For every , the convexity of the function yields It remains to prove . Using first and then the convexity of the functions , we get This finishes the proof.

Using \eqref{eq:saddle} with first and then with , we have for every , The existence of a saddle point implies the strong duality since which is the converse inequality of \eqref{eq:weak}. This finally implies:

Note that shows that .

The upper bound in \eqref{eq:saddle} regardless the value of shows that for every , that is . Now taking \eqref{eq:saddle} with yields for , which shows that . Combining with \eqref{eq:all} yields so that and since and , this implies for all .

Set The condition is equivalent to saying that and since is clearly a convex set, by a separation argument, there exists a non-null vector such that Take an arbitrary . If then where and . The previous inequality implies for all , therefore . Now, since for all , a simple limiting argument yields for all , To conclude, and since we already know that for all , it only remains to prove that and to set in that case . The rest of the argument is by contradiction. If , then the previous inequality on a Slater point yields but since for all , we finally get for all . Finally all the components of are null and we are face to a contradiction. The proof is completed.

Assume now that for some . Then, so that there exists satisfying for all , . And this implies that is a saddle point since: for all and , so that is a saddle point and this implies that the strong duality holds (as we have seen in the previous section). This concludes the proof.