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Let $\{c_i\}_{i=1}^n$ be a sequence of real numbers such that $c_i \geq 0$ for each $i$ and $\sum_{i=1}^n c_i = 1$. Let $\omega_i \in [\delta, \Delta]$ for each $i$, where $\delta$ and $\Delta$ are strictly positive reals. Define $$f(t) = \sum_{i=1}^n c_i \sin(\omega_i t)$$ for $t \in [0, \infty)$, and let $t'$ be the smallest non-zero value of $t$ for which $f(t) = 0$.

  • How can we show that there exists a constant $T > 0$ such that $t' \leq T$ for all possible choices of $c_i$ and $\omega_i$?
  • Can we give upper/lower bounds on $T$ in terms of any of the other variables?
  • In particular, how can we determine what, if any, is the dependence of $T$ on $n$?
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  • $\begingroup$ For $[\delta,\Delta]=[1,2]$, is it possible to have $t’>\pi$? $\endgroup$ – Matt F. Apr 17 '19 at 11:53
  • $\begingroup$ I am sure it is possible to have $t'>\pi$. The $t'$ must depend on $n$. THe example is somewhat complicated, and I will post it later. $\endgroup$ – Alexandre Eremenko Apr 17 '19 at 19:05
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    $\begingroup$ For n=2 and the smallest omega_i = 1, I get 3pi/4 as a rough upper bound for T. It might be of interest when n=2 to determine t exactly for all allowed combinations of c and omega. You might then be able to say something when n=3 and omega_3 is the biggest of the omegas. You might get T of order ( log n ). Gerhard "Maybe Apply Surfing To This?" Paseman, 2019.04.17. $\endgroup$ – Gerhard Paseman Apr 17 '19 at 19:43
  • $\begingroup$ @l2j6 It is not obvious to me that $t' < \infty$. As far as I can see, Alexandre Eremenko's answer (through Fedja's argument) is the only indisputable proof of this fact. $\endgroup$ – Luc Guyot Apr 18 '19 at 8:42
  • $\begingroup$ There is a lower bound which is essentially the time-energy uncertainty principle. Look for quantum speed limit. $\endgroup$ – lcv Apr 20 '19 at 0:59
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That one cannot guarantee a zero on $(0,\pi/\delta]$ or on any interval $(0,c]$ with $c$ independent on $n$ follows from the example of B. Logan, (Theorem 5.5.1 of his thesis Properties of high-pass signals, Columbia Univ., 1965). He constructs a bounded $L^2$ function whose Fourier transform is supported on any given set $[-b,-a]\cup [a,b]$ with $0<a<b$, and which is positive on any given interval $(0,c)$. This function can be approximated by functions of your class with large $n$.

Logan's example is reproduced as Example 1 here.

Here is the solution suggested by user Fedja.

Theorem. $f$ must have a zero on the interval $(0,\pi n/\delta)$. (This estimate is unlikely to be exact for all $n$).

Proof. Define the linear operators $f\mapsto A_kf$, where $(A_kf)(t)=f(t)+f(t+\pi/\omega_k)$. When we apply $A_k$ to our $f$ it kills all summands of the form $\sin\omega_k(t+\alpha)$. So $$A_nA_{n-1}\ldots A_1f=0.$$ On the other hand $$(A_nA_{n-1}\ldots A_1f)(t)=f(t)+f(t+\pi/\omega_1)+f(t+\pi/\omega_2)+f(t+\pi/\omega_1+\pi/\omega_2)+\ldots,$$ all summands are of the form $f(t+\alpha)$ where $\alpha$ is a sum of some $\pi/\omega_j$, with at most $n$ such summands, so $\alpha\leq \pi n/\delta$. Thus if our $f(t)>0$ on $(0,\pi n/\delta+\epsilon)$ this sum will be positive on $(0, \epsilon)$ - contradiction.

Notice that the estimate does not depend on $\Delta$.

