Does midpoint-convexity at a point imply midpoint-convexity at other points? This question is a follow-up of this one.
Let $f:\mathbb R \to \mathbb [0,\infty)$ be a $C^{\infty}$ function satisfying $f(0)=0$.
Suppose that $f$ is strictly decreasing on $(-\infty,0]$ and strictly increasing on $[0,\infty)$, and that $f$ is strictly convex in some neighbourhood of $0$.

Given $c \in \mathbb R$, we say that $f$ is midpoint-convex at the point $c$ if
$$
f((x+y)/2) \le (f(x) + f(y))/2,
$$
whenever $(x+y)/2=c$, $x,y \in \mathbb R$.

Question: Let $r<s<0$, and suppose that $f$ is midpoint-convex at $r$. Is $f$ midpoint-convex at $s$?

In the example given here (essentially $(f(x)=-x^3$) $f$ is concave after its global minimum point, while here I assume that there is a convex neighbourhood.
 A: No, there are counterexamples for each such $r,s$.
Before I start with the proof and deal with the technicalities, I'd like to present the basic idea: Take a convex function that fullfills all convexity conditions (here $x^2$), then pertube it a small bit on a short interval after $s$, such that it becomes concave there, but still keeps the monotonicity. This makes it not mid-convex arund $s$, while the pertubation is too small to affect the mid-convexity at $r$, which is "far away".

Define for $\delta > 0$
$$b_\delta(x)=
\begin{cases}
x^2,                                   & \text {if } 0 \le x \le \frac{\delta}4;\\
\frac{\delta^2}8-(x-\frac{\delta}2)^2, & \text {if } \frac{\delta}4 \le x \le \frac{3\delta}4;\\
(x-\delta)^2,                          & \text {if } \frac{3\delta}4 \le x \le \delta;\\
0,                                     & \text {otherwise.}\\
\end{cases}
$$
It's easy to see that this is a $C^1$ function, the parabolas have been "stuck together" such that their values and first derivates agree on the points where the piece-wise definitions change, we have
$$b_\delta(0)=b_\delta(\delta)=0, b_\delta(\frac{\delta}4)=b_\delta(\frac{3\delta}4)=\frac{\delta^2}{16}.$$
We also have
$$b'_\delta(x)=
\begin{cases}
2x,                                   & \text {if } 0 \le x \le \frac{\delta}4;\\
-2(x-\frac{\delta}2), & \text {if } \frac{\delta}4 \le x \le \frac{3\delta}4;\\
2(x-\delta),                          & \text {if } \frac{3\delta}4 \le x \le \delta;\\
0,                                     & \text {otherwise.}\\
\end{cases}
$$
and again we see that
$$b'_\delta(0)=b'_\delta(\delta)=0, b_\delta(\frac{\delta}4)=\frac\delta2, b_\delta(\frac{3\delta}4)=-\frac\delta2$$
by using definitions on both sides of the piece-wise definition.
Since $b'_\delta$ is a piece-wise linear function, the minimum value is easily seen to be taken at $\frac{3\delta}4$, so we get
$$\forall x \in \mathbb R: b'_\delta(x) \ge -\frac\delta2.$$
We now define a counterexample to the proposition, as
$$f_\delta(x):=x^2-4b_\delta(x-s)$$
First we see that $f_\delta(x)=x^2$ outside $[s,s+\delta]$, so the conditions on monotony of $f$ are fullfilled outside $[s,s+\delta]$.
Inside this interval, we have
$$f'_\delta (x)=2x-4b'_\delta(x-s) \le 2(s+\delta) + 4\frac\delta2 = 2s+4\delta,$$
where we used $2x \le 2(s+\delta)$ for $x\in [s,s+\delta]$ and the lower bound for $b'_\delta$ given above.
That means for $\delta$ small enough ($\delta < -\frac{s}2$) we also have $f'_\delta (x) < 0$ in the interval $[s,s+\delta]$ and since with the above upper bound $s+\delta < 0$ that is exactly what is needed to prove that our $f_\delta$ has the correct monoticity. In addition since $f_\delta(x)=x^2$ around $x=0$, it is strictly convex there.
I'll show that $f$ is not midpoint-convex at $s$ for any $\delta$.
We have $f_\delta(s)=s^2, f_\delta(s-\frac\delta4) = (s-\frac\delta4)^2, f_\delta(s+\frac\delta4) = (s+\frac\delta4)^2-4b_\delta(\frac\delta4) = (s+\frac\delta4)^2 -4 \frac{\delta^2}{16}$
and so $2f_\delta(s) = 2s^2$, but
$$f_\delta(s-\frac\delta4) + f_\delta(s+\frac\delta4) = (s-\frac\delta4)^2 + (s+\frac\delta4)^2 -4 \frac{\delta^2}{16} = 2s^2 +2\frac{\delta^2}{16} - 4 \frac{\delta^2}{16} = 2s^2 -2\frac{\delta^2}{16} < 2f_\delta(s)$$
in contradiction to midpoint-convexity at $s$.
Now the only thing left to do is to prove midpoint convexity at $r$. That can only be broken if one of $x,y$ lies in the interval $[s,s+\delta]$, as otherwiese $f_\delta(x)=x^2$ which is convex everywhere and thus midpoint convex at any point.
So let's assume $x=r-\alpha$, $y=r+\alpha \in [s,s+\delta]$, we get similar to the above calculation $2f_\delta(r)=2r^2$ and
$$f_\delta(r-\alpha) + f_\delta(r+\alpha) = (r-\alpha)^2 + (r+\alpha^2) - 4b_\delta(r+\alpha-s) = 2r^2 + 2\alpha^2 - 4b_\delta(r+\alpha-s).$$
We have $r+\alpha \ge s \Rightarrow \alpha \ge s-r > 0$, so finally we get
$$f_\delta(r-\alpha) + f_\delta(r+\alpha) \ge 2r^2 + 2(s-r)^2 - 4b_\delta(r+\alpha-s)$$
But since $b_\delta(x) $ can be made as small as desired by decreasing $\delta$, we can find a suitable $\delta$ such that $2(s-r)^2 - 4b_\delta(x) >0$ for any $x$. That proves finally that the defined $f_\delta$ is a counteraxample for a sufficiently small $\delta$.
A: Problem: Let $r < s < 0$ be given. Find a $C^\infty$ function $f : \mathbb{R} \to [0, \infty)$ with $f(0)=0$ such that:
i) $f$ is strictly decreasing on $(-\infty, 0)$ and strictly increasing on $(0, \infty)$;
ii) $f(x) + f(2r - x) \ge 2f(r)$ for all $x\in \mathbb{R}$;
iii) $f(y) + f(2s - y) < 2f(s)$ for some $y \in \mathbb{R}$;
iv) $f$ is strictly convex in some neighbourhood of $0$.
Solution: Let $f(x) = x^4 + bx^3 + cx^2$. Clearly $f(0)=0$.
We claim that $f$ satisfies i), ii), iii) and iv) if
\begin{align}
32c - 9b^2 &> 0, \tag{1}\\
6r^2 + 3br + c &> 0, \tag{2}\\
6s^2 + 3bs + c &< 0. \tag{3}
\end{align}
Indeed, first, we have $f'(x) = x(4x^2 + 3bx+2c)$, and if $32c - 9b^2 > 0$,
then $4x^2 + 3bx+2c > 0$ for all $x\in \mathbb{R}$, so i) is satisfied;
second, we have $f(x) + f(2r-x) - 2f(r) = 2(x-r)^2((x-r)^2 + 6r^2 + 3br + c)$,
and if $6r^2 + 3br + c > 0$, then ii) is satisfied;
third, we have $f(y) + f(2s-y) - 2f(s) = 2(y-s)^2((y-s)^2 + 6s^2 + 3bs + c)$,
and if $6s^2 + 3bs + c < 0$, then iii) is satisfied.
fourth, we have $f''(x) = 12x^2 + 6bx + 2c$, and if $c > 0$ (follows from $32c - 9b^2 > 0$),
then $f''(0) = 2c > 0$ and $f$ is strictly convex in some neighbourhood of $0$ (due to continuity of $f''(x)$).
Then, we prove that there exist $b, c$ such that (1), (2) and (3) are satisfied.
We simply choose $b = -3s$ and
$$c = 3s^2 - \frac{1}{2}\min\left(3(2r-s)(r-s), \ \frac{15}{32}s^2\right).$$
Indeed, first, since $c \ge 3s^2 - \frac{1}{2}\cdot \frac{15}{32}s^2$,
we have $32c - 9b^2 = \frac{15}{2}s^2 > 0$;
second,
since $c \ge 3s^2 - \frac{1}{2} \cdot 3(2r-s)(r-s)$, we have
$6r^2 + 3br + c \ge \frac{3}{2}(2r-s)(r-s) > 0$;
third, since $c < 3s^2$,
we have $6s^2 + 3bs + c < 6s^2 + 3(-3s)s + 3s^2 = 0$.
