Convexity and concavity

Optimization: Extreme points

Convexity and concavity

The first two facts of the partial derivatives test can be somewhat explained by the following geometric interpretation of the conditions.

Let $f(x,y)$ be a bivariate function.

If no point of the line segment between any two points on the graph of $f$ lies below the graph, then $f$ is said to be convex.
If no point of the line segment between any two points on the graph of $f$ lies above the graph, then $f$ is said to be concave.

The same definitions can be used for functions of a single variable. The graph of the function $f(x)=x^2$ is a parabola opening upward and so is convex. In the bivariate version, the functions $g(x,y)=x^2$ and $h(x,y)=x^2+y^2$ are both convex, but the function $k(x,y)=x^2-y^2$ is not.

A function $f$ is convex if and only $-f$ is concave.

Let $f(x,y)$ be a bivariate function all of whose partial derivatives of first and second order exist and are continuous. Assume that $D$ is an open disk that is contained in the domain of $f$ . Denote the Hessian of $f$ by $H$ . Then

$f$ is convex on $D$ if and only if, for each point $\rv{a,b}$ in $D$ , we have $H(a,b)\ge0$ and $f_{xx}(a,b)\ge0$ and $f_{yy}(a,b)\ge0$ .
$f$ is concave on $D$ if and only if, for each point $\rv{a,b}$ in $D$ , we have $H(a,b)\ge0$ and $f_{xx}(a,b)\le0$ and $f_{yy}(a,b)\le0$ .

In algebraic terms, the interpretation of convexity involving the line segment can be stated as follows:

For each pair of distinct points $\rv{a,b}$ and $\rv{c,d}$ of $D$ , and for each real number $\lambda$ with $0\le \lambda\le1$ , the following inequality holds $f\left((1-\lambda)\cdot a+\lambda\cdot c,(1-\lambda)\cdot b+\lambda\cdot d)\right)\le (1-\lambda)\cdot f(a,b)+\lambda\cdot f(c,d)\tiny.$

Similarly, the geometric interpretation of concavity can be stated as:

For each pair of distinct points $\rv{a,b}$ and $\rv{c,d}$ of $D$ , and for each real number $\lambda$ with $0\le \lambda\le1$ , the following inequality holds $f\left((1-\lambda)\cdot a+\lambda\cdot c,(1-\lambda)\cdot b+\lambda\cdot d)\right)\ge (1-\lambda)\cdot f(a,b)+\lambda\cdot f(c,d)\tiny.$

In the first case of the partial derivatives test, the conditions for concavity are satisfied on a small disk around the maximum.

Consider the following function $f(x,y)=A\cdot x^{a}\cdot y^{b}-q\cdot y-p\cdot x-r\tiny,$ where

$A$ , $a$ , $b$ are positive constants with $a+b \le 1$ ,
$p$ , $q$ , and $r$ are arbitrary constants, and
the domain of $f$ is restricted to $x \ge 0\land y \ge 0$ .

What is the nature of $f$ ?

You can use the following expressions for the second derivatives and the Hessian of $f$ :
$\begin{array}{rcl} f_{xx} &=& \left(A\cdot a^2-A\cdot a\right)\cdot x^{a-2}\cdot y^{b} \\ f_{xy}&=& A\cdot a\cdot b\cdot x^{a-1}\cdot y^{b-1} \\ f_{yy}&=& \left(A\cdot b^2-A\cdot b\right)\cdot x^{a}\cdot y^{b-2} \\ f_{xx} \cdot f_{yy} - f_{xy}^2 &=& -A^2\cdot a\cdot b\cdot \left(b+a-1\right)\cdot x^{2\cdot a-2}\cdot y^{2\cdot b-2} \end{array}$

$f$ is a concave function.

To show that a function $f(x,y)$ is concave, we check the following conditions:
$\begin{array}{rcl} f_{xx} &\le& 0 \\ f_{yy} &\le& 0 \\ f_{xx}\cdot{}f_{yy}-(f_{xy})^2 &\ge& 0 \end{array}$ We check the signs of $f_{xx}$ , $f_{yy}$ , and the Hessian $f_{xx}\cdot{}f_{yy}-(f_{xy})^2$ .
$\begin{array}{rcl} f_{xx} &=&\left(A\cdot a^2-A\cdot a\right)\cdot x^{a-2}\cdot y^{b} \\ &\le& 0 \\ &&\phantom{xxx}\color{blue}{\text{by assumption, }0\lt a\lt a+b\le 1\text{, so }a^2-a\le 0}\\ \\ f_{yy} &=& \left(A\cdot b^2-A\cdot b\right)\cdot x^{a}\cdot y^{b-2} \\ &\le& 0 \\ &&\phantom{xxx}\color{blue}{\text{by assumption, }0\lt b\lt a+b\le 1\text{, so }b^2-b\le0} \end{array}$
$\begin{array}{rcl} f_{xx}\cdot{}f_{yy}-(f_{xy})^2 &=& \left( (a^2-a)\cdot A\cdot x^{a-2} \cdot y^b\right)\cdot \left( (b^2-b)\cdot A\cdot x^a y^{b-2}\right)\\&&\phantom{xx} - \left(a\cdot b\cdot A\cdot x^{a-1} y^{b-1}\right)^2\\ &=& a\cdot b\cdot(a-1)\cdot(b-1)\cdot A^2\cdot x^{2a-2}\cdot y^{2b-2}\\ &&\phantom{xx}-a^2\cdot b^2\cdot A^2\cdot x^{2a-2}\cdot y^{2b-2} \\ &=& a\cdot b\cdot A^2\cdot x^{2a-2}\cdot y^{2b-2}\cdot \left((a-1)(b-1)-a\cdot b\right) \\ &=& a\cdot b\cdot A^2\cdot x^{2a-2}\cdot y^{2b-2}\cdot \left(a\cdot b-a-b+1-a\cdot b\right) \\ &=& a\cdot b\cdot A^2\cdot x^{2a-2}\cdot y^{2b-2}\cdot (1-a-b)\\& \ge & 0\\ &&\phantom{xxx}\color{blue}{\text{by assumption, } a+b\le 1} \end{array}$ Thus, by the concavity theorem, the function is concave on any open disk in the domain.

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