I'm going through an article lately and there is one point which is very confusing. So, we have the following original constrained binary quadratic problem as the following. The pre-assumption of certain parameters are $Q\in Z^{n*n}$, $A\in Z^{n*n}$, and $b\in Z^{n*1}$.
$\text{min}$ $x^{T}Qx$,
$s.t.,$ $Ax\leq b$ $\text{and}$ $x\in \{0,1\}^{n}$. Let's call this optimization problem the problem $P$.
Now, by the classical lagrangian reduction, this can be relaxed as
$\text{min}$ $x^{T}Qx+\lambda^{T}(Ax-b)$,
$s.t.,$ $x\in \{0,1\}^{n}$ and $\lambda >0$. Let's call this optimization problem the problem $L_{\lambda}$.
Since the relaxation serves as a lower bound, we ideally plan to make it as maximum as possible. So, let $d(\lambda) = x^{T}Qx + \lambda^{T}(Ax-b)$, we have the other optimization,
$\text{max}$ $d(\lambda)$
$s.t.,$ $\lambda \geq 0$. Let's call this optimization problem the problem $L$.
So far so good. Now, here is the confusing point.
The article mentioned that the problem $L$ can be rewritten as the following.
$max$ $\mu$
$s.t.,$ $\mu \leq x^{T}Qx + \lambda^{T}(Ax-b) $ for any $x\in \{0,1\}^{n}$ and $\lambda \geq 0$.
Let's call the last optimization problem the problem $W$.
I do not quite understand why this is a rewritten. So, suppose $\mu_{0}$ is the optimal solution of problem $W$, then it follows that $\mu_{0}\leq x^{T}Qx+\lambda^{T}(Ax-b)$ for any $x\in \{0,1\}^{n}$ and $\lambda \geq 0$. While on the other hand, for the problem $L$, what I understand it is like this: for each given $\lambda$, there is one optimization problem $L_{\lambda}$, and one optimal solution (which we take a minimum by the definition of $L_{\lambda}$). Then, finally, among all the optimum solutions, one maximum value is chosen. Somehow, I can not see that in the formulation of the optimization problem $W$.
Or did I overthink? Any comments would be greatly appreciated.