<p>Take a dataset <span class="math-container">$D = (B^t, x^t)_{t \in T}$</span> such that for all <span class="math-container">$t$</span>, <span class="math-container">$x^t \in B^t$</span>. I'll say that <span class="math-container">$D$</span> is <strong>rationalisable</strong> by the utility function <span class="math-container">$u$</span> if for all <span class="math-container">$t$</span> and all <span class="math-container">$x \in B$</span>: <span class="math-container">$u(x^t) \ge u(x)$</span>.</p>
<p>If you only impose convexity on <span class="math-container">$u$</span>, then any dataset <span class="math-container">$D$</span> is rationalisable by the constant (convex) utility function <span class="math-container">$u(x) = k$</span>. So this has not testable implications.</p>
<p>Assume now that you impose <strong>strict monotonicity</strong> on <span class="math-container">$u$</span> in the sense that <span class="math-container">$x &gt; y$</span> implies <span class="math-container">$u(x) &gt; u(y)$</span>.</p>
<p>Consider a budget set <span class="math-container">$\overline{y^t z^t}$</span> and take the comprehensive hull:
<span class="math-container">$$
CH(\overline{y^t z^t}) = \left\{x \in \mathbb{R}^n: x \le w \text{ for some } w \in \overline{y^t z^t}\right\}.
$$</span></p>
<p>Then we can show that the dataset <span class="math-container">$D = (\overline{y^t z^t}, x^t)_{t \in T}$</span> is rationalisable by a montone utility function <span class="math-container">$u$</span> <strong>if and only if</strong> the dataset <span class="math-container">$D' = (CH(\overline{y^t z^t}), x^t)_{t \in T}$</span> is rationalisable by the same monontone utility function <span class="math-container">$u$</span>.</p>
<blockquote>
<p>For the proof: <span class="math-container">$(\to)$</span> Let <span class="math-container">$D$</span> be rationalisable by the montone function <span class="math-container">$u$</span>. We want to show that for all <span class="math-container">$t$</span> and all <span class="math-container">$x \in CH(\overline{y^t z^t})$</span> we have <span class="math-container">$u(x^t) \ge u(x)$</span>. Indeed, if <span class="math-container">$x \in CH(\overline{y^t z^t})$</span>, we know that <span class="math-container">$x \le w$</span> for some <span class="math-container">$w \in \overline{y^t z^t}$</span>. As such:
<span class="math-container">$$
u(x) \le u(w) \le u(x^t),
$$</span>
where the first inequality comes from the monotonicity conditionon <span class="math-container">$u$</span> and the last one from rationalisability of D.</p>
</blockquote>
<blockquote>
<p><span class="math-container">$(\leftarrow)$</span> Let <span class="math-container">$D'$</span> be rationalisable by <span class="math-container">$u$</span>. Let us first show that for all <span class="math-container">$t$</span>, <span class="math-container">$x^t \in \overline{y^t z^t}$</span>. If not, then as <span class="math-container">$x^t \in CH(\overline{y^t z^t})$</span> there must exist a <span class="math-container">$w \in \overline{y^t z^t}$</span> such that <span class="math-container">$x^t &lt; w$</span>; Then, by monotonicity of <span class="math-container">$u$</span> and rationalisability of <span class="math-container">$D'$</span>:
<span class="math-container">$$
u(x^t) &lt; u(w) \le u(x^t),
$$</span>
a contradiction. Now for any <span class="math-container">$z \in \overline{y^t z^t}$</span> it then follows from rationalisability of <span class="math-container">$D'$</span> that <span class="math-container">$u(z) \le u(x^t)$</span>, so <span class="math-container">$D$</span> is also rationalisable by <span class="math-container">$u$</span>.</p>
</blockquote>
<p>This shows the following equivalence:</p>
<ol>
<li>The dataset <span class="math-container">$D = (\overline{y^t, z^t}, x^t)_{t \in T}$</span> is rationalisable by a convex and montone utility function <span class="math-container">$u$</span>.</li>
<li>The dataset <span class="math-container">$D = (CH(\overline{y^t, z^t}), x^t)_{t \in T}$</span> is rationalisable by a convex and monotone utility function <span class="math-container">$u$</span>.</li>
</ol>
<p>As the sets <span class="math-container">$CH(\overline{y^t z^t})$</span> 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)</p>

