This is a follow-up on my previous question that turned out to be almost-trivial.
Let $\varphi(t)=\sin(t)+\sin(t\sqrt{2})+\sin(t\sqrt{3})$. 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:
- 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;
- 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 fractions of $\sqrt{2}$ and/or $\sqrt{3}$?
- It might by practical to consider the winding number of the curve $\gamma:[0,T]\to \mathbb{C}$ given by $\gamma(t) = e^{it}+e^{it\sqrt{2}}+e^{it\sqrt{3}}$.
Addendum: an explicit proof of $E<+\infty$ through Diophantine Approximation. Let $R\subset\mathbb{R}^+$ the set of real numbers such that $r,r\sqrt{2},r\sqrt{3}$ are simultaneously close to integer multiples of $\pi$. For any $r\in R$ we have that $\varphi(r)$ is close to zero: since $\varphi$ is bounded and Lipschitz-continuous, it is enough to show that $R$ is syndetic to have that $E$ is finite. If we consider a cube with side length $\varepsilon>0$ centered at $\frac{m}{\pi}\left(1,\sqrt{2},\sqrt{3}\right)\pmod{1}$ we easily get than for some integer $m\leq\frac{1}{\varepsilon^3}$ the numbers $m,m\sqrt{2},m\sqrt{3}$ are simultaneously at most $\pi\varepsilon$-apart from an integer multiple of $\pi$. By choosing $\varepsilon=\frac{1}{3\pi}$ the ridiculous bound $E\leq 6+27\pi^3$ can be easily derived.
The inequality $E\leq 12$ can be deduced from my approach below, however the optimal bound for $E$ seems to be around $4.5$, so there still is some work to be done.
Here there are some extra thoughts. It is not difficult to show by the sum formulas that the roots of $\sin(t)+\sin(t\sqrt{2})=2\sin\left(t\frac{\sqrt{2}+1}{2}\right)\cos\left(t\frac{\sqrt{2}-1}{2}\right)$ are located at $$ 2\pi(\sqrt{2}-1)\mathbb{Z} \cup \left(\pi(\sqrt{2}+1) + 2\pi(\sqrt{2}+1)\mathbb{Z}\right) $$
hence $\sin(t)+\sin(t\sqrt{2})$ has a sign change on any interval $[a,b]\subset\mathbb{R}^+$ whose length exceeds $2\pi(\sqrt{2}-1)$. By a similar argument, $\sin(t\sqrt{2})+\sin(t\sqrt{3})$ has a sign change on any interval $[a,b]\subset\mathbb{R}^+$ whose length exceeds $2\pi(\sqrt{3}-\sqrt{2})\leq 2$.
Let $g(t)=\sin(t\sqrt{2})+\sin(t\sqrt{3})$ and $h(t)=g(t)g(t+2\pi(\sqrt{3}-\sqrt{2}))$.
$h$ can be factored through the sum formulas and has the following behaviour:
Let us consider the set $H=\{t\in\mathbb{R}^+: h(t)\leq -2\}$. Since $|g(t)|\leq 2$, for any $t\in H$ we have that $g(t)$ and $g(t+2\pi(\sqrt{3}-\sqrt{2}))$ have opposite signs and absolute values $\geq 1$. In particular:
and $E$ is bounded by the length of the largest interval over which $h(t)\geq -2$.
This proves $\color{blue}{E\leq 12}$, roughly.
And that can be probably improved up to $E\leq 10$ by directly considering the largest interval over which $g(t)^2\leq 1$, since $h(t)$ is rarely positive.