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Problem statement:

Define $p(t) = \sum\limits_{j=-N}^{N}c_{j}e^{ijt}$ be a real-valued trigonometric polynomial. Suppose there exists an $x_{0}\in\mathbb{R}$ such that $p(x_{0}) = \|p\|_{\infty}$.

I need to show that $\|p'\|_{\infty}\leq N\cdot \|p\|_{\infty}$.

But this is actually the last part of a larger problem.

Letting $M=\|p\|_{\infty}$, part one was to show that $p(x_{0} + t)\geq M\cdot \cos(Nt)$ for every $t\in \left[-\frac{\pi}{N},\frac{\pi}{N}\right]$.

This was finally solved (with much rejoicing) in roughly the following way:

Step 1: For any choice of constants $c_{j}$, the equation $p(t) = 0$ can have at most $2^{N}$ solutions (counting multiplicity).

Step 2: The function $g(t) = p(x_{0} + t) - M\cdot \cos(Nt)$ can be written in the same form as $p(t)$ by adjusting $c_{N}$ and $c_{-N}$. Thus the equation $g(t) = 0$ can have at most $2^{N}$ roots counting multiplicity.

Step 3: Assuming the desired inequality fails for some $t\in\left[-\frac{\pi}{N},\frac{\pi}{N}\right]$, we obtain $2^N + 1$ roots to the equation $g(t) = 0$, which is a contradiction.

So to summarize more precisely, the question says to use the result

$p(x_{0} + t)\geq M\cdot \cos(Nt)$ for every $t\in \left[-\frac{\pi}{N},\frac{\pi}{N}\right]$

to obtain

$\|p'\|_{\infty}\leq N\cdot \|p\|_{\infty}$.

Now what has left me and my peer banging our heads against the wall is the fact that we can't seem to get any insight on $\|p'\|_{\infty}$. By differentiating $p$ term by term (and then taking the absolute value) we could not get a factor of $N$ to appear without losing too much information about the size of $|p'(t)|$.

Any insight or suggestions on this would be greatly appreciated!

share|cite|improve this question
Assuming that you consider the uniform norm over $[-\pi,\pi],$ this is simply Bernstein's inequality for the derivatives. A good reference would be P.Borwein and T.Erdelyi Polynomial and Polynomial inequalities. – leshik Oct 14 '11 at 5:23
Yes the question mentions later that this is called Bertein's inequality and that it holds also for complex valued trigonometric polynomials. I will see if I can find that reference. Thanks. – roo Oct 14 '11 at 13:25

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