# Orthogonality relation in $L^2$ implying periodicity

Let $$\theta(t)$$ and $$\phi(t)$$ be two real $$C^1$$ functions $$[0,2\pi]\rightarrow \mathbb{R}$$. Let us assume $$\theta$$ has the properties $$\int_0^{2\pi} e^{i\theta(t)} dt=0.$$ Geometrically this means that the curve obtained by integrating the (tangent) vector function $$(\cos(\theta),\sin(\theta))$$ over $$(0,2\pi)$$ is closed. I would like to prove the following claim

If the integral $$\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt=0$$ for all $$\lambda \in \mathbb{R}$$ than $$\phi$$ is periodic of period $$2\pi/l$$ with $$1\neq l\in\mathbb{N}$$.

Note that the function $$F(\lambda):=\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt$$ is analytic in $$\lambda$$ and therefore it is constantly equal to $$0$$ iff its derivatives $$F^{(n)}(0)=\int_0^{2\pi} e^{i(\theta(t))} (\phi(t))^n dt$$ vanish for all $$n\in \mathbb{N}$$.

OBSERVATION. If $$\frac{d}{dt}\theta$$ 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 $$0$$.

• Is it not true that if $\phi$ is constant then $\theta$ can be any function satisfying the original conditions? – Nik Weaver Jun 13 at 7:56
• Or if $\phi$ has even period, any $\theta$ with $\theta(t+\pi) = -\theta(t)$ would work. – Nik Weaver Jun 13 at 8:05
• For the first comment, yes. This is providing a rotation of the original curve by changing the parameter I start measuring the turning angle from. You are right, my formulation include also this trivial case, which is something I do not want to consider. I am going to edit it. – Leonardo Jun 13 at 8:08
• Concerning the second comment, good point. Can one at least deduce the periodicity of $\phi$ of some non-trivial period $2\pi/l$? – Leonardo Jun 13 at 9:00
• Yes, creating a new post is generally preferred over changing the question after getting the answer you didn't want. But before you do that you might spend a little time thinking about the problem on your own, now that you understand why the original question had a negative answer... – Nik Weaver Jun 13 at 9:57