Karamata's inequality

In mathematics, Karamata's inequality,^[1] named after Jovan Karamata,^[2] also known as the majorization inequality, is a theorem in elementary algebra for convex and concave real-valued functions, defined on an interval of the real line. It generalizes the discrete form of Jensen's inequality, and generalizes in turn to the concept of Schur-convex functions.

Statement of the inequality

Let $I$ be an interval of the real line and let $f$ denote a real-valued, convex function defined on $I$ . If $x 1, . . . , x n$ and $y 1, . . . , y n$ are numbers in $I$ such that $(x 1, . . . , x n)$ majorizes $(y 1, . . . , y n)$ , then

f(x_{1})+\cdots +f(x_{n})\geq f(y_{1})+\cdots +f(y_{n}).

(1)

Here majorization means that $x 1, . . . , x n$ and $y 1, . . . , y n$ satisfies

x_{1}\geq x_{2}\geq \cdots \geq x_{n}

and

y_{1}\geq y_{2}\geq \cdots \geq y_{n},

(2)

and we have the inequalities

x_{1}+\cdots +x_{i}\geq y_{1}+\cdots +y_{i}

for all

i \in {1, . . . , n - 1

}.

(3)

and the equality

x_{1}+\cdots +x_{n}=y_{1}+\cdots +y_{n}

(4)

If $f$ is a strictly convex function, then the inequality (1) holds with equality if and only if we have $x i = y i$ for all $i \in {1, . . . , n$ }.

Remarks

If the convex function $f$ is non-decreasing, then the proof of (1) below and the discussion of equality in case of strict convexity shows that the equality (4) can be relaxed to

x_{1}+\cdots +x_{n}\geq y_{1}+\cdots +y_{n}.

(5)

The inequality (1) is reversed if $f$ is concave, since in this case the function $- f$ is convex.

Example

The finite form of Jensen's inequality is a special case of this result. Consider the real numbers $x 1, . . . , x n \in I$ and let

a:={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}

denote their arithmetic mean. Then $(x 1, . . . , x n)$ majorizes the $n$ -tuple $(a, a, . . . , a)$ , since the arithmetic mean of the $i$ largest numbers of $(x 1, . . . , x n)$ is at least as large as the arithmetic mean $a$ of all the $n$ numbers, for every $i \in {1, . . . , n - 1$ }. By Karamata's inequality (1) for the convex function $f$ ,

f(x_{1})+f(x_{2})+\cdots +f(x_{n})\geq f(a)+f(a)+\cdots +f(a)=nf(a).

Dividing by $n$ gives Jensen's inequality. The sign is reversed if $f$ is concave.

Proof of the inequality

We may assume that the numbers are in decreasing order as specified in (2).

If $x i = y i$ for all $i \in {1, . . . , n$ }, then the inequality (1) holds with equality, hence we may assume in the following that $x i \neq y i$ for at least one $i$ .

If $x i = y i$ for an $i \in {1, . . . , n - 1$ }, then the inequality (1) and the majorization properties (3) and (4) are not affected if we remove $x i$ and $y i$ . Hence we may assume that $x i \neq y i$ for all $i \in {1, . . . , n - 1$ }.

It is a property of convex functions that for two numbers $x \neq y$ in the interval $I$ the slope

{\frac {f(x)-f(y)}{x-y}}

of the secant line through the points $(x, f (x))$ and $(y, f (y))$ of the graph of $f$ is a monotonically non-decreasing function in $x$ for $y$ fixed (and vice versa). This implies that

c_{i+1}:={\frac {f(x_{i+1})-f(y_{i+1})}{x_{i+1}-y_{i+1}}}\leq {\frac {f(x_{i})-f(y_{i})}{x_{i}-y_{i}}}=:c_{i}

(6)

for all $i \in {1, . . . , n - 1$ }. Define $A 0 = B 0 = 0$ and

A_{i}=x_{1}+\cdots +x_{i},\qquad B_{i}=y_{1}+\cdots +y_{i}

for all $i \in {1, . . . , n$ }. By the majorization property (3), $A i \geq B i$ for all $i \in {1, . . . , n - 1$ } and by (4), $A n = B n$ . Hence,

{\begin{aligned}\sum _{i=1}^{n}{\bigl (}f(x_{i})-f(y_{i}){\bigr )}&=\sum _{i=1}^{n}c_{i}(x_{i}-y_{i})\\&=\sum _{i=1}^{n}c_{i}{\bigl (}\underbrace {A_{i}-A_{i-1}} _{=\,x_{i}}{}-(\underbrace {B_{i}-B_{i-1}} _{=\,y_{i}}){\bigr )}\\&=\sum _{i=1}^{n}c_{i}(A_{i}-B_{i})-\sum _{i=1}^{n}c_{i}(A_{i-1}-B_{i-1})\\&=c_{n}(\underbrace {A_{n}-B_{n}} _{=\,0})+\sum _{i=1}^{n-1}(\underbrace {c_{i}-c_{i+1}} _{\geq \,0})(\underbrace {A_{i}-B_{i}} _{\geq \,0})-c_{1}(\underbrace {A_{0}-B_{0}} _{=\,0})\\&\geq 0,\end{aligned}}

(7)

which proves Karamata's inequality (1).

To discuss the case of equality in (1), note that $x 1 > y 1$ by (3) and our assumption $x i \neq y i$ for all $i \in {1, . . . , n - 1$ }. Let $i$ be the smallest index such that $(x i, y i) \neq (x i +1, y i +1)$ , which exists due to (4). Then $A i > B i$ . If $f$ is strictly convex, then there is strict inequality in (6), meaning that $c i +1 < c i$ . Hence there is a strictly positive term in the sum on the right hand side of (7) and equality in (1) cannot hold.

If the convex function $f$ is non-decreasing, then $c n \geq 0$ . The relaxed condition (5) means that $A n \geq B n$ , which is enough to conclude that $c n (A n - B n) \geq 0$ in the last step of (7).

If the function $f$ is strictly convex and non-decreasing, then $c n > 0$ . It only remains to discuss the case $A n > B n$ . However, then there is a strictly positive term on the right hand side of (7) and equality in (1) cannot hold.

References

^ Kadelburg, Zoran; Đukić, Dušan; Lukić, Milivoje; Matić, Ivan (2005), "Inequalities of Karamata, Schur and Muirhead, and some applications" (PDF), The Teaching of Mathematics, 8 (1): 31–45, ISSN 1451-4966
^ Karamata, Jovan (1932), "Sur une inégalité relative aux fonctions convexes" (PDF), Publ. Math. Univ. Belgrade (in French), 1: 145–148, Zbl 0005.20101

External links

An explanation of Karamata's inequality and majorization theory can be found here.

[1] Kadelburg, Zoran; Đukić, Dušan; Lukić, Milivoje; Matić, Ivan (2005), "Inequalities of Karamata, Schur and Muirhead, and some applications" (PDF), The Teaching of Mathematics, 8 (1): 31–45, ISSN 1451-4966

[2] Karamata, Jovan (1932), "Sur une inégalité relative aux fonctions convexes" (PDF), Publ. Math. Univ. Belgrade (in French), 1: 145–148, Zbl 0005.20101

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