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I want to show following.

Fix $n$ a positive integer. Suppose that $z_1, \cdots z_n$ are complex numbers satisfying \begin{align} \left| \sum_{j=1}^n z_j w_j \right| \leq 1 \end{align} for all $w_1, \cdots, w_n\in \mathbb{C}$ such that $\sum_{j=1}^n | w_j|^2 \leq 1$. then, $\sum_{j=1}^n | z_j|^2 \leq 1$.

Any idea or hint will be helpful.

adopting from hint i got for \begin{align} |\sum_{j=1}^n z_j w_j | = |\sum_{j=1}^n \lambda |z_j|^2| = |\lambda| \sum_{j=1}^n |z_j|^2 \leq1 \end{align} and form \begin{align} \sum_{j=1}^n |w_j|^2 = \sum_{j=1}^n |\lambda|^2 |z_j|^2 = |\lambda|^2 \sum_{j=1}^n |z_j|^2 \leq1 \end{align} Then the equality seems to be depend on value of $\lambda$. $i.e$, i have

\begin{align} \sum_{j=1}^n |z_j|^2 \leq \frac{1}{|\lambda|}, \qquad \sum_{j=1}^n |z_j|^2 \leq \frac{1}{|\lambda|^2} \end{align} which i don't know how to interpret

Did
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phy_math
  • 6,448

2 Answers2

3

Suppose that $$ \sum_{j=1}^n|z_j|^2=\lambda^2\gt1\tag{1} $$ for real $\lambda\gt1$. Define $$ w_j=\frac{\overline{z}_j}\lambda\tag{2} $$ Then $$ \sum_{j=1}^n|w_j|^2=1\tag{3} $$ However, $$ \left|\,\sum_{j=1}^nz_jw_j\,\right|=\lambda\gt1\tag{4} $$ Thus, if $$ \left|\,\sum_{j=1}^nz_jw_j\,\right|\le1\tag{5} $$ for all $w_j$ so that $(3)$ holds, we must have $$ \sum_{j=1}^n|z_j|^2\le1\tag{6} $$

robjohn
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Hint: Try $w_j=\lambda \overline{z_j}$.

Hagen von Eitzen
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    adopting your hint i got for $|\sum_{j=1}^n z_j w_j | = |\sum_{j=1}^n \lambda |z_j|^2| \leq1$, and form $\sum_{j=1}^n |w_j|^2 = \sum_{j=1}^n |\lambda|^2 |z_j|^2 \leq1$, Then the equality seems to be depend on value of $\lambda$. – phy_math Sep 18 '16 at 04:11