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Do you know proofs or references for the following inequality:

There exists a positive constant $C>0$ such that for any complex numbers $a_1,\ldots,a_n$

$$ |a_1|+\cdots+|a_n| \leq C\sup_{z_1^3=1,\ldots,z_n^3=1 } |a_1z_1+\cdots + a_n z_n| $$ where the supremum is taken over the complex numbers $z_1,\ldots,z_n$ such that $z_1^3=1,\ldots,z_n^3=1$?

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2 Answers 2

up vote 5 down vote accepted

The strategy is to rotate the $a$ values so that they are roughly pointing in the same direction. More formally, for each complex $a$ choose a third root of unity $z$ so that the argument $\theta $ of $az$ lies between plus and minus $\pi/3$.

Then \begin{eqnarray*} |a_1z_1+\cdots +a_nz_n|&\geq&|\mbox{Re}(a_1z_1+\cdots +a_nz_n)| \cr &=&|a_1z_1|\cos(\theta_1)+\cdots +|a_nz_n|\cos(\theta_n) \cr &\geq &(|a_1|+\cdots +|a_n|)\cos(\pi/3)\cr &=&{1\over 2}\ (|a_1|+\cdots +|a_n|)\cr \end{eqnarray*}

This gives your result with $C=2$.

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It seems to me that we must take the the "furthest away from the argument" and not the "closest argument". What do you think about this? –  Zouba Nov 11 '11 at 15:46
    
You are right, I didn't describe it well. I will modify my answer. –  Byron Schmuland Nov 11 '11 at 15:51
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Notice that the mapping $\|\cdot\|_* \colon \mathbb{C}^n \to \mathbb{R}$ given by $$\|(a_1,\dots,a_n)\|_* = \sup_{z_1^3=z_2^3=\dots=1} |a_1z_1 + \dots + a_n z_n|$$ defines a norm. The result then follows by the equivalence of norms in finite dimensional vector spaces.

The triangle inequality is easily seen to be true: Just use the ordinary triangle inequality. To prove that if $\|(a_1,\dots,a_n)\|_* = 0$ then $(a_1,\dots,a_n) = 0$ notice that if one of the coordinates (let's say $a_1$) wasn't $0$, then there would be three different numbers: $a_1 z_1 + a_2 + \dots + a_n$, $a_1 z_2 + a_2 + \dots + a_n$ and $a_1 z_3 + a_2 + \dots + a_n$ one of which is non-zero and thus the sup we get would be non-zero also.

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Does the op want a constant independent of $n$, though? –  David Mitra Nov 11 '11 at 15:14
    
Yes, the constant $C$ must be independent of $n$. –  Zouba Nov 11 '11 at 15:21
    
@David Mitra: A good point. –  J. J. Nov 11 '11 at 15:23
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