For given $\sum_{j=1}^n | w_j|^2 \leq 1$ and $\sum_{j=1}^n | z_j|^2 \leq 1$ implies $ \left| \sum_{j=1}^n z_j w_j \right| \leq 1 $

Asked Sep 18 '16 at 05:08

Active Sep 18 '16 at 05:08

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Now little bit twist the problem ( I want to do with converse way of my previous question If $\sum\limits_{j=1}^n | w_j|^2 \leq 1$ implies $ \left| \sum\limits_{j=1}^n z_j w_j \right| \leq 1$, then $\sum\limits_{j=1}^n | z_j|^2 \leq 1$)

Situation is similar.

Fix $n$ a positive integer. Suppose that $z_1, \cdots z_n$, $w_1, \cdots w_n$ are complex numbers satisfying $\sum_{j=1}^n | w_j|^2 \leq 1$.and $\sum_{j=1}^n | z_j|^2 \leq 1$. then is following true \begin{align} \left| \sum_{j=1}^n z_j w_j \right| \leq 1 \end{align}

It seems that above equality true. $i.e$, pick $w_j = \bar{z}_j$ then it became trivial, but i am not sure for proving such inequality in this fashion.

edited Apr 13 '17 at 12:20

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asked Sep 18 '16 at 05:08

phy_math

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It's not using the Cauchy-Schwarz inequality? – iam_agf Sep 18 '16 at 05:09
4

It is exactly the Cauchy-Schwarz inequality. – Robert Israel Sep 18 '16 at 05:25

For given $\sum_{j=1}^n | w_j|^2 \leq 1$ and $\sum_{j=1}^n | z_j|^2 \leq 1$ implies $ \left| \sum_{j=1}^n z_j w_j \right| \leq 1 $

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