Is KKT conditions necessary and sufficient for any convex problems?

Question

In Boyd's Convex Optimization, pp. 243,

for any optimization problem ... for which strong duality obtains, any pair of primal and dual optimal points must satisfy the KKT conditions

i.e. $\mathrm{strong ~ duality} \implies \mathrm{KKT ~ is ~ necessary ~ condition ~ for ~ optimal ~ solution}$

and in pp. 244,

(When the primal problem is convex) if $\tilde{x}, \tilde{\lambda}, \tilde{\mu}$ are any points that satisfy the KKT conditions, then $\tilde{x}$ and $(\tilde{\lambda}, \tilde{\mu})$ are primal and dual optimal, with zero duality gap.

If duality gap = 0, the problem satisfies strong duality, and in the 3rd paragraph:

If a convex optimization problem ... satisfies Slater’s condition, then the KKT conditions provide necessary and sufficient conditions for optimality

For me it means: (for any convex problems KKT is already sufficient for optimal)

$$\mathrm{KKT} \implies \mathrm{optimal ~ with ~ zero ~ duality ~ gap} \implies \mathrm{strong ~ duality} \implies \mathrm{KKT ~ is ~ also ~ necessary}$$

so KKT is necessary and sufficient for any convex problems? (Because Slater's condition can be automatically satisfied for the zero duality gap)

Answer 1

Boyd and Vandenberghe considers convex optimization problems of the form \begin{align} \text{minimize} &\quad f_0(x) \\ \text{subject to} & \quad f_i(x) \leq 0 \quad \text{for } i = 1,\ldots, m \\ &\quad a_i^T x = b_i \quad \text{for } i = 1,\ldots, p, \end{align} where $f_0,\ldots, f_m$ are convex functions. The optimization variable is $x \in \mathbb R^n$ and the domain of the optimization problem is $$ \mathcal D = \bigcap_{i=0}^m \textbf{dom} \,f_i. $$ (See equation (4.15), p. 136 in Boyd and Vandenberghe.)

Let $x \in \mathbb R^n$, $\lambda \in \mathbb R^m$, and $\nu \in \mathbb R^p$. Then the following two statements are equivalent:

1. $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.
2. $x$ and $(\lambda,\nu)$ are primal and dual optimal, and strong duality holds.

If Slater's condition is satisfied, then strong duality is guaranteed to hold, and so we can make a simpler and more useful statement. In this case, the following are equivalent:

1. $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.
2. $x$ and $(\lambda,\nu)$ are primal and dual optimal.

Warning: If strong duality does not hold, then it is possible for $x$ and $(\lambda,\nu)$ to be primal and dual optimal without satisfying the KKT conditions. An example where this occurs is given below.

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By the way, if Slater's condition holds, then dual optimal variables $(\lambda,\nu)$ are guaranteed to exist. So if $x$ is primal optimal, then $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.

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Example: [Problem 5.21 in Boyd and Vandenberghe] Consider the convex optimization problem \begin{align} \text{minimize} &\quad e^{-x} \\ \text{subject to} & \quad \frac{x^2}{y}\leq 0 \end{align} with variables $x$ and $y$ and domain $$ \mathcal D = \{ (x,y) \mid y > 0 \}. $$ The primal optimal value is $1$. The Lagrangian is $$ L(x,y,\lambda) = e^{-x} + \lambda \frac{x^2}{y}. $$ The dual function is $$ g(\lambda) = \inf_{(x,y) \in \mathcal D} L(x,y,\lambda) =\begin{cases} 0 & \quad \text{if } \lambda \geq 0, \\ -\infty & \quad \text{otherwise.} \end{cases} $$ The dual problem is \begin{align} \text{maximize} &\quad 0 \\ \text{subject to} &\quad \lambda \geq 0. \end{align} The optimization variable in the dual problem is $\lambda$. We can see that any $\lambda \geq 0$ is dual optimal.

So, any $(x,y)$ with $x = 0$ and $y > 0$ is primal optimal, and any $\lambda \geq 0$ is dual optimal. Although the primal and dual optimal values are both attained, strong duality does not hold.

Answer 2

The KKT conditions are not necessary for optimality even for convex problems. Consider $$ \min x $$ subject to $$ x^2\le 0. $$ The constraint is convex. The only feasible point, thus the global minimum, is given by $x=0$. The gradient of the objective is $1$ at $x=0$, while the gradient of the constraint is zero. Thus, the KKT system cannot be satisfied.