One of the better properties of convex set is called convex set separation theorem, it tells us, you can choose a hyperplane to separate the two convex disjoint sets! We'll also see the application of this theorem in convex optimization problems, such as Slater theorem.
Convex Sets separation theorem (Euclidean space case): A collecting \ (of S_ {. 1} \) , \ (of S_ {2} \) is \ (\ mathbb {R & lt} ^ {n-} \) ( \ (n-\ GEQ 1 \) ) of the two non-empty disjoint convex sets, there is a separating hyperplane \ (of S_ {1} \) , \ (of S_ {2} \) , there are both \ (v \ in \ mathbb { R & lt} ^ {n-} \) , \ (V \ NEQ 0 \) and \ (b \ in \ mathbb { R} \) such that:
\ the begin {Equation} V \ CDOT X + B \ GEQ 0, \ End { equation} for any \ (x \ in S_ {1 }, \) and:
\ the begin {equation} V \ CDOT X + B \ Leq 0, \ End {equation} for any \ (x \ in S_ {2 } \ ) .
Proof:
a \ (. 1 of S_ {} \) , \ (of S_ {2} \) disjoint set of convex sets easily verified: \ (of S_. 1} {2} {-S_ \ triangleq \ lbrace XY \ MID X \ in S_ {1}, y \ in S_ {2} \ rbrace \) does not contain \ (0 \) a convex set, so we only need to prove the presence of \ (v \ in \ mathbb { R} ^ {n} \ ) , \ (V \ NEQ 0 \) such that for any \ (X \ in -S_ of S_ {{2}}. 1 \) , there are: \ (V \ CDOT X \ GEQ 0 \) , because when for any the \ (X \ in. 1 of S_ {} \) , \ (Y \ {2} of S_ in \) , then \ (XY \ in -S_ of S_ {{2}}. 1 \) , \ (V \ CDOT (XY ) = V \ CDOT XV \ CDOT Y \ GEQ 0 \) , so we make the \ (B \ triangleq - \ sup_ {Y \ in of S_ {2}} \ lbrace V \ CDOT Y \ rbrace \) , at this time \ ( V \) , \ (B \) just to meet the above inequality (1), (2).
So you may wish to prove the following lemma.
Lemma 1: set \ (S \) is \ (\ mathbb {R} ^ {n} \) is a closed convex subset \ (0 \) is not set \ (S \) point inside, then presence \ (V \ in \ mathbb {R & lt} ^ {n-} \) , \ (V \ NEQ 0 \) , ST \ (V \ CDOT Y \ GEQ 0 \) , for any \ (y \ in S \) .
Note that if the above lemma 1 exists, this time \ (S \ triangleq \ overline { S_ {1} -S_ {2}} \) is a closed convex set, and since the \ (0 \ notin S1-S2 \) readily known by convex sets \ (0 \) does not belong to \ (S \) inside, and from the above can be found in the corresponding Lemma 1 \ (v \ in \ mathbb { R} ^ {n} \) so original proposition is true, then we only need to prove more than lemma.
Proof of Lemma 1: First we prove \ (0 \ notin S \) situations. This case, since \ (S \) is a closed set, the presence \ (x ^ {\ ast} \ in S \) such that: (. 1) \ (\ Vert ^ {X \ AST} \ Vert = \ {inf_ Y \ in S} \ lbrace \ Vert the y-\ Vert \ rbrace \) . we conclude that this time \ (the X-^ {\ AST} \ cdot the y-\ geq 0 \) , for any \ (the y-\ in S \) . otherwise, there is \ ( Y_ {0} \ in S \) , so \ (X ^ {\ AST} \ CDOT Y_ {0} <0 \) , at this time we let: \ [T = \ FRAC {Y_ {0} \ CDOT (Y_ the -X-^ {0} {\ AST})} {\ Vert the -X-Y_ {0} {^ \ AST} \ Vert ^ {2}} \] , by the \ (y_ {0} \ cdot x ^ {\ ast } <0 \) readily known . \ (t \ in (0,1 ) \) we let:
\ [Z \ {^ triangleq TX \ AST} + (. 1-T) Y_ {0}, \]
is \ ( z \ in S \) and:
\ [Z \ PERP (X ^ {\} the -Z AST), \]
then:
\ [\ Vert z \ Vert ^ {2} = \ Vert x ^ {\ ast} \ Vert ^ {2} - \ Vert x ^ {\ ast} -z \ Vert ^ {2} <\ Vert x ^ {\ ast} \ Vert ^ {2}
, \] which (1) contradicted case then \ (0 \ notin S \) is proved.
We now consider \ (0 \ in \ partial S \) case, then the presence of the sequence \ (X_ {K} \ notin S \ rightarrow 0 \) , when the \ (K \ rightarrow \ infty \) . Seen from the case of the above has been demonstrated, the presence \ (V_ {K} \ in \ mathbb {R & lt} ^ {n-} \) , \ (\ Vert V_ {K} \ Vert =. 1 \) such that for any \ (Y \ in S \) have \ (V_ {K} \ CDOT (Y-X_ {K}) \ GEQ 0 \) . and because the sequence \ (v_ {k} \) bounded, then there must be a convergent subsequence \ (v_ {k_ {i}} \) converges to a \ (V \ in \ mathbb {n-R & lt} ^ {} \) , time \ (\ Vert v \ Vert = 1 \) and for any \ (Y \ in S \) we have \ (V_ {K_ {I}} \ CDOT (Y-X_ {K_ {I}}) \ GEQ 0 \) , we let \ (k \ rightarrow \ infty \ ) may naturally be obtained \ (V \ CDOT Y \ GEQ 0 \) , the lemma is proved this time, Theorem proof.