We now an optimization problem for any (not necessarily a convex optimization problem): \
Split the begin {} \ text min {} \ {0} Quad & F_ (X) \ NEWLINE \ {text Subject to:} \ {F_ & Quad i} (x) \ leq 0 , i = 1, ..., m \ newline & h_ {i} (x) = 0, = 1, ..., p \ end {split}
if all occurring function are first order differentiable function, and assuming that \ (X ^ {\ AST} \) , \ ((\ ^ {the lambda \} AST, \ NU ^ {\ AST}) \) respectively (), and issues its the optimal solution of the dual problem, gap is zero, then we naturally have the following conditions:
\ {Split} the begin F_ {i} (the X-^ {\} AST) & \ Leq 0, i = 1, ... m; \ newline h_ {i} (x ^ {\ ast}) & = 0, i = 1, ..., p \ newline \ lambda_ {i} ^ {\ ast} & \ geq 0, i = 1. ..m \ newline \ lambda_ {i} ^ {\ ast} f_ {i} (x ^ {\ ast}) & = 0, i = 1, ... m \ newline \ nabla f_ {0} (x ^ {\ ast}) + \ sum_ {i = 1} ^ {m} \ lambda_ {i} ^ {\ ast} \ nabla f_ {i} (x ^ {\ ast}) + \ sum_ {i = 1} ^ {p} \ nu_ {i} ^ {\ ast} \ nabla h_ {i} (x ^ {\ ast}) & = 0, \ end {split}
We now consider the case of convex optimization problem, this time, we have the inverse proposition easy to get set up as follows:
Theorem If the optimization problem (), \ (F_ {I} \) (I = 0,1, ..., m) are convex function may be a first order, \ (H_ {I} \) (I =. 1 , ..., p) are affine function, \ (\ tilde are {X} \ in D \) , \ (\ {tilde are \} the lambda \ in \ R & lt mathbb {m} ^ {} \) , \ ( \ tilde {\ nu} \ in \ mathbb {R} ^ {p} \) satisfy the KKT condition :
\begin{split}f_{i}(\tilde{x})&\leq 0, i=1,...m; \newline h_{i}(\tilde{x})&=0, i=1,...,p \newline \tilde{\lambda_{i}}&\geq 0, i=1...m \newline \tilde{\lambda_{i}}f_{i}(\tilde{x})&=0, i=1,...m\newline \nabla f_{0}(\tilde{x})+\sum_{i=1}^{m}\tilde{\lambda_{i}}\nabla f_{i}(\tilde{x})+\sum_{i=1}^{p}\tilde{\nu_{i}}\nabla h_{i}(\tilde{x})&=0,\end{split}
The \ (\ tilde are {X} \) , \ ((\ {tilde are \ the lambda}, \ {tilde are \ NU}) \) are the original optimization problem, the solution to the dual problem.
Proof :
考察 Lagrange 函数：\[L: D \times\mathbb{R}^{m}\times\mathbb{R}^{p}\rightarrow \mathbb{R},\]
\(L(x,\lambda,\nu)=f_{0}(x)+\sum_{i=1}^{m}\lambda_{i}f_{i}(x)+\sum_{i=1}^{p}\nu_{i}h_{i}(x)\).
For convenience, we use the same notation before, note: \ (P ^ {\ AST} = \ inf_ {X \ in D (F, H)} F_ {0} \) , \ (G (\ tilde are {\ } the lambda, \ {tilde are \ NU}) = \ {inf_ X \ D} in L (X, \ the lambda, \ NU) \) .
Now fixed \ (\ the lambda = \ tilde are {\ the lambda} \) , \ (\ NU = \ tilde are {\ NU} \) , since \ (\ tilde are {\ lambda_ {I}} \ GEQ 0 \) , \ ( h_ {i} \) is an affine function, easy to know \ (L (x, \ tilde {\ lambda}, \ tilde {\ nu}) \) is about \ (X \) is a convex function, all time \ (\ nabla_ {x} L ( \ tilde {x}, \ tilde {\ lambda}, \ tilde {\ nu}) = 0 \) ensures \ (L (x, \ tilde {\ lambda}, \ tilde { \ nu}) \) at (x = \ tilde {x} \) \ in addition to a minimum value, so this time, for any \ (X \ in D \) , we have:
\ begin {split} L (x , \ tilde {\ lambda}, \ tilde {\ nu}) & \ geq L (\ tilde {x}, \ tilde {\ lambda}, \ tilde {\ nu}) \ newline & = f_ {0} (\ tilde {x}) \ end {split}
noted by the KKT conditions, the equal sign in the above formula is clearly established. This time on both sides of the equation removed supremum of \ (the X-\ in D \) , we naturally are:
\ [G (\ {tilde are \ the lambda}, \ {tilde are \ NU}) \ {0} GEQ F_ (\ tilde are {X}) \ P ^ {GEQ \ AST} \] , but we note that \ (G ( \ {tilde are \ the lambda}, \ {tilde are \ NU}) \ P ^ {GEQ \ AST} \) , the formula can be seen two equal sign are true, the proposition have proved.