Fejer-Riesz theorem

Let $x_0, x_1, x_2, \ldots , x_n$ be arbitrary complex numbers, $x_0 \neq 0$, $x_n \neq 0$. The expression

represents a real non-negative trigonometric polynomial of order precisely $n$. Calculate the coefficients of this trigonometric polynomial.

If the real trigonometric polynomial of $n$th order

can assume only non-negative values for all real $\vartheta$, then it may be represented in the following form:

where $h(z) = x_0 + x_1 z + x_2 z^2 + \cdots + x_n z^n$ is a polynomial of $n$th degree.

If $g(\vartheta )$ ($\vartheta$ real) denotes a real non-negative cosine polynomial, then one can always find a polynomial $h(z) = x_0 + x_1 z + x_2 z^2 + \cdots + x_n z^n$ with only real coefficients such that we have

Show that the representation of a real non-negative trigonometric polynomial as given in problem 40 for real $\vartheta$ is in general possible in several ways.

Show that in problem 40 there always exists a representation such that $h(z)$ satisfies the following conditions:

Every polynomial of degree $n$ which can assume only real and non-negative values for $-1 \leqq x \leqq 1$ may be represented in the form

where $A(x)$ and $B(x)$ are polynomials of $n$th and $(n-1)$th degrees, respectively, with only real coefficients.

Let

($\vartheta$ real) be a real non-negative trigonometric polynomial of $n$th order, and let the constant term in $g(\vartheta )$ be equal to $1$, i.e.

Then

With the notation of problem 50, we have

With the notation of problem 50, we have

Obtain an analogous description of those rational functions which are positive on $T$. Hint: Such a function must have the same number of zeros as poles in $U$. Consider products of factors of the form

where $\left|\alpha \right| < 1$ and $\left|\beta \right| < 1$.

Suppose $f$ is a trigonometric polynomial,

and $f(\theta ) > 0$ for all real $\theta$. Prove that there is a polynomial $P(z) = c_{0}+c_{1} z+\cdots +c_{n} z^{n}$ such that

Hint: Apply Exercise 4 to the rational function $\Sigma a_{k} z^{k}$. Is the result still valid if we assume $f(\theta ) \geq 0$ instead of $f(\theta ) > 0$?