Does convoluting $f$ with a unimodal function not increase the number of extremums of $f$?

Question

This question can be formulated in two environments: discrete and continuous. For discrete case let $G=\mathbb{Z}$ and for continuous let $G=\mathbb{R}$. We can think of $G$ as of group so that we understand what convoluting on $G$ means.

Let $f:G\rightarrow[0,\infty)$ be a bounded function with exactly $n$ local extremums and let $g:G\rightarrow[0,\infty)$ be a unimodal $L_1$ function. Does then the convolution $f*g$ have at most $n$ local extremums?

My attempt:
In summary, I really didn't make any meaningful progress. I tried proving the case $n=1$ first and then decomposing $f$ into a sum of $n$ unimodal functions, but the sum of two unimodal functions need not be bimodal (or uni-) so this wouldn't work. Then I tried Defining the function $H:G\times G\rightarrow[0,\infty)^2:(x,y)\mapsto f(x)g(y)$ which would have exactly $n$ extremums and then "projecting" it (integrating along perpendicular lines), hoping that "projections" would not increase the number of extremums, but that turned out to be false as well. Maybe requiring the $H$ to have at most $n$ extremums when restricted to any affine line in $G\times G$ would suffice for a proof, but I doubt that my definition of $H$ satisfies that condition.

Answer 1

There are three rules for convolution between unimodal / log-concave functions.

1. The convolution of two log-concave functions is log-concave.
2. The convolution of one log-concave and one unimodal function is unimodal.
3. However, the convolution of two unimodal functions are not unimodal.

To illustrate the third point, consider functions on $\mathbb{Z}$ $$f = g = \{\cdots, 0, 4, 5, 11, 0, \cdots\}$$ Then $$f * g = \{\cdots, 0, 16, 40, 113, 110, 121, 0, \cdots\}$$ which is no longer unimodal.

In particular, your conjecture doesn't hold even with $n = 1$.