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.