Understanding Holder's Inequality: A Key Step in Proving Minkowski's Inequality

[pbandjay]

I am actually attempting the proof for Minkowski's inequality, but have not gotten that far yet. I am stuck on a step in Holder's inequality, and I have a feeling it's something very simple that I am just overlooking...

I have easily been able to show [tex]ab \leq \frac{a^p}{p} + \frac{b^q}{q}[/tex]

And if [tex]a,b[/tex] are normalized vectors then:

[tex]\sum_k a_k b_k \leq \frac{1}{p}\sum_k {a_k}^p + \frac{1}{q}\sum_k {b_k}^q[/tex]

And I am aware that through normalizing the vectors, I am supposed to be able to deduce the formula for Holder's inequality:

[tex]\sum_k a_k b_k \leq (\sum_k {a_k}^p)^{\frac{1}{p}}(\sum_k {b_k}^q)^{\frac{1}{q}}[/tex]

But I just cannot figure this step out for some reason! Please give me at least a hint...

Thank you in advance!

 

[maze]

This inequality:
[tex]\sum_k a_k b_k \leq \frac{1}{p}\sum_k {a_k}^p + \frac{1}{q}\sum_k {b_k}^q[/tex]

holds for all vectors, not just those that are normalized.

Whenever there are several things being added and you want to turn them into a single term, a good strategy is to do whatever it takes to turn the sum into a constant 1. Then after things simplify, bring back in the complexity that you threw away, but now the sum is gone. This happens all the time, and I think there is a deep meaning why it works so much, but I don't know what it is.

In this case, if you restrict ||a||_p = 1, and ||b||_q = 1, then the equation becomes

[tex]\sum_k a_k b_k \leq 1[/tex]

for a and b normalized like so. Now bring back the complexity for arbitrary vectors x and y,

[tex]\sum_k \frac{x_k}{||x||_p} \frac{y_k}{||y||_q} \leq 1[/tex]

[tex]\sum_k x_k y_k \leq ||x||_p ||y||_q[/tex]

 

[pbandjay]

Oh wow, okay. That's pretty nice and slick. Thank you!

 

[maze]

Also note that the same exact strategy will prove the integral form.