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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 x1, . . . , xn and y1, . . . , yn are numbers in I such that (x1, . . . , xn) majorizes (y1, . . . , yn), then

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

Here majorization means that x1, . . . , xn and y1, . . . , yn satisfies

$${\displaystyle x_{1}\geq x_{2}\geq \cdots \geq x_{n}}$$ and $${\displaystyle y_{1}\geq y_{2}\geq \cdots \geq y_{n},}$$ (2)

and we have the inequalities

$${\displaystyle x_{1}+\cdots +x_{i}\geq y_{1}+\cdots +y_{i}}$$ for all i ∈ {1, . . . , n − 1}. (3)

and the equality

$${\displaystyle x_{1}+\cdots +x_{n}=y_{1}+\cdots +y_{n}}$$ (4)

If f  is a strictly convex function, then the inequality (1) holds with equality if and only if we have xi = yi 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

$${\displaystyle x_{1}+\cdots +x_{n}\geq y_{1}+\cdots +y_{n}.}$$ (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, . . . , xn ∈ I and let

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

denote their arithmetic mean. Then (x1, . . . , xn) majorizes the n-tuple (a, a, . . . , a), since the arithmetic mean of the i largest numbers of (x1, . . . , xn) 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,

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

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 xi = yi for all i ∈ {1, . . . , n}, then the inequality (1) holds with equality, hence we may assume in the following that xiyi for at least one i.

If xi = yi for an i ∈ {1, . . . , n − 1}, then the inequality (1) and the majorization properties (3) and (4) are not affected if we remove xi and yi. Hence we may assume that xiyi 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

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

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

$${\displaystyle c_{i+1}:={\frac {f(x_{i+1})-f(y_{i+1})}{x_{i+1}-y_{i+1}}}\leq {\frac {f(x_{i})-f(y_{i})}{x_{i}-y_{i}}}=:c_{i}}$$ (6)

for all i ∈ {1, . . . , n − 1}. Define A0 = B0 = 0 and

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

for all i ∈ {1, . . . , n}. By the majorization property (3), AiBi for all i ∈ {1, . . . , n − 1} and by (4), An = Bn. Hence,

{\displaystyle {\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 xiyi for all i ∈ {1, . . . , n − 1}. Let i be the smallest index such that (xi, yi) ≠ (xi+1, yi+1), which exists due to (4). Then Ai > Bi. If f  is strictly convex, then there is strict inequality in (6), meaning that ci+1 < ci. 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 cn ≥ 0. The relaxed condition (5) means that AnBn, which is enough to conclude that cn(AnBn) ≥ 0 in the last step of (7).

If the function f  is strictly convex and non-decreasing, then cn > 0. It only remains to discuss the case An > Bn. 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