Suppose we have a primal problem


$
\min\_x f(x), s.t. h\_i(x)=0,
$


where $h\_i$ are all affine, and $f$ is convex. 


Then its Lagrangian is


$\min\_x \max\_{z\_i} f(x) + \sum\_i z\_i h\_i(x)$


and the dual problem is


$\max\_{z\_i} \min\_x f(x) + \sum\_i z\_i h\_i(x)$


The KKT condition (sufficient and necessary here) only says for any optimal solution x\*, there exists $z^\*\_i$ such that


$\nabla f(x^\*) + \sum\_i z^\*\_i \nabla h\_i(x^\*) = 0$


But is the following claim correct?


For any optimal $z^\*\_i$ of the dual problem, there must be an optimal $x^\*$ of the primal problem, such that


$\nabla f(x^\*) + \sum\_i z^\*\_i \nabla h\_i(x^\*) = 0$.



Lagrangian duality 
At the level of generality you asked about, the answer is **no**, the claim is not correct. Of course, your case of interest may rule out counterexamples like the one below.


It can happen that the primal is bounded below but does not achieve its optimum, whereas the dual does. For example take $f(x\_1,x\_2) = \exp(x\_1)$ and $h\_1(x\_1,x\_2)=x\_2$. Then the primal is $\min\_{x\_1\in\mathbb{R}} \exp(x\_1)$ and the dual is $\max\_{z\_1 = 0} 0$. Both have optimal value zero, but in the case of the primal the infimum is not achieved. The KKT equation 
\[
\begin{bmatrix} \exp(x\_1) & 0\end{bmatrix} + z\_1\begin{bmatrix} 0 & 1\end{bmatrix} = 0
\]
has no solutions.


