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, 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

Here majorization means that $x_{1}, …, x_{n}$ and $y_{1}, …, y_{n}$ satisfies

and we have the inequalities

and the equality

If $i ∈ {1, …, n − 1}$&thinsp; is a strictly convex function, then the inequality ($$) holds with equality if and only if we have $f$ for all $x_{i} = y_{i}$.

Remarks
 If the convex function $i ∈ {1, …, n}$&thinsp; is non-decreasing, then the proof of ($$) below and the discussion of equality in case of strict convexity shows that the equality ($$) can be relaxed to 

 The inequality ($$) is reversed if $f$&thinsp; is concave, since in this case the function $f$&thinsp; is convex. 

Example
The finite form of Jensen's inequality is a special case of this result. Consider the real numbers $−f$ and let


 * $$a := \frac{x_1+x_2+\cdots+x_n}{n}$$

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


 * $$f(x_1)+f(x_2)+ \cdots +f(x_n) \ge f(a)+f(a)+\cdots+f(a) = nf(a).$$

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

Proof of the inequality
We may assume that the numbers are in decreasing order as specified in ($$).

If $f$ for all $x_{i} = y_{i}$, then the inequality ($$) holds with equality, hence we may assume in the following that $i ∈ {1, …, n}$ for at least one $x_{i} ≠ y_{i}$.

If $i$ for an $x_{i} = y_{i}$, then the inequality ($$) and the majorization properties ($$) and ($$) are not affected if we remove $i ∈ {1, …, n}$ and $x_{i}$. Hence we may assume that $y_{i}$ for all $x_{i} ≠ y_{i}$.

It is a property of convex functions that for two numbers $i ∈ {1, …, n}$ in the interval $x ≠ y$ the slope


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

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

for all $y$. Define $i ∈ {1, …, n − 1}$ and


 * $$A_i=x_1+\cdots+x_i,\qquad B_i=y_1+\cdots+y_i$$

for all $A_{0} = B_{0} = 0$. By the majorization property ($$), $i ∈ {1, …, n}$ for all $A_{i} ≥ B_{i}$ and by ($$), $i ∈ {1, …, n − 1}$. Hence,

which proves Karamata's inequality ($$).

To discuss the case of equality in ($$), note that $A_{n} = B_{n}$ by ($$) and our assumption $x_{1} > y_{1}$ for all $x_{i} ≠ y_{i}$. Let $i ∈ {1, …, n − 1}$ be the smallest index such that $i$, which exists due to ($$). Then $(x_{i}, y_{i}) ≠ (x_{i+1}, y_{i+1})$. If $A_{i} > B_{i}$&thinsp; is strictly convex, then there is strict inequality in ($$), meaning that $f$. Hence there is a strictly positive term in the sum on the right hand side of ($$) and equality in ($$) cannot hold.

If the convex function $c_{i+1} < c_{i}$&thinsp; is non-decreasing, then $f$. The relaxed condition ($$) means that $c_{n} ≥ 0$, which is enough to conclude that $A_{n} ≥ B_{n}$ in the last step of ($$).

If the function $c_{n}(A_{n}−B_{n}) ≥ 0$&thinsp; is strictly convex and non-decreasing, then $f$. It only remains to discuss the case $c_{n} > 0$. However, then there is a strictly positive term on the right hand side of ($$) and equality in ($$) cannot hold.