Does convexity of $f(x)$ imply convexity of $f(e^x)$? Let $0<\epsilon<1$, and let $f:[1-\epsilon,1] \to \mathbb{R}$ be a strictly decreasing, continuous, and strictly convex function. Suppose that $f_-'(1)=0$.
Define $g:[\log(1-\epsilon),0] \to \mathbb{R}$ by setting $g(x)=f(e^x)$. Is $g$ convex on some half-neighbourhood of $0$?

I specifically don't want to assume stronger differentiability assumptions .
The differentiability assumptions do matter here:

*

*If we assume $f,g \in C^2$ and $f''(1)>0$, then $g''(0)=f''(1)>0$ and $g$ is convex.


*If we remove the assumption $f_-'(1)=0$, $g$ might be non-convex. Take $f(x)=1-x$. Then $g(x)=1-e^x$, so $g''(x)=-e^x<0$.
 A: No, the given conditions do not imply that $f(e^x)$ is convex in some interval $(-\delta, 0]$. We will give a concrete counterexample below.
Preliminaries/motivation
If $f$ is differentiable then $g(u) = f(e^u)$ is convex iff $h(x) = xf'(x)$ is increasing. We have that $h$ is negative and $h(1) = 0$. Of course that does not imply that $h$ is increasing near $x=1$. But on the other hand, $h$ can not be decreasing near $x=1$ either. Therefore a counterexample must be such that $xf'(x)$ “oscillates” arbitrarily close to $x=1$.
The counterexample
We define $f:(0, 1] \to \Bbb R$ as
$$
 f(x) = -\int_x^1 \frac{h(t)}{t} \, dt
$$
where the function $h:[0, 1] \to \Bbb R$ has the following properties:

*

*$h$ is continuous on $[0, 1]$.

*$h$ is negative on $[0, 1)$ and $h(1) = 0$.

*$h(x)/x$ is strictly increasing on $(0, 1)$.

*There are intervals arbitrary close to $1$ on which $h$ is strictly decreasing.

Then $f'(x) = h(x)/x$ on $(0, 1)$, so that $f$ is strictly decreasing and  strictly convex on $(0, 1]$, with $f_-'(1) = h(1) = 0$. For $g(u) = f(e^u)$ we have
$$
 g'(u) = f'(e^u) e^u = h(e^u)
$$
which is strictly decreasing on intervals arbitrary close to $u=0$, so that $g$ is not convex on any interval $(-\delta, 0]$.
Constructing the function $h$
The idea is to construct $h$ as a piecewise linear function which is alternatingly increasing and decreasing on intervals arbitrarily close to $1$. The slope of the decreasing parts is chosen such that $h(x)/x$ is still increasing.
First we define two sequences recursively:
$$
 x_0 = 0 \, ,\,
 y_n = \frac{x_n+1}{2} \,, \,
 x_{n+1} = \frac{y_n(y_n+3)}{3y_n + 1} \, .
$$
It can be verified that both sequences converge to $1$ and
$$
 0 = x_0 < y_0 < x_1 < y_1 < \ldots < x_n < y_n < x_{n+1} < \ldots
$$
Then $h$ is defined as a linear function on each interval $[x_n, y_n]$ and $[y_n, x_{n+1}]$ with
$$
 h(x_n) = x_n - 1 \, , \, h(y_n) = \frac{y_n-1}{2} \, .
$$
It can be verified that $h(y_n) > h(x_{n+1})$ so that $h$ is strictly decreasing on intervals arbitrary close to $1$.
Also $x-1 \le h(x) \le (x-1)/2$ for all $x \in [0, 1)$, so that $h$ is negative on $[0, 1)$ and can be continuously extended with $h(1) = 0$.
On each interval $[x_n, y_n]$ is $h$ strictly increasing, so that $h(x)/x$ is strictly increasing as well. On each interval $[y_n, x_{n+1}]$ is
$$
 h(x) = \frac{y_n-1}{2} + (x-y_n)\frac{y_n-1}{4y_n}
= \frac{y_n-1}{4} + \frac{y_n-1}{4y_n} x
$$
so that $h(x)/x$ is strictly increasing as well.
Remark
By “smoothing” the function $h$ at all $x_n$ and $y_n$ we can make $f$ infinitely differentiable.
