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Given $p \in (1,\infty)$ and $\frac{1}{p}+\frac{1}{q}=1$ and $x,y \in \mathbb{R}^n$, show that:

$$\sum_{j=1}^n |x_jy_j| \leq |x|_p|y|_q$$ where $|x|_p :=\left(\sum_{j=1}^n |x|^p \right)^{1/p}$

I believe that Young's inequality can be used here. If I can reduce the given inequality somehow to Young's inequality then I would have proved Holder's inequality. Since log function is monotonically increasing:

$$\log{\sum_{j=1}^n |x_jy_j|} \leq \log{|x|_p|y|_q}$$ $$\log{\sum_{j=1}^n |x_jy_j|} \leq \frac{1}{p}\log{\sum_{j=1}^n |x_j|^p} + \frac{1}{q}\log{\sum_{j=1}^n |y_j|^q}$$

Now Young's inequality is: $uv \leq \frac{1}{p}u^p+\frac{1}{q}v^q$. So if I can show that when (define): $u^p := \log{\sum_{j=1}^n |x_j|^p}$ multiplied by $v^q:=\log{\sum_{j=1}^n |y_j|^q}$ is $\log{\sum_{j=1}^n |x_jy_j|}=uv$, then I have proved Holder inequality. But no matter how I approach this later part, I get to here:

$$uv=\frac{1}{p}\frac{1}{q}\log{\sum_{j=1}^n |x_j|^p}\log{\sum_{j=1}^n |y_j|^q}$$

And I cannot proceed further.

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  • $\begingroup$ HINT: start from the Young inequality and in some way you must end in the Holder inequality, the other direction to construct a proof (as you are trying) seems more complicated. $\endgroup$ – Masacroso Mar 11 '17 at 14:48
  • $\begingroup$ good point! I have not thought about that. $\endgroup$ – i squared - Keep it Real Mar 11 '17 at 14:49
  • $\begingroup$ I must advice that the proof that I know is not so easy after all... its needed some specific setup in the Young inequality with each coordinate. After, when you sum all these Young inequalities you get the Holder inequality. $\endgroup$ – Masacroso Mar 11 '17 at 15:05
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Hint: Assume $x$ and $y$ non-zero. Let $\widehat{x}_i=x_i/|x|_p$ and $\widehat{y}_i=y_i/|y|_q$. Then apply Young's inequality: $$ \sum_i |\widehat{x}_i \widehat{y}_i| \leq \frac{1}{p} \sum_i |\widehat{x}_i|^p + \frac{1}{q} \sum_i |\widehat{y}_i|^q = \frac1p + \frac1q =1$$ and you get the desired inequality for $\sum_i |x_iy_i|$.

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