analysisconvex-analysisfunctionsmonotone-functionsreal-analysis
Example of a convex and injective function $f:I \to \mathbb{R}$ on an interval $I$, such that $f(I)$ is an interval and $f^{-1}:f(I)\to \mathbb{R}$ is not concave.
Attempt. Our function $f$ cannot be monotone (in that case, since $f(\mathbb{R})$ is an interval, $f$ would be continuous and it that case $f^{-1}$ would be concave if $f$ is strictly increasing and convex if $f$ is strictly decreasing).
Thanks for the help.
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)
If $g(x) = y$, I get $$ h''(x) = \dfrac{f''(y/2) f'(y) - 2 f'(y/2) f''(y)}{4 f'(y)^3} $$ so for $h$ to be convex requires $$ \dfrac{f''(y/2)}{f'(y/2)} \ge 2 \dfrac{f''(y)}{f'(y)}$$ For a counterexample, take $f$ that is strictly concave on $[0,1/2]$ but linear on $[1/2, 1]$, so that if $1/2 \le y < 1$ we have $f''(y) = 0$ but $f''(y/2) < 0$.
Best Answer
How about $x \longmapsto e^{-x}$?