# Does $\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt=0$ $\forall \; n\in\mathbb{N}_0$ imply $\phi$ periodic?

PROBLEM. Let $$\theta(t)$$ and $$\phi(t)$$ be two real analytic non-constant functions $$[0,2\pi]\rightarrow \mathbb{R}$$. I am trying to prove the following claim

If the integral $$\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt=0$$ for all $$n\in\mathbb{N}_0$$ than the first derivative $$\theta'$$ and $$\phi$$ are periodic of common period $$2\pi/l$$ with $$1\neq l\in\mathbb{N}$$.

Note that this is equivalent to $$F(\lambda):=\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt=0$$ for all $$\lambda \in \mathbb{R}$$. In fact, $$F(\lambda)$$ is analytic in $$\lambda$$ and its being constantly equal to 0 is equivalent to the vanishing of all its derivatives $$F^{(n)}(0)=\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt$$. Geometrically this means that the curve obtained by integrating the (tangent) vector function $$(\cos(\theta+\lambda\phi),\sin(\theta+\lambda\phi))$$ over $$[0,2\pi]$$ is closed $$\forall \lambda$$.

Just in case, a back-up less general claim for which I would like to see a clean solution is

If, in the hypotesis above, $$\phi$$ is a polynomial, then $$\phi$$ is constantly $$0$$.

OBSERVATION. If $$\theta'$$ and $$\phi$$ are periodic of common period $$\frac{2\pi}{l}$$ with $$1\neq l \in \mathbb{N}$$ and $$\int_0^{\frac{2\pi}{l}} e^{i\theta}\neq 0$$ then the converse implication is true. In fact, in this setting $$\theta=c\cdot t+\theta_p(t)$$ with $$c=\frac{2\pi}{l}(\theta(\frac{2\pi}{l})-\theta(0))$$ and $$\theta_p$$ periodic of period $$\frac{2\pi}{l}$$. Then \begin{align} \int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt &=& \sum_{j=0}^{l-1} \int_{j \frac{2\pi}{l}}^{(j+1) \frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt \\ &=& \sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt, \end{align} where the last equality is obtained by repetedly applying the substitution $$t'=t-\frac{2\pi}{l}$$. Since we know $$\sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i\theta(t)}dt=\int_0^{2\pi} e^{i\theta(t)} dt=0$$ then also the integral above must be $$0$$. In the following picture the curve associated to $$\theta(t)=t + \cos( 12 t)$$ deformated in the direction $$\cos(3 t)$$. In this case $$l=3$$ and the curve is closed $$\forall \lambda$$.

$\theta(t)=t + \cos( 12 t)$ deformated in the direction $$\cos( 3 t)$$. In this case $$l=3$$ and the curve is closed $$\forall \lambda$$.">

IDEA. If $$\theta$$ monotone one can substitute $$s=\theta(t)$$ in the integral and get $$\int_{\theta(0)}^{\theta(2\pi)} e^{i s} \frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))} ds=0.$$ In this case the idea behind the hypotesis becomes apperent: $$\phi(\theta^{-1}(s))$$ is periodic of non-trivial period iff $$\phi$$ and $$\theta'$$ have the common period property. It seems here that looking at the Fourier expansion of our functions on $$[\theta(0),\theta(2\pi)]$$ could be a good idea: the condition we have means indeed that, $$\forall n$$, the first harmonic of the function $$\frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))}$$ is $$0$$. Fourier coefficients of a product are obtained by convolutions and therefore the condition above becomes, $$\forall n$$: $$\sum_{k_n=-\infty}^{+\infty} \sum_{k_{n-1}} ... \sum_{k_{2}}\sum_{k_{1}} \widehat{\frac{1}{\theta'}}(1-\sum_{i=1}^{n} k_i) \prod_{i=1}^{n} \widehat{\phi}(k_i)=0.$$ Is this approach viable? Can one from here exploit the fact that a function is periodic of non-trivial period iff there exists $$k$$ such that only harmonics multiple of $$k$$ are different from 0? Other way round, do non-zero harmonics of coprime orders imply a contradiction with our constraints? As for a toy example, if $$\theta(t)=t$$,$$\theta'(s)=1$$ and $$\phi(s)=\cos(2s)+\cos(3s)$$ already $$\widehat{f^2}(1)= 2 \widehat{f}(3)\widehat{f}(-2) \neq 0$$; in the general setting interaction of coefficients is not straightforward.

