Convex rationalization when the budget sets are segments? Backgroud:
SARP can be defined on general budget set.

SARP: Assume for all $B$ the choice $c(B)$ is only one element. If $x_i,x_{i+1}\in B_i$, and $x_i = c(B_i)$, for all $i\in \{1,N-1\}$, then $x_1=c(B_1)\notin B_N$.

I learnt that, if the budget is this: $B_i=\{x\mid p_ix\leq p_ix_i\}$, then, SARP implies that the finite data is rationalized by convex preferece.
Motivation: most real life problems have more complicated budgets. For example, if the agent is allocating her money between risky security and riskless cash, then her budget will be a segment rather than a triangle.
My question: Let $y^i$ and $z^i$ be two points, and $\overline{ y^iz^i}$ is the segment connecting $y^i$ and $z^i$.
Let $B^i=\overline{ y^iz^i}$ be the segment budget set.
Let the data be $(x^i,B^i)^{i\in\{1,2,...,n\}}$ where $n$ is a natural number.
Can we still have a convex rationalization of the data?
That is, we are trying to find an Axiom, similar to SARP, such that the following two conditions are equivalent:

The data satisfies the Axiom,
The data is rationalized by a convex preference.

If we could also get monotonicity, then it will be better. The focus is on convexity of preference.
My try: I've gone through the paper &quot;Revealed preference analysis for convex
rationalizations on nonlinear budget sets&quot;. However, this paper requires that budget set $B$ is monotone, and a segment budget set is not monotone.
Any comments, long or short, will help. Any related reference will help, too. Thanks in advance.
Take a dataset $D = (B^t, x^t)_{t \in T}$ such that for all $t$, $x^t \in B^t$. I'll say that $D$ is rationalisable by the utility function $u$ if for all $t$ and all $x \in B$: $u(x^t) \ge u(x)$.
If you only impose convexity on $u$, then any dataset $D$ is rationalisable by the constant (convex) utility function $u(x) = k$. So this has not testable implications.
Assume now that you impose strict monotonicity on $u$ in the sense that $x &gt; y$ implies $u(x) &gt; u(y)$.
Consider a budget set $\overline{y^t z^t}$ and take the comprehensive hull:
$$
CH(\overline{y^t z^t}) = \left\{x \in \mathbb{R}^n: x \le w \text{ for some } w \in \overline{y^t z^t}\right\}.
$$
Then we can show that the dataset $D = (\overline{y^t z^t}, x^t)_{t \in T}$ is rationalisable by a montone utility function $u$ if and only if the dataset $D' = (CH(\overline{y^t z^t}), x^t)_{t \in T}$ is rationalisable by the same monontone utility function $u$.

For the proof: $(\to)$ Let $D$ be rationalisable by the montone function $u$. We want to show that for all $t$ and all $x \in CH(\overline{y^t z^t})$ we have $u(x^t) \ge u(x)$. Indeed, if $x \in CH(\overline{y^t z^t})$, we know that $x \le w$ for some $w \in \overline{y^t z^t}$. As such:
$$
u(x) \le u(w) \le u(x^t),
$$
where the first inequality comes from the monotonicity conditionon $u$ and the last one from rationalisability of D.


$(\leftarrow)$ Let $D'$ be rationalisable by $u$. Let us first show that for all $t$, $x^t \in \overline{y^t z^t}$. If not, then as $x^t \in CH(\overline{y^t z^t})$ there must exist a $w \in \overline{y^t z^t}$ such that $x^t &lt; w$; Then, by monotonicity of $u$ and rationalisability of $D'$:
$$
u(x^t) &lt; u(w) \le u(x^t),
$$
a contradiction. Now for any $z \in \overline{y^t z^t}$ it then follows from rationalisability of $D'$ that $u(z) \le u(x^t)$, so $D$ is also rationalisable by $u$.

This shows the following equivalence:

The dataset $D = (\overline{y^t, z^t}, x^t)_{t \in T}$ is rationalisable by a convex and montone utility function $u$.
The dataset $D = (CH(\overline{y^t, z^t}), x^t)_{t \in T}$ is rationalisable by a convex and monotone utility function $u$.

As the sets $CH(\overline{y^t z^t})$ are comprehensive, the revealed preference conditions for answering the condition 2) can be found in the JET paper that you refer to (Revealed preference analysis for convex rationalizations on nonlinear budget sets)