Relation to other types of "convexity" Every convex function is pseudoconvex, but the converse is not true. For example, the function f(x) = x + x^{3} is pseudoconvex but not convex. Similarly, any pseudoconvex function is
quasiconvex; but the converse is not true, since the function f(x) = x^{3} is quasiconvex but not pseudoconvex. This can be summarized schematically as: : convex \Rightarrow pseudoconvex \Rightarrow quasiconvex To see that f(x) = x^{3} is not pseudoconvex, consider its derivative at x = 0: f^{\prime}(0) = 0. Then, if f(x) = x^{3} was pseudoconvex, we should have: : f^{\prime}(0) (y-0) = 0 \geq 0 \Rightarrow f(y) \geq f(0), \quad \forall \, y \in \mathbb{R}. In particular it should be true for y=-1. But it is not, as: f(-1) = (-1)^{3} = -1 .
Sufficient optimality condition For any differentiable function, we have the
Fermat's theorem necessary condition of optimality, which states that: if f has a local minimum at x^{*} in an
open domain, then x^{*} must be a
stationary point of f (that is: \nabla f(x^{*}) = 0). Pseudoconvexity is of great interest in the area of
optimization, because the converse is also true for any pseudoconvex function. That is: if x^{*} is a
stationary point of a pseudoconvex function f, then f has a global minimum at x^{*}. Note also that the result guarantees a global minimum (not only local). This last result is also true for a convex function, but it is not true for a quasiconvex function. Consider for example the quasiconvex function: : f(x)=\frac{e^{x}}{x^{2}+1} + \frac{1}{e^{x}}. This function is not pseudoconvex, but it is quasiconvex. Also, the point x=0 is a critical point of f, as f^{\prime}(0)=0. However, f does not have a global minimum at x=0 (not even a local minimum). Finally, note that a pseudoconvex function may not have any critical point. Take for example the pseudoconvex function: f(x)=x^{3}+x, whose derivative is always positive: f^{\prime}(x)=3x^{2}+1>0, \, \forall \, x \in \mathbb{R}. ==Examples==