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I'm having trouble proving the following:

Suppose that $\frac{1}{p} + \frac{1}{q} = 1$. Then for any $\mathbf{x} \in \mathbb{R}^n$, $$ \lVert \mathbf{x} \rVert_p = \sup_{\mathbf{y} \in \mathbb{R}^n} \frac{\lvert \mathbf{x} \cdot \mathbf{y} \rvert}{\lVert \mathbf{y} \rVert_q} , \qquad \mathbf{y} \neq \mathbf{0} . $$

This post hints at the intuition behind how a proof might work, but I have no idea how to flesh out the details.

Definition of $p$-norm:

Let $1 \leq p < \infty$. For $\mathbf{x} = (x_1, \ldots, x_n) \in \mathbb{R}^n$, define $$ \lVert \mathbf{x} \rVert_p = \left( \sum_{k = 1}^n \lvert x_k \rvert^p \right)^{1/p} . $$

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What you need is a special case of Hölder's Inequality: $$ |x\cdot y|\leq\|x\|_p\,\|y\|_q. $$ This gives you immediately that $$ \lVert \mathbf{x} \rVert_p \geq \sup_{\mathbf{y} \in \mathbb{R}^n} \frac{\lvert \mathbf{x} \cdot \mathbf{y} \rvert}{\lVert \mathbf{y} \rVert_q} , \qquad \mathbf{y} \neq \mathbf{0} . $$ For the reverse inequality, note that if $a_j$ is such that $x_ja_j=|x_j|$, defining $y$ by $$ y_j=a_j\,|x_j|^{p-1} $$ gives us $$ \frac{|x\cdot y|}{\|y\|_q} =\frac{\displaystyle\sum_j |x_j|^p}{\Big(\sum_j |x_j|^{q(p-1)} \Big)^{1/q}} =\frac{\|x\|_p^p}{\|x\|_p^{p/q}}=\|x\|_p. $$

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  • $\begingroup$ 1. Why is $(\sum_j \lvert x_j \rvert^{q(p-1)})^{1/q} = \lVert \mathbf{x} \rVert_p^{1/q}$? $\endgroup$ Commented Oct 1, 2020 at 7:10
  • $\begingroup$ 2. For the reverse inequality, why are we allowed define $\mathbf{y}$? Shouldn't we be considering all $\mathbf{y} \in \mathbb{R}^n$? In fact I don't understand what the reverse inequality (equality) is for. $\endgroup$ Commented Oct 1, 2020 at 7:13
  • $\begingroup$ There was a $p$ missing on the right. And $q(p-1)=p$. $\endgroup$ Commented Oct 1, 2020 at 13:27

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