If $\frac{1}{p}+\frac{1}{q}=1$ and $x,y \in \mathbb{F}^m, m>1$, then how to use Hölder's Inequality to prove $\lVert {x}\rVert_p = \max \{ |\langle x,y \rangle|: \lVert y \rVert_q \leq 1\}$？
$\lVert {x}\rVert_p = \max \{ |\langle x,y \rangle|: \lVert y \rVert_q \leq 1\}$
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1 Answers
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I assume $x,y$ are finite dimensional vectors, if not, you need to specify what they exactly are.
Firstly, Holder's inequality gives $$\|x\|_p\ge\|x\|_p\|y\|_q\ge \langle x,y\rangle,\forall \|y\|_q \le 1$$ Then, simply take $y = \frac{x}{\|x\|_q}$.
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0If you take $y=\frac{x}{\lVert x \rVert_q}$, then $\lVert y \rVert_q=1$, but how you achieve the equality in $\lVert x \rVert_p \lVert y \rVert_q \geq \langle x,y \rangle$? – 2017-01-28