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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<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]] |
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Revision as of 00:11, 18 June 2023
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:
![$\left(1 + \frac {x}{y}\right)^k + \left(1 + \frac {y}{x}\right)^k\geq 2^{k+1}$](http://latex.artofproblemsolving.com/4/0/6/406c5de6ae650893aa91bcdcb3865a0ee7d6b308.png)