# Difference between revisions of "Hölder's Inequality"

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− | + | == 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 | ||

+ | <cmath> \begin{align*} | ||

+ | 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} \\ | ||

+ | \le{}& (a_1 + \dotsb + a_n)^{\lambda_a} (b_1 + \dotsb + b_n)^{\lambda_b} \dotsm (z_1 + \dotsb + z_n)^{\lambda_z} . | ||

+ | \end{align*}</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]]. | ||

+ | |||

+ | We can state the inequality more concisely thus: Let <math>\{ \{a_{ij}\}_{i=1}^n \} _{j=1}^m</math> be several sequences of nonnegative reals, and let <math>\{ \lambda_i \}_{i=1}^n</math> be a sequence of nonnegative reals such that <math>\sum \lambda = 1</math>. Then | ||

+ | <cmath> \sum_j \prod_i a_{ij}^{\lambda_i} \le \prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} . </cmath> | ||

+ | |||

+ | == Proof of Elementary Form == | ||

+ | We will use weighted [[AM-GM]]. We will disregard sequences <math>\{ a_{ij} \}_{i=1}^n</math> 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 <math>1 \le k \le m</math>, let us define | ||

+ | <cmath> \beta_k = \frac{\prod_i a_{ik}^{\lambda_i}}{\sum_j \prod_i a_{ij}^{\lambda_i}} .</cmath> | ||

+ | Evidently, <math>\sum \beta_j = 1</math>. Then for all integers <math>1\le i \le n</math>, by weighted AM-GM, | ||

+ | <cmath> \sum_j a_{ij} = \sum_j \beta_j \left(\frac{a_{ij}}{\beta_j} \right) \ge \prod_j \left( \frac{a_{ij}}{\beta_j} \right)^{\beta_j} . </cmath> | ||

+ | Hence | ||

+ | <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>, | ||

+ | <cmath> \prod_i \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \beta_k} = \frac{\prod_j a_{ij}^{\lambda_i}}{ \prod_j a_{ij}^{\lambda_i} / \sum_j \prod_i a_{ij}^{\lambda_i}} = \sum_j \prod_i a_{ij}^{\lambda_i} . </cmath> | ||

+ | 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> | ||

+ | since the sum of the <math>\beta_k</math> is one. Hence in summary, | ||

+ | <cmath> \prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} \ge \sum_j \prod_i a_{ij}^{\lambda_i} , </cmath> | ||

+ | as desired. Equality holds when <math>a_{ij}/\beta_j = a_{ij'}/\beta_{j'}</math> for all integers <math>i,j,j'</math>, i.e., when all the sequences <math>\{a_{ij}\}_{j=1}^m</math> are proportional. <math>\blacksquare</math> | ||

+ | |||

+ | == Statement == | ||

+ | If <math>p,q>1</math>, <math>1/p+1/q=1</math>, <math>f\in L^p, g\in L^q</math> then <math>fg\in L^1</math> and <math>||fg||_1\leq ||f||_p||g||_q</math>. | ||

+ | |||

+ | == Proof == | ||

+ | If <math>||f||_p=0</math> then <math>f=0</math> a.e. and there is nothing to prove. Case <math>||g||_q=0</math> is similar. On the other hand, we may assume that <math>f(x),g(x)\in\mathbb{R}</math> for all <math>x</math>. Let <math>a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q</math>. [[Young's Inequality]] gives us | ||

+ | <cmath> \frac{|f(x)|}{||f||_p}\frac{|g(x)|}{||g||_q} \leq \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p} + \frac{1}{q}\frac{|g(x)|^q}{||g||_q^q}. </cmath> | ||

+ | These functions are measurable, so by integrating we get | ||

+ | <cmath> \frac{||fg||_1}{||f||_p||g||_q}\leq \frac{1}{p}\frac{||f(x)||^p}{||f||_p^p} + \frac{1}{q}\frac{||g(x)||^q}{||g||_q^q} = \frac{1}{p}+\frac{1}{q}=1 . </cmath> | ||

+ | |||

+ | == Examples == | ||

+ | * Prove that, for positive reals <math>x,y,k</math>, the following inequality holds: | ||

+ | <center><math>\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}</math></center> | ||

+ | |||

+ | [[Category:Inequality]] | ||

+ | [[Category:Definition]] | ||

+ | [[Category:Theorems]] |

## Latest revision as of 13:15, 29 November 2018

## 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: