Master Holder's Inequality with Expertly Guided Solutions

[Cairo]

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

A proof would be great.

 

[sylas]

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

A proof would be great.

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
$$\left( \sum_{n=1}^\infty\left| a_n \right|^p \right)^{1/p}<br /> \left( \sum_{n=1}^\infty\left| b_n \right|^q \right)^{1/q}$$
​

My question. What does l_p mean?

 

[Cairo]

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

Thanks in advance.

 

[sylas]

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

Thanks in advance.

Define

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

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

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

as required.

Felicitations -- sylas

 

[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?

 

[sylas]

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

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.

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

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

 

[Cairo]

Thanks.