# Given $\theta$, find $f$ such that $\int_{\mathbb{T}} \text{e}^{i\theta} \cos(h \cdot f) = 0,$ for all $h \in \mathbb{N}$

Let $$\theta$$ be a $$C^{\infty}$$ (resp. analytic) real-valued function on $$\mathbb{T}=[0,2\pi]/\{0,2\pi\}$$.

When can one find $$f \neq 0$$, $$C^{\infty}$$ (resp. analytic) real-valued function on $$\mathbb{T}$$ such that $$\tag{1} \int_{[0,2\pi]} \text{e}^{i\theta(s)} \cos(h \cdot f(s)) \text{d}s = 0, \text{ for all } h\in \mathbb{N}\cup \{0\}?$$

Setting $$n=0$$, one obvious necessary condition is $$\int_{[0,2\pi]} \text{e}^{i\theta(s)} \text{d}s = 0$$. A sufficient condition I found, which is far from being explicit, is the existence of a reparametrization $$v:\mathbb{T}\rightarrow \mathbb{T}$$ that switches off all the harmonics multiple of a certain $$k$$. More precisely, given the Fourier expansion $$v'(t) \text{e}^{i\theta(v(t))}= \sum_{j\in\mathbb{Z}} c_j e^{i(jt)},$$ if there exists $$k\in \mathbb{N}$$ such that $$c_{k \cdot h}=0$$ for all $$h\in \mathbb{Z}$$, then $$\int_{[0,2\pi]} \text{e}^{i\theta(s)} \cos(h \cdot k \cdot v^{-1}(s)) \text{d}s=\int_{[0,2\pi]} v'(t) \text{e}^{i\theta(v(t))} \cos(h \cdot k \cdot t) \text{d}t=0.$$ How can one more explicitely characterize/construct (or fail to construct) such $$f$$'s in the $$C^\infty$$ (resp. analytic) case? Does this problem connect to something you know from the literature?

MORE EQUIVALENT CONDITIONS

The existence of a function as above is equivalent to the existence of a function $$\phi$$ such that $$\tag{2} \int_{[0,2\pi]} \text{e}^{i\theta(s)} \phi(s)^n \text{d}s = 0, \text{ for all } n\in \mathbb{N}\cup \{0\}.$$

One direction follows from the Weierstrass approximation theorem: if (2) holds, then it must hold for all the composition of $$g\circ \phi$$, with $$g$$ an $$L^2$$ function, and in particular for $$g(x)=\cos(h\cdot x)$$. The other direction is obtained by observing that $$\cos(f)^n$$ is a linear combination of $$\cos(h\cdot f), h \in \mathbb{N}$$, and therefore if (1) holds then (2) holds for $$\phi=\cos(f)$$.

Knowing that, I attempted, without success for the moment, to construct such a $$\phi$$ by finding suitable Fourier coefficients. If $$\phi(t)=\sum_{j\in\mathbb{Z}} \widehat{\phi}(j) e^{i(jt)}$$, then, by convolution, $$\phi(t)^n=\sum_{j}\;\sum_{j_{n-1}}\sum_{j_{n-2}} ... \sum_{j_{1}} \widehat{\phi}(j-\sum_{r=1}^{n-1} j_r) \prod_{r=1}^{n-1} \widehat{\phi}(j_r) \; \text{e}^{i (j t)},$$ and hence (2) is translated into a sequence of homogeneus equations in $$l^2$$. The regularity of $$\phi$$ should then be deduced by bounding the decay of its smartly chosen Fourier coefficients.

Do you see how this other approach could work?

The two conditions have very interesting geometrical implications. In fact, another equivalent characterization is the existence of $$\phi$$ such that
$$\int_{[0,2\pi]} e^{i(\theta(s)+\lambda\phi(s))} ds=0, \text{ for all } \lambda \in \mathbb{R}.$$ The equivalence can be obtained by observing that the term on the left is analytic in $$\lambda$$ and by computing its $$n$$-th derivative at $$\lambda=0$$. This would provide a one parameter family of closed curve associated to an affine line in the space of turning angle functions.

Note. The first part of this question had been already posted without receiving an answer. Since edits were major I just decided to repost it from scratch. Thank you