# majorization of polynomials over the unit circle

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 $$\sum_i a_i\overline{a_{i-n}} z^{-n} + \overline{a_i}{a_{i-n}}z^n \ge \sum_i b_i\overline{b_{i-n}} z^{-n} + \overline{b_i}{b_{i-n}}z^n$$

$$\newcommand{scal}[2]{\left\langle {#1};{#2}\right\rangle}$$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 $$\Xi(x)=\begin{cases}0&\text{if }x\le 0\lor x\ge1\\ \exp\frac1{x^2-x}&\text{if }0 and call $$\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.
• 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 Mar 18, 2022 at 20:15