# proof of Clarkson inequality

Suppose $2\leq p<\infty\textrm{ and }f,g\in L^{p}$.

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

By the triangle inequality, we have the following two inequalities

 $|f+g|^{p}\leq|f|^{p}+|g|^{p}\qquad\mbox{and}\qquad|f-g|^{p}\leq|f|^{p}+|g|^{p},$

and summing the two inequalities we get

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

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

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

Hence it follows that

 $\displaystyle\left\|\frac{f+g}{2}\right\|_{p}^{p}+\left\|\frac{f-g}{2}\right\|% _{p}^{p}$ $\displaystyle\leq$ $\displaystyle\frac{1}{2^{p-1}}\left(\int|f|^{p}\,d\mu+\int|g|^{p}\,d\mu\right)$ $\displaystyle=$ $\displaystyle\frac{1}{2^{p-1}}\left(\left\|f\right\|_{p}^{p}+\left\|g\right\|_% {p}^{p}\right),$

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

Title proof of Clarkson inequality ProofOfClarksonInequality 2013-03-22 16:24:46 2013-03-22 16:24:46 CWoo (3771) CWoo (3771) 8 CWoo (3771) Proof msc 28A25