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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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    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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    You are very welcome!2012-12-29