\section{Proof of Lemma~\ref{alpha00}}\label{GS}
Recall from Section~\ref{A} the definition of an active point $v$ and the related concepts including its set of active constraints $B_v$, $A^{B_v}$, the set of active points $\operatorname{Act}(\cD)$, and $\rho_{m i n}({\cD})$. Before we start the proof of Lemma~\ref{alpha00}, we mention some observations regarding these concepts. 
Letting $b^{B_v}=[b^{i_1},\ldots,b^{i_d}]^{\top}\in\mathbb{R}^d$, we have $A^{B_v}v=b^{B_v}$. Based on the fact that $A^{B_v} \in \mathbb{R}^{d \times d}$ is invertible, we can write $v=(A^{B_v})^{-1} b^{B_v}$. Moreover, note that an active point of ${\cD}$ may or may not belong to ${\cD}$. Indeed, vertices of ${\cD}$ are those active points of ${\cD}$ which belong to ${\cD}$ and hence $\operatorname{Vert}({\cD})\subseteq \operatorname{Act}({\cD})$. Finally, it is important to note that the compactness of ${\cD}$ results in $\rho_{\min}({\cD})>0$. The proof of Lemma~\ref{alpha00} can now be presented as follows.

\begin{proof}
	Since the faces of $\cD_{\tau}$ are parallel to the faces of $\cD$, the maximum distance between a point $x\in \cD$ and its projection $\pi_{\cD_{\tau}} (x)\in \cD_{\tau}$ obviously occurs at some vertex $v\in \operatorname{Vert}(\cD)$. In other words, we have $\arg \max_{x\in \cD} \left\|x-\pi_{\cD_{\tau}} (x)\right\| \in \operatorname{Vert}(\cD)$.  Note that $B_v$ consists of $d$ independent hyper-plains that intersect at $v$. Now consider their shifted versions in $\cD_{\tau}$. These shifted hyper-plains are also independent. Denote their intersection by $u$. Note that  $B_v=B_u$. Therefore, for any $v\in \operatorname{Vert}(\cD)$, it follows that
	\begin{align*}
	&A^{B_v}v-b^{B_v} = 0 \quad \Rightarrow \quad v = \left(A^{B_v}\right)^{-1}b^{B_v},\\
	&A^{B_u}u-b^{B_u} +\tau\mathbb{1}_{\abs{B_u}}= 0 \quad \Rightarrow \quad u = \left(A^{B_u}\right)^{-1}\left(b^{B_u}-\tau\mathbb{1}_{\abs{B_u}}\right).
	\end{align*}
	By subtracting two previous lines and noting that $B_v=B_u$, for any $v\in\operatorname{Vert}(\cD)$, we conclude that
	\begin{align*}
	\left\|v-\pi_{\cD_{\tau}} (v)\right\| &\leqslant \left\|v-u\right\|= \tau \left\|\left(A^{B_v}\right)^{-1}\mathbb{1}_{\abs{B_v}}\right\| \leqslant \tau \,\sqrt{d} \left\|\left(A^{B_v}\right)^{-1}\right\|= \frac{\tau \,\sqrt{d}}{\rho_{\min}\left(A^{B_v}\right)}\leqslant \frac{\tau \,\sqrt{d}}{\rho_{\min}\left(\cD\right)},
	\end{align*}
	where in the last inequality, we used the fact that $\operatorname{Vert}(\cD)\subseteq \operatorname{Act}(\cD)$. Hence, for all $x\in\cD$, $\|x-\pi_{\cD_{\tau}} (x)\| \leqslant \alpha_{\cD} \tau$, where $\alpha_{\cD}=\sqrt{d}/\rho_{\min}(\cD)$.
	
	Moreover, note that $\pi_{\cD_{\tau}} (\pi_{\cD}(x))\in \cD_{\tau}$. Thus, for every $x\in\mathbb{R}^d$, we have $\|x-\pi_{\cD_{\tau}} (x)\|  \leqslant \|x-\pi_{\cD_{\tau}} (\pi_{\cD}(x))\|$ based on the definition of projection on $\cD_{\tau}$. Therefore, for every $x\in\mathbb{R}^d$,
	\begin{equation}\label{e1}
		 \|x-\pi_{\cD_{\tau}} (x)\|  \leqslant \|x-\pi_{\cD_{\tau}} (\pi_{\cD}(x))\|\leqslant \|x-\pi_{\cD} (x)\|+ \|\pi_{\cD} (x)-\pi_{\cD_{\tau}} (\pi_{\cD}(x))\|.
	\end{equation}
	Based on the earlier argument, since $\pi_{\cD} (x)\in\cD$, we have $\|\pi_{\cD} (x)-\pi_{\cD_{\tau}} (\pi_{\cD}(x))\|\leqslant \alpha_\cD\tau$. This together with \eqref{e1} results in $\|x-\pi_{\cD_{\tau}} (x)\|\leqslant \|x-\pi_{\cD} (x)\|+\alpha_\cD\tau$.
\end{proof}
\begin{remark}
	Note that $\alpha_{\cD}$ is a parameter of the polytope ${\cD}$ and does not depend on $\tau$. Figure~\ref{alpha_shr} illustrates the geometrical meaning of $\alpha_{\cD}$ for a simple polytope (here a triangle) in two dimensions. As it can be seen, the faces of ${\cD}_{\tau}$ are parallel to the faces of ${\cD}$ with a distance less than $\tau$ but not necessarily equal to $\tau$. 
\end{remark}
{"config": "arxiv", "file": "2006.13326/ShrinkageLemma.tex"}