Suppose $f: \mathbb R \longrightarrow \mathbb R$ is differentiable and convex. Define $\alpha: \mathbb R \times \{0,1\} \longrightarrow \mathbb R$ by letting $$\alpha(x,i) = f(x) + (i-x)f'(x),$$ and let $$F(x,y) = x\alpha_1(y) + (1-x)\alpha_0(y).$$
So, in general, we know that $F$ has a critical point at $y = x$.
From what I understand about an often cited result on the literature of scoring rules (see e.g. Gneiting and Raftery, "Strictly Proper Scoring Rules, Prediction, and Estimation"), it is known that for a fixed $x$, $F(x,y)$ restricted to $[0,1]\times[0,1]$ is maximized at $y=x$.
That all makes me think $F$ must be concave when restricted to $[0,1]\times[0,1]$. Is that so?
(Updated: I added the restriction to the unit square in response to some of the comments below.)
Last update: So the answer to this is indeed 'no'. But if I understand lemma 14 of this paper, though, $\alpha_i$ is quasi-concave.
The function $f(x)=\sin x$ is convex on $[\pi,2\pi]$. The functions $\sin x-x\cos x$ and $\sin x+(1-x)\cos x$ are neither convex nor concave on this interval.
For the sine function, If $F$ were concave, then $F(1,y)=2\sin y-2y\cos y+\cos y$ were also concave as a function of one variable, but this is not the case.