Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Hlder's Inequality"

 
m (proofreading)
Line 1: Line 1:
 
Hölder's inequality: If <math>p,q>1</math>, <math>1/p+1/q=1</math>, <math>f\in L^p, g\in L^q</math> then <math>fg\in L^1</math> and <math>||fg||_1\leq ||f||_p||g||_q</math>.
 
Hölder's inequality: If <math>p,q>1</math>, <math>1/p+1/q=1</math>, <math>f\in L^p, g\in L^q</math> then <math>fg\in L^1</math> and <math>||fg||_1\leq ||f||_p||g||_q</math>.
  
−
Proof: If <math>||f||_p=0</math> then <math>f=0</math> a.e. and there is nothing to prove. Case <math>||g||_q=0</math> is similar. On the other hand, we may assume that <math>f(x),g(x)\in\mathbb{R}</math> for all <math>x</math>. Let <math>a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q</math>. [[Young's inequality]] gives us
+
Proof: If <math>||f||_p=0</math> then <math>f=0</math> a.e. and there is nothing to prove. Case <math>||g||_q=0</math> is similar. On the other hand, we may assume that <math>f(x),g(x)\in\mathbb{R}</math> for all <math>x</math>. Let <math>a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q</math>. [[Young's Inequality]] gives us
 
<center><math>\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}.</math></center> These functions are measurable, so by integrating we get
 
<center><math>\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}.</math></center> These functions are measurable, so by integrating we get
 
<center><math>\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</math></center>
 
<center><math>\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</math></center>

Revision as of 12:18, 27 June 2006

Hölder's inequality: If $p,q>1$, $1/p+1/q=1$, $f\in L^p, g\in L^q$ then $fg\in L^1$ and $||fg||_1\leq ||f||_p||g||_q$.

Proof: If $||f||_p=0$ then $f=0$ a.e. and there is nothing to prove. Case $||g||_q=0$ is similar. On the other hand, we may assume that $f(x),g(x)\in\mathbb{R}$ for all $x$. Let $a=\frac{|f(x)|^p}{||f||_p^p}, b=\frac{|g(x)|^q}{||g||_q^q},\alpha=1/p,\beta=1/q$. Young's Inequality gives us

$\frac{|f(x)|}{||f||_p}\frac{|g(x)|}{||g||_q}\leq \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p}+\frac{1}{q}\frac{|g(x)|^q}{||g||_q^q}.$

These functions are measurable, so by integrating we get

$\frac{||fg||_1}{||f||_p||g||_q}\leq\frac{1}{p}\frac{||f(x)||^p}{||f||_p^p}+\frac{1}{q}\frac{||g(x)||^q}{||g||_q^q}=\frac{1}{p}+\frac{1}{q}=1$
Retrieved from ""