How can I show that $g(r) = (a - be^{-rt})^{+}$ is not necessarily convex nor concave in $r$ when $b > 0$?
By the way,
$$x^{+} = \begin{cases} 0 & \text{ if } x < 0 \\[.7em] x & \text{ if } x \geq 0 \end{cases} $$
I tried plotting in Desmos to get an understanding of the function. It looks really strange at some points. You can also get a plot by entering
$$f\left(r\right)=\left\{r\ge-\frac{1}{t}\ln\left(\frac{b}{a}\right):0,\left(a-be^{-rt}\right)\right\}$$
on https://www.desmos.com/calculator.
Now, I don't know a great way to explain it. A result that the book gives is $g(r) = (ae^{-rt} - b)^{+}$ is, for $a > 0$, decreasing and convex in $r$. So I can use this result, I guess, but I don't see much relation between the two functions.
Any help would be appreciated.
First assume that $t>0$. Observe the following: $g=0$ for $r$ less then some number $r_0$. [ $\lim_{r \to -\infty} (a-be^{-rt})=a-b\lim_{r \to -\infty} e^{-rt}=a-b\infty=-\infty$]. (Hence $g'(r)=0$ for $r<r_0$. For $r$ sufficiently large $g'(r)=bte^{-rt} >0$. For a function which is convex or concave the derivative is monotonic. It follows that $g'$ is increasing. Hence $g'' \geq 0$. But $g''(r)=-bt^{2}e^{-rt} <0$ for $r$ sufficiently large. Hence $g$ is neither concave nor convex.
The case $t<0$ is similar.