How to prove Clarkson's inequality?

Question

I do not know how to prove one of the four Clarkson's inequalities: let $u,v \in L^p(\Omega)$, if $1 < p < 2$, then $$ \bigg\lVert \frac{u+v}{2} \bigg\rVert_p^p + \bigg\lVert \frac{u-v}{2} \bigg\rVert_p^p \geq \frac{1}{2}\lVert u \rVert_p^p + \frac{1}{2}\lVert v \rVert_p^p $$ Could your please help provide a detailed proof? Thanks a lot!

Answer 1

Since both sides are just integrals over $\Omega$, it is enough to prove the inequality for real numbers. This can be found here.

Alternatively, you can do the following: \begin{align*}\newcommand\abs[1]{|#1|} \newcommand\biggnorm[1]{\big\|#1\big\|} \abs{x+y}^p + \abs{x - y}^p &= \biggnorm{ \begin{pmatrix} x+y \\ x-y \end{pmatrix} }_{p}^p \le 2^{1 - \frac p2} \, \biggnorm{ \begin{pmatrix} x+y \\ x-y \end{pmatrix} }_{2}^p \\&= 2 \, \biggnorm{ \begin{pmatrix} x \\ y \end{pmatrix} }_{2}^p \le 2 \, \biggnorm{ \begin{pmatrix} x \\ y \end{pmatrix} }_{p}^p = 2 \, \big( \abs{x}^p + \abs{y}^p \big) . \end{align*} Here, one uses the equivalence between the $2$-norm and the $p$-norm on $\mathbb{R}^2$ and the parallelogram identity.