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

Immediately from Holder's Inequality I get that: $$\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}=1}\left\le\Vert a\Vert_{q}$$ 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 $\left=\Vert a\Vert_{p}$ but I can't seem to manage to do that. Since it's also easy to show that: $$\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}=1}\left=\max_{x\in\mathbb{R}^{n},\,\Vert x\Vert_{p}\leq1}\left$$ 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.2012-12-28
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    Try $x_k = \text{sgn } a_k |a_k|^{\frac{q}{p}}$.2012-12-28
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    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?2012-12-28
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    I added an answer below.2012-12-28

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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!2012-12-29
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    You are very welcome!2012-12-29