Suppose you are given a set of functions $f_1, \ldots, f_n$. Every function is defined as follows
$$f_i(x) = \sqrt{1+C^2_i-2C_i\cos (x-D_i)}$$
where $0<C_i<1$ and $0\leq D_i<2\pi$ are real-valued constants and $x\in \mathbb R$. Now we restrict $x$ to an interval $]a,b[$, s.t.,
\begin{align} \forall i, j \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i(x)<f_j(x) &~~\lor \forall x\!\in]a, b[:& ~~f_i(x)>f_j(x) \\ \forall i, j \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i'(x)<f_j'(x) &~~\lor \forall x\!\in]a, b[:& ~~f_i'(x)>f_j'(x) \\ \forall i, j \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i''(x)<f_j''(x) &~~\lor \forall x\!\in]a, b[:& ~~f_i''(x)>f_j''(x) \\ \forall i \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i(x)<0 &~~\lor \forall x\!\in]a, b[:& ~~f_i(x)>0 \\ \forall i \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i'(x)<0 &~~\lor \forall x\!\in]a, b[:& ~~f_i'(x)>0 \\ \forall i \in \{1, \ldots, n\} ~~\forall x\!\in]a, b[:& ~~f_i''(x)<0 &~~\lor \forall x\!\in]a, b[:& ~~f_i''(x)>0 \end{align}
In words within the interval $]a,b[$ the following holds:
- there is a strict total order over all functions
- there is a strict total order of their first derivative (although it does not need to be the same order)
- there is a strict total order of their second derivative (although it does not need to be the same order)
- no function intersects the $x$-axis
- no function has an extremum
- no function has an inflection point
Question: How many minima can the function $f_{\Sigma}(x) = \sum \limits_{i=1}^{n} f_i(x)$ have within the interval $]a,b[$?
My hope is that the answer to the question is that there can only be one or a constant number of minima, but I would also be happy with any polynomial upper OR exponential lower bound.
What I've tried: Initially I hoped that the minimum of the sum function would necessarily coincide with a point of the summands, of which there are only $O(n)$, like extrema and inversion points. This does not seems to be true. Then I tried: "Between any two minima of the sum, there has to be a minimum/maximum/inversion point of a summand function", but no such statement seems to be true either. I am not entirely convinced that a charging argument will not work, but I have not been able to find the right angle.
Next I have tried to accumulate the functions into a constant number of functions by grouping them into four sets $F^+_+, F^+_-, F^-_+$ and $F^-_-$, s.t., any function in the set $F^+_-$ has positive first derivative and negative second derivative (and similar for the other three sets). Summing up all of the functions in these sets yields four functions $$ f^+_+(x) =\!\! \sum\limits_{f\in F^+_+} f(x),~~~ f^+_-(x) =\!\! \sum\limits_{f\in F^+_-} f(x),~~~ f^-_+(x) =\!\! \sum\limits_{f\in F^-_+} f(x),~~~ f^-_-(x) =\!\! \sum\limits_{f\in F^-_-} f(x) $$ and we can observe that we retain that for, e.g., $f^+_-$ the first derivative is entirely positive and the second is entirely negative within $]a,b[$ (and similar for $f^+_+, f^-_-, f^-_+$). Note that these function do not necessarily have a total order anymore, nor do their derivatives. The hope was that reducing the number of functions to just 4 would be helpful, but I could not figure out how to proceed from here.
I have also tried to plot these functions in Mathematica and I have not encountered any instance with more than $n$ minima, which is where my suspicion comes from that the answer is $O(n)$.
Alternative follow up question: If it is for some reason not possible to give any upper bound on the number of minima, what would be a sufficient property of these functions to obtain one?
This question has been asked a couple of days ago on the Mathematics Stack Exchange, but there haven't been any answers yet.