Proving domination phenomena of convexity gaps Let $\phi:\mathbb [0,\infty) \to [0,\infty)$ be affine on $[0,a]$ and strictly convex on $[a,\infty)$, for some  $a >0$.
Let $x<a<c<y$, where $c=t x + (1-t) y$, and define 
$$G_1= t\phi(x) + (1-t)\phi(y)-\phi(t x + (1-t) y). \tag{1}.$$
Now, rewrite $c$ as a convex combination of $a$ and $y$, i.e. $c=s a + (1-s) y$, and set
$$G_2= s\phi(a) + (1-s)\phi(y)-\phi(s a + (1-s) y)  \tag{2}$$.
Does $G_2 \le G_1$? 
 A: Unless I'm missing something, this is true, and there a lot of extraneous assumptions involved here: we don't need the function to be positive, affine on any interval, or strict convexity. For any convex function $\phi$ over a domain including $a < b < c < d$ such that $p:= t a + (1 - t)d = sb + (1 - s)c$, we have
$$t\phi(a) + (1 - t)\phi(d) \ge s\phi(b) + (1 - s)\phi(c), \tag{$\star$}$$
which implies domination of the gaps:
$$t\phi(a) + (1 - t)\phi(d) - \phi(p) \ge s\phi(b) + (1 - s)\phi(c) - \phi(p).$$
To prove $(\star)$, let $A$ be the half-space of $\Bbb{R}^2$ below the line generated by the points $(a, \phi(a))$ and $(d, \phi(d))$, i.e.
$$A = \left\{(x, y) \in \Bbb{R}^2 : y \le \frac{\phi(d) - \phi(a)}{d - a}(x - a) + \phi(a)\right\}.$$
Note that $A$ is convex.
I claim that $(b, \phi(b)), (c, \phi(c)) \in A$. Note that
\begin{align*}
\frac{\phi(d) - \phi(a)}{d - a}(b - a) + \phi(a) &= \frac{b - a}{d - a} \phi(d) + \frac{d - b}{d - a}\phi(a) \\
&\ge \phi\left(\frac{b - a}{d - a}d + \frac{d - b}{d - a}a\right) \\
&= \phi(b),
\end{align*}
hence $(b, \phi(b)) \in A$. Similarly, $(c, \phi(c)) \in A$. Thus, the entire line segment between them is contained in $A$. Therefore,
$$s(b, \phi(b)) + (1 - s)(c, \phi(c)) = (p, s\phi(b) + (1 - s)\phi(c)) \in A,$$
i.e.
\begin{align*}
s\phi(b) + (1 - s)\phi(c) &\le \frac{\phi(d) - \phi(a)}{d - a}(p - a) + \phi(a) \\
&= \frac{\phi(d) - \phi(a)}{d - a}(ta + (1 - t)d - a) + \phi(a) \\
&= (1 - t)\frac{\phi(d) - \phi(a)}{d - a}(d - a) + \phi(a) \\
&= t\phi(a) + (1 - t)\phi(d),
\end{align*}
proving $(\star)$.
