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Hölder's Inequality

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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_... 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_... 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_... 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... 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... 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^... 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...

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}


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