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Is it true that $$ \left|\sum_{j=1}^n \lambda_j e^{i\theta_j}\right|^2 = 1, $$ where $\sum_{i=j}^n |\lambda_j|^2 = 1$ if, and only if $|\lambda_{j^{\star}}| = 1$ for some $j^{\star}$?

Jean Marie
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SRichoux
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    Consider your LHS as the squared modulus of a (hermitian) dot product, then apply Cauchy-Schwarz. – Jean Marie Jan 26 '23 at 14:48
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    Actually I cannot work this out, are you considering the following dot product $\sum_j \overline{v_j} e^{i\theta_j} w_j$? – SRichoux Jan 26 '23 at 15:25
  • No, this one : $\sum_{j=1}^n \lambda_j \overline{e^{-i\theta_j}}$ – Jean Marie Jan 26 '23 at 17:19
  • But then applying Cauchy-Schwarz only yields the inequality: $\left|\sum_{j=1}^n \lambda_j e^{i\theta_j}\right|^2 \le (\sum_{j=1}^n |\lambda_j|^2) (\sum_{j=1}^n |e^{i\theta_j}|^2)$ – SRichoux Jan 27 '23 at 11:43

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In fact, the "only if" condition is false.

Here is a counterexample.

Let $\omega:=e^{i \pi/3}$. We have

$$1+\omega+\overline{\omega}=2 \ \ \iff \ \ \frac{1}{2\sqrt{2}}i+\frac{1}{2\sqrt{2}}(-i)+\frac12.1+\frac12.\omega+\frac12.\overline{\omega}=1,$$

otherwise said :

$$\frac{1}{2\sqrt{2}}e^{i \pi/2}+\frac{1}{2\sqrt{2}}e^{-i \pi/2}+\frac12.e^{i 0\pi}+\frac12.e^{i \pi/3}+\frac12.e^{-i \pi/3}=1,$$

none of the coefficients being with modulus $1$ (though condition $\sum_j \lambda_j^2 =1$ is verified).

Remark : Cauchy-Schwarz was in fact a deadend (sorry for that) because it gives an evident relationship :

$$\underbrace{\left|\sum_{j=1}^n \lambda_j e^{i\theta_j}\right|^2}_1 \le \underbrace{(\sum_{j=1}^n |\lambda_j|^2)}_1 (\sum_{j=1}^n |e^{i\theta_j}|^2)$$

is equivalent to $\sum_{j=1}^n |e^{i\theta_j}|^2 \ge 1$

which is always verified.

Jean Marie
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