# About the zeroes of an almost-periodic function

Recently, when providing an answer to a question asking to prove the boundedness of an integral function, I invoked the following Lemma:

Lemma. Let $\varphi(t)=\sin(t)+\cos(t\sqrt{2})$. Such function is not periodic, but it is bounded, Lipshitz-continuous and with mean zero, i.e. $\lim_{b\to +\infty}\frac{1}{b-a}\int_{a}^{b}\varphi(t)\,dt = 0$. Its real zeroes are simple, hence by denoting as $\zeta_0<\zeta_1<\zeta_2<\zeta_3<\ldots$ the real positive zeroes we have that $$E = \sup_{n\in\mathbb{N}}\left(\zeta_{n+1}-\zeta_n\right) < +\infty.$$

Now my actual question:

Q: How can we improve the previous inequality and show, for instance, that $E\leq 2\pi$?

My thoughts:

1. If one is able to produce accurate bounds for $\frac{1}{2\pi i}\oint_{\gamma}\frac{\varphi'(t)}{\varphi(t)}\,dt$, with $\gamma$ being the boundary of a thin rectangle in the complex plane enclosing the real interval $[a,b]$, is also able to estimate the density of real zeroes;
2. If for some non-negative function $\psi(t)$ over the interval $[a,b]$ the integrals $\int_{a}^{\frac{a+b}{2}}\varphi(t)\psi(t)\,dt$ and $\int_{\frac{a+b}{2}}^{b}\varphi(t)\psi(t)\,dt$ have opposite signs, $\varphi(t)$ has a zero in $[a,b]$. But what is an efficient way for constructing such weigth functions $\psi$? Can we exploit the convergents of the continued fraction of $\sqrt{2}$?
3. It might by practical to consider the winding number of the curve $\gamma:[0,T]\to \mathbb{C}$ given by $\gamma(t) = e^{it}+e^{it\sqrt{2}}$.

## 1 Answer

I would be tempted to try to solve directly $\varphi(t)=0$.

Let $\zeta$ such that $\phi(\zeta)=0$. Then $\cos(\zeta \sqrt{2})=-\sin(\zeta)=\cos(\zeta+\pi/2)$

This gives $\pm \zeta \sqrt{2} = \zeta+\pi/2 + 2n\pi$ for $n \in \mathbb{Z}$

Therefore you have two types of solutions: $\alpha_n=\dfrac{\sqrt{2}+1}{2}\pi(1+4n)$ and $\beta_n=-\dfrac{\sqrt{2}-1}{2}\pi(1+4n)$ for $n \in \mathbb Z$.

You can see that $\beta_{n}-\beta_{n+1}=2(\sqrt{2}-1)\pi<\pi$.

Of course since you can have $\alpha$ solutions between the $\beta$ ones you might be able to improve this bound.

• The simplest ways are always the best. So we just need to understand how two arithmetic progressions interleave (interlace? I am not sure about the correct wording), and that should not be difficult. Many thanks and (+1). – Jack D'Aurizio Aug 5 '17 at 20:04
• yes exactly! not sure about the wording either though – fonfonx Aug 5 '17 at 20:15
• Answer accepted. If you are interested, I have just proposed a non-trivial version of the same question: math.stackexchange.com/questions/2383844/… – Jack D'Aurizio Aug 5 '17 at 20:20