Suppose we have an (not necessarily convex) optimization problem : <mtable columnalign="right l

Carolyn Beck 2022-06-24 Answered
Suppose we have an (not necessarily convex) optimization problem :
min x f 0 ( x ) f 1 ( x ) 0.
Let L ( x , λ ) = f 0 ( x ) + λ ( f 1 ( x ) ). Then the above problem can be equivalently written as:
min x max λ 0 L ( x , λ ) .
The dual of the above problem can be written as:
max λ 0 min x L ( x , λ ) .
We say that strong duality holds at a point when
min x max λ 0 L ( x , λ ) = max λ 0 min x L ( x , λ ) .
By weak duality, the inequality
min x max λ 0 L ( x , λ ) max λ 0 min x L ( x , λ )
always holds true. My doubt is: suppose there exists an x that minimizes L ( x , λ ) for a fixed λ, can i say that the following inequality holds true:
max λ 0 ( min x L ( x , λ ) ) = max λ 0 L ( x , λ ) min x max λ 0 L ( x , λ ) .
If the above is true, then are we saying that if there is a primal variable that attains its minimum in the dual problem, then strong duality holds true? Somewhere this does not seem to add up.
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Answers (1)

Harold Cantrell
Answered 2022-06-25 Author has 21 answers
No, because the last inequality, max λ 0 L ( x , λ ) min x max λ 0 L ( x , λ ), does not hold in general. Note that x is a function of λ. It would have been clearer to write x∗(λ) instead of just x .
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