The Book States 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 $f(\theta_{o})=0$ whenever $f$ is continuous at the point $\theta_{o}$.
I do not understand exactly why he choose $\delta \leq \frac{\pi}{2}$ and why this leads to $f(\theta) > \frac{f(0)}{2}$ and what is the importance of that $f(\theta) > \frac{f(0)}{2}$?Also why he choose the trigonometric polynomial in that way and what is the importance of choosing $\eta < \delta$, so that $p(\theta) \geq 1 + \frac{\epsilon}{2}$? why he let $p_{k}(\theta) = [p(\theta)]^{k}$ in part $2$ and $3$ of the proof. Could anyone explain this for me please?$
They choose $\delta \leq \frac{\pi}{2}$ to make sure that $p(\theta)$ is non-negative whenever $|\theta|<\delta$.
All they want is a neighborhood of $0$ on which $f>0$. The value $\frac{f(0)}{2}$ is just a simple choice.
That they can choose such a $\delta$ is a consequence of the continuity of $f$ at $0$ : $$ \exists\, \delta > 0, \forall\, \theta : |\theta| < \delta \implies f(0)-\frac{f(0)}{2} =\frac{f(0)}{2} < f(\theta) $$
They would like to choose $p$ such that $p>1$ on $(-\delta,\delta)$. But that's impossible because $p$ is continuous and they chose $\epsilon$ such that $|p(\theta)| < 1$ whenever $\delta \leq \theta \leq \pi$. That's why they need to restrict to $(-\eta,\eta) \subset (-\delta,\delta)$.
The value $1+\frac{\epsilon}{2}$ is unimportant as long as it is $> 1$. They could not simply choose $1+\epsilon$ because $p(0) = 1+\epsilon$.
They saw that for any trigonometric polynomial $p$, $p^k$ is also a trigonometric polynomial. So they really started with $p_k := p^k$ in mind and chose $p$ so that $p^k$ peaks at $0$. To do so they used the fact that $a^k$ tends to $\infty$ if $a > 1$ and to $0$ if $|a| < 1$.
While the proof is a bit technical, the idea is simple enough:
In the proof they take care of the details on how to construct $p_k$ precisely and how to show rigorously $(\star)$ happens.