【Derivation】Convex Optimization

Separation theorems and supporting hyperplanes（分离定理与支撑超平面）

       Inner and outer polyhedral approximations.（内部与外部多面体逼近）

       Let C belongs to Rn be a closed convex set.and suppose that x1,...xk are on the boundary of C.Suppose that for each i,Ati(x-xi)=0 defines a supporting hyperplane for C at xi,.e.,C belongs to {x|aAti(x-xi)<=0}.consider the two polyhedra

                     Pinner = conv{ x1,...xk },  Pouter = { x | Ati(x-xi)<=0 , i=1,...,K}.

       Show that Pinner belongs to C belongs to Pouter.Draw a picture illustrating this.

      令C包含于Rn 为闭凸集，并设x1,...xk 在C的边界上。设对于每个i,Ati(x-xi)=0定义了C在xi处的一个支撑超平面，即C包含于 {x|aAti(x-xi)<=0}。考虑两个多面体

                    Pinner = conv{ x1,...xk },  Pouter = { x | Ati(x-xi)<=0 , i=1,...,K}.

      证明Pinner 包含于 C 包含于 Pouter 并画出图像进行说明。

     Solution.

           The points xi are in  C because C is closed.Any point in Pinner = conv{x1,...xk },  is also in C because C is convex .Therefore Pinner belongs to C. If x belongs to C then Ati(x-xi)<=0 for  i=1,...,K,i.e,x belongs to pouter.therefore C belongs to Pouter.

          Pinner = conv(x1,...,xk).

          C = { x | Ati(x-xi)<=0 }.

          Pouter = { x | Ati(x-xi)<=0 , i=1,...,K}.

          The figure shows an example with k=4.

     解：

          由于C是闭的所以xi都在C中。由于C为凸集， 所以Pinner = conv{x1,...xk } 中的任意点也在C中，因此Pinner包含于C。如果x包含于C 及 Ati(x-xi)<=0 for  i=1,...,K,即x包含于Pouter, 因此C包含于Pouter。