Difference between revisions of "Hölder's Inequality"

Revision as of 15:41, 29 December 2021

Elementary Form

If $a_1, a_2, \dotsc, a_n, b_1, b_2, \dotsc, b_n, \dotsc, z_1, z_2, \dotsc, z_n$ are nonnegative real numbers and $\lambda_a, \lambda_b, \dotsc, \lambda_z$ are nonnegative reals with sum of 1, then $\begin{align*} a_1^{\lambda_a}b_1^{\lambda_b} \dotsm z_1^{\lambda_z} + \dotsb &+ a_n^{\lambda_a} b_n^{\lambda_b} \dotsm z_n^{\lambda_z} \\ \le{}& (a_1 + \dotsb + a_n)^{\lambda_a} (b_1 + \dotsb + b_n)^{\lambda_b} \dotsm (z_1 + \dotsb + z_n)^{\lambda_z} . \end{align*}$ Note that with two sequences $\mathbf{a}$ and $\mathbf{b}$ , and $\lambda_a = \lambda_b = 1/2$ , this is the elementary form of the Cauchy-Schwarz Inequality.

We can state the inequality more concisely thus: Let $\{ \{a_{ij}\}_{i=1}^n \} _{j=1}^m$ be several sequences of nonnegative reals, and let $\{ \lambda_i \}_{i=1}^n$ be a sequence of nonnegative reals such that $\sum \lambda = 1$ . Then $\sum_j \prod_i a_{ij}^{\lambda_i} \le \prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} .$

Proof of Elementary Form

We will use weighted AM-GM. We will disregard sequences $\{ a_{ij} \}_{i=1}^n$ 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 $1 \le k \le m$ , let us define $\beta_k = \frac{\prod_i a_{ik}^{\lambda_i}}{\sum_j \prod_i a_{ij}^{\lambda_i}} .$ Evidently, $\sum \beta_j = 1$ . Then for all integers $1\le i \le n$ , by weighted AM-GM, $\sum_j a_{ij} = \sum_j \beta_j \left(\frac{a_{ij}}{\beta_j} \right) \ge \prod_j \left( \frac{a_{ij}}{\beta_j} \right)^{\beta_j} .$ Hence $\prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} \ge \prod_i \prod_j \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i \beta_j} = \prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} .$ But from our choice of $\beta_j$ , for all integers $1 \le j \le m$ , $\prod_i \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \beta_k} = \frac{\prod_j a_{ij}^{\lambda_i}}{ \prod_j a_{ij}^{\lambda_i} / \sum_j \prod_i a_{ij}^{\lambda_i}} = \sum_j \prod_i a_{ij}^{\lambda_i} .$ Therefore $\prod_j \biggl[ \prod_i \Bigl( \frac{a_{ij}}{\beta_j} \Bigr)^{\lambda_i} \biggr]^{\beta_j} = \prod_k \biggl( \sum_j \prod_i a_{ij}^{\lambda_i} \biggr)^{\beta_k} = \sum_j \prod_i a_{ij}^{\lambda_i},$ since the sum of the $\beta_k$ is one. Hence in summary, $\prod_i \biggl( \sum_j a_{ij} \biggr)^{\lambda_i} \ge \sum_j \prod_i a_{ij}^{\lambda_i} ,$ as desired. Equality holds when $a_{ij}/\beta_j = a_{ij'}/\beta_{j'}$ for all integers $i,j,j'$ , i.e., when all the sequences $\{a_{ij}\}_{j=1}^m$ are proportional. $\blacksquare$

Statement

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

Proof

If $||f||_p=0$ then $f=0$ a.e. and there is nothing to prove. Case $||g||_q=0$ is similar. On the other hand, we may assume that $f(x),g(x)\in\mathbb{R}$ for all $x$ . Let $a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q$ . Young's Inequality gives us $\frac{|f(x)|}{||f||_p}\frac{|g(x)|}{||g||_q} \leq \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p} + \frac{1}{q}\frac{|g(x)|^q}{||g||_q^q}.$ These functions are measurable, so by integrating we get $\frac{||fg||_1}{||f||_p||g||_q}\leq \frac{1}{p}\frac{||f(x)||^p}{||f||_p^p} + \frac{1}{q}\frac{||g(x)||^q}{||g||_q^q} = \frac{1}{p}+\frac{1}{q}=1 .$

Examples

Prove that, for positive reals $x,y,k$ , the following inequality holds:

$\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}$

@@ Line 40: / Line 40: @@
 <center><math>\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}</math></center>
+[[Category:Algebra]]
 [[Category:Inequality]]
-[[Category:Definition]]
-[[Category:Theorems]]

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