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$$\frac{\| x \|_1}{\| x \|_2} \le E$$

$$\frac{\| x \|_1}{\| x \|_2} \le \frac{n\| x \|_{\infty}}{\| x \|_2}$$

Considering :

$$\frac{\| x \|_{\infty}}{\| x \|_2} \le 1$$

Because:

$$\| x \|_{\infty} \le \cdots \le \| x \|_3 \le \| x \|_2 \le \| x \|_1 $$

So:

$$ \frac{n\| x \|_{\infty}}{\| x \|_2} \le n$$

$$E=n$$

Is it correct?

Thanks

  • 3
    Yes, it is! But I guess, the question is not properly stated. For instance, the inequality still holds even if I say, something like $E=n^2$? Think about it! – 2017-02-27
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    It depends how you define $E$. If you define $E:=\inf\{|y|\in\Bbb R:\|x\|_1\le \|x\|_2\quad\forall x\in\Bbb R\}$ then your answer is not enough. But of course is true that if you want some constant $K$ such that $\|x\|_1\le K\|x\|_2,\forall x\in\Bbb R$ then $K=n$ holds. – 2017-02-27
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    Think about how you could employ the Cauchy-Schwarz inequality in this situation. – 2017-02-27

1 Answers 1

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Assume $x\in{\mathbb R}^n$. Then $$\|x\|_1=\sum_{i=1}^n 1\cdot |x_i|\leq\sqrt{\sum_{i=1}^n 1^2}\cdot\sqrt{\sum_{i=1}^n x_i^2}=\sqrt{n}\>\|x\|_2$$ with equality sign iff all $|x_i|$ are equal. The "optimal" value for $E$ therefore is $\sqrt{n}$.