}}{\sigma_{min}(A)}$ and the lower bound $\varepsilon \geq \dfrac{\delta \sqrt{m}}{\sigma_{max}(A)}$ follow.
(*) Drawback: is it justified to consider only the points in the faces? (for me it is intuitive if I draw it, but how do you go on to formalizing this?) 
 A: The upper bound is correct, but you can't have $\sqrt{m}$ in the lower bound. Counterexample: $n=1$,  $A=(1,-1,0,0,\dots,0)$. Here the sole singular value of $A$ is $\sqrt{2}$, and the set $\{Ax\le b\}$ is $\{-b_2\le x\le b_1\}$, implying that $d_H(X,X_\delta)=\delta$ regardless of $m$.
Preliminaries: metric spaces
Some such things are better discussed in the general context of Lipschitz and coLipschitz maps between metric spaces. A map $F:Y\to Z$ is $L$-Lipschitz if $\|F(x)-F(y)\|\le L\|x-y\|$ for all $x,y\in Y$. Such a map satisfies $$d_H(F(S), F(T))\le Ld_H(S, T)\tag1$$ for any two nonempty  sets $S,T\subset Y$. Indeed, any point $F(s)$, $s\in S$, has a neighbor $F(t)$, where $d(s,t)$ can be arbitrarily close to $d_H(S,T)$. 
Furthermore, a map $F:Y\to Z$ is $\ell$-coLipschitz if $F(B(x,r))\supset B(F(x), \ell r)$ for all $x\in Y$ and $r>0$; here $B(x,r)$ means the ball centered at $x$ with radius $r$. Such a map satisfies $$d_H(S, T)\ge \ell d_H(F^{-1}(S), F^{-1}(T))\tag2$$ for any two nonempty  sets $S,T\in Z$.  Indeed, if a point $x\in F^{-1}(S)$ is at distance $>r$ from set $F^{-1}(T)$, that means $F(B(x,r))$ is disjoint from $T$. Hence, $B(F(x), \ell r)$ is disjoint from $T$, giving a lower bound on $d_H(S,T)$.
Notice that $S,T$ don't have to be bounded here; the distances may be infinite in some cases, and the claims still hold. 
Singular values
Let $M$ be the range of $A$; this is an $n$-dimensional subspace of $\mathbb{R}^m$. With respect to the Euclidean distance, the linear map $A: \mathbb{R}^n\to M$ is $\sigma_\max(A)$-Lipschitz and $\sigma_\min(A)$-coLipschitz. Hence, 
$$
\frac{1}{\sigma_\max(A)}d_H(AX,AX_\delta) \le d_H(X,X_\delta)\le \frac{1}{\sigma_\min(A)}d_H(AX,AX_\delta) \tag3
$$
Slicing a box
The set $AX$ is the intersection of subspace $M$ with the box $Q=\{y\in \mathbb{R}^m:y\le b\}$. Similarly, $AX_\delta$ is the intersection of   $M$ with the box $Q_\delta=\{y\in \mathbb{R}^m:y\le b+\delta u\}$. The Hausdorff distance between the two boxes is $\sqrt{m}\delta$, but this is not the distance we want: our goal is to estimate the distance between their intersections with $M$. Since $Q_\delta$ is obtained by vector addition of $Q$ and the $\delta$-ball in $\ell_\infty$ norm, it follows that 
$$
d_H(M\cap Q, M\cap Q_\delta) = \delta\sup_{x\in M}\frac{\|x\|_2}{\|x\|_\infty} \tag4
$$
The right hand side of (4) is at most $\sqrt{m}$, which yields the upper bound. The lower bound, according to A lower bound for the ratio of $2$- and $\infty$-norms within a linear subspace, is $\sqrt{n}$. Final conclusion:
$$
\frac{\delta \sqrt{n}}{\sigma_\max(A)}  \le d_H(X,X_\delta)\le \frac{\delta\sqrt{m}}{\sigma_\min(A)} \tag5
$$
