Difference between revisions of "Karamata's Inequality"
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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: | 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} | + | <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}} | {{stub}} | ||
==See also== | ==See also== | ||
− | [[Category: | + | |
− | [[Category: | + | [[Category:Algebra]] |
+ | [[Category:Inequalities]] |
Revision as of 16:48, 29 December 2021
Karamata's Inequality states that if majorizes and is a convex function, then
Proof
We will first use an important fact:
This is proven by taking casework on . If , then
A similar argument shows for other values of .
Now, define a sequence such that:
Define the sequences such that and similarly.
Then, assuming and similarily with the 's, we get that . Now, we know: .
Therefore,
Thus we have proven Karamata`s Theorem This article is a stub. Help us out by expanding it.