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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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    but you can't. Use it.2012-06-28
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    seriously? I was looking for a "pure proof", i wanted to use just the basic tools.2012-06-28
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    What the hell does "pure proof" means?2012-06-28
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    Seriously. The fact is that $\max_{x\in S}T(x)=(\sum |y_i|^q)^{1/q}$ with $q=p/(p-1)$, and once you know this fact, you know Hölder's inequality.2012-06-28
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    its just a joke, I mean a proof using just the basics. I think that the Holder's inequality would be too much.2012-06-28
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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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    Should have $p-1$ in the formula for the gradient.2012-06-28
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    @LeonidKovalev: The $\frac{p}{x_k}$ before the $|x_k|^p$ should take care of that and handle the sign.2012-06-28
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    I did not... Anyway, I would not put $x_k$ in the denominator, since it can be zero.2012-06-28
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    @LeonidKovalev: I could use $\frac{|y_k|^p}{y_k}=\mathrm{sgn}(y_k)|y_k|^{p-1}$, but I think simplicity is nice. Besides, the $y_k$ in the denominator and the $\mathrm{sgn}(y_k)$ both get canceled in the final answer.2012-06-28
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    @LeonidKovalev: I changed them to prevent any complaints about division by $0$.2012-06-28
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    What means $\nabla F\Vert y$ ?2012-06-29
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    @Integral: That means that the gradient of $F$ is parallel to $y$. That means that the surface where $F(x)=1$ is perpendicular to $y$.2012-06-29
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    Ok. Thank you very much!2012-06-29