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# 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.

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## Comments

## proof of Clarkson inequality is not correct!?

Hello,

in my opinion this proof is not correct. I think you are not allowed to use

|a+b|^p <= |a|^p+|b|^p.

Counter-example:

a,b=1 p = 2

=> |1 + 1|^2=4 is NOT smaller then 1^2+1^2=2.

Iam very interested in an alternative proof.

greets flowwsen

## Re: proof of Clarkson inequality is not correct!?

I agree, the proof is wrong.

For a correct proof see e.g. Adams, Fournier "Sobolev Spaces" (in the 2nd edition it's 2.38 in the section on Uniform Convexity of L^p spaces).

darthporter