Holder's Inequality: Proving (17) is Sufficient

[sakodo]

Hi guys, I am reading a proof on Holder's inequality. There is a line I don't understand.

Here is the extract from Kolmogorov & Fomin, Introductory Real Analysis.

"The proof of [Minkowski's inequality] is in turn based on Holder's inequality
\sum_{k=1}^n |a_k b_k|\leq (\sum_{k=1}^n|a_k\mid^p)^{\frac{1}{p}}(\sum_{k=1}^n|b_k\mid^q)^{\frac{1}{q}}. \ \ \ (15) , where \frac<br /> {1}{p}+\frac{1}{q}=1.

We begin by observing that the inequality (15) is homogeneous,i.e., if it holds for two points (a_1,...,a_n) \ and \ (b_1,...,b_n), then it holds for any two points
(\lambda a_1,...,\lambda a_n) \ and \ (\mu b_1,...,\mu b_n) where \lambda \ and \ \mu are arbitrary real numbers. Therefore we need only prove (15) for the case
\sum_{k=1}^n|a_k\mid^p = \sum_{k=1}^n|b_k\mid^q = 1. \ \ \ (17)

Thus, assuming that (17) holds, we now prove that \sum_{k=1}^n |a_k b_k|\leq 1. \ \ \ (18)"

I understand the inequality being homogeneous, but I don't understand how he got to the assumption of (17). Why is proving the case of (17) sufficient?

Any help would be appreciated.

Thanks.

 

[henry_m]

Any general a_n and b_n can be rescaled so it satisfies the condition. Suppose you've proved it for the case
<br /> \sum_{k=1}^n |a_k|^p = \sum_{k=1}^n |b_k|^p = 1<br />
and you'd like to prove it for
<br /> \sum_{k=1}^n |a_k|^p = A,\quad \sum_{k=1}^n |b_k|^p = B<br />
All you need to do is consider the sequences a_k/A^{1/p} and b_k/B^{1/p}, which satisfy the special case condition, since homogeneity means it's equivalent.