CC BY-SA 4.0
2021-04-23T20:54:30.153Z
Very cool answer tdm! I truly appreciate you. I am also sorry that I forgot to emphasize that I want to use SARP (or similar axioms), such that there is only one unique demand for each budget set. The JET paper we are using focus only on GARP.
High GPA
CC BY-SA 4.0
2021-04-23T08:45:01.380Z
2021-04-23T08:45:01.380Z
tdm
<p>The idea is to consider the hyperplane tangent at the indifference curve through <span class="math-container">$x^t$</span> as a &quot;linear budget&quot;. These linear budgets have to include the set <span class="math-container">$\overline{y^t, z^t}$</span>. Then making use of these these linear budgets, we can leverage the results of Matzkin &amp; Richter (JET,1991, Testing strictly concave rationality) to obtain a strictly convex rationalisation..</p>
<p><strong>Afriat style Theorem</strong> Let <span class="math-container">$D = (\overline{y^t z^t}, x^t)_{t \in T}$</span> be a dataset. Then the following are equivalent:</p>
<ol>
<li><p><span class="math-container">$D$</span> is rationalisable by a strictly monotone utility function with strictly convex indifference curves.</p>
</li>
<li><p>For all <span class="math-container">$t \in T$</span>, there exists a vector <span class="math-container">$p^t \in \mathbb{R}^n_{++}$</span>, and a number <span class="math-container">$u^t \in \mathbb{R}$</span> such that (i) for all <span class="math-container">$t, v, \in T$</span>, if <span class="math-container">$x^v \ne x^t$</span>:
<span class="math-container">$$
u^v - u^t &lt; p^t(x^v - x^t),
$$</span>
(ii) for all <span class="math-container">$t, v \in T$</span> with <span class="math-container">$x^v = x^t$</span>,
<span class="math-container">$$
u^t = u^v,
$$</span>
and (iii) for all <span class="math-container">$t \in T$</span>:
<span class="math-container">$$
p^t x^t \ge p^t y^t \text{ and } p^t x^t \ge z^t.
$$</span></p>
</li>
<li><p>For all <span class="math-container">$t \in T$</span>, there exists a vector <span class="math-container">$\tilde p^t \in \mathbb{R}^n_{++}$</span> such that (i) for all <span class="math-container">$t \in T$</span>:
<span class="math-container">$$
\tilde p^t x^t \ge \tilde p^t y^t \text{ and } \tilde p^t x^t \ge \tilde p^t z^t
$$</span>
and (ii):
<span class="math-container">$$
(\tilde p^t, x^t)_{t \in T} \text{ satisfies SARP}.
$$</span></p>
</li>
<li><p><span class="math-container">$D$</span> is rationalisable by a strictly monotone and strictly concave utility function.</p>
</li>
</ol>
<blockquote>
<p>(<span class="math-container">$1 \to 3$</span>) The set <span class="math-container">$\overline{y^t z^t}$</span> is convex and the upper countour sets <span class="math-container">$UC(x^t) = \{y \in \mathbb{R}^n: u(y) \ge u(x^t)\}$</span> are strictly convex. As such, using a suitable supporting hyperplane theorem, we can find vectors <span class="math-container">$\tilde p^t$</span> tangent to <span class="math-container">$UC(x^t)$</span> at <span class="math-container">$x_t$</span> that separates <span class="math-container">$UC(x^t)$</span> from <span class="math-container">$\overline{y^t z^t}$</span> which means that <span class="math-container">$\tilde p^t x^t \ge \tilde p^t y^t$</span> and <span class="math-container">$\tilde p^t x^t \ge \tilde p^t z^t$</span>. This shows (i). Also from strict monotonicity of <span class="math-container">$u$</span>, we must have that <span class="math-container">$\tilde p^t \in \mathbb{R}^n_{++}$</span>. Next, for all <span class="math-container">$w \ne x^t$</span> with <span class="math-container">$\tilde p^t w \le \tilde p^t x^t$</span>, we have <span class="math-container">$u(x^t) &gt; u(w)$</span>.</p>
<p>Now, define <span class="math-container">$x^t S x^v$</span> if <span class="math-container">$x^t \ne x^v$</span> and <span class="math-container">$\tilde p^t x^t \ge \tilde p^t x^v$</span>. Then it follows that <span class="math-container">$x^t S x^v$</span> implies <span class="math-container">$u(x^t) &gt; u(x^v)$</span>, so <span class="math-container">$S$</span> has to be acyclic, which shows that <span class="math-container">$(\tilde p^t, x^t)_{t \in T}$</span> satisfies SARP.</p>
</blockquote>
<blockquote>
<p>(<span class="math-container">$2 \iff 3$</span>) Follows from Theorem 2 in Matzkin &amp; Richter (Testing strictly concave rationality, 1991, JET), where we define <span class="math-container">$p^t = \lambda^t \tilde p^t$</span>.</p>
</blockquote>
<blockquote>
<p>(<span class="math-container">$3 \to 4$</span>) From Matzkin and Richter it follows that 3 implies the existence of a strictly monotone and strictly concave utility function such that for all <span class="math-container">$w$</span> with <span class="math-container">$\tilde p^t w \le \tilde p^t x^t$</span>, we have <span class="math-container">$u(x^t) \ge u(w)$</span>. Then if <span class="math-container">$w \in \overline{y^t z^t}$</span> we immediately have that <span class="math-container">$\tilde p^t w \le \tilde p^t x^t$</span> (as <span class="math-container">$w$</span> is a convex combination of <span class="math-container">$y^t$</span> and <span class="math-container">$z^t$</span>) so <span class="math-container">$u(x^t) \ge u(w)$</span> as we wanted to show.</p>
