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!