When has weighted sum of two specific unimodal functions two peaks?

Question

Consider the functions $f(x) = x e^{-x}$ and $g(x) = xe^{-kx}$ for $x \geq 0$, where $e$ is the Euler constant and $k > 1$. It is easy to check that $f(x)$ is unimodal and has its global maximum at $x_f^* = 1$, while $g(x)$ is also unimodal and has its global maximum at $x_g^* = \dfrac{1}{k} \in (0,1)$. Notice further that both functions are strictly concave up to $\tilde{x}_f = 2$ (for $f(x)$) and $\tilde{x}_g = \dfrac{2}{k}$ (for $g(x)$), after which they are strictly convex.

What I am interested in is the following: Take the weighted sum of the functions $$h_{\lambda}(x) := (1-\lambda)f(x) + \lambda g(x),$$ where $0 < \lambda < 1$. For which values of $k$ can we guarantee that there exists a range of weights $\lambda$ such that $h_{\lambda}(x)$ has two peaks?

What we can easily deduce is that for every $\lambda$, $h_\lambda(x)$ is strictly increasing on $[0, 1/k]$ and strictly decreasing on $[1, \infty)$. Moreover, due to the concavity property of the individual functions, $h_{\lambda}(x)$ is clearly strictly concave on $[0, 2/k]$. Taken together, this implies that for $k \leq 2$, $h_{\lambda}(x)$ is unimodal and has a single peak somewhere in $(1/k, 1)$.

What about the case where $k > 2$? Using Mathematica, it appears that two peaks can only emerge for $k \gtrapprox 6$. But can we somehow show this analytically? Or at least prove that for $k$ sufficiently large, the function has two peaks when $\lambda$ is set appropriately? Many thanks in advance!

Answer 1

$\def\e{\mathrm{e}}$Note that between any two adjacent local maximum points, there exists exactly one local minimum point, and $f(x) = (1 - λ)x\e^{-x} + λx\e^{-kx}$ is increasing on $\left[ 0, \dfrac{1}{k} \right]$ and decreasing on $[1, +\infty]$, thus$$ f \text{ has two peaks} \Longleftrightarrow f' \text{ has three zeros}. $$ Also, the zeros of $f'$ lie on $\left( \dfrac{1}{k}, 1 \right)$, and $f'(x) = (1 - λ)(1 - x) \e^{-x} + λ(1 - kx) \e^{-kx}$, thus\begin{align*} \mathrel{\phantom{\Longleftrightarrow}}{} f' \text{ has three zeros} &\Longleftrightarrow \frac{kx - 1}{1 - x} \e^{-(k - 1)x} = \frac{1 - λ}{λ} \text{ has three roots}\\ &\Longleftrightarrow \ln(kx - 1) - \ln(1 - x) - (k - 1)x = \ln\frac{1 - λ}{λ} \text{ has three roots}. \end{align*} Define $g(x) = \ln(kx - 1) - \ln(1 - x) - (k - 1)x$. Because the range of $\dfrac{1 - λ}{λ}$ is $(0, +\infty)$ given that $λ \in (0, 1)$, then the range of $\ln\dfrac{1 - λ}{λ}$ is $(-\infty, +\infty)$. Therefore,\begin{align*} &\mathrel{\phantom{\Longleftrightarrow}}{} \exists λ \in (0, 1): g(x) = \ln\frac{1 - λ}{λ} \text{ has three roots}\\ &\Longleftrightarrow \exists c \in \mathbb{R}: g(x) = c \text{ has three roots}\\ &\Longleftrightarrow g' \text{ has two zeros}. \end{align*} Now, because\begin{align*} g'(x) &= \frac{k}{kx - 1} + \frac{1}{1 - x} - (k - 1)\\ &= (k - 1) \frac{kx^2 - (k + 1)x + 2}{(kx - 1)(1 - x)}, \quad x \in \left( \frac{1}{k}, 1\right) \end{align*} thus\begin{align*} &\mathrel{\phantom{\Longleftrightarrow}}{} g' \text{ has two zeros} \Longleftrightarrow kx^2 - (k + 1)x + 2 = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right). \end{align*} Define $h(x) = kx^2 - (k + 1)x + 2$. Note that the symmetry axis of the graph of $y = h(x)$ is $x = \dfrac{k + 1}{2k}$ and $\dfrac{1}{k} < \dfrac{k + 1}{2k} < 1$, also $h\left( \dfrac{1}{k} \right) = h(1) = 1 > 0$, thus\begin{align*} h(x) = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right) &\Longleftrightarrow Δ = (k + 1)^2 - 8k > 0\\ &\stackrel{k > 1}{\Longleftrightarrow} k > 3 + 2\sqrt{2} ≈ 5.828427. \end{align*}

Answer 2

We have the familiar unimodal function $f(x):=xe^{-x}$ $\>(x\geq0)$ and the same function at another time scale. The question is whether suitable convex combinations of the two can have two peaks. In the following I consider $f$ as the "fast" function and add to it a suitable multiple of the "slower" function $$g_s(x):= f(sx)=s x \>e^{-s x},\qquad 0<s<1\ .$$ Beginning with $h(x):=f(x)+g_{1/12}(x)$ (this corresponds to $k=12$ in your question) we see that $$h(1)>f(1)={1\over e},\qquad h(12)>g_{1/12}(12)={1\over e}\ ,$$ but $$h(4)=4 e^{-4}+{1\over 3}e^{-1/3}=0.312<0.368={1\over e}\ .$$ This shows that the $h$ considered here is not unimodal.

In order to get an overview over the set of all combinations of $f$ and $g_s$ which are not unimodal I consider the family $$h_{s,t}(x):=x\bigl(e^{-x}+ t e^{-sx}\bigr),\quad s>0, \quad t>0\ .$$ The $h$ considered above corresponds to $s=t={1\over12}$, see the following figure.

Unfortunately the analysis of the $h_{s,t}$ and their possible local extrema leads to transcendental equations. I therefore let Mathematica do the checking for which $(s,t)$ of a fine lattice the function $h_{s,t}$ has more than one peak. Here is the result (the $s$-axis is horizontal, the $t$-axis vertical):

The characteristic peak in the figure is at $(s,t)=(0.169,0.335)$. (This corresponds to your $k=5.917$). The corresponding function $h_{s,t}$ has a single local maximum of type $y(x)=y_0-c(x-x_0)^4$, $\,c>0$.