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An invex function $f$ is a differentiable function from $\Bbb R^n$ to $\Bbb R$ that for some function $\eta : \Bbb R^n \times \Bbb R^n \to \Bbb R^n$ satisfies for all $x, u$, $f(x) - f(u) \geq \eta(x, u) \nabla f(u)$ .

The Wikipedia article on invex functions can be found here.

Since there are no restrictions on $\eta$, give a pair of vectors $x$ and $u$, I can always find a vector $z$ such that the dot product of this vector with the gradient vector at $u$ takes any real value (assuming the gradient vector is not degenerate or unbounded). Thus, the only bite this definition has is at points where the gradient is $0$. In fact, any stationary point i.e. point with gradient of $0$ must be a global minimum. So invex functions are just those functions that have all stationary points as global minima.

Have I misunderstood something?

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up vote 10 down vote accepted

You're right that if $f$ is invex then any stationary point is a global minimum. Other classes of functions, such as convex and quasiconvex functions, also share this property, though.

However, a key aspect of invex functions is that the relationship goes the other direction, too; i.e., if $f$ is differentiable and all stationary points of $f$ are global minima then $f$ must be invex. (See here for a proof.) So invex functions are exactly that class of functions that behave like our calculus students tend to think functions ought to behave when minimizing: Just find where $\nabla f = {\bf 0}$, and you're done. No need to test for local minima or saddle points or maxima or anything else. :)

Another key property of invex functions has to do with the Karush-Kuhn-Tucker conditions. These give necessary conditions for a solution to a nonlinear programming problem to be optimal. Invex functions are the class of functions for which the KKT conditions are sufficient as well as necessary (assuming you're minimizing), in the sense that if the objective and the inequality constraints are invex and continuously differentiable, then the KKT conditions are sufficient. (The equality constraints must be affine.)

In terms of the question in your title, I would say that the usefulness of invex functions lies in these two properties.

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"if the objective and the inequality constraints are invex and continuously differentiable, then the KKT conditions are sufficient." That's not correct since linear combination of invex functions is not guaranteed to preserve the global optimality property. The correct statement is: The KKT conditions are sufficient if the Lagrangian function of the nonlinear program is invex. – user130142 Feb 20 '14 at 4:20

There has in fact been some controversy over the usefulness of invex functions at all.

This article is interesting A critical view on invexity

As far as I can tell, the criticism is justified and the definition is vacuous. Please can someone prove me wrong by providing a concrete example of a novel, interesting theorem which cannot be easily proven without using the concept of invexity!

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