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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?

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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 $n$.)

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  • $\begingroup$ I did not understand how you changed the limits of integration from pi to infinity. $\endgroup$ – erdoswiles May 11 '17 at 15:26
  • $\begingroup$ @erdoswiles Sorry, that was a typo! $\endgroup$ – Kenny Wong May 11 '17 at 15:41
  • $\begingroup$ Oh ok! Thanks for the answer. $\endgroup$ – erdoswiles May 11 '17 at 15:43

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