Strictly Quasiconvex Function

Let $f:S ightarrow mathbb{R}^n$ and S be a non-empty convex set in $mathbb{R}^n$ then f is said to be strictly quasicovex function if for each $x_1,x_2 in S$ with $fleft ( x_1 ight ) eq fleft ( x_2 ight )$, we have $fleft ( lambda x_1+left ( 1-lambda ight )x_2 ight )< max :left { fleft ( x_1 ight ),fleft ( x_2 ight ) ight }$

Remarks

Every strictly quasiconvex function is strictly convex.

Strictly quasiconvex function does not imply quasiconvexity.

Strictly quasiconvex function may not be strongly quasiconvex.

Pseudoconvex function is a strictly quasiconvex function.

Theorem

Let $f:S ightarrow mathbb{R}^n$ be strictly quasiconvex function and S be a non-empty convex set in $mathbb{R}^n$.Consider the problem: $min :fleft ( x ight ), x in S$. If $hat{x}$ is local optimal solution, then $ar{x}$ is global optimal solution.

Proof

Let there exists $ ar{x} in S$ such that $fleft ( ar{x} ight )leq f left ( hat{x} ight )$

Since $ar{x},hat{x} in S$ and S is convex set, therefore,

$$lambda ar{x}+left ( 1-lambda ight )hat{x}in S, forall lambda in left ( 0,1 ight )$$

Since $hat{x}$ is local minima, $fleft ( hat{x} ight ) leq fleft ( lambda ar{x}+left ( 1-lambda ight )hat{x} ight ), forall lambda in left ( 0,delta ight )$

Since f is strictly quasiconvex.

$$fleft ( lambda ar{x}+left ( 1-lambda ight )hat{x} ight )< max left { fleft ( hat{x} ight ),fleft ( ar{x} ight ) ight }=fleft ( hat{x} ight )$$

Hence, it is contradiction.

Strictly quasiconcave function

Let $f:S ightarrow mathbb{R}^n$ and S be a non-empty convex set in $mathbb{R}^n$, then f is saud to be strictly quasicovex function if for each $x_1,x_2 in S$ with $fleft (x_1 ight ) eq fleft (x_2 ight )$, we have

$$fleft (lambda x_1+left (1-lambda ight )x_2 ight )> min left { f left (x_1 ight ),fleft (x_2 ight ) ight }$$.

Examples

$fleft (x ight )=x^2-2$

It is a strictly quasiconvex function because if we take any two points $x_1,x_2$ in the domain that satisfy the constraints in the definition $fleft (lambda x_1+left (1- lambda ight )x_2 ight )< max left { f left (x_1 ight ),fleft (x_2 ight ) ight }$ As the function is decreasing in the negative x-axis and it is increasing in the positive x-axis (since it is a parabola).

$fleft (x ight )=-x^2$

It is not a strictly quasiconvex function because if we take take $x_1=1$ and $x_2=-1$ and $lambda=0.5$, then $fleft (x_1 ight )=-1=fleft (x_2 ight )$ but $fleft (lambda x_1+left (1- lambda ight )x_2 ight )=0$ Therefore it does not satisfy the conditions stated in the definition. But it is a quasiconcave function because if we take any two points in the domain that satisfy the constraints in the definition $fleft ( lambda x_1+left (1-lambda ight )x_2 ight )> min left { f left (x_1 ight ),fleft (x_2 ight ) ight }$. As the function is increasing in the negative x-axis and it is decreasing in the positive x-axis.