Efficient and strategy-proof allocation mechanisms in many-agent economies

Abstract

In this paper, we show that in pure exchange economies, any Pareto-efficient and strategy-proof allocation mechanism is alternately dictatorial; that is, it always allocates the total endowment to a single agent even if the receivers vary. While many studies have shown that such an allocation mechanism is dictatorial in two-agent economies, it has long remained an open question whether such a characterization can be obtained in many-agent economies.

Appendix

1.1 Proof of Lemma 1

For example, we construct $\hat{R}$ as follows. Figure 5 depicts the construction. Fix a vector $\bar{x}\in Q'$. For each $y\in \partial Q\bigcap I(\bar{x};\bar{R})$, we consider the hyperplane $y+p(\bar{R},y)^\perp$ that supports the upper contour set $UC(\bar{x};\bar{R})$ at y and the space $y+p(\bar{R},y)^\perp -R_{++}^L$ strictly on the lower left-hand side of the hyperplane. We define V as

$$\begin{aligned} V=UC(\bar{x};\bar{R})\bigcup Q\setminus \bigcup _{y\in \partial Q \bigcap I(\bar{x};\bar{R})} \{y+p(\bar{R},y)^\perp -R^L_{++}\}. \end{aligned}$$

That is, V is $U(\bar{x};\bar{R})\bigcup Q$ whose subsets in the lower left-hand sides of the hyperplanes $y+p(\bar{R};y)^\perp$ for $y\in \partial Q\bigcap I(\bar{x};\bar{R})$ are cut off. Note that $\partial V$ equals $I(\bar{x};\bar{R})$ outside of Q. Further, $\partial V\bigcap Q'$ is away from $UC(\bar{x};\bar{R})$; that is, they have no intersection.

Fig. 5

Proof of Lemma 1

We modify V into a smooth set by truncating its edge as follows. We let $\bar{D}_\epsilon$ denote a closed ball with sufficiently small radius $\epsilon$ and define $V'$ as the union of such closed balls included in V:

$$\begin{aligned} V'=\bigcup _{\bar{D}_\epsilon \subset V}\bar{D}_\epsilon . \end{aligned}$$

See Lemma A2 in Momi (2017) for the proof that $V'$ is a smooth set as desired. Clearly, $\partial V'$ equals $\partial V$ outside Q

However, $V'$ is not a strictly convex set. We modify $\partial V'$ into the surface of a strictly convex set so that it is an indifference set of a preference in ${{\mathcal{R}}}$. We pick any positive vector $\alpha \in R_{++}^L$ and consider the hyperplane $\alpha ^\perp$ passing through the origin and perpendicular to the vector $\alpha$. We let L(z) denote the half line starting from z and extending in the direction of $\alpha$: $L(z)=\{x\in R^L: x=z+t\alpha , t\ge 0\}$. With a sufficiently small positive scalar s, we define

$$\begin{aligned} I=\bigcup _{z\in \alpha ^\perp }\left\{ s\left( L(z)\bigcap I(\bar{x};\bar{R})\right) +(1-s)\left( L(z)\bigcap \partial V'\right) \right\} \end{aligned}$$

We let $\tilde{R}\in{{\mathcal{R}}}$ be the preference that has I as its indifference set. See Lemma A1 in Momi (2017) for the proof that I is well defined as an indifference set. As $I(\bar{x};\bar{R})$ equals $\partial V'$ outside Q, I equals $I(\bar{x};\bar{R})$ outside Q. In addition, $I\bigcap \overline{Q'}$ is away from $UC(\bar{x};\bar{R})$. We fix a vector $\tilde{x}\in I$.

We pick $\hat{x}=t\bar{x}$, where $t<1$, so that $I\bigcap \overline{Q'}$ has no intersection with $UC(\hat{x};\bar{R})$. Clearly, we can pick such $\hat{x}$ with t sufficiently close to 1 because $I(\hat{x};\bar{R})\bigcap \overline{Q'}$ converges to $I(\bar{x};\bar{R})\bigcap \overline{Q'}$ as $t\rightarrow 1$ with respect to the standard metric introduced into the set of compact subsets of $R^L$.⁶

When a preference R is sufficiently close to $\bar{R}$, $I(\hat{x};R)\bigcap K$ is sufficiently close to $I(\hat{x}; \bar{R})\bigcap K$ with any compact set K. Therefore, we set the preference R so close to $\bar{R}$ that $I(\hat{x}, R)\bigcap \overline{Q}$ has no intersection with $UC(\bar{x};\bar{R})$ and $I\bigcap \overline{Q'}$ has no intersection with $UC(\hat{x};R)$.

