Mathematics Prelims

July 14, 2008

Holder’s Inequality

Filed under: Analysis,Functional Analysis — cjohnson @ 6:38 pm

If (x_n)_{n \in \mathbb{N}} and (y_n)_{n \in \mathbb{N}} are sequences of real numbers where the following series converge, for a given p > 1 and q \in \mathbb{R} such that 1/p + 1/q = 1,

\sum_{n = 1}^\infty |x_n|^p < \infty   and  \sum_{n=1}^\infty |y_n|^q < \infty

Then we have the following inequality, know an Holder’s inequality.

\sum_{n=1}^\infty |x_n y_n| \leq \left( \sum_{n=1}^\infty |x_n|^p \right)^{1/p} \left(\sum_{n=1}^\infty |y_n|^q \right)^{1/q}

Proof: Suppose \sum_{n=1}^\infty |x_n|^p = \sum_{n=1}^\infty |y_n|^q = 1 and note the following, due to Young’s Inequality.

|x_n y_n| \leq \frac{1}{p} |x_n|^p + \frac{1}{q} |y_n|^q

Summing over n, we have

\sum_{n=1}^\infty |x_n y_n| \leq \frac{1}{p} \sum_{n=1}^\infty |x_n|^p + \frac{1}{q} \sum_{n=1}^\infty |y_n|^q = 1/p + 1/q = 1

Now, suppose we’re given sequences (x_n) and (y_n) where the associated series converge, but not to one.  In that case we construct two new sequences, (a_n) and (b_n) as follows.

a_n = \frac{x_n}{ \left( \sum_{n = 1}^\infty |x_n|^p \right)^{1/p}}

b_n = \frac{y_n}{ \left( \sum_{n = 1}^\infty |y_n|^q \right)^{1/q}}

Now note that \sum |a_n|^p = \sum |b_n|^q = 1, so we can apply the previous result to obtain the following.

\sum_{n=1}^\infty |a_n b_n| = \sum_{n=1}^\infty \frac{|x_n y_n|}{\left(\sum_{n=1}^\infty |x_n|^p\right)^{1/p} \left(\sum_{n=1}^\infty |y_n|^q\right)^{1/q}} \leq 1

Multiplying both sides by the denominator on the left (which we can pull out of the sum since it’s constant), we obtain the desired inequality.

(It’s worth mentioning that there is another, more general, version of Holder’s Inequality that uses Lebesgue integrals.  The Lebesgue integral version actually implies the version we’ve mentioned here, and I’ll post about that once I start posting my notes from measure theory, specifically Lebesgue integration.)


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