</blockquote>
<blockquote>
<p>(<span class="math-container">$4 \to 1$</span>) Straighforward.</p>
</blockquote>
<p><strong>Remark 1</strong>: The conditions in 2 provide a set of linear inequalities which can be efficiently  verified (except maybe for the strict inequality which will require some <span class="math-container">$\varepsilon$</span>-tweaking).</p>
<p><strong>Remark 2</strong>: To get some more intuition behind condtion 3-(i), consider the two dimensional case, <span class="math-container">$y^t, z^t, x^t \in \mathbb{R}^2$</span>. See figure below. there are 3 cases. Here the dashed line gives the slope <span class="math-container">$\tilde p^t$</span></p>
<ol>
<li>If <span class="math-container">$y^t &gt; z^t$</span>. Then, due to strict monotonicity, <span class="math-container">$x^t$</span> has to be equal to <span class="math-container">$y^t$</span>. There are (a priori) no restrictions on <span class="math-container">$\tilde p^t$</span> (see top left panel of figure). A similar reasoning holds if <span class="math-container">$z^t &gt; y^t$</span>.</li>
<li>If <span class="math-container">$y^t$</span> and <span class="math-container">$z^t$</span> are not ordered and <span class="math-container">$x^t = y^t$</span> or <span class="math-container">$x^t = z^t$</span>. Then the slope <span class="math-container">$\tilde p^t$</span> is either flatter, steeper, or equal to the slope from the line <span class="math-container">$\overline{y^t z^t}$</span>. (panel top right or bottom left).</li>
<li>If <span class="math-container">$y^t$</span> and <span class="math-container">$z^t$</span> are not ordered and <span class="math-container">$x^t$</span> is between <span class="math-container">$y^t$</span> and <span class="math-container">$z^t$</span>, then <span class="math-container">$\tilde p^t$</span> has to equal the slope from the line <span class="math-container">$\overline{y^t z^t}$</span>. (panel bottom right).</li>
</ol>
<p><a href="https://i.sstatic.net/UVfA2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UVfA2.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-04-24T12:32:45.157Z
I am quite interested in your condition (3.i). For a 2-dimensional problem, does (3.i) simply mean that the budget segment has negative slope? Suppose that everything is in the first quadrant.
High GPA
CC BY-SA 4.0
2021-04-24T12:00:36.607Z
2021-04-26T06:07:24.213Z
2021-04-26T06:07:24.213Z
tdm
<p><strong>Backgroud:</strong></p>
<p>SARP can be defined on general budget set.</p>
<blockquote>
<p>SARP: Assume for all <span class="math-container">$B$</span> the choice <span class="math-container">$c(B)$</span> is only one element. If <span class="math-container">$x_i,x_{i+1}\in B_i$</span>, and <span class="math-container">$x_i = c(B_i)$</span>, for all <span class="math-container">$i\in \{1,N-1\}$</span>, then <span class="math-container">$x_1=c(B_1)\notin B_N$</span>.</p>
</blockquote>
<p>I learnt that, if the budget is this: <span class="math-container">$B_i=\{x\mid p_ix\leq p_ix_i\}$</span>, then, SARP implies that the finite data is rationalized by convex preferece.</p>
<p><strong>Motivation:</strong> 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.</p>
<p><strong>My question</strong>: Let <span class="math-container">$y^i$</span> and <span class="math-container">$z^i$</span> be two points, and <span class="math-container">$\overline{ y^iz^i}$</span> is the segment connecting <span class="math-container">$y^i$</span> and <span class="math-container">$z^i$</span>.</p>
<p>Let <span class="math-container">$B^i=\overline{ y^iz^i}$</span> be the segment budget set.</p>
<p>Let the data be <span class="math-container">$(x^i,B^i)^{i\in\{1,2,...,n\}}$</span> where <span class="math-container">$n$</span> is a natural number.</p>
<p><strong><em>Can we still have a convex rationalization of the data</em>?</strong></p>
<p>That is, we are trying to find an Axiom, similar to SARP, such that the following two conditions are equivalent:</p>
<ol>
<li>The data satisfies the Axiom,</li>
<li>The data is rationalized by a convex preference.</li>
</ol>
<p>If we could also get monotonicity, then it will be better. The focus is on convexity of preference.</p>
<p><strong>My try:</strong> 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 <span class="math-container">$B$</span> is monotone, and a segment budget set is not monotone.</p>
<p>Any comments, long or short, will help. Any related reference will help, too. Thanks in advance.</p>