EDIT. For $f(x)=(2/3)\sin x+(1/3)\sin(2.5 x)$ the smallest positive zero is about $3.48>\pi$,

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    $\begingroup$ Very interesting! And thanks for the link! $\endgroup$ – paul garrett Apr 17 '19 at 21:25
  • $\begingroup$ It seems to me that you only showed the following: $f^{\ast}(t) \ge 0$ for every $t \ge 0$. It is very likely to be impossible (and hence the desired contradiction), but I would be glad to know a proof of it. $\endgroup$ – Luc Guyot Apr 17 '19 at 21:59
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    $\begingroup$ @Luc Guyot: This is not relevant enymore, but $f^*\not\equiv 0$ because $\sum c_j^*=1$, and I used an elementary theorem of Hurwitz that if a con-constant analytic function has a zero, then every sufficiently close analytic function has a zero nearby. $\endgroup$ – Alexandre Eremenko Apr 17 '19 at 23:41
  • $\begingroup$ @Luc Guyot: I must have added that every function of the class not only has zeros but sign changes (zeros of odd multiplicity). This is needed when applying Hurwitz to ensure that the nearby function has a real zero not just a pair of complex ones nearby. $\endgroup$ – Alexandre Eremenko Apr 17 '19 at 23:55
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    $\begingroup$ @Luc Guyot: The existence of sign changes (points where $f(t)<0$) for every such function follows from the result in the paper linked to my answer. $\endgroup$ – Alexandre Eremenko Apr 18 '19 at 10:31
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The upper bound $\frac{\pi n}{\delta}$, with $\delta = \min_i(\omega_i)$, can be improved when $n = 2$.

Claim. Let $f(t) = c_1 \sin(\omega_1 t) + c_2 \sin(\omega_2 t)$, with $c_i, \omega_i > 0$ for $i = 1, 2$. Then there is $0 < \theta < \frac{3\pi}{2 \delta}$ such that $f(\theta) = 0$.

Proof. If $\omega_1 = \omega_2$, set $\theta = \frac{\pi}{\omega_1}$. Swapping $\omega_1$ and $\omega_2$ if needed, we can hence assume, without loss of generality, that $\omega_1 < \omega_2$. If $\frac{\pi}{\omega_2} \ge \frac{\pi}{2\omega_1}$, then we have $f(\frac{\pi}{\omega_1}) = c_2 \sin(\frac{\omega_2}{\omega_1} \pi) \le 0$. Since $f(\epsilon) = (c_1 \omega_1 + c_2 \omega_2) \epsilon + o(\epsilon^2)$ is positive for all sufficiently small $\epsilon > 0$, the claim follows from the intermediate value theorem. Let us assume now that $\frac{\pi}{\omega_2} < \frac{\pi}{2\omega_1}$. As $\sin(\omega_2 t)$ cannot be positive on any interval of length larger than $\frac{\pi}{\omega_2}$, it takes a negative value at some $\theta$ in $(\frac{\pi}{\omega_1}, \frac{3\pi}{2\omega_1})$, an interval on which $\sin(\omega_1 t)$ is negative. Hence $f(\theta) < 0$, which completes the proof.

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    $\begingroup$ One can scale t and assume the smallest frequency is 1, possibly pushing the zero out as far as $x=\pi$. Then either this is a bound for the first zero, or else the other frequency is at least 2, which would give an upper bound of $3\pi/2$. By using the c's, you can probably get an upper bound of $\pi$. Gerhard "Likes Picture Form Of Proof" Paseman, 2019.04.17. $\endgroup$ – Gerhard Paseman Apr 17 '19 at 23:05
  • $\begingroup$ @GerhardPaseman Thanks for your comment. I simplified the proof and improved the bound according to your suggestion. $\endgroup$ – Luc Guyot Apr 20 '19 at 16:28
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    $\begingroup$ f(0) is actually 0, so use $f(\epsilon) \gt 0$ for a wise choice of $\epsilon \gt 0$. (Maybe you need to use $\omega_2 \lt 2\omega_1$ for the first case too.) I like seeing this proof, which is concise and easy to follow. Thanks for adding rigor. Gerhard "And For Doing The Work" Paseman, 2019.04.20. $\endgroup$ – Gerhard Paseman Apr 20 '19 at 19:03
  • $\begingroup$ @GerhardPaseman Many thanks again, it is now corrected. $\endgroup$ – Luc Guyot Apr 20 '19 at 19:49
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Below is a contour plot (courtesy of Wolfram Alpha) of sin t + sin ct. A better plot would be of sin t + A(c)*sin ct to maximize the light regions. This plot confirms the $3\pi/2$ bound for the case of equal amplitudes. Maybe someone could make a better plot with different amplitudes?

enter image description here

Gerhard "Use Pretty Pictures To Bound" Paseman, 2019.04.19.

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