Suppose we have the optimization problem in standard form:
minimize$ f_0(x)$
subject to
$f_i(x)\le 0,i=1,..,m$
$h_i(x)= 0,i=1,..,p$
And we indicate $dom \ f_0=dom \ f_i=dom \ h_i =D$. Whereas the feasible set is $F \subset D$. $p^*$ is instead the value of the minimum.
The theory of duality proceeds definining the lagrangian $L(x,\lambda,\nu):$
$L(x,\lambda,\nu)=f_0(x) + \sum_{i=1}^m \lambda_i f_i(x)+\sum_{i=1}^p \nu_ih_i(x)$
and the function:
$g(\lambda,\nu)=\inf_{x \in D}L(x,\lambda,\nu)$
Now one central observation is that in this way:
$p^* \ge g(\lambda,\nu)$ if $\lambda \ge 0 \ [1]$
And this is useful because:
1- It is a variational estimate of the minimum.
2- $g$ is concave even if the $f$ are not.
My question is this. If we define the modified $g$ with $\tilde g$:
$\tilde g(\lambda,\nu)=\inf_{x \in F}L(x,\lambda,\nu)$
still I think we would have:
$p^* \ge \tilde g(\lambda,\nu) \ge g(\lambda,\nu), \lambda \ge 0$
and we would have therefore a stronger bound on the minimum.
Why the theory than proceeds with using $g$ rather than $\tilde g$. What is the reason ?