Which of the following functions is quasiconvex?

Question

a) $F(x)=x^{2}$

b) $F(x)=e^{-x}$

c) $F(x)=\cos (x)$

d) $F(x)=x^{-1}$ if $f\neq 0$ and $f=0$ if $x=0$

Hi, i am finding quasi concavity and convexity very difficult to understand. Any help with this will be appreciated.

Answer 1

All convex functions are surely quasiconvex so for $f(x)=x^2$ the second derivative is 2 >0 so it is convex. Giving us that A is quasiconvex. C) cosx Changes from being convex to concave and back every $\pi$ period starting at an integer multiple of $\pi$.

Answer 2

The function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is said to be Quasi-convex function if its domain and all its sublevel sets as:

$$S_\alpha = \{\mathbf{x}| \mathbf{x}\in \text{dom}f, f(x) \leq \alpha\}$$

are convex sets for every $\alpha$. Means, if you draw a horizontal line for any $\alpha$ then the corresponding domain for $f(x) \leq \alpha$ should be a convex set.

Notice: Any convex function is quasiconvex.

Here:

$x^2$ is a convex function and hence is a quasiconvex function.

$e^{-x}$ is a convex function and hence is a quasiconvex function.

$\cos x$ is not a quasi convex function for $x \in [-\pi, \pi]$.

$f(x) = \frac{1}{x}$ with $f(0) = 0$ is not a quasiconvex function.

Notice that from optimization view point the $e^{-x}$ is unsolvable objectve function where: the optimal value is $p^{\star} = 0$ but no exact optimum point $x$ for that.

Answer 3

Note that any convex (concave) function is quasiconvex (quasiconcave), but the converse may or may not be true.

Let $f(x_1,x_2,...,x_n) \in C^2$. Then $f$ is quasiconvex (or quasiconcave) if $|B_r|\leq0$ (or$(-1)^r|B_r|\geq0$) for $r=1,2,...,n,$ where $B_r$ is the bordered Hessian matrix: $$B_r=\begin{pmatrix} 0 & f_{x_1} & f_{x_2} & ... & f_{x_r} \\ f_{x_1} & f_{x_1x_1} & f_{x_1x_2} & ... & f_{x_1x_r} \\ . & . & . & ... & . \\ f_{x_r} & f_{x_rx_1} & f_{x_rx_2} & ... & f_{x_rx_r} \\ \end{pmatrix}.$$

Note: Sometimes the above rules do not work. Then one must check the other rule: $f$ is quasiconvex (quasiconcave) for $x\in(a,b)$ if $f(x)\le \max(f(a),f(b)) \ \ (f(x)\ge \min (f(a),f(b))$.

a) For $f(x)=x^2$,

$|B_1|=\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = \begin{vmatrix} 0 & 2x \\ 2x & 2 \\ \end{vmatrix} =-4x^2\le0$. Thus, it is quasiconvex.

$(-1)^1|B_1|=-\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = -\begin{vmatrix} 0 & 2x \\ 2x & 2 \\ \end{vmatrix} =4x^2\ge0$. Thus, it is quasiconcave for $x\in(a,b)$ if $f(x)$ is monotonous.

b) For $f(x)=e^{-x}$,

$|B_1|=\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = \begin{vmatrix} 0 & -e^{-x} \\ -e^{-x} & e^{-x} \\ \end{vmatrix} =-e^{-2x}\le0$. Thus, it is quasiconvex.

$(-1)^1|B_1|=-\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = -\begin{vmatrix} 0 & -e^{-x} \\ -e^{-x} & e^{-x} \\ \end{vmatrix} =e^{-2x}\ge0$. Thus, it is also quasiconcave as well.

c) For $f(x)=\cos{x}$,

$|B_1|=\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = \begin{vmatrix} 0 & -\sin{x} \\ -\sin{x} & -\cos{x} \\ \end{vmatrix} =-\sin^2{x}\le0$. Thus, it is quasiconvex for $x\in(a,b)$ if $f(x)\le \max(f(a),f(b)) $.

$(-1)^1|B_1|=-\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = -\begin{vmatrix} 0 & -\sin{x} \\ -\sin{x} & -\cos{x} \\ \end{vmatrix} =\sin^2{x}\ge0$. Thus, it is quasiconcave for $x\in(a,b)$ if $f(x)\ge \min(f(a),f(b)) $.

d) For $f(x)=\begin{cases} \frac{1}{x}, \ if \ x\ne0 \\ 0, \ if \ x=0 \end{cases} $,

$|B_1|=\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = \begin{vmatrix} 0 & -\frac{1}{x^2} \\ -\frac{1}{x^2} & \frac{2}{x^3} \\ \end{vmatrix} =-\frac{1}{x^4}\leq0$. Thus, it is quasiconvex.

$(-1)^1|B_1|=-\begin{vmatrix} 0 & f_{x_1} \\ f_{x_1} & f_{x_1x_1} \\ \end{vmatrix} = -\begin{vmatrix} 0 & -\frac{1}{x^2} \\ -\frac{1}{x^2} & \frac{2}{x^3} \\ \end{vmatrix} =\frac{1}{x^4}\geq0$. Thus, it is quasiconcave.