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Suppose you have two sequences of complex numbers $a_i$ and $b_i$ indexed over the integer numbers such that they are convergent in $l^2$ norm and $a$ has norm greater than $b$ in the sense $$\infty>\sum_i|a_i|^2 \ge \sum_i|b_i|^2.$$ Suppose moreover they are uncorrelated over any time delay, meaning $$\sum_i a_i\overline{b_{i-n}} = 0 \quad \forall n\in \mathbb Z.$$

Is it true that the polinomial $a(z) = \sum_i a_iz^{-i}$ is greater in absolute value than $b(z) = \sum_i b_iz^{-i}$ for any unit norm complex number $z$?

I thought some algebraic trick would lead me to the answer, but I get stuck at proving $$ \sum_i a_i\overline{a_{i-n}} z^{-n} + \overline{a_i}{a_{i-n}}z^n \ge \sum_i b_i\overline{b_{i-n}} z^{-n} + \overline{b_i}{b_{i-n}}z^n $$

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$\newcommand{scal}[2]{\left\langle {#1};{#2}\right\rangle}$Since we're interested only in $z\in S^1$, I'll call $z=e^{ix}$ and I'll work in $L^2_{\Bbb C}[0,2\pi]$ (since you also don't seem to be actually interested in polynomials). Call $\Xi(x)=\begin{cases}0&\text{if }x\le 0\lor x\ge1\\ \exp\frac1{x^2-x}&\text{if }0<x<1\end{cases}$ and call $\Phi(x)=\left(\frac1{2\pi}\int_0^1\lvert\Xi(t)\rvert^2\,dt\right)^{-1/2}\Xi(x)$. Then consider the Fourier series in $L^2[0,2\pi]$ \begin{align}&a(e^{ix}):=5\Phi(x)=\sum_{k\in\Bbb Z} a_k e^{ikx}\\ &b(e^{ix}):=\Phi(x-1)=\sum_{k\in\Bbb Z} b_k e^{ikx}\end{align}

You have \begin{align}&\sum_{k\in\Bbb Z} a_k\overline b_{k-n}=\frac5{2\pi}\int_0^{2\pi}\overline{e^{nix}\Phi(x-1)}\Phi(x)\,dx=0\\ &\sum_{k\in\Bbb Z}\lvert a_k\rvert^2=\frac{25}{2\pi}\int_0^{2\pi}\lvert \Phi(x)\rvert^2\,dx=25\\ &\sum_{k\in\Bbb Z}\lvert b_k\rvert^2=\frac{1}{2\pi}\int_0^{2\pi}\lvert \Phi(x-1)\rvert^2\,dx=1\end{align}

However, $\lvert a(e^{ix})\rvert=0<\lvert b(e^{ix})\rvert$ for $1< x<2$.

Notice that inequalities of norm and absolute value hold eventually for the partial sums, because these Fourier series converge uniformly on $[0,2\pi]$. However, the condition of "uncorrelation over any time delay" doesn't hold and, in fact, it can't hold for non-zero trigonometric polynomials.

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  • $\begingroup$ yeah, it's enough to take two functions with disjoint supports over the unit circle. And yes, both function must have infinitely many non zero Fourier coefficient, otherwise $b$ is zero $\endgroup$
    – Exodd
    Mar 18, 2022 at 20:15

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