Difference between revisions of "Hölder's Inequality"
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<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> | <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>, | 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}}{ \ | + | <cmath> \prod_i \left( \frac{a_{ij}}{\beta_j} \right)^{\lambda_i} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \beta_j} = \frac{\prod_i a_{ij}^{\lambda_i}}{ \prod_i a_{ij}^{\lambda_i} / \sum_j \prod_i a_{ij}^{\lambda_i}} = \sum_j \prod_i a_{ij}^{\lambda_i} . </cmath> |
Therefore | 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> | <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> |
Latest revision as of 02:33, 21 July 2024
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)