# Constraint from level sets of an analytic function

Let $$\theta$$ and $$f$$ be two real analytic non-constant functions defined on $$[0,2\pi]$$. For simplicity we assume $$f$$ has just two critical values $$m (in the picture $$-1$$ and $$1$$); we index as $$\{f_j\}_{j \in\{1,2,...,N\}}$$ its invertible branches. Our hypotesis is that $$\theta$$ and $$f$$ satisfy $$\forall \;a\in(m,M)$$ $$\sum_j \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{|f'(f_j^{-1}(a))|}=0.$$ This is a balancing condition for 2-dimensional vectors represented with complex numbers and involving all points in a given level set (which is finite since $$f$$ is analytic).

EDIT. As pointed out in the comments it was probably unclear that the hypotesis is about a pair of functions $$\theta$$ and $$f$$ that satisfy the condition above. I am not claiming this relation holds for any $$\theta$$. Such pair of functions can for example be constructed by considering a closed regular curve on $$[0,2\pi]$$ of the form $$\gamma(t)=\sum_{j=1}^N a_j \sin(j\cdot t)+i\sum_{j=1}^N b_j \cos(j\cdot t),$$ and observing that for example $$\int_0^{2\pi}\gamma(t) \Bigl(\cos((N+1)\cdot t)\Bigl)^n=0$$, $$\forall n$$. Reparametrizing by arc-length $$t=v(s)$$ and denoting with $$\theta(s)$$ the turning angle of $$\gamma$$ we get $$\int \text{e}^{i\theta(s)} \Bigl(\cos((N+1)\cdot v(s))\Bigl)^n=0.$$ From here one can show that if such property holds for $$\cos((N+1)\cdot v(s))$$ then it must hold for any composition of this function with $$g \in L^1$$. Picking $$g$$ as the characteristic function of $$[a,a+\delta]$$ and derivating in $$\delta$$, we get the equality above for $$f=\cos((N+1)\cdot v(s))$$. You can read more about that in this other question of mine: We observe

$$0=\lim_{a \to m^+} \sum_j \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{|f'(f_j^{-1}(a))|} = - 2 \lim_{a \to m^+} \sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))},$$ where the second equality follows from the analiticity of $$f$$ at inner points. Similarly, derivating in $$a$$, one obtains $$\forall \; n \in \mathbb{N}$$ $$\lim_{a \to m^+} \frac{d^n}{da^n} \sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}=0.$$

I would like to prove from these hypoteses that also the sum over odd indices is constantly $$0$$, that is $$\forall \;a\in(m,M)$$. $$\sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}=0.$$

Note that the fact that all the derivatives are $$0$$ in the limit is not enough. We could theoretically have something like $$\text{e}^{-1/x}$$, whose derivatives go to $$0$$ when $$x$$ approches $$0$$, but the function is not constant. Nevertheless, I have experimental evidence that this does not happen in our setting. I tried to exploit the fact that here we are dealing with a single analytic function and that its global behaviour is determined by its Taylor expansion at any point but I could not conclude the implication this question is about.

EDIT 2. Calling our function $$F(a):=\sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}$$, proving that it is constantly $$0$$ is equivalent to prove the existence of $$C_\varepsilon>0$$ such that $$\| F^{(n)}(a) \| \leq C_\varepsilon^{n+1} n!, \; \forall n \in \mathbb{N}, \; a \in (m,m+\varepsilon).$$ In this case, the series $$0 \equiv \sum F^{(n)}(a) \frac{a^n}{n!}$$ would converge uniformly to $$F(s)$$ in a neighbohood of $$0$$, providing the desired identity. This implication is explained for example in One-Sided Analyticity Condition Guarantees Analytic Function?.

Although this approach looks reasonable I am still stuck. Does anyone have some hints?

The question is pretty specific but I guess an answer in the positive could connect to properties of analytic functions interesting also from a more general perspective. Thanks.

• Something is crazy as written: if $\theta$ is an arbitrary analytic function, you can just interpolate it to be anything you want at the evaluation points, which will definitely destroy the suggested identity. – fedja Jan 27 at 16:22
• @fedja $\theta$ and $f$ are arbitrary as long as they satisfy the condition on the level sets on top, which must hold for all the a. They are a specific pair of functions, which satisfy the given constraint. Is that unclear? – Leonardo Jan 27 at 17:01
• Question has been edited. Hope now it makes more sense. Thanks for pointing that out. – Leonardo Jan 29 at 15:22

Something is still strange. Let's take $$f=\cos(3x)$$. Then the "odd" level set is $$t,t+\frac{2\pi}3,t+\frac{4\pi}3$$ and the "even" one is the same with $$2\pi-$$ everywhere for not too large $$t$$. The derivatives at the pre-images are all the same in absolute value, so we are just asking if the function $$g(t)=e^{i\theta(t)}+e^{i\theta(t+\frac{2\pi}3)}+e^{i\theta(t+\frac{2\pi}3)}$$ can satisfy $$g(t)+g(2\pi-t)=0$$ without being identically $$0$$. Let $$\theta(t)=t+\psi(t)$$ where $$\psi$$ is real-analytic, $$2\pi$$-periodic. We are interested in the Fourier coefficients of $$e^{i\theta(t)}$$ with indices divisible by $$3$$. It is enough to obtain some non-trivial sequence satisfying $$a_{-k}=-a_k$$ for $$3\mid k$$. This means that we should get some prescribed sequence for the Fourier coefficients of $$e^{i\psi(t)}$$ with indices $$k\equiv -1\mod 3$$. The restriction that $$\psi$$ must be real relates only the pairs of indices $$m$$ and $$-m$$, so we have no relation for the indices in our set and the differential (in any reasonable Banach algebra of real-analytic $$2\pi$$-periodic functions) of the mapping $$\psi(\cdot)\mapsto e^{i\psi(\cdot)}$$ is just $$i$$ times the identity, so as a mapping from real functions to the sequences of Fourier coefficients with indices congruent to $$-1$$ modulo 3 is onto with bounds, which means that we can get any sequence that is sufficiently small, thus refuting your conjecture. Am I missing anything?
• Thanks for your answer. I guess you are on point. It is just not completely apparent to me how one handles the condition $\widehat{\psi}(m)=\overline{\widehat{\psi}(-m)}$ when it comes to construct $\text{e}^{i\psi}$ with the desired coefficients using your invertibility result. Your answer in the negative is a bit sad for me ;) since I hoped to use my equality to prove certain periodicity properties of $\theta$ and $f$, in particular that, under the sum condition on level sets, if $\text{e}^{i\theta}$ is $2\pi$-periodic then also $f$ must be. Do you have maybe an intuition about that? – Leonardo Feb 3 at 13:05
• @Leonardo "It is just not completely apparent to me how one handles the condition..." You just force it at every iteration step. If you want to solve $\widehat{e^{i\psi}}(k)=a_k$ for $k\equiv -1\mod 3$ and $\psi_m$ is an approximate solution giving you discrepancies $b_k$, then you just add a real-valued function whose Fourier coefficients with the indices $k$ you are interested in are $-ib_k$ and the coefficients with indices $-k$ are $i\bar b_k$. This function adds just a little to the norm but decreases the discrepancy fixed number of times, so you have a geometric convergence. – fedja Feb 3 at 14:21