Holder's Inequality Help

1. Jan 26, 2010

Cairo

I'm currently reading a book and stuck on an exercise with no solutions.

A proof would be great.

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2. Jan 26, 2010

sylas

I have a bit of an interest in Holders inequality, so I'm going to make the effort to convert your word docoument into something readable for the forums. In general, posting a word document is not adequate for readers. I have also converted an "a" to a "b" in the last formula.

The problem is:

Let $1 < p < \infty$ and 1/p + 1/q = 1

Show that if $a \in l_p$ and $b \in l_q$ then the series
$$S(a,b) = \sum_{n=1}^{\infty} a_n b_n$$
converges absolutely, and is bounded above by
$$\left( \sum_{n=1}^\infty\left| a_n \right|^p \right)^{1/p} \left( \sum_{n=1}^\infty\left| b_n \right|^q \right)^{1/q}$$

My question. What does lp mean?

3. Jan 27, 2010

Cairo

Lp = {a = a(n) | Sigma |a(n)|^p < infinity}.

4. Jan 27, 2010

sylas

Define
\begin{align*} X & = \sum_{n=0}^\infty \left| a_n \right| ^p \\ Y & = \sum_{n=0}^\infty \left| b_n \right| ^q \\ x_n & = | a_n | / X^{1/p} \\ y_n & = | b_n | / Y^{1/q} \\ \intertext{Hence} \sum_{n=0}^\infty x_n^p & = \frac{\sum_{n=0}^\infty \left| a_n \right|^p}{X} & = 1 \\ \sum_{n=0}^\infty y_n^q & = \frac{\sum_{n=0}^\infty \left| b_n \right|^q}{Y} & = 1 \end{align*}​

I will also use Young's inequality. Given 1/p + 1/q = 1, for p,q > 0, a,b >= 0, we have
$$ab \leq \frac{a^p}{p} + \frac{b^q}{q}$$​

Hence
\begin{align*} \sum_{n=0}^\infty a_n b_n & \leq X^{1/p}Y^{1/q} \sum_{n=0}^\infty x_n y_n \\ & \leq X^{1/p}Y^{1/q} \sum_{n=0}^\infty \left( \frac{x_n^p}{p} + \frac{y_n^q}{q} \right) \\ & \leq \frac{X^{1/p}Y^{1/q}}{p} \sum_{n=0}^\infty x_n^p + \frac{X^{1/p}Y^{1/q}}{q} \sum_{n=0}^\infty y_n^q \\ & = \frac{X^{1/p}Y^{1/q}}{p} + \frac{X^{1/p}Y^{1/q}}{q} \\ & = X^{1/p}Y^{1/q} \left( \frac{1}{p} + \frac{1}{q} \right) \\ & = X^{1/p}Y^{1/q} \end{align*}​
as required.

Felicitations -- sylas

5. Jan 27, 2010

Cairo

Thanks Sylas. I can see from the proof that the series is bounded above. But where does the absolute convergence come in?

Would you have to use the MCT?

Also, do I not need to consider finite sums for Holder's Inequality?

Last edited: Jan 27, 2010
6. Jan 27, 2010

sylas

I am not sure what you mean. The given condition
$$\sum_{n=0}^\infty |a_n|^p < \infty$$​
is simply saying that the sum is finite. So it has a finite value, and I give it a name, X.

There are many forms of Holder's inequality. I gave the proof for the case you presented. A finite sum follows directly as a special case of the infinite sum, where all but an infinite number of the terms are zero.

And by the way, I see you started all series at n=1, whereas I started at n=0. There's no difference; it is still just an infinite series.

Cheers -- sylas

7. Jan 28, 2010

Thanks.