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Motivated by this question asked earlier, I was wondering whether one can prove easily that the local maxima of the sum of Gaussians:

$$f_n(x):= \sum_{i=1}^{n}e^{-(x-x_i)^2}, \quad x_1 < x_2 < \dots < x_n$$

are always strict local maxima, i.e. there's a punctured neighborhood $U^*$ around each local maximum $x^*$, so that the $\forall x \in U^*, f(x)< f(x^*),$ (instead of $f(x)\le f(x^*).$)

I don't know of any analytical expression for the local maxima of the sum of Gaussians.

Of course when $n=1$, one can show that the unique local and global maximum is $x_1$. For $n=3$, I took $x_1= 1, x_2=5, x_3=8,$ and I got the following graph:

enter image description here

which clearly shows that the local maxima happen near the $x_i$'s (but not necessarily at the $x_i$'s). It also seems from the graph that my intuition is correct: all local maxima of the sum of Gaussians are indeed strict local maxima. But I was wondering whether there's a rigorous proof?

I guess one way to proceed might be by first observing that when $x< x_1$ or $x> x_n, f_n(x) > 0, f'(x) < 0$ respectively. So this means that we just need to show that when $x \in [x_1, x_n], f(x)$ is at most a $2n$-to-$1$ function, i.e. the preimage of any point can have at most $2n$ elements (this is intuitively thought out by the fact that for $n=1, f_1(x)$ is even, and for $n=3, f_3(x)$ is at most a $6$-to-$1$ function, and same behavior is observed for other values of $n$. Once proved, this will show that all local maxima of $f$ are indeed strict, as otherwise, it'd contradict that the preimage $f_n^{-1}(y), y \in \mathbb{R}$ of any $y$ can have at most $2n$ elements. But I'm not sure how to show this...Next step will be to generalize it to multiple dimensions.

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The derivative $f'_n$ of $f_n$ is a nonzero real-analytic (and even entire) function. Therefore, $f'_n$ can have only finitely many zeroes in any bounded interval. So, all the local maxima of $f_n$ are isolated and therefore strict.

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  • $\begingroup$ Hello and thank you for your answer. Just to understand this part better: "Therefore, $f_n'$ can have only finitely many zeroes in any bounded interval", is this the explanation? $f_n'$ is analytic, so if it vanishes infinite number of times on a closed interval $[a,b],$ it must have an accumulation point. But that means $f_n'$ is identically zero by identity theorem for real analytic functions - en.wikipedia.org/wiki/Identity_theorem. So $f_n$ is a constant. But since $f_n$ is not constant, this can't happen. Hence $f_n'$ must have only finite number of zero on $[x_1, x_n].$ $\endgroup$
    – Learning math
    Commented Feb 6, 2023 at 18:23
  • $\begingroup$ It seems to me that in higher dimensions so $x\in \mathbb{R}^p$, if we replace $(x-x_i)^2$ by $||x-x_i||^2,$ $f_n$ will still have all its local maxima strict local maxima. But I think the identity theorem may not apply to higher dimensions, as the zeros don't need to be isolated there... $\endgroup$
    – Learning math
    Commented Feb 6, 2023 at 18:28
  • $\begingroup$ Is there a reason to believe it to be true in higher dimensions? If we have a single Gaussian, it is radial, and therefore your property fails. This argument no longer works for a non-trivial sum, but the behavior does seem somewhat different. $\endgroup$
    – Mark Schultz-Wu
    Commented Feb 6, 2023 at 19:31
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    $\begingroup$ @Learningmath : I suggest you ask the question about higher dimensions in a separate post. $\endgroup$
    – Iosif Pinelis
    Commented Feb 6, 2023 at 22:24
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    $\begingroup$ @Learningmath : I will have this in mind. At this point, I don't have good ideas about higher dimensions. $\endgroup$
    – Iosif Pinelis
    Commented Feb 7, 2023 at 3:26

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