Difference between revisions of "Karamata's Inequality"

Latest revision as of 02:39, 28 March 2024

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: If $f(x)$ is convex over the interval $(a, b)$ , then $\forall a\leq x_1\leq x_2 \leq b$ 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: $\sum_{i=1}^{n}f(a_i) - \sum_{i=1}^{n}f(b_i)=\sum_{i=1}^{n}f(a_i)-f(b_i)=\sum_{i=1}^{n}c_i(a_i-b_i)=\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})$ $\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})=\sum_{i=1}^{n}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)$ $\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=1}^{n}(c_i-c_{i+1})(A_i-B_i)\geq 0$ .

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

Thus, we have proven Karamata's Theorem.

This article is a stub. Help us out by expanding it.

@@ Line 1: / Line 1: @@
-'''Karamata's Inequality''' states that if <math>(x_i)</math> [[Majorization|majores]] <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==
-{{incomplete|proof}}
+We will first use an important fact:
+If <math>f(x)</math> is convex over the interval <math>(a, b)</math>, then <math>\forall a\leq x_1\leq x_2 \leq b</math> and <math>\Gamma(x, y):=\frac{f(y)-f(x)}{y-x}</math>, <math>\Gamma(x_1, x)\leq \Gamma (x_2, x)</math>
+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})</cmath><cmath>\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=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)</cmath><cmath>\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=1}^{n}(c_i-c_{i+1})(A_i-B_i)\geq 0</cmath>.
+Therefore, <cmath>\sum_{i=1}^{n}f(a_i) \geq \sum_{i=1}^{n}f(b_i)</cmath>
+Thus, we have proven Karamata's Theorem.
+{{stub}}
 ==See also==
-[[Category:Theorems]]
-[[Category:Inequality]]
+[[Category:Algebra]]
+[[Category:Inequalities]]

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Difference between revisions of "Karamata's Inequality"

Latest revision as of 02:39, 28 March 2024

Proof

See also