Does strong duality hold when I dualize only a subset of the constraints?

If strong duality holds, then it also holds when only a subset of the constraints is dualized.

We define the following three problems: the original, the partially dualized, and the dual.

Problem (P1): begin{align}min_x&quad f(x)\text{s.t.}&quad g_i(x)leq 0, i in Cend{align}

Problem (P2): begin{align}max_{lambdage0} min_x&quad f(x) + sum_{i in A}lambda_ig_i(x)\text{s.t.}&quad g_i(x)leq 0, i in Csetminus Aend{align}

Problem (P3): begin{align}max_{muge0} max_{lambdage0} min_x&quad f(x) + sum_{i in A}lambda_ig_i(x) + sum_{i in Csetminus A}mu_ig_i(x)end{align}

It is given that strong duality holds, which means that (P1) and (P3) have the same objective value. For convenience, denote this by f(P1) = f(P3).

Using weak duality, we will show that f(P1) $ge$ f(P2) $ge$ f(P3). Because we know that f(P1) = f(P3), it must be that f(P1) = f(P2) = f(P3).

From (P1) to (P2): let $bar{x}$ be an optimal solution to (P1). Because $bar{x}$ is feasible for (P1), we have $g_i(bar{x})le0$ for all $iin C$. Next, plug $bar{x}$ into (P2), which is feasible. Due to the non-negativity of the multipliers, it follows for any $lambda ge 0$ that $f(bar{x}) ge f(bar{x}) + sum_{i in A}lambda_ig_i(bar{x})$. Hence, f(P1) $ge$ f(P2).

From (P2) to (P3): let $bar{lambda} ge 0$ be optimal multipliers for (P2) and let $bar{x}$ be corresponding optimal primal variables. Using a similar argument, $bar{lambda}$ and $bar{x}$ can be plugged into (P3). Because $mu ge 0$ and $g_i(bar{x})le0$ for all $iin Csetminus A$, we have for all $mu ge 0$ that $$quad f(bar{x}) + sum_{i in A}bar{lambda}_ig_i(bar{x}) ge f(bar{x}) + sum_{i in A}bar{lambda}_ig_i(bar{x}) + sum_{i in Csetminus A}mu_ig_i(bar{x}).$$ It follows that f(P2) $ge$ f(P3), which completes the proof.