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Let $S= \{(x_1,\ldots, x_n)\in \mathbb{R}^n$; $|x_1|^p+\ldots+|x_n|^p=1\}$, where $p>1$ is real(and fixed), consider a fixed $y\in\mathbb{R}^n$ and $T:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $T(x) = x\cdot y$, where $x\cdot y = x_1y_1+\ldots+x_ny_n$.

I'm having a hard time to find $\max_{x\in S}\ T(x)$.

I already noticed a few things but its still really difficult to do something useful.

1) $\forall (x_1\ldots, x_n)\in S, |x_1|^p+\ldots+|x_n|^p\leq |x_1|+\ldots+|x_n|$;

2) Taking the norm $\Vert(x_1,\dots, x_n)\Vert=(|x_1|^p+\ldots+|x_n|^p)^{\frac{1}{p}}$ and the ball $B(0,1)$, with center $0\in\mathbb{R}^n$ and radius $1$, we have $S=\partial B$;

3) $\forall i=1\ldots n, T(e_i) = y_i$.

Also, I'm trying to avoid Holder's inequality.

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    I changed max$_{x\in S}$ to $\max_{x\in S}$. That is standard usage. When you write$\max$with$a$backslash, then (1) $\max$ is not italicized (as if it were the product of three variables $m$, $a$, and $x$), and (2) proper spacing appears before and after it when one write something like "5\max A", and (3) when it is in "display" style rather than "inline" style, then the subscript appears directly under it, thus: $\displaystyle\max_{x\in S}$.2012-06-28

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Define $\newcommand{\sgn}{\operatorname{sgn}}$ $ F(x)=\sum_{k=1}^n|x_k|^p\tag{1} $ then $ \nabla F(x)=\left(p\sgn(x_k)|x_k|^{p-1}\right)_{k=1}^n\tag{2} $ You want to find a point on the surface where $\nabla F\,||\,y$. Therefore, $ x=\left(\sum_{k=1}^n|y_k|^{\frac{p}{p-1}}\right)^{-\frac{1}{p}}\left(\sgn(y_k)|y_k|^{\frac{1}{p-1}}\right)_{k=1}^n\tag{3} $ should be the point where $T(x)$ is the greatest. Computing $T(x)$ yields $ T(x)=\left(\sum_{k=1}^n|y_k|^{\frac{p}{p-1}}\right)^{\frac{p-1}{p}}\tag{4} $

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    Ok. Thank you very much!2012-06-29