Karamata's Inequality: Difference between revisions

Revision as of 13:26, 14 August 2018

Karamata's Inequality states that if $(a_i)$ majorizes $(b_i)$ and $f$ is a convex function, then

$\sum_{i=1}^{n}f(a_i)\geq \sum_{i=1}^{n}f(b_i)$

Proof

We will first use an important fact: $\text{If }f(x)\text{ is convex over the interval }(a, b)\text{, then }\forall \hspace{2mm} a\leq x_1\leq x_2 \leq b \text{ and } \Gamma(x, y):=\frac{f(y)-f(x)}{y-x}, \Gamma(x_1, x)\leq \Gamma (x_2, x)$

This is proven by taking casework on $x\neq x_1,x_2$ . If $x<x_1$ , then $\Gamma(x_1, x)=\frac{f(x_1)-f(x)}{x_1-x}\leq \frac{f(x_2)-f(x)}{x_2-x}=\Gamma(x_2, x)$

A similar argument shows for other values of $x$ .

Now, define a sequence $C$ such that: $c_i=\Gamma(a_i, b_i)$

Define the sequences $A_i$ such that $A_i=\sum_{j=1}^{i}a_j, A_0=0$ and $B_i$ similarly.

Then, assuming $a_i\geq a_{i+1}$ and similarily with the $b_i$ 's, we get that $c_i\geq c_{i+1}$ . Now, we know: $\sum_{i=1}^{n}f(a_i) - \sum_{i=1}^{n}f(b_i)=\sum_{i=1}^{n}f(a_i)-f(b_i)=sum_{i=1}^{n}c_i(a_i-b_i)=\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})=\sum_{i=1}^{n}c_i(A_i-B_i) - \sum_{i=1}^{n}f(b_i)$

Now, we know that This article is a stub. Help us out by expanding it.

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-'''Karamata's Inequality''' states that if <math>(x_i)</math> [[Majorization|majorizes]] <math>(y_i)</math> and <math>f</math> is a [[convex function]], then
+'''Karamata's Inequality''' states that if <math>(a_i)</math> [[Majorization|majorizes]] <math>(b_i)</math> and <math>f</math> is a [[convex function]], then
-<center><math>\sum_{i=1}^{n}f(x_i)\geq \sum_{i=1}^{n}f(y_i)</math></center>
+<cmath>\sum_{i=1}^{n}f(a_i)\geq \sum_{i=1}^{n}f(b_i)</cmath>
 ==Proof==
+We will first use an important fact:
+<cmath>\text{If }f(x)\text{ is convex over the interval }(a, b)\text{, then }\forall \hspace{2mm} a\leq x_1\leq x_2 \leq b \text{ and } \Gamma(x, y):=\frac{f(y)-f(x)}{y-x}, \Gamma(x_1, x)\leq \Gamma (x_2, x)</cmath>
+This is proven by taking casework on <math>x\neq x_1,x_2</math>. If <math>x<x_1</math>, then
+<cmath>\Gamma(x_1, x)=\frac{f(x_1)-f(x)}{x_1-x}\leq \frac{f(x_2)-f(x)}{x_2-x}=\Gamma(x_2, x)</cmath>
+A similar argument shows for other values of <math>x</math>.
+Now, define a sequence <math>C</math> such that:
+<cmath>c_i=\Gamma(a_i, b_i)</cmath>
+Define the sequences <math>A_i</math> such that
+<cmath>A_i=\sum_{j=1}^{i}a_j, A_0=0</cmath>
+and <math>B_i</math> similarly.
+Then, assuming <math>a_i\geq a_{i+1}</math> and similarily with the <math>b_i</math>'s, we get that <math>c_i\geq c_{i+1}</math>. Now, we know:
+<cmath>\sum_{i=1}^{n}f(a_i) - \sum_{i=1}^{n}f(b_i)=\sum_{i=1}^{n}f(a_i)-f(b_i)=sum_{i=1}^{n}c_i(a_i-b_i)=\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})=\sum_{i=1}^{n}c_i(A_i-B_i) - \sum_{i=1}^{n}f(b_i)</cmath>
+Now, we know that
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Karamata's Inequality: Difference between revisions

Revision as of 13:26, 14 August 2018

Proof

See also