Let $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$. Then $$|\sum\limits_{i = 1}^n x_i y_i| \leq ||x||_p ||x||_q, \;\; \forall x, y \in \mathbb{R}^n.$$ I have to prove it considering $$u = \frac{x}{||x||_p} \;\; \text{and} \;\; v = \frac{y}{||y||_q}$$ and using $\bf{Young's}$ inequality. Can someone, please, give me a hint?
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Hint: proofs of this inequality are everywhere online. – Jun 20 '15 at 06:54
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I didn't find a proof that considers $u = \frac{x}{||x||_p}, u = \frac{y}{||y||_q}, $ – g.pomegranate Jun 20 '15 at 06:56
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I have to use Young's inequality, not Jensen's inequality! – g.pomegranate Jun 20 '15 at 06:57
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I don't know how. I need a hint – g.pomegranate Jun 20 '15 at 06:58
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1Here's a relevant Wiki link. – Viktor Vaughn Jun 20 '15 at 07:04
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Hint: Write $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$.
- By Young's inequality, $$|u_i \cdot v_i| \leq \frac{|u_i|^p}{p} + \frac{|v_i|^q}{q}. \tag{1}$$ Rewrite this inequality using the definition of $u$ and $v$.
- Sum $(1)$ over $i=1,\ldots,n$. Deduce that $$\sum_{i=1}^n |u_i v_i| \leq 1.$$
- Conclude.
saz
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1I don't understand why $\sum\limits_{i=1}^n |u_i v_i| \leq 1$. We have from (1) $$\sum\limits_{i=1}^n |u_i v_i| \leq \frac{1}{p} | \frac{x_i}{||x||_p}|^p + \frac{1}{q} |\frac{y_i}{||y||_q}|^q.$$ And then? – g.pomegranate Jun 20 '15 at 08:15
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Yes. In the right side we have $$\frac{1}{p} ||u||_p^p + \frac{1}{q} ||v||_q^q = \frac{1}{p} + \frac{1}{q} = 1$$. Thank you! – g.pomegranate Jun 20 '15 at 08:23