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} .
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>
<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>
<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>

Revision as of 01:47, 18 December 2020

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