NOTE: This question originated from Orthogonality relation in $L^2$ implying periodicity. As suggested in the comments to the previous post, since the target of the question changed over time and edits were major, here I hope I gave a clearer and more consistent presentation of my problem.

• You do not need to move 6 points in a highly nontrivial way. Just rotate 3 opposite pairs in some independent ways for a while recovering the original star after 30 degree rotation and then move the 2 equilateral triples in some independent ways arriving at the 60 degree rotation. From pairs $\ell$ can be only $2$, but from triples it can be only $3$, so no $\ell$ exists in the end. – fedja Jun 17 at 22:03

I missed the real analyticity condition (my comment makes perfect sense for $$C^\infty$$ though), so let's move points in a fancy way to satisfy it.

First, observe that if $$a_0,a_1,a_2$$ are positive reals close to $$1$$, then there exist unique $$\theta_1\approx \frac {2\pi} 3$$ and $$\theta_2\approx \frac {4\pi}3$$ such that $$a_0+a_1e^{i\theta_1}+a_2e^{i\theta_2}=0$$. Moreover, $$\theta_{1,2}$$ are real analytic functions of $$a_{0,1,2}$$ in some neighborhood of $$1$$. This is just the implicit function theorem.

Now choose your favorite $$2\pi$$-periodic real analytic function $$F(\tau)$$ with uniformly small derivative that is not periodic with any smaller period (say, $$\varepsilon\cos\tau$$) and put $$t(\tau)=\tau+F(\tau)$$. Then $$\tau$$ is uniquely determined by $$t$$ and the dependence is real analytic as well.

Next define $$\theta_{1,2}(\tau)$$ by $$\theta_j(\tau)\approx \frac {2\pi j}3$$ such that $$t'(\tau)+t'(\tau+\tfrac{2\pi}3)e^{i\theta_1(\tau)}+t'(\tau+\tfrac{4\pi}3)e^{i\theta_2(\tau)}=0$$ Everything is real analytic so far.

By uniqueness, we must have the relations $$\theta_1(\tau+\frac{2\pi}{3})=\theta_2(\tau)-\theta_1(\tau)$$ and $$\theta_1(\tau+\frac{4\pi}{3})=2\pi -\theta_2(\tau)$$. Thus $$\theta_1(\tau)+\theta_1(\tau+\frac{2\pi}{3})+\theta_1(\tau+\frac{4\pi}{3})=2\pi$$. This implies that there exists a real analytic $$\Theta(\tau)$$ such that $$\Theta(\tau+\frac{2\pi}3)=\Theta(\tau)+\theta_1(\tau)$$ (just divide the Fourier coefficients by appropriate numbers to get the periodic part and add $$\tau$$; note that the identity for $$\theta_1$$ implies that $$\widehat\theta_1(3k)=0$$ for $$k\ne 0$$, so no division by $$0$$ will be encountered). Then, automatically, $$\Theta(\tau+2\pi)=\Theta(\tau)+2\pi$$ and $$\Theta(\tau+\frac{4\pi}3)=\Theta(\tau+\frac{2\pi}3)+\theta_1(\tau+\frac{2\pi}3) \\ =\Theta(\tau)+\theta_1(\tau)+\theta_1(\tau+\frac{2\pi}3)=\Theta(\tau)+\theta_2(\tau)$$
Hence, we have the identity $$\sum_{j=0}^2 t'(\tau+\tfrac{2\pi j}3)e^{i\Theta(\tau+\frac{2\pi j}3)}=0$$ We can now pick up any $$\frac{2\pi}3$$-periodic real analytic function $$\Psi(\tau)$$, multiply the terms by the corresponding values of $$\Psi^n$$ (they are equal), integrate in $$\tau$$ from $$0$$ to $$\frac{2\pi}3$$, and use the standard change of variable formula to get $$\int_0^{2\pi} e^{i\Theta(\tau(t))}\Psi(\tau(t))^n\,dt=0$$ but $$\psi(t)=\Psi(\tau(t))$$ is no longer $$\frac{2\pi}3$$ periodic in $$t$$ because the composition kills periodicity.

As I said from the beginning, "there are many fancy ways to move six (well, even three) points around the circle and keep the sum balanced".

• Cool and nicely explained. Thank you very much! I am gonna update the post with some drawings that present your solution :) – Leonardo Jun 19 at 16:41
• @Leonardo You are cordially welcome. :-) – fedja Jun 19 at 16:50