Suppose we have an (not necessarily convex) optimization problem :

$\begin{array}{r}\underset{x}{min}{f}_{0}(x)\\ {f}_{1}(x)\le 0.\end{array}$

Let $L(x,\lambda )={f}_{0}(x)+\lambda ({f}_{1}(x))$. Then the above problem can be equivalently written as:

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda ).$

The dual of the above problem can be written as:

$\underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda ).$

We say that strong duality holds at a point when

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda )=\underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda ).$

By weak duality, the inequality

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda )\ge \underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda )$

always holds true. My doubt is: suppose there exists an ${x}^{\ast}$ that minimizes $L(x,\lambda )$ for a fixed $\lambda $, can i say that the following inequality holds true:

$\underset{\lambda \ge 0}{max}{\textstyle (}\underset{x}{min}L(x,\lambda ){\textstyle )}=\underset{\lambda \ge 0}{max}L({x}^{\ast},\lambda )\ge \underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda ).$

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.

$\begin{array}{r}\underset{x}{min}{f}_{0}(x)\\ {f}_{1}(x)\le 0.\end{array}$

Let $L(x,\lambda )={f}_{0}(x)+\lambda ({f}_{1}(x))$. Then the above problem can be equivalently written as:

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda ).$

The dual of the above problem can be written as:

$\underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda ).$

We say that strong duality holds at a point when

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda )=\underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda ).$

By weak duality, the inequality

$\underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda )\ge \underset{\lambda \ge 0}{max}\underset{x}{min}L(x,\lambda )$

always holds true. My doubt is: suppose there exists an ${x}^{\ast}$ that minimizes $L(x,\lambda )$ for a fixed $\lambda $, can i say that the following inequality holds true:

$\underset{\lambda \ge 0}{max}{\textstyle (}\underset{x}{min}L(x,\lambda ){\textstyle )}=\underset{\lambda \ge 0}{max}L({x}^{\ast},\lambda )\ge \underset{x}{min}\underset{\lambda \ge 0}{max}L(x,\lambda ).$

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.