Difference between revisions of "Minkowski Inequality"

(Equivalence with the standard form)
m (Equivalence with the standard form)
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== Equivalence with the standard form ==
 
== Equivalence with the standard form ==
  
For <math>r>s>0</math>, putting <math>x_{ij}:=a_{ij}^s</math> and <math>p:=\frac rs>1</math>, the 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}
 
<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>.
  
Put <math>m=2, a_i:=x_{i1},b_i:=x_{i2}</math> and 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}
 
<math>\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p}

Revision as of 14:06, 12 November 2010

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

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