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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, im trying to avoid Holder's inequality.

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but you can't. Use it. – user31373 Jun 28 '12 at 20:24
seriously? I was looking for a "pure proof", i wanted to use just the basic tools. – Integral Jun 28 '12 at 20:33
What the hell does "pure proof" means? – D. Thomine Jun 28 '12 at 20:35
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. – user31373 Jun 28 '12 at 20:37
its just a joke, I mean a proof using just the basics. I think that the Holder's inequality would be too much. – Integral Jun 28 '12 at 20:38
up vote 3 down vote accepted

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. – user31373 Jun 28 '12 at 20:40
@LeonidKovalev: The $\frac{p}{x_k}$ before the $|x_k|^p$ should take care of that and handle the sign. – robjohn Jun 28 '12 at 20:42
I did not... Anyway, I would not put $x_k$ in the denominator, since it can be zero. – user31373 Jun 28 '12 at 20:43
@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. – robjohn Jun 28 '12 at 20:50
@LeonidKovalev: I changed them to prevent any complaints about division by $0$. – robjohn Jun 28 '12 at 20:56

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