- Convex Optimization Tutorial
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- Jensen's Inequality
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- Sufficient & Necessary Conditions for Global Optima
- Quasiconvex & Quasiconcave functions
- Differentiable Quasiconvex Function
- Strictly Quasiconvex Function
- Strongly Quasiconvex Function
- Pseudoconvex Function
- Convex Programming Problem
- Fritz-John Conditions
- Karush-Kuhn-Tucker Optimality Necessary Conditions
- Algorithms for Convex Problems
- Convex Optimization Resources
- Convex Optimization - Quick Guide
- Convex Optimization - Resources
- Convex Optimization - Discussion
Strictly Quasiconvex Function
Let $f:S\rightarrow \mathbb{R}^n$ 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 $f\left ( x_1 \right ) \neq f\left ( x_2 \right )$, we have $f\left ( \lambda x_1+\left ( 1-\lambda \right )x_2 \right )< max \:\left \{ f\left ( x_1 \right ),f\left ( x_2 \right ) \right \}$
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 $f:S\rightarrow \mathbb{R}^n$ be strictly quasiconvex function and S be a non-empty convex set in $\mathbb{R}^n$.Consider the problem: $min \:f\left ( x \right ), x \in S$. If $\hat{x}$ is local optimal solution, then $\bar{x}$ is global optimal solution.
Proof
Let there exists $ \bar{x} \in S$ such that $f\left ( \bar{x}\right )\leq f \left ( \hat{x}\right )$
Since $\bar{x},\hat{x} \in S$ and S is convex set, therefore,
$$\lambda \bar{x}+\left ( 1-\lambda \right )\hat{x}\in S, \forall \lambda \in \left ( 0,1 \right )$$
Since $\hat{x}$ is local minima, $f\left ( \hat{x} \right ) \leq f\left ( \lambda \bar{x}+\left ( 1-\lambda \right )\hat{x} \right ), \forall \lambda \in \left ( 0,\delta \right )$
Since f is strictly quasiconvex.
$$f\left ( \lambda \bar{x}+\left ( 1-\lambda \right )\hat{x} \right )< max \left \{ f\left ( \hat{x} \right ),f\left ( \bar{x} \right ) \right \}=f\left ( \hat{x} \right )$$
Hence, it is contradiction.
Strictly quasiconcave function
Let $f:S\rightarrow \mathbb{R}^n$ 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 $f\left (x_1\right )\neq f\left (x_2\right )$, we have
$$f\left (\lambda x_1+\left (1-\lambda\right )x_2\right )> min \left \{ f \left (x_1\right ),f\left (x_2\right )\right \}$$.
Examples
$f\left (x\right )=x^2-2$
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 $f\left (\lambda x_1+\left (1- \lambda\right )x_2\right )< max \left \{ f \left (x_1\right ),f\left (x_2\right )\right \}$ As the function is decreasing in the negative x-axis and it is increasing in the positive x-axis (since it is a parabola).
$f\left (x\right )=-x^2$
It is not a strictly quasiconvex function because if we take take $x_1=1$ and $x_2=-1$ and $\lambda=0.5$, then $f\left (x_1\right )=-1=f\left (x_2\right )$ but $f\left (\lambda x_1+\left (1- \lambda\right )x_2\right )=0$ 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 $f\left ( \lambda x_1+\left (1-\lambda\right )x_2\right )> min \left \{ f \left (x_1\right ),f\left (x_2\right )\right \}$. As the function is increasing in the negative x-axis and it is decreasing in the positive x-axis.
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