Difference between revisions of "Minkowski Inequality"

m (Equivalence with the standard form)
m (Equivalence with the standard form)
Line 8: Line 8:
 
For <math>r>s>0</math>, putting <math>x_{ij}:=a_{ij}^s</math> and <math>p:=\frac rs>1</math>, the symmetrical form given above becomes
 
For <math>r>s>0</math>, putting <math>x_{ij}:=a_{ij}^s</math> and <math>p:=\frac rs>1</math>, the symmetrical form given above becomes
  
<math> \sum_{j=1}^{m}\biggl(\sum_{i=1}^{n}x_{ij}^p\biggr)^{1/p}
+
<center><math> \sum_{j=1}^{m}\biggl(\sum_{i=1}^{n}x_{ij}^p\biggr)^{1/p}
\geq\left(\sum_{i=1}^{n}\biggl(\sum_{j=1}^{m}x_{ij}\biggr)^p\right)^{1/p}</math>.
+
\geq\left(\sum_{i=1}^{n}\biggl(\sum_{j=1}^{m}x_{ij}\biggr)^p\right)^{1/p}</math>.</center>
  
 
Putting <math>m=2</math> and <math>a_i:=x_{i1},b_i:=x_{i2}</math>, we get the form in which the Minkowski Inequality is given most often:
 
Putting <math>m=2</math> and <math>a_i:=x_{i1},b_i:=x_{i2}</math>, we get the form in which the Minkowski Inequality is given most often:
  
<math>\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p}
+
<center><math>\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p}
\geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}</math>
+
\geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}</math></center>
  
As the latter can be iterated, there is no loss of generality by putting <math>m=2</math> .
+
As the latter can be iterated, there is no loss of generality by putting <math>m=2</math>.
  
 
== Problems ==
 
== Problems ==

Revision as of 14:50, 11 March 2011

The Minkowski Inequality states that if $r>s$ is a nonzero real number, then for any positive numbers $a_{ij}$, the following holds: $\left(\sum_{j=1}^{m}\left(\sum_{i=1}^{n}a_{ij}^r\right)^{s/r}\right)^{1/s}\geq \left(\sum_{i=1}^{n}\left(\sum_{j=1}^{m}a_{ij}^s\right)^{r/s}\right)^{1/r}$

Notice that if either $r$ or $s$ is zero, the inequality is equivalent to Holder's Inequality.

Equivalence with the standard form

For $r>s>0$, putting $x_{ij}:=a_{ij}^s$ and $p:=\frac rs>1$, the symmetrical form given above becomes

$\sum_{j=1}^{m}\biggl(\sum_{i=1}^{n}x_{ij}^p\biggr)^{1/p} \geq\left(\sum_{i=1}^{n}\biggl(\sum_{j=1}^{m}x_{ij}\biggr)^p\right)^{1/p}$.

Putting $m=2$ and $a_i:=x_{i1},b_i:=x_{i2}$, we get the form in which the Minkowski Inequality is given most often:

$\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p} \geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}$

As the latter can be iterated, there is no loss of generality by putting $m=2$.

Problems

This article is a stub. Help us out by expanding it.