Proving that $\sum_{j=1}^{n}|z_{j}|^{2}\le 1$ when $|\sum_{j=1}^{n}z_{j}w_{j}| \le 1$ for all $\sum_{j=1}^{n}|w_{j}|^{2}\le 1$

The problem is like this:

Fix $n$ a positive integer. Suppose that $z_{1},\cdots,z_{n} \in \mathbb C$ are complex numbers satisfying $|\sum_{j=1}^{n}z_{j}w_{j}| \le 1$ for all $w_{1},\cdots,w_{n} \in \mathbb C$ such that $\sum_{j=1}^{n}|w_{j}|^{2}\le 1$. Prove that $\sum_{j=1}^{n}|z_{j}|^{2}\le 1$.

For this problem, I so far have that $|z_{i}|^{2}\le 1$ for all $i$ by plugging $(0, \cdots,0,1,0,\cdots,0)$ for $w=(w_{1},\cdots,w_{n} )$

Also, by plugging $(1/\sqrt{n},\cdots,1/\sqrt{n})$ for $w=(w_{1},\cdots,w_{n} )$ we could have $|z_{1}+\cdots+z_{n}|\le \sqrt{n}$

I wish we can conclude that $|z_{i}|\le 1/\sqrt{n}$ for each $i$.

Am I in the right direction?

Any comment would be grateful!

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3 Answers

To have a chance of success, one must choose a family $(w_j)_j$ adapted to the input $(z_j)_j$. If $z_j=0$ for every $j$, the result holds. Otherwise, try $w_j=c^{-1}\bar z_j$ with $c^2=\sum\limits_{k=1}^n|z_k|^2$.

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Thank you! That makes it precise! :-) –  Emily Sep 10 '12 at 15:49

Let $w_j:=\frac{\bar z_j}{\sqrt{\sum_{k=1}^n|z_k|^2}}$ (when $\sum_j|z_j|^2\neq 0$). Then $|w_j|^2=\frac{|z_j|^2}{\sum_{k=1}^n|z_k|^2}$ and we get $\sqrt{\sum_{j=1}^n|z_j|^2}\leq 1$, hence the result.

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Thank you! This gives me even more details! :-) –  Emily Sep 10 '12 at 15:50

If $z =0$ there is nothing to be done, so suppose $z \neq 0$.

Let $w =\frac{1}{\|z\|} z$. Note that $\|w\| = 1$. Then $|\sum_{j=1}^{n}z_{j}w_{j}| = \frac{1}{\|z\|}|\sum_{j=1}^{n}z_{j}^2| = \|z\| \leq 1$.

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