# Mathematics Prelims

## July 14, 2008

### Young’s Inequality

Filed under: Analysis,Functional Analysis — cjohnson @ 3:18 pm

Suppose that $p$ is a real number with $p > 1$.  Let $q = 1/(1 - 1/p)$.  Then we have that $1/p + 1/q = 1$.  For any positive $x$ and $y$, the following inequality holds:

$xy \leq \frac{x^p}{p}+ \frac{y^q}{q}$

Proof: Note that $x = (x^p)^\frac{1}{p}$ and $y = (y^q)^\frac{1}{q}$.  Taking the log xy we have

$\log xy = \log x + \log y = \log ((x^p)^\frac{1}{p}) + \log ((y^q)^\frac{1}{q}) \leq \log (\frac{x^p}{p} + \frac{y^q}{q})$

If we then exponentiate, we have $xy \leq \frac{x^p}{p} + \frac{y^q}{q}$.

A measure theory book of mine (Measure, Integral and Probability, by Capinski) gives another proof, but I like this one, from PlanetMath, more.