Difficulty in understanding a part in a proof from Stein and Shakarchi Fourier Analysis book.

Theorem 2.1 : Suppose that $f$ is an integrable function on the circle with $\hat f(n)=0$ for all $n \in \Bbb Z$. Then $f(\theta_0)=0$ whenever $f$ is continuous at the point $\theta_0$.

Proof : We suppose first that $f$ is real-valued, and argue by contradiction. Assume, without loss of generality, that $f$ is defined on $[-\pi,\pi]$, that $\theta_0=0$, and $f(0) \gt 0$.

Since $f$ is continuous at $0$, we can choose $ 0\lt \delta \le \frac \pi2$, so that $f(\theta) \gt \frac {f(0)}2$ whenever $|\theta| \lt \delta$. Let $$p(\theta)=\epsilon + \cos\theta,$$ Where $\epsilon \gt 0$ is chosen so small that $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$. Then, choose a positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1 + \frac \epsilon2$, for $|\theta| \lt \eta$. Finally, let $p_k(\theta)=|p(\theta)|^k$, and select $B$ so that $|f(\theta)| \le B$ for all $\theta$. This is possible since $f$ is integrable, hence bounded.

By construction, each $p_k$ is a trigonometric polynomial, and since $\hat f(n)=$ for all $n$, we must have $\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\,d\theta=0$ for all $k$.

I understood the first paragraph clearly. But the rest is not making it's way into my head.

1. In the beginning of second paragraph, how does the given range works for choosing $\delta$? If the continuity is used to get the range, then how?
2. How can we choose $\epsilon$ so small such that, $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$?
3. How can we choose positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1+ \frac \epsilon2$, for $|\theta| \lt \eta$.
4. Why do we must have $\int_{-\pi}^{\pi}f(\theta)p_k(\theta)\,d\theta=0$ for all $k$?

Solution 1:

1. Continuity tells us we can choose $\delta>0$ so that $|f(\theta) - f(0)|< \frac{f(0)}{2}$ if $|\theta - 0| < \delta$ (which in particular implies $f(\theta) > \frac{f(0)}{2}$). Once the existence of such a $\delta$ is established, we can assume it is as small as we need; in particular, we're free to take it to be less than $\pi/2$, without affecting the inequality $f(\theta) > \frac{f(0)}{2}$. If you want to see this more concretely, then take the first $\delta$ that we obtained through continuity, and set $\delta' = \min(\delta,\frac{\pi}{2})$. Then $0<\delta < \frac{\pi}{2}$, and $|\theta|<\delta'$ implies $|\theta|<\delta$ implies $f(\theta) > \frac{f(0)}{2}$.

2. Here we're using the fact that $\cos\theta < 1$ when $\theta$ is away from $0$. To make this quantitative, $\cos\theta$ ranges from a maximum of $\cos\delta$ to a minimum of $-1 = \cos\pi$ on the set $\{\delta \leq |\theta|\leq\pi\}$. Therefore $$ \epsilon - 1 \leq p(\theta) \leq \epsilon + \cos\delta $$ when $\delta \leq |\theta| \leq \pi$. On the left, we can throw out an extra $\epsilon/2$: $$ \frac{\epsilon}{2} - 1 \leq \epsilon - 1 \leq p(\theta). $$ On the right, we have $$ \epsilon + \cos\delta = \epsilon + 1 - (1-\cos\delta) = \epsilon + 1 - \lambda. $$ Note $\lambda > 0$. If we choose $\frac{3\epsilon}{2}< \lambda$, then $-\lambda < -\frac{3\epsilon}{2}$, and $$ \epsilon + 1 - \lambda < 1 - \frac{\epsilon}{2}. $$ Therefore if we choose $\epsilon < \frac{2}{3}(1-\cos\delta)$, and obtain $\delta$ as above, then $$ \frac{\epsilon}{2} - 1 \leq p(\theta) \leq 1 - \frac{\epsilon}{2}, $$ or equivalently $$ |p(\theta)| \leq 1 - \frac{\epsilon}{2} $$ whenever $\delta \leq |\theta| \leq \pi$.

3. This is similar to both 1 and 2. Now we're using the fact that near $\theta = 0$, $p(\theta) \sim \epsilon + 1$. By continuity of $\cos\theta$ at $\theta = 0$, there exists $\eta>0$ such that if $|\theta| < \eta$, then $|1 - \cos\theta| < \frac{\epsilon}{2}$. This inequality implies $\cos\theta > 1 - \frac{\epsilon}{2}$. Therefore $$ p(\theta) = \epsilon + \cos\theta > 1 + \frac{\epsilon}{2}. $$ Again, once the existence of such $\eta>0$ is established, then we are free to take it to be as small as we want; in particular we may specify that $\eta<\delta$.

   4. For example, look at $p_1(\theta) = p(\theta) = \epsilon + \cos\theta$. Then $$ \int_{-\pi}^\pi f(\theta)p_1(\theta) d\theta = \epsilon\int_{-\pi}^\pi f(\theta) d\theta + \int_{-\pi}^\pi f(\theta)\cos(\theta) d\theta . $$ Note that the first integral on the RHS is just $\epsilon\hat{f}(0)$, so this is $0$. The second integral is just the first Fourier cosine coefficient: in fact, $$ \int_{-\pi}^\pi f(\theta)\cos\theta d\theta = \int_{-\pi}^\pi \text{Re}(f(\theta)e^{i\theta} )d\theta = \text{Re}\left(\int_{-\pi}^\pi f(\theta)e^{i\theta} d\theta\right) = \text{Re}\hat{f}(1) = 0. $$ (I may be off by a constant prefactor of $\frac{1}{2\pi}$, but this is unimportant.) Here I'm using the assumption that $f$ is real-valued to take $f(\theta)$ under the real part. For higher powers of $p(\theta)$, you'll see the other Fourier coefficients come up just like this, because $p(\theta)^k$ is a trigonometric polynomial (i.e. linear combinations of $\sin(k\theta)$ and $\cos(k\theta)$ for all $k$. So all of these integrals vanish.