Difference between revisions of "Hlder's Inequality"
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m (Hölder's inequality moved to Hölder's Inequality: capitalization change) |
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Revision as of 13:36, 11 July 2006
Hölder's inequality: 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
![$\frac{|f(x)|}{||f||_p}\frac{|g(x)|}{||g||_q}\leq \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p}+\frac{1}{q}\frac{|g(x)|^q}{||g||_q^q}.$](http://latex.artofproblemsolving.com/2/5/6/2560a53826cfcf80c81290ea65b647b40e55e9e3.png)
These functions are measurable, so by integrating we get
![$\frac{||fg||_1}{||f||_p||g||_q}\leq\frac{1}{p}\frac{||f(x)||^p}{||f||_p^p}+\frac{1}{q}\frac{||g(x)||^q}{||g||_q^q}=\frac{1}{p}+\frac{1}{q}=1$](http://latex.artofproblemsolving.com/2/8/9/289991c7864bbe6d91b07bef65e347358dbb93cd.png)