Bound on truncated trigonometric polynomial

Question

Let $p(\theta)$ be a real trigonometric polynomial of degree $N>0$ $$ p(\theta) = \sum_{j=-N}^N a_{|j|}e^{\text ij\theta} $$ and for any $0\le s<N$ define the left trucation of $p$ as $$ p_s(\theta):= \sum_{j=-s}^N a_{|j|}e^{\text ij\theta}. $$

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..

Answer 1

No, it fails for $p(\theta) = P(e^{i\theta})$ 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.

Answer 2

Take $N=3$ and a polynomial $$p(\theta)=a+be^{i\theta}+be^{-i\theta}+ce^{2i\theta}+ce^{-2i\theta}=a+2b\cos\theta+2c\cos2\theta$$ $$p_1(\theta)=a+be^{i\theta}+be^{-i\theta}+ce^{2i\theta}=a+2b\cos\theta+c\cos2\theta+ic\sin2\theta$$

Basically, the question is asking whether by transferring a $\cos2\theta$ to $i\sin2\theta$, 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, $$p(\theta)=\alpha(1+\cos\theta-0.2\cos2\theta)$$ $$p_1(\theta)=\alpha(1+\cos\theta-0.1(\cos2\theta+i\sin2\theta))$$ where $\alpha$ is the a normalizing factor, in this case $\alpha=1/1.85$.

Here is a plot of the two functions against $\theta$: