Bound on truncated trigonometric polynomial

Question

Let $p(theta)$ be a real trigonometric polynomial of degree $N>0$
$$
p(theta) = sum_{j=-N}^N a_{|j|}e^{text ijtheta}
$$
and for any $0le s<N$ define the left trucation of $p$ as
$$
p_s(theta):= sum_{j=-s}^N a_{|j|}e^{text ijtheta}.
$$
Is it true that, for any $s$,
$$
|p(theta)|_infty le 1 implies |p_s(theta)|_infty le 1?
$$
(Here the infinity norm is just the sup of the absolute value of the funtion on $[-pi,pi]$)
I verified by hand that it is true for $N=1$, but I am quite sure there exists a counterexample..

Chrystomath · Answer

Take $N=3$ and a polynomial $$p(theta)=a+be^{itheta}+be^{-itheta}+ce^{2itheta}+ce^{-2itheta}=a+2bcostheta+2ccos2theta$$
$$p_1(theta)=a+be^{itheta}+be^{-itheta}+ce^{2itheta}=a+2bcostheta+ccos2theta+icsin2theta$$
Basically, the question is asking whether by transferring a $cos2theta$ to $isin2theta$, is it possible to increase the maximum value of the polynomial. Simplifying drastically for intuition, let's seek real numbers $x,y$, such that $|x+y|le1$, $|x+iy|>1$. Expanding out shows that we need $y<0$, $x^2+y^2>1>(x-y)^2$.
A quick search with $a=1=b$ and negative $c$ gives this counterexample, $$p(theta)=alpha(1+costheta-0.2cos2theta)$$
$$p_1(theta)=alpha(1+costheta-0.1(cos2theta+isin2theta))$$ where $alpha$ is the a normalizing factor, in this case $alpha=1/1.85$.
Here is a plot of the two functions against $theta$:

Joshua P. Swanson · Answer

No, it fails for $p(theta) = P(e^{itheta})$ where $P(x) = 1/x^3 + 1/x^2 - 1/x + 1 - x + x^2 + x^3$ and $s=2$. Expecting it to be rarely true, I just did trial and error a few times.

Bound on truncated trigonometric polynomial

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