CC BY-SA 4.0
2021-04-21T13:44:43.373Z
This question is difficult to answer without knowing the full framework. What kind of (additional) assumptions are you imposing on the utility function (e.g. montonicity?).

For example, if the underlying utiliity function is strictly monotone, then whether the budget set is monotone or not will not really matter, as any rational choice will have to be on the frontier of the budget set. In this case, you can simply replace the budget set by its comprehensive hull.
tdm
CC BY-SA 4.0
2021-04-21T21:29:09.723Z
@tdm Thanks for your comment. The question is updated thanks to your suggestions.
High GPA
<microeconomics><reference-request><preferences><theory><choice-theory>
Convex rationalization when the budget sets are segments?
High GPA
# Convex rationalization when the budget sets are segments?

(Asked by: username_0 on 2021-04-19)

**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:

1. The data satisfies the Axiom,
2. 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 "Revealed preference analysis for convex
rationalizations on nonlinear budget sets". 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.

**Tags:** microeconomics, reference-request, preferences, theory, choice-theory
> **Comments:**
> **username_1**: This question is difficult to answer without knowing the full framework. What kind of (additional) assumptions are you imposing on the utility function (e.g. montonicity?).  For example, if the underlying utiliity function is strictly monotone, then whether the budget set is monotone or not will not really matter, as any rational choice will have to be on the frontier of the budget set. In this case, you can simply replace the budget set by its comprehensive hull.  
> **username_0**: @tdm Thanks for your comment. The question is updated thanks to your suggestions.  

---

## 2 answers

### Answer by username_1 on 2021-04-24. Score: 3 | Accepted answer

The idea is to consider the hyperplane tangent at the indifference curve through $x^t$ as a "linear budget". These linear budgets have to include the set $\overline{y^t, z^t}$. Then making use of these these linear budgets, we can leverage the results of Matzkin & Richter (JET,1991, Testing strictly concave rationality) to obtain a strictly convex rationalisation..

**Afriat style Theorem** Let $D = (\overline{y^t z^t}, x^t)\_{t \in T}$ be a dataset. Then the following are equivalent:

1. $D$ is rationalisable by a strictly monotone utility function with strictly convex indifference curves.
2. For all $t \in T$, there exists a vector $p^t \in \mathbb{R}^n\_{++}$, and a number $u^t \in \mathbb{R}$ such that (i) for all $t, v, \in T$, if $x^v \ne x^t$:
   $$
   u^v - u^t < p^t(x^v - x^t),
   $$
   (ii) for all $t, v \in T$ with $x^v = x^t$,
   $$
   u^t = u^v,
   $$
   and (iii) for all $t \in T$:
   $$
   p^t x^t \ge p^t y^t \text{ and } p^t x^t \ge z^t.
   $$
3. For all $t \in T$, there exists a vector $\tilde p^t \in \mathbb{R}^n\_{++}$ such that (i) for all $t \in T$:
   $$
   \tilde p^t x^t \ge \tilde p^t y^t \text{ and } \tilde p^t x^t \ge \tilde p^t z^t
   $$
   and (ii):
   $$
   (\tilde p^t, x^t)\_{t \in T} \text{ satisfies SARP}.
   $$
4. $D$ is rationalisable by a strictly monotone and strictly concave utility function.

