Is the sum of a bounded unimodal function and a bounded concave increasing function still unimodal or concave increasing?

Question

Suppose we have a differentiable unimodal function $f(x),x \in R$ and a differentiable bounded concave increasing function $g(x), x \in R$. $f(x)$ is bounded and converges to $c_1$. $g(x)$ is bounded and converges to $c_2$. Is $h(x)=f(x)+g(x)$ still unimodal or convace increasing?

Answer 1

Try $f(x) = \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-(x-1)^2/2}$, the PDF of the normal distribution with mean $1$ and variance $1$, which is unimodal, infinitely differentiable, and bounded and $g(x) = \arctan(x)$, which is infinitely differentiable, bounded, concave (on $[0,\infty)$, which seems to be the intention, based on your graph) and increasing. Their sum is bounded and infinitely differentiable, but is not increasing or concave. Here $c_1 = 0$ and $c_2 = \pi/2$.

(Put the peak of the unimodal function where the bounded function is still small.)

With more effort than I'm willing to expend, we can find scalars $a,b$ such that $af(x) + bg(x)$ has a local maximum near $x = 3/2$ whose height is the limit $\lim_{x \rightarrow \infty} af(x) + bg(x)$, defeating unimodality. The above has $a = b =1$. The below has $a = 2$, $b=1$, so there is an $a$ in $(1,2)$ that produces this outcome with $b = 1$. ($c_1$ and $c_2$ are unchanged.)