Karamata's inequality

In mathematics, Karamata's inequality, named after Jovan Karamata, 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.

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 x₁, . . . , x_n and y₁, . . . , y_n are numbers in I such that (x₁, . . . , x_n) majorizes (y₁, . . . , y_n), then

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

Here majorization means that x₁, . . . , x_n and y₁, . . . , 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 ∈ {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 ∈ {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 x₁, . . . , x_n ∈ I and let

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

denote their arithmetic mean. Then (x₁, . . . , x_n) majorizes the n-tuple (a, a, . . . , a), since the arithmetic mean of the i largest numbers of (x₁, . . . , x_n) is at least as large as the arithmetic mean a of all the n numbers, for every i ∈ {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 ∈ {1, . . . , n}, then the inequality (1) holds with equality, hence we may assume in the following that x_i ≠ y_i for at least one i.

If x_i = y_i for an i ∈ {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 ≠ y_i for all i ∈ {1, . . . , n − 1}.

It is a property of convex functions that for two numbers x ≠ 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 ∈ {1, . . . , n − 1}. Define A₀ = B₀ = 0 and

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

for all i ∈ {1, . . . , n}. By the majorization property (3), A_i ≥ B_i for all i ∈ {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 x₁ > y₁ by (3) and our assumption x_i ≠ y_i for all i ∈ {1, . . . , n − 1}. Let i be the smallest index such that (x_i, y_i) ≠ (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 ≥ 0. The relaxed condition (5) means that A_n ≥ B_n, which is enough to conclude that c_n(A_n−B_n) ≥ 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.

External links

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