Remark 1: If $f: \ \mathbb{R} \to \mathbb{R}$ is a twice differentiable function
with $f'' < 0$ and finite maximum $f(x_0)$ at $x_0 > 0$. We may analyze the condition by using Danskin's theorem. See https://en.wikipedia.org/wiki/Danskin%27s_theorem

Remark 2: If $\alpha = 0$, we have $g(x, \beta) = \beta f(0) + (1-\beta)f(x)$
and $g^{\ast}(\beta) = \beta f(0) + (1-\beta) f(x_0)$.
Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = f(0) - f(x_0) \le 0$.

Remark 3: If $\alpha = \frac{1}{2}$, we have $g(x, \beta) = f(\frac{x}{2})$
and $g^{\ast}(\beta) = f(x_0)$. Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = 0$.

$\phantom{2}$

According to Remarks 2 and 3, we may restrict to $\alpha \in (0, \frac{1}{2})$.

Clearly, for any $\beta \in [0, \frac{1}{2}]$, $g(x, \beta)$ has a unique maximizer denoted by $x^\ast(\beta)$,
which is the unique solution of
$$\alpha\beta f'(\alpha x) + (1-\alpha)(1-\beta)f'((1-\alpha)x) = 0.$$
It implies that $\frac{x_0}{1-\alpha} \le x^\ast(\beta) \le \frac{x_0}{\alpha}$.
Thus, we have $g^{\ast}(\beta) = \max_{x\in S} g(x, \beta)$ for some compact set $S$ containing $[\frac{x_0}{1-\alpha}, \frac{x_0}{\alpha}]$, which satisfies the requirement of Danskin's theorem.

By using Danskin's theorem, we have
$$\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta)
= f(\alpha x^\ast(\beta)) - f((1-\alpha) x^\ast(\beta)).$$

A sufficient and necessary condition
for $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta)\le 0, \ \forall \beta \in [0, \frac{1}{2}]$
given $\alpha \in (0, \frac{1}{2})$ is that
\begin{align}
&\alpha\beta f'(\alpha x) = - (1-\alpha)(1-\beta)f'((1-\alpha)x)\\
\Longrightarrow\quad & f(\alpha x) \le f((1-\alpha) x), \quad \forall \alpha \in (0, \tfrac{1}{2}), \beta \in [0, \tfrac{1}{2}].
\end{align}
Maybe we can obtain some sufficient conditions.