Difference between revisions of "Hölder's Inequality"
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== Elementary Form == | == Elementary Form == | ||
If <math>a_1, a_2, \dotsc, a_n, b_1, b_2, \dotsc, b_n, \dotsc, z_1, z_2, \dotsc, z_n</math> are [[nonnegative]] [[real number]]s and <math>\lambda_a, \lambda_b, \dotsc, \lambda_z</math> are nonnegative reals with sum of 1, then | If <math>a_1, a_2, \dotsc, a_n, b_1, b_2, \dotsc, b_n, \dotsc, z_1, z_2, \dotsc, z_n</math> are [[nonnegative]] [[real number]]s and <math>\lambda_a, \lambda_b, \dotsc, \lambda_z</math> are nonnegative reals with sum of 1, then | ||
− | <cmath> \ | + | <cmath>(a_1 + \dotsb + a_n)^{\lambda_a} (b_1 + \dotsb + b_n)^{\lambda_b} \dotsm (z_1 + \dotsb + z_n)^{\lambda_z} |
− | + | \geq a_1^{\lambda_a}b_1^{\lambda_b} \dotsm z_1^{\lambda_z} + \dotsb + a_n^{\lambda_a}b_n^{\lambda_b} \dotsm z_n^{\lambda_z}.</cmath> | |
− | + | ||
− | |||
Note that with two sequences <math>\mathbf{a}</math> and <math>\mathbf{b}</math>, and <math>\lambda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]]. | Note that with two sequences <math>\mathbf{a}</math> and <math>\mathbf{b}</math>, and <math>\lambda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]]. | ||
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<cmath> \prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} \ge \prod_i \prod_j \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i \beta_j} = \prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} . </cmath> | <cmath> \prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} \ge \prod_i \prod_j \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i \beta_j} = \prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} . </cmath> | ||
But from our choice of <math>\beta_j</math>, for all integers <math>1 \le j \le m</math>, | But from our choice of <math>\beta_j</math>, for all integers <math>1 \le j \le m</math>, | ||
− | <cmath> \prod_i \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \ | + | <cmath> \prod_i \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \beta_j} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \prod_i a_{ij}^{\lambda_i} / \sum_j \prod_i a_{ij}^{\lambda_i}} = \sum_j \prod_i a_{ij}^{\lambda_i} . </cmath> |
Therefore | Therefore | ||
<cmath> \prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} = \prod_k \biggl( \sum_j \prod_i a_{ij}^{\lambda_i} \biggr)^{\beta_k} = \sum_j \prod_i a_{ij}^{\lambda_i}, </cmath> | <cmath> \prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} = \prod_k \biggl( \sum_j \prod_i a_{ij}^{\lambda_i} \biggr)^{\beta_k} = \sum_j \prod_i a_{ij}^{\lambda_i}, </cmath> | ||
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<center><math>\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}</math></center> | <center><math>\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}</math></center> | ||
− | [[Category: | + | [[Category:Algebra]] |
− | [[Category: | + | [[Category:Inequalities]] |
+ | [[Category:Mathematics]] |
Latest revision as of 17:42, 28 September 2024
Elementary Form
If are nonnegative real numbers and are nonnegative reals with sum of 1, then
Note that with two sequences and , and , this is the elementary form of the Cauchy-Schwarz Inequality.
We can state the inequality more concisely thus: Let be several sequences of nonnegative reals, and let be a sequence of nonnegative reals such that . Then
Proof of Elementary Form
We will use weighted AM-GM. We will disregard sequences for which one of the terms is zero, as the terms of these sequences do not contribute to the left-hand side of the desired inequality but may contribute to the right-hand side.
For integers , let us define Evidently, . Then for all integers , by weighted AM-GM, Hence But from our choice of , for all integers , Therefore since the sum of the is one. Hence in summary, as desired. Equality holds when for all integers , i.e., when all the sequences are proportional.
Statement
If , , then and .
Proof
If then a.e. and there is nothing to prove. Case is similar. On the other hand, we may assume that for all . Let . Young's Inequality gives us These functions are measurable, so by integrating we get
Examples
- Prove that, for positive reals , the following inequality holds: