I'm reading through Stein & Shakarchi's book on Fourier Analysis on my own, and have a question about the proof of the following theorem: Suppose that $f$ is an integrable function on the circle with $\hat{f}(n)=0$ for all $n \in \mathbb{Z}$. Then $f(\theta_{0})=0$ whenever f is continuous at the point $\theta_{0}$.
They prove the theorem by contradiction – assuming that $f$ satisfies the hypotheses, $\theta_{0}=0$, and $f(0)>0$. They then construct a family of trigonometric polynomials $\{p_{k}\}$ that peak at $0$, resulting in $$\int p_{k}(\theta)f(\theta)\; d\theta \rightarrow \infty \;\; \text{as}\;\; k \rightarrow \infty.$$ Here, the peak functions are constructed in the following manner: Since $f$ is continuous at $0$, we can choose $0 < \delta \leq \pi/2$, so that $f(\theta)>f(0)/2$ whenever $|\theta|< \delta$. Let $p(\theta) = \epsilon + \cos\theta$ where $\epsilon >0$ is chosen so small that $|p(\theta)|< 1-\epsilon/2$, whenever $\delta \leq |\theta| \leq \pi$. Then choose a positive $\eta$ with $\eta < \delta$, so that $p(\theta) \geq 1 + \epsilon/2$, for $|\theta| < \eta$. Finally, we define $p_{k}(\theta) = [p(\theta)]^k$.
It's not exactly clear to me how this gives a contradiction, since the Fourier coefficients of $f$ are given by $$\int f(\theta)e^{-in\theta}\; d\theta.$$ Could someone explain how the first integral implies there is a nonzero Fourier coefficient?
Best Answer
If the Fourier coefficients of f are all 0 then the integral of f multiplied by any trigonometric polynomial is 0. This follows by just taking linear combinations.