# Is it Necessary to Assign Dual Variables to Sign Constraints in the Lagrangian?

Consider an optimization problem with sign constraints on the variables: \begin{align} \min_{x\in\mathbb{R}^n}&\quad f(x) \\ \text{s.t.}&\quad g_i(x)\le 0,\quad i=1,...,m \\ &\quad h_j(x)=0,\quad j=1,....,\ell \\ &\quad x\le 0. \end{align} When forming the Lagrangian for this problem, I wonder if it is necessary to assign dual variables to the constraint $$x\le 0$$?

To be more specific, I know for certain that I can treat $$x\le 0$$ as general inequality constraints, which yields the Lagrangian $$L(x,\lambda,\mu,\nu)=f(x)+\sum_{i=1}^m\lambda_ig_i(x)+\sum_{j=1}^{\ell}\mu_jh_j(x)+\sum_{i=1}^n\nu_ix_i$$ and the dual function $$d(\lambda,\mu,\nu)=\inf_{x\in\mathbb{R}^n}L(x,\lambda,\mu,\nu),\quad\forall\,(\lambda,\mu,\nu)\in\mathbb{R}^m_+\times\mathbb{R}^{\ell}\times\mathbb{R}^n_+,$$ so the dual problem is \begin{align} (D)\quad\max&\quad d(\lambda,\mu,\nu) \\ \text{s.t.}&\quad \lambda\ge 0,\quad\nu\ge 0. \end{align}

What if I do not assign dual variables to the constraints $$x\le 0$$? That is, let $$\widetilde{L}(x,\lambda,\mu)=f(x)+\sum_{i=1}^m\lambda_ig_i(x)+\sum_{j=1}^{\ell}\mu_jh_j(x)$$ and, when forming the dual function, perform minimization over $$x\le0$$ instead of $$x\in\mathbb{R}^n$$, which gives $$\widetilde{d}(\lambda,\mu)=\inf_{x\le 0}L(x,\lambda,\mu),\quad\forall\,(\lambda,\mu)\in\mathbb{R}^m_+\times\mathbb{R}^{\ell}.$$ This yields \begin{align} (\widetilde{D})\quad\max&\quad \widetilde{d}(\lambda,\mu) \\ \text{s.t.}&\quad \lambda\ge 0. \end{align} Are $$(D)$$ and $$(\widetilde{D})$$ equivalent?

## 1 Answer

It turns out that $$(D)$$ and $$(\widetilde{D})$$ are not equivalent. As an example, let $$n=m=\ell=1$$, $$g_1(x)=h_1(x)=0$$, and $$f(x)=-x^3$$. Then $$(D)$$ is given by \begin{align} (D)\quad\max_{\lambda,\mu,\nu}&\quad\min_{x\in\mathbb{R}}(-x^3+\nu x)=-\infty\\ \text{s.t.}&\quad\lambda,\nu\ge 0, \end{align} and has optimal value $$-\infty$$. In contrast, $$(\widetilde{D})$$ is \begin{align} (\widetilde{D})\quad\max_{\lambda,\mu}&\quad\min_{x\le 0}(-x^3)=0 \\ \text{s.t.}&\quad\lambda\ge 0 \end{align} and has optimal value $$0$$.