Difference between revisions of "Karamata's Inequality"

Revision as of 13:31, 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: $\sum_{i=1}^{n}f(a_i) - \sum_{i=1}^{n}f(b_i)=\sum_{i=1}^{n}f(a_i)-f(b_i)=\sum_{i=1}^{n}c_i(a_i-b_i)=\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})$ $\sum_{i=1}^{n}c_i(A_i-A_{i-1}-B_i+B_{i+1})=\sum_{i=1}^{n}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)$ $\sum_{i=0}^{n-1}c_i(A_i-B_i) - \sum_{i=0}^{n-1}c_{i+1}(A_i-B_i)=\sum_{i=1}^{n}(c_i-c_{i+1})(A_i-B_i)\geq 0$ .

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

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

@@ Line 20: / Line 20: @@
 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>
+<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>
 Now, we know that

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Difference between revisions of "Karamata's Inequality"

Revision as of 13:31, 14 August 2018

Proof

See also