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I have an equality constraints as

$\qquad \qquad \min_x f(x) \\ \qquad \qquad s.t. \quad Ax \leq b \\\qquad \qquad \qquad x = h(x)$.

The KKT conditions of the minimization without the equality constraints is

$\qquad \qquad \nabla f(x) + A^T u = 0, Ax < b \\ \qquad \qquad u \geq 0 \quad \&\quad u^TAx = 0$.

Can I replace $x = H(x)$ in the above KKT conditions, and assume that the equality constraints are also satisfied?

Majid Mohammadi
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    After replacing $x=h(x)$ you still have $x$ left in the objective that must satisfy the constraint, so no. – A.Γ. Jan 03 '19 at 20:23

1 Answers1

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KKT conditions have another expression when $h(x)=0$ restrictions are included.

Theorem.(General KKT conditions) Let (P) be the convex optimization problem ($f,\ g_{i}$ convex and $h_{j}$ affine) as it follows,

\begin{equation} \begin{split} \min&\ \ f(\textbf{x})\\ s.t.&\ \ g_{i}(\textbf{x})\leq0\ \ i=1,\dots ,m\\ &\ \ h_{j}(\textbf{x})=0\ \ j=1,\dots,l. \end{split} \end{equation}

Let $\bar{\textbf{x}}$ a factible point such that

  • Functions $f$ and $g_{i}$ for all $i\in\{1,\dots,m\}$ are differentiable in $\bar{\textbf{x}}$.
  • $h_{j}(\bar{\textbf{x}})\in \mathcal{C}^{1}$, with $j\in\{1,\dots,l\}$.
  • Vectors $\nabla g_{i}(\bar{\textbf{x}})$, with $i\in\{i:g_{i}(\bar{\textbf{x}})=0\}$ and $\nabla h_{j}(\bar{\textbf{x}})$ with $j\in\{1,\dots,l\}$ are linear independent.

If $\bar{\textbf{x}}$ is a local minimum, then exists $u_{i}$ unique scalars for all $i$ and $v_{j}$ for all $j\in\{1,\dots,l\}$ such that

\begin{equation} \begin{split} \nabla f(\bar{\textbf{x}})+\sum_{i=1}^{m}u_{i}\nabla g_{i}(\bar{\textbf{x}})+\sum_{j=1}^{l}v_{j}\nabla h_{j}(\bar{\textbf{x}})&=\textbf{0}\\ u_{i}g_{i}(\bar{\textbf{x}})&=0\ \ \ \textrm{ for all }i\in\{1,\dots,m\}\\ u_{i}&\geq0\ \ \ \textrm{ for all }i\in\{1,\dots,m\}. \end{split} \end{equation}

Theorem from Bazaraa M. S. (2006). Nonlinear Programming: Theory and Algorithms. 3rd edition.

Lancet S.
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