0
$\begingroup$

Let $z_1, z_2, \ldots, z_n$ and $w_1, w_2,\ldots,w_n$ be some complex numbers, and $n$ a positive integer greater than $1$. If $$|z_1 \pm z_2 \pm \ldots \pm z_n| \le |w_1 \pm w_2 \pm \ldots \pm w_n|$$ Prove $$|z_1|^2 + |z_2|^2 + \ldots + |z_n|^2 \le |w_1|^2 + |w_2|^2 + \ldots + |w_n|^2$$

Let $z_k = a_k + ib_k$ and $w_k = c_k + id_k$. The two relations become: $$\left(a_1 + a_2 + \ldots + a_n\right)^2 + \left(b_1 + b_2 + \ldots + b_n\right)^2 \le \left(c_1 + c_2 + \ldots + c_n\right)^2 + \left(d_1 + d_2 + \ldots + d_n\right)^2$$ $$a_1^2 + b_1^2 + a_2^2 + b_2^2 + \ldots + a_n^2 + b_n^2 \le c_1^2 + d_1^2 + c_2^2 + d_2^2 + \ldots + c_n^2 + d_n^2$$

Can you give me a hint, please? Thanks!

Maybe this helps you (it's the first part of the problem): $$|z + w|^2 + |z - w|^2 = 2|z|^2 + 2|w|^2$$

$\endgroup$
  • $\begingroup$ Induction should be better in general. $\endgroup$ – Nosrati Jul 23 '18 at 11:40
  • $\begingroup$ @mrtaurho Your inequality is false: take $n=2,z_1=1,z_2=i$ $\endgroup$ – Kavi Rama Murthy Jul 23 '18 at 11:55
  • $\begingroup$ Yeah, I just realized for myself. I am sorry for the wrong hint. $\endgroup$ – mrtaurho Jul 23 '18 at 11:58
  • 3
    $\begingroup$ Hint: $$\sum\limits_{\epsilon_2,\cdots,\epsilon_n \in \{-1,1\}}|z_1 + \epsilon_2 z_2 + \cdots + \epsilon_n z_n|^2 = 2^{n-1}\left(|z_1|^2 + \cdots + |z_n|^2\right)$$ $\endgroup$ – achille hui Jul 23 '18 at 12:00
2
$\begingroup$

For $n=2$ we have $|z_1+z_2|^{2}+|z_1-z_2|^{2}=2|z_1|^{2}|+2|z_2|^{2}$. Using a similar formula for the $w$'s we get the result immediately. The above formula for $n=2$ generalizes to any $n$ and (with the constant $2$ on the right side replaced by $2^{n}$) and so the same proof works for any $n$. In all cases what you do is write down the inequality for all possible plus minus signs and add them up.

$\endgroup$
  • $\begingroup$ The constant on the right side is replaced by $2^n$, not by $n$. $\endgroup$ – uniquesolution Jul 23 '18 at 11:57
  • $\begingroup$ You are right. I have edited the answer. $\endgroup$ – Kavi Rama Murthy Jul 23 '18 at 11:58
  • $\begingroup$ the factor in RHS is $2^{n-1}$, not $2^n$. $\endgroup$ – achille hui Jul 23 '18 at 12:03
  • $\begingroup$ @achillehui - You are wrong. The generalized parallelogram identity says that the average of $\sum_{i=1}^n\|\pm v_i\|^2$ over all possible signs $\pm 1$ is exactly $\sum_i\|v_i\|^2$. There are $2^n$ possible choices of signs in $n$ dimensions. $\endgroup$ – uniquesolution Jul 23 '18 at 12:07
  • $\begingroup$ @uniquesolution The first $z_1$ doesn't have a -ve sign attached to it. $\endgroup$ – achille hui Jul 23 '18 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.