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 13:29, 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
where
is a positive real number, and also
.
Then the following inequality holds for all
and
with equality iff
.
Proof
The logarithm is concave and we know that , so by [Jensen's Inequality], we have
Young's Inequality then follows by exponentiation of both sides.