Consider the primal problem $$ \min_x \{f(x):G(x)\in Y\}\tag{$P$} $$ rewritten as $$ \nu(P^\prime) = \min_x \{f(x):y\in Y,\;y=G(x)\}\tag{$P^\prime$} $$ where $G:\mathbb{R}^n\to \mathbb{R}^m$ with $G(x)=\{g_1(x),\ldots,g_m(x)\}$ where $Y\subseteq\mathbb{R}^m$. For simplicity, assume that $g_i$, $f$, and $Y$ are convex. Is there a convenient way to write a dual problem of ($P^\prime$) if we know: $g_i^*$, $f^*$, and $\delta_Y^*$? I tried the following:
- Lagrangian $$ L(x,y;\lambda)=f(x)+\sum_{i=1}^m\lambda_ig_i(x)+\delta_Y(y)-\lambda^\top y $$
- Optimal value of dual program \begin{align*} \nu(D^\prime)&=\max_\lambda\min_{x} \{f(x)+\sum_{i=1}^m\lambda_ig_i(x)\}-\delta_Y^*(\lambda)\\ &=\max_{\lambda,\mu}\min_{x,x^i} \{f(x)+\sum_{i=1}^m\lambda_ig_i(x^i) : x^i=x\}-\delta_Y^*(\lambda)\\ &=\max_{\lambda,\mu}\min_{x,x^i} \{f(x)+\sum_{i=1}^m\lambda_ig_i(x^i) + (\mu^i)^\top(x^i - x)\}-\delta_Y^*(\lambda)\\ &=\max_{\lambda,\mu} -f^*\Bigl(-\sum_{i=1}^m \mu^i\Bigr) - \sum_{i=1}^m \lambda_i g_i^*(-\mu_i/\lambda_i) - \delta_Y^*(\lambda)\\ \end{align*} but am not sure if there is a nicer way to simplify the "$\min_x$" part than introducing $x^i=x$ constraints.