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.

Second order derivative test for convexity 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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