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I'm trying show that if $p,q$ are Holder Conjugates then: $$\forall\, a\in\mathbb{R}^{n}:\,\Vert a\Vert_{q}=\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}=1}\left<a,x\right>$$ Where $\left<a,x\right>$ is the Euclidian Inner-Product on $\mathbb{R}^{n}$ .

Immediately from Holder's Inequality I get that: $$\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}=1}\left<a,x\right>\le\Vert a\Vert_{q}$$ To show the other direction of the inequality I wanted to pick a $v\in\mathbb{R}^{n}$ such that $\Vert v\Vert_{p}=1$ and $\left<a,v\right>=\Vert a\Vert_{p}$ but I can't seem to manage to do that. Since it's also easy to show that: $$\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}=1}\left<a,x\right>=\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}\leq1}\left<a,x\right>$$ It would also suffice to find a $v$ with $\Vert v\Vert_{p}\leq1$.

Help would be appreciated!

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Look at how Hölder's inequality is usually proved, ie, Young's inequality. This gives a condition for equality, namely there are constants $\alpha>0, \beta>0$ such that $\alpha|a_k|^q = \beta|x_k|^p$ for all $k$. You could simplify life by assuming $\|a\|_q= 1$ to start with. –  copper.hat Dec 28 '12 at 20:01
    
Try $x_k = \text{sgn } a_k |a_k|^{\frac{q}{p}}$. –  copper.hat Dec 28 '12 at 20:15
    
That choice of vector doesn't seem to yield the right result unless I'm making some error in calculation, did you check it out? –  Serpahimz Dec 28 '12 at 21:10
    
I added an answer below. –  copper.hat Dec 28 '12 at 23:50

1 Answer 1

up vote 1 down vote accepted

Let $x_k = \text{sgn } a_k |a_k|^{\frac{q}{p}}$. Then $x_k a_k = (\text{sgn } a_k) a_k |a_k|^{\frac{q}{p}} = |a_k|^{\frac{q}{p}+1} = |a_k|^q$, and $\sum x_k a_k = \|a\|_q^q$. In addition, we have $\sum_k |x_k|^p = \sum_k |a_k|^q = \|a\|_q^q$, and so $\|x\|_p = \|a\|_q^{\frac{q}{p}}$.

Hence we have $\sum_k x_k a_k =\|a\|_q^{\frac{q}{p}} \|a\|_q^{q-\frac{q}{p}} = \|x\|_p \|a\|_q$. Then choosing $v = \frac{1}{\|x\|_p} x$ will produce the desired result.

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Indeed, I made a calculation error before but I've sorted it out since and reached the same result. I also took your previous advice and simplified it by assuming $∥a∥q$=1. Thanks for the help! –  Serpahimz Dec 29 '12 at 0:04
    
You are very welcome! –  copper.hat Dec 29 '12 at 0:10

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