About the number of critical points of a function

Question

Suppose that $f$ is a totally monotone function on $(0,\infty)$, so that $(-1)^n f^{(n)}\ge0$ for all $n=0,1,\dots$, $f(0+)\in(0,\infty)$, and $f(t)\sim\frac{1}{t^{\frac{3}{2}}}$ as $t\to\infty$. Can we prove that $g(t)=tf(t)$ has only one critical point over $(0,\infty)$?

Answer 1

Here is a counterexample:
$$f(t)=10^{10} \left(8 e^{-10 t}+e^{-t}\right)+\frac{1}{(t+1)^{3/2}}$$ for real $t\ge0$.

Then $(-1)^n f^{(n)}\ge0$ on $[0,\infty)$ for all $n=0,1,\dots$. Also, $f(t)\sim \frac{1}{t^{3/2}}$ as $t\to\infty$.

Also, $g(0)=0$, $g(\frac1{10})\approx3.85\times10^9$, $g(\frac4{10})\approx3.27\times10^9$, $g(\frac{10}{10})\approx3.68\times10^9$, and $g(\infty-)=0$, so that $g(\frac1{10})>\max(g(0),g(\frac4{10}))$ and $g(\frac{10}{10})>\max(g(\frac4{10}),g(\infty-))$. So, there is a critical point of $g$ in each of the intervals $(0,\frac4{10})$ and $(\frac4{10},\infty)$. There also is a third critical point of $g$ in the interval $(\frac1{10},\frac{10}{10})$.

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For an illustration, here is the graph $\{(t,g(t))\colon 0.05\le t\le1.5\}$:

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A simpler version of the previous example:

The idea here is the same as before: $8\approx10\approx\infty$.