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.


1 Comment »

  1. Hi! Interesting blog. I do have a suggestion, WordPress can parse latex code 🙂

    Comment by chuckie — July 15, 2008 @ 7:26 am | Reply

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