> ($1 \to 3$) The set $\overline{y^t z^t}$ is convex and the upper countour sets $UC(x^t) = \{y \in \mathbb{R}^n: u(y) \ge u(x^t)\}$ are strictly convex. As such, using a suitable supporting hyperplane theorem, we can find vectors $\tilde p^t$ tangent to $UC(x^t)$ at $x\_t$ that separates $UC(x^t)$ from $\overline{y^t z^t}$ which means that $\tilde p^t x^t \ge \tilde p^t y^t$ and $\tilde p^t x^t \ge \tilde p^t z^t$. This shows (i). Also from strict monotonicity of $u$, we must have that $\tilde p^t \in \mathbb{R}^n\_{++}$. Next, for all $w \ne x^t$ with $\tilde p^t w \le \tilde p^t x^t$, we have $u(x^t) > u(w)$.
>
> Now, define $x^t S x^v$ if $x^t \ne x^v$ and $\tilde p^t x^t \ge \tilde p^t x^v$. Then it follows that $x^t S x^v$ implies $u(x^t) > u(x^v)$, so $S$ has to be acyclic, which shows that $(\tilde p^t, x^t)\_{t \in T}$ satisfies SARP.

> ($2 \iff 3$) Follows from Theorem 2 in Matzkin & Richter (Testing strictly concave rationality, 1991, JET), where we define $p^t = \lambda^t \tilde p^t$.

> ($3 \to 4$) From Matzkin and Richter it follows that 3 implies the existence of a strictly monotone and strictly concave utility function such that for all $w$ with $\tilde p^t w \le \tilde p^t x^t$, we have $u(x^t) \ge u(w)$. Then if $w \in \overline{y^t z^t}$ we immediately have that $\tilde p^t w \le \tilde p^t x^t$ (as $w$ is a convex combination of $y^t$ and $z^t$) so $u(x^t) \ge u(w)$ as we wanted to show.

> ($4 \to 1$) Straighforward.

**Remark 1**: The conditions in 2 provide a set of linear inequalities which can be efficiently verified (except maybe for the strict inequality which will require some $\varepsilon$-tweaking).

**Remark 2**: To get some more intuition behind condtion 3-(i), consider the two dimensional case, $y^t, z^t, x^t \in \mathbb{R}^2$. See figure below. there are 3 cases. Here the dashed line gives the slope $\tilde p^t$

1. If $y^t > z^t$. Then, due to strict monotonicity, $x^t$ has to be equal to $y^t$. There are (a priori) no restrictions on $\tilde p^t$ (see top left panel of figure). A similar reasoning holds if $z^t > y^t$.
2. If $y^t$ and $z^t$ are not ordered and $x^t = y^t$ or $x^t = z^t$. Then the slope $\tilde p^t$ is either flatter, steeper, or equal to the slope from the line $\overline{y^t z^t}$. (panel top right or bottom left).
3. If $y^t$ and $z^t$ are not ordered and $x^t$ is between $y^t$ and $z^t$, then $\tilde p^t$ has to equal the slope from the line $\overline{y^t z^t}$. (panel bottom right).

[![enter image description here](https://i.sstatic.net/UVfA2.png)](https://i.sstatic.net/UVfA2.png)

> **Comments:**
> **username_0**: I am quite interested in your condition (3.i). For a 2-dimensional problem, does (3.i) simply mean that the budget segment has negative slope? Suppose that everything is in the first quadrant.  

---

### Answer by username_1 on 2021-04-23. Score: 3

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 $x > y$ implies $u(x) > u(y)$.

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 $x^t < w$; Then, by monotonicity of $u$ and rationalisability of $D'$:
> $$
> u(x^t) < u(w) \le u(x^t),
> $$
> 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:

1. The dataset $D = (\overline{y^t, z^t}, x^t)\_{t \in T}$ is rationalisable by a convex and montone utility function $u$.
2. 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)

> **Comments:**
> **username_0**: Very cool answer tdm! I truly appreciate you. I am also sorry that I forgot to emphasize that I want to use SARP (or similar axioms), such that there is only one unique demand for each budget set. The JET paper we are using focus only on GARP.  

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