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 $k$?
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    $\begingroup$ @takecare $\theta_0=0$ just makes sure that the mess is less. If you take any value of $\theta_0$, then the proof will go exactly by the same steps as outlined in this post. Have a look at the theme of the proof and convince yourself about this. :) $\endgroup$ – Error 404 Sep 16 '16 at 13:56
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    $\begingroup$ @VikrantDesai How does the proof go exactly the same for other values? In the proof we make use of cos0=1 and it's values near 0. $\endgroup$ – nomadicmathematician Sep 16 '16 at 21:03
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    $\begingroup$ @takecare For any other $\theta_0$, define $f$ as $f(\theta_0) \gt 0$ and domain of $f$ is $[\theta_0 - \pi,\theta_0 + \pi]$. By continuity, we can choose $0 \lt \delta \le \frac {\pi}{2}$ so that $f(\theta) \gt \frac {f(\theta_0)}2$ whenever $|\theta - \theta_0| \lt \delta$. Now from here onwards the messier part will occur (defining $p$) but nevertheless will dance around the steps of the given proof. Try to figure it out. I'll post that if you want but currently I don't have time so have patience. At least go through the given proof. $\endgroup$ – Error 404 Sep 17 '16 at 11:00
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    $\begingroup$ @takecare Remember we have to construct a family of trigonometric polynomials that peak at $\theta_0$. So defining $p$ is the important part of the proof. $\endgroup$ – Error 404 Sep 17 '16 at 11:01
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    $\begingroup$ @VikrantDesai In that case, we can't use the fact that $\cos |\theta| <1$ when $\theta$ is away from $n\pi$, so I don't really see how the proof can be adapted. $\endgroup$ – nomadicmathematician Sep 17 '16 at 15:12
  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 $\eta<\delta$.

    1. 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.
  • $\begingroup$ Thanks for putting so much efforts in answering this! I am now clear with 1,3 & 4. But in 2, you have wrote $\epsilon + 1 - \lambda \le 1 - \lambda$. How is that possible? I suggest a fix here for choosing our $\epsilon$ to be $\epsilon \lt \frac 23 (1-\cos\delta)$. Then $p(\theta) \le \epsilon + \cos\delta \lt 1 - \frac \epsilon2$. $\endgroup$ – Error 404 Feb 25 '16 at 16:01
  • $\begingroup$ @VikrantDesai Made a mistake, you're right. I've fixed it in the answer. $\endgroup$ – Gyu Eun Lee Feb 27 '16 at 6:07
  • $\begingroup$ Possibly in a hurry you forgot to fix $\epsilon \lt 2(1-\cos \delta)$.Divide by $3$ in RHS. $\endgroup$ – Error 404 Feb 29 '16 at 14:02

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