TITLE: sufficient condition for KKT problems
QUESTION [11 upvotes]: For the Karush-Kuhn Tucker optimsation problem, Wikipedia notes that: 
"The necessary conditions are sufficient for optimality if the objective function f and the inequality constraints g_j are continuously differentiable convex functions and the equality constraints h_i are affine functions." 
link
Could someone please show me how this result is derived? That is, given a convex objective function, convex inequality constraints and affine equality constraints, how can we show that any point in the feasible set that satisfies the KKT conditions must be a minimizer of the function over the feasible set?

REPLY [8 votes]: The primal problem is
\begin{align}
\operatorname{minimize}_x & \quad f_0(x) \\
\text{subject to} & \quad f_i(x) \leq 0 \quad \text{for } i = 1,\ldots, m\\
& \quad Ax = b.
\end{align}
The functions $f_i, i = 0,\ldots,m$ are differentiable and convex.
Assume $x^*$ is feasible for the primal problem (so $f_i(x^*) \leq 0$ for $i = 1,\ldots,m$ and $A x^* = b$) and that there exist vectors $\lambda \geq 0$ and $\eta$ such that 
$$
\tag{$\spadesuit$}\nabla f_0(x^*) + \sum_{i=1}^m \lambda_i \nabla f_i(x^*) + A^T \eta = 0  
$$
and
$$
\lambda_i f_i(x_i) = 0 \quad \text{for } i = 1,\ldots,m.
$$
Because the functions $f_i$ are convex, equation ($\spadesuit$) implies that $x^*$ is a minimizer (with respect to $x$) of the Lagrangian
$$
L(x,\lambda,\eta) = f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \eta^T(Ax - b).
$$
Thus, if $x$ is feasible for the primal problem, then
\begin{align*}
f_0(x) & \geq f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \eta^T(Ax - b) \\
& \geq f_0(x^*) + \sum_{i=1}^m \lambda_i f_i(x^*) + \eta^T(Ax^* - b) \\
& = f_0(x^*).
\end{align*}
This shows that $x^*$ is a minimizer for the primal problem.