We define

$$\begin{aligned} W=\left\{ UC(\hat{x};R)\bigcap UC(\tilde{x};\tilde{R})\bigcap Q\right\} \bigcup \left\{ UC(\tilde{x};\tilde{R})\bigcap Q^c\right\} . \end{aligned}$$

We observe W in Q and outside Q separately. In Q, W equals $UC(\hat{x};R)\bigcap UC(\tilde{x};\tilde{R})$. In particular, $I=I(\tilde{x}; \tilde{R})$ is away from $UC(\hat{x}; R)$ in $Q'$. Therefore, $\partial W$ equals $I(\hat{x};R)$ in $Q'$; that is $\partial W\bigcap Q'=I(\hat{x}; R)\bigcap Q'$. Outside Q, W equals $UC(\tilde{x};\tilde{R})$. Therefore, $\partial W$ equals $I(\tilde{x};\tilde{R})$ outside Q. Thus, we have $\partial W\bigcap Q^c=I(\tilde{x};\tilde{R})\bigcap Q^c=I(\bar{x};\bar{R})\bigcap Q^c$.

As this set W has the edge at the intersection of $I(\hat{x}; R)$ and $I(\tilde{x},\tilde{R})$ in $Q\setminus Q'$, we slightly modify it into a smooth set by truncating the edge as we did for V. We let $\bar{D}_\epsilon$ denote a closed ball with sufficiently small radius $\epsilon$ and define $W'$ as the union of such closed balls included in W:

$$\begin{aligned} W'=\bigcup _{\bar{D}_\epsilon \subset V}\bar{D}_\epsilon . \end{aligned}$$

We let $\hat{R}\in{{\mathcal{R}}}$ be a preference that has $W'$ as its upper contour set. When $\epsilon$ is sufficiently small, $W'$ equals W in $Q'\bigcup Q^c$, and hence $W'$ has the above-mentioned properties of W: $\partial W'\bigcap Q'=I(\tilde{x};\tilde{R})\bigcap Q'$ and $\partial W'\bigcap Q^c=I(\bar{x};\bar{R})\bigcap Q^c$. Thus, $\hat{R}$ is identical to R in $Q'$ and identical to $\bar{R}$ in $Q^c$ as desired.

1.2 Proof of Proposition 1

The proof is essentially the same as the proof in Momi (2017, Section 4). Although a slight logical difference is only in the proof of Lemma A3, we rewrite the whole proof. As explained after the statement of Proposition 1, we focus on the determinacy of consumption among agents 1 and 2, and transform the preferences in Momi’s proof into preferences in our domains using Lemma 1.

Throughout this section, we suppose the following conditions of the proposition: $\bar{\mathbf{R }}=(\bar{R}^1,\ldots ,\bar{R}^N)$ is a preference profile such that $\bar{R}^i$, $i\ge 3$, are identical to $\bar{R}^3$ in a cone $Q\in{\mathcal{Q}}$; $g^3(\bar{\mathbf{R }}, f)\in Q$; $Q^i\in{\mathcal{Q}}$, $i=1,2$, are cones such that $g^i(\bar{\mathbf{R }},f)\in Q^i$; f is a social choice function pseudo-efficient and strategy-proof on ${{\mathcal{R}}}(\bar{R}^1, Q^{1 c})\times{{\mathcal{R}}}(\bar{R}^2,Q^{2c})\times \{\bar{R}^3\} \times \cdots \times \{\bar{R}^N\}$; the consumption-direction vectors $g^i(\bar{\mathbf{R }},f)$, $i=1,2,3$, are different; and $f^i(\bar{\mathbf{R }})\in int A$ for both $i=1,2$.

The next lemma, which is a counterpart of Lemma 2 in Momi (2017), is an immediate consequence of the difference of consumption-direction vectors.

Lemma A1

Consider preferences$\tilde{R}^i, R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$, $i=1,2$. Suppose$g^i(\tilde{R}^1,\tilde{R}^2,\bar{\mathbf{R }}^{-\{1,2\}})$, $i=1,2,3$, are different and$g^i(\tilde{R}^1,\tilde{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=g^3(\tilde{R}^1,\tilde{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})$for$i\ge 3$. For$i,j \in \{1,2\}$, if$f^i(\tilde{R}^1,\tilde{R}^2, \bar{\mathbf{R }}^{-\{1,2\}}) =f^i(\hat{R}^1,\hat{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})\ne 0$and$\tilde{R}^i$and$\hat{R}^i$have the same gradient vector at the consumption bundle, then$f^j(\tilde{R}^1,\tilde{R}^2, \bar{\mathbf{R }}^{-\{1,2\}}) =f^j(\hat{R}^1,\hat{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})$.

Proof

Without loss of generality, we let $i=1$ and $j=2$. For simplicity, we write $\tilde{\mathbf{R }}=(\tilde{R}^1,\tilde{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})$ and $\hat{\mathbf{R }} =(\hat{R}^1,\hat{R}^2,\bar{\mathbf{R }}^{-\{1,2\}})$. As $f^1(\tilde{\mathbf{R }})=f^1(\hat{\mathbf{R }})\ne 0$, and $\tilde{R}^1$ and $\hat{R}^1$ have the same gradient vector at the consumption bundle, we have $g^i(\tilde{\mathbf{R }},f) =g^i(\hat{\mathbf{R }},f)$ for all agents.

As the consumption of all agents at the preference profiles $\tilde{\mathbf{R }}$ and $\hat{\mathbf{R }}$, respectively, sum to $\Omega$, we have $\sum _{i=1}^Nf^i(\tilde{\mathbf{R }}) =\sum _{i=1}^N\parallel f^i(\tilde{\mathbf{R }})\parallel g^i(\tilde{\mathbf{R }},f)=\Omega$ and $\sum _{i=1}^Nf^i (\hat{\mathbf{R }})=\sum _{i=1}^N\parallel f^i(\hat{\mathbf{R }}) \parallel g^i(\hat{\mathbf{R }},f)=\Omega$. Considering the difference between these equations, we have $(\parallel f^2(\tilde{\mathbf{R }})\parallel -\parallel f^2(\tilde{\mathbf{R }})\parallel )g^2(\tilde{\mathbf{R }},f)+(\sum _{i=3}^N\parallel f^i(\tilde{\mathbf{R }})\parallel -\sum _{i=3}^N\parallel f^i(\hat{\mathbf{R }})\parallel )g^3(\tilde{\mathbf{R }},f)=0$ because $f^1(\tilde{\mathbf{R }})=f^1(\hat{\mathbf{R }})$, $g^2(\tilde{\mathbf{R }},f)=g^2(\hat{\mathbf{R }},f)$ and $g^i(\tilde{\mathbf{R }},f)=g^i(\tilde{\mathbf{R }},f)=g^3(\tilde{\mathbf{R }},f)$ for $i\ge 3$. As the consumption-direction vectors of agents 2 and 3 are independent, we have $\parallel f^2(\tilde{\mathbf{R }}) \parallel = \parallel f^2(\hat{\mathbf{R }})\parallel$, and hence $f^2(\tilde{\mathbf{R }})=f^2(\hat{\mathbf{R }})$. $\square$

By the definition of the option set, $f^i(\bar{\mathbf{R }})$ is the most preferred consumption bundle in the option set $G^i(\bar{\mathbf{R }}^{-i})$ with respect to $\bar{R}^i$ for $i=1,2$. We want to establish the reverse: the most preferred consumption bundle in $G^i(\bar{\mathbf{R }}^{-i})$ with respect to $\bar{R}^i$ is $f^i(\bar{\mathbf{R }})$. For tractability, we initially work with the closure of the option set, $\overline{G^i(\bar{\mathbf{R }}^{-i})}$. By considering the closure, we ensure the existence of the most preferred consumption vector in $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ with respect to any preference $R^i$. The next lemma, which is a counterpart of Lemma 3 in Momi (2017), shows that if a consumption bundle is the most preferred in the closure of the option set with respect to some preference, then the consumption bundle is actually an element of the option set.

Lemma A2

For$i=1,2$, if$x\in Q^i$is the most preferred consumption bundