proof of Clarkson inequality

Suppose .

${\parallel {\displaystyle \frac{f+g}{2}}\parallel }_p^p+{\parallel {\displaystyle \frac{f-g}{2}}\parallel }_p^p$ $=$ ${\displaystyle \int {\left|\frac{f+g}{2}\right|}^p𝑑\mu }+{\displaystyle \int {\left|\frac{f-g}{2}\right|}^p𝑑\mu }$ (1)

$=$ ${\displaystyle \frac{1}{2^p}}\left({\displaystyle \int {\left|f+g\right|}^p𝑑\mu }+{\displaystyle \int {\left|f-g\right|}^p𝑑\mu }\right).$ (2)

By the triangle inequality, we have the following two inequalities

$${|f+g|}^p\le {|f|}^p+{|g|}^p\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}{|f-g|}^p\le {|f|}^p+{|g|}^p,$$

and summing the two inequalities we get

$${|f+g|}^p+{|f-g|}^p\le 2({|f|}^p+{|g|}^p).$$

This means that expression (2) above is less than or equal to

${\displaystyle \frac{1}{2^{p-1}}}{\displaystyle \int ({|f|}^p+{|g|}^p)𝑑\mu }.$ (3)

Hence it follows that

${\parallel {\displaystyle \frac{f+g}{2}}\parallel }_p^p+{\parallel {\displaystyle \frac{f-g}{2}}\parallel }_p^p$ $\le$ ${\displaystyle \frac{1}{2^{p-1}}}\left({\displaystyle \int {|f|}^p𝑑\mu }+{\displaystyle \int {|g|}^p𝑑\mu }\right)$

$=$ ${\displaystyle \frac{1}{2^{p-1}}}\left({\parallel f\parallel }_p^p+{\parallel g\parallel }_p^p\right),$

which since $p\ge 2$ directly implies the desired inequality.

Title	proof of Clarkson inequality
Canonical name	ProofOfClarksonInequality
Date of creation	2013-03-22 16:24:46
Last modified on	2013-03-22 16:24:46
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	8
Author	CWoo (3771)
Entry type	Proof
Classification	msc 28A25