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Strongly Quasiconvex Function
Let $f:S\rightarrow \mathbb{R}^n$ and S be a non-empty convex set in $\mathbb{R}^n$ then f is strongly quasiconvex function if for any $x_1,x_2 \in S$ with $\left ( x_1 \right ) \neq \left ( x_2 \right )$, we have $f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )< max \:\left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \},\forall \lambda \in \left ( 0,1\right )$
Theorem
A quasiconvex function $f:S\rightarrow \mathbb{R}^n$ on a non-empty convex set S in $\mathbb{R}^n$ is strongly quasiconvex function if it is not constant on a line segment joining any points of S.
Proof
Let f is quasiconvex function and it is not constant on a line segment joining any points of S.
Suppose f is not strongly quasiconvex function.
There exist $x_1,x_2 \in S$ with $x_1 \neq x_2$ such that
$$f\left ( z \right )\geq max\left \{ f\left ( x_1 \right ), f\left ( x_2 \right ) \right \}, \forall z= \lambda x_1+\left ( 1-\lambda \right )x_2, \lambda \in \left ( 0,1 \right )$$
$\Rightarrow f\left ( x_1 \right )\leq f\left ( z \right )$ and $f\left ( x_2 \right )\leq f\left ( z \right )$
Since f is not constant in $\left [ x_1,z \right ]$ and $\left [z,x_2 \right ] $
So there exists $u \in \left [ x_1,z \right ]$ and $v=\left [ z,x_2 \right ]$
$$\Rightarrow u= \mu_1x_1+\left ( 1-\mu_1\right )z,v=\mu_2z+\left ( 1- \mu_2\right )x_2$$
Since f is quasiconvex,
$$\Rightarrow f\left ( u \right )\leq max\left \{ f\left ( x_1 \right ),f \left ( z \right ) \right \}=f\left ( z \right )\:\: and \:\:f \left ( v \right ) \leq max \left \{ f\left ( z \right ),f\left ( x_2 \right ) \right \}$$
$$\Rightarrow f\left ( u \right )\leq f\left ( z \right ) \:\: and \:\: f\left ( v \right )\leq f\left ( z \right )$$
$$\Rightarrow max \left \{ f\left ( u \right ),f\left ( v \right ) \right \} \leq f\left ( z \right )$$
But z is any point between u and v, if any of them are equal, then f is constant.
Therefore, $max \left \{ f\left ( u \right ),f\left ( v \right ) \right \} \leq f\left ( z \right )$
which contradicts the quasiconvexity of f as $z \in \left [ u,v \right ]$.
Hence f is strongly quasiconvex function.
Theorem
Let $f:S\rightarrow \mathbb{R}^n$ and S be a non-empty convex set in $\mathbb{R}^n$. If $\hat{x}$ is local optimal solution, then $\hat{x}$ is unique global optimal solution.
Proof
Since a strong quasiconvex function is also strictly quasiconvex function, thus a local optimal solution is global optimal solution.
Uniqueness − Let f attains global optimal solution at two points $u,v \in S$
$$\Rightarrow f\left ( u \right ) \leq f\left ( x \right ).\forall x \in S\:\: and \:\:f\left ( v \right ) \leq f\left ( x \right ).\forall x \in S$$
If u is global optimal solution, $f\left ( u \right )\leq f\left ( v \right )$ and $f\left ( v \right )\leq f\left ( u\right )\Rightarrow f\left ( u \right )=f\left ( v\right )$
$$f\left ( \lambda u+\left ( 1-\lambda\right )v\right )< max \left \{f\left ( u \right ),f\left ( v \right ) \right \}=f\left ( u \right )$$
which is a contradiction.
Hence there exists only one global optimal solution.
Remarks
- A strongly quasiconvex function is also strictly quasiconvex fucntion.
- A strictly convex function may or may not be strongly quasiconvex.
- A differentiable strictly convex is strongly quasiconvex.