intersection of convex regions

Question

Consider $n$ vectors in $R^n$ given by $a_1=(1,1,\ldots,1)$ and $a_2=(-1,1,\ldots,1)$, $a_{3}=(-1,-1,1,\ldots,1)$,..., $a_n=(-1,-1,\ldots,-1,1).$ Suppose we are given $2n$ positive numbers $\lambda_i$, $1\leq i\leq 2n.$ For $1\leq i\leq n$ we define $P_i=\{x:~ \langle x,a_i\rangle\leq \lambda_i\}$, and for $n+1\leq i\leq 2n$, we define $P_i=\{x:~ \langle x,-a_{i-n}\rangle\leq \lambda_i\}.$ Is the intersection of $P_i,$ $\cap_{i=1}^{2n}P_i$, a compact region?

Answer 1

Hint take $ x \in \cap_{i=1}^{2n}P_i $ looking at first component of $x$ i.e, $x_1 $ you can show that this $x_1 \in R$ running over a bounded set as $x$ running a over $\cap_{i=1}^{2n}P_i$! To show this , let assume WLOG $x_1 \rightarrow -\infty$, by extracting suitable subsequence of $x \in \cap_{i=1}^{2n}P_i$. since $x \in P_2 \cap P_n$ we have $\langle a_2, x \rangle \leq \lambda_2$ and $\langle - a_1 , x \rangle\leq \lambda_{n+1} $ therefore $$ -2 x_1 =\langle a_2-a_1 , x \rangle \leq \lambda_1 + \lambda_{n+1} $$ Which is a contradiction, similar idea works for other components of $x$ (some of them requires more P_i s).

Answer 2

Hint: None of the $P_i$ Are compact. Neither are the $P_i \cap P_{i+n}, i=1,\dots,n,$ but they are bounded in the direction of $a_i$ and are essentially closed tubular neighborhoods of $\{a_i\}^\perp$. What is the intersection of these orthogonal complements? Just $\vec{0}$, so we should expect $\bigcap_{i=1}^{2n}P_i$ to be a compact neighborhood of $\vec{0}$.

Answer 3

Here is another approach. Define $$A = \begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}^T,\Lambda_1 = \begin{pmatrix}\lambda_1&\cdots&\lambda_n\end{pmatrix}^T, \Lambda_2 = \begin{pmatrix}\lambda_{n+1}&\cdots&\lambda_{2n}\end{pmatrix}^T,$$ $$ \cal{C} = \{y\in\mathbb{R}^n: -\Lambda_2\leq y\leq\Lambda_1 \ \ (\mathrm{entrywise})\}.$$ Clearly $\cal{C}$ is compact and $A$ is nonsingular, so $\bigcap_{i=1}^{2n}P_i = A^{-1}\cal{C}$ is compact.