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.)

Advertisements

1 Comment »


RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: