# Fourier coefficients zero implies integral with trigonometric polynomial zero

I was studying this theorem :

Suppose that $f$ is an integrable function on the circle with $\widehat{f}(n) = 0$ for all n ∈ Z. Then $f(θ_0)$ = 0 whenever $f$ is continuous at the point $θ_{0}$.

from the Princeton Lectures in Analysis, Fourier Series. In the proof the author then says that since the Fourier coefficients of $f$ are zero for all n and $p_k$ is a trigonometric polynomial, we must have $\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\ d\theta = 0$ for all $k$.

I have not been able to understand why this must be so. I started with the condition:

$\widehat{f}(n) = \frac{1}{L}\int_{a}^{b} f(\theta)e^{\frac{-2\pi inx}{L}} \ d\theta = 0 \ \forall n \in \mathbb{Z}$.

But I couldn't prove that this implies that any integral with trigonometric polynomial is also zero. How can this be seen?

I'm afraid I don't have the book, so I don't know what $p_k(\theta)$ is!
Fortunately, this doesn't matter. By definition, a trigonometric polynomial is a finite sum of the form, $$p(\theta) = \sum_{n = -N}^N c_n e^{i n \theta}.$$ Given that $\int_{-\pi}^\pi d\theta \ f(\theta) e^{i n \theta} = 2\pi \hat f(-n) = 0$ for all $n \in \mathbb N$, we have $$\int_{-\pi}^\pi d\theta \ f (\theta) p(\theta) = \int_{-\pi}^\pi d\theta \sum_{n = -N}^N c_n f(\theta) e^{in \theta} = \sum_{n = -N}^Nc_n \int_{-\pi}^\pi d\theta \ f(\theta) e^{in\theta} =0.$$ (It is legitimate to take the sum out of the integral because the sum is over finitely many values of $n$.)