sets $S(x_i)$ covers $\partial B.$ Let $v_i\mathrel{\mathop{:}}=v(x_i).$ For any natural number $d$, we define [^6] $$p_d(y)=\sum_{i=1}^N \left(\frac{v_i^Ty}{v_i^Tx_i}\right)^{2d} \text{ and } S_d=\{y \in \mathbb{R}^n|~ p_d(y)\leq 1\}.$$ Note that $p_d$ is convex as the sum of even powers of linear forms and homogeneous. We first show that $$(\rho + \frac12(1-\rho)) B \subseteq int(S_d).$$ As $(\rho + \frac12(1-\rho)) B \subset int(C)$, for all $y \in (\rho + \frac12(1-\rho)) B$ and for all $i=1,\ldots,N$, we have $v_i^Ty < v_i^Tx_i$. Hence there exists a positive integer $d$ such that $$\left(\max_i \max_{y \in (\rho +\frac12 (1-\rho))B} \frac{v_i^Ty}{v_i^Tx_i}\right)^{2d}<\frac{1}{N}$$ and so $p_d(y)<1$ for all $y \in (\rho + \frac12(1-\rho)) B$.

We now show that $S_d \subseteq B.$ Let $y \in S_d$, and so $p_d(y)\leq 1.$ This implies that $$\frac{v_i^Ty}{v_i^Tx_i} \leq 1, \forall i=1,\ldots,N.$$ From this, we deduce that $y \notin \partial B.$ Indeed if $y \in \partial B,$ there exists $i \in \{1,\ldots,N\}$ such that $y \in S(x_i)$, which implies that $v_i^T y>v_i^Tx_i$ and contradicts the previous statement. Hence $\partial B \cap S_d =\emptyset$. As both $B$ and $S_d$ contain the zero vector, we conclude that $S_d \subseteq int(B) \subseteq B$. Note that this guarantees positive definiteness of $p_d$ as $p_d$ is homogeneous and its 1-sublevel set is 