Let $g:\mathbb R \to \mathbb R$ be a continuous function, satisfying
$$
|g(\frac{x + y}{2}) |\le \left|\frac{g(x) + g(y)}{2}\right|,
$$
for every $x,y \in \mathbb R$.
Is it true that
$$
\left |g\big(\lambda x + (1-\lambda)y\big)\right|\le \left|\lambda g(x) + (1-\lambda)g(y) \right|
$$
for every $x,y$ and $\lambda \in [0,1]$?
Note that the situation here is slightly different from the classical one, when there are no absolute values.
In that case, midpoint-convexity does imply full convexity. However, trying to adapt the proof, I hit an obstacle:
The proof (for the standard unsigned convexity) begins by showing
$$
g((x_1+\dots+x_m)/m)\leq (g(x_1)+\dots+g(x_m))/m$$ for any $m=2^k$
by applying midpoint convexity $k$ times.
However, this first step seems to fail in our context:
$$
|g|(\frac{x + y+z+w}{4})=\left|g\left(\frac{1}{2}(\frac{x + y}{2})+\frac{1}{2}(\frac{z + w}{2})\right) \right| \le \left|\frac{g(\frac{x + y}{2}) + g(\frac{z + w}{2})}{2}\right| \le \frac{|g|(\frac{x + y}{2}) + |g|(\frac{z + w}{2})}{2} \le \frac{1} {4}\big(|g(x)+g(y)|+|g(z)+g(w)|\big).
$$
We lost something, since we wanted $|g|(\frac{x + y+z+w}{4}) \le \frac{1} {4}\big(|g(x)+g(y)+g(z)+g(w)|\big).$
Best Answer
A function $|g|$ is continuous and mid-point convex, so it is convex (see, for instance, [L, 5.1]). It implies the required inequality $$|g(\frac{x + y}{2}) |\le \left|\frac{g(x) + g(y)}{2}\right|$$ for every $x,y \in \mathbb R$ such that $f$ does not change its sign on $[x,y]$. The condition implies that $g$ is zero between any two its zeroes, so it either does not change its sign (and then the required claim holds) or changes its sign only once. In the latter case the required claim can fail. For instance, let $g(x)=x|x|$ for each $x\in\Bbb R$. It is easy to check that $g$ satisfies the required condition, but for $x=-4/3$, $y=3$, and $\lambda=8/9$ we have $$\left |g\big(\lambda x + (1-\lambda)y\big)\right|=\frac {529}{729}>\frac {423}{729}=\left|\lambda g(x) + (1-\lambda)g(y) \right|.$$
References
[L] Hojoo Lee, Topics in Inequalities - Theorems and Techniques, (February 25, 2006).