Difference between revisions of "Young's Inequality"
(Created page with "== Form for Hölder exponents == If <math>a, b</math> are non-negative reals, and <math>p,q</math> are positive reals that satisfy <math>\frac{1}{p}+\frac{1}{q}=1</math>, then...") |
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with equality iff <math>f(a) = b</math>. | with equality iff <math>f(a) = b</math>. | ||
== Proof == | == Proof == | ||
| − | The logarithm is concave and we know that <math>\frac{1}{p}+\frac{1}{q}=1</math>, so by Jensen's Inequality, we have | + | The logarithm is concave and we know that <math>\frac{1}{p}+\frac{1}{q}=1</math>, so by [Jensen's Inequality], we have |
<cmath>\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)</cmath> | <cmath>\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)</cmath> | ||
<cmath>\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \log{a}+\log{b} = \log{ab}</cmath> | <cmath>\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \log{a}+\log{b} = \log{ab}</cmath> | ||
Young's Inequality then follows by exponentiation of both sides. | Young's Inequality then follows by exponentiation of both sides. | ||
Revision as of 12:29, 18 August 2024
Form for Hölder exponents
If 
are non-negative reals, and 
are positive reals that satisfy 
, then the following inequality holds for all possible values of 
and 
.
with equality iff 
Form for definite integrals
Suppose 
is a strictly increasing and continuous function on the interval ![$[0,t]$](http://latex.artofproblemsolving.com/9/3/9/9398cf678fee12444e20e80c8ec38bb2803749fb.png)
where 
is a positive real number, and also 
.
Then the following inequality holds for all ![$a \in [0,c]$](http://latex.artofproblemsolving.com/e/2/0/e20820c2c463d957998fcbec3591370a1fda4448.png)
and ![$b \in [0,f(c)]$](http://latex.artofproblemsolving.com/d/2/3/d238f6a4d2e9dbfb6774aa36b634bbcb5a6d016a.png)
with equality iff 
.
Proof
The logarithm is concave and we know that , so by [Jensen's Inequality], we have
![\[\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)\]](http://latex.artofproblemsolving.com/b/1/e/b1e1c08c3759dfff12f8fb7c02c712e81f5fca1a.png)
![\[\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \log{a}+\log{b} = \log{ab}\]](http://latex.artofproblemsolving.com/3/4/b/34b370c0294f687b7e6d39223f45f43f5876b5d7.png)
Young's Inequality then follows by exponentiation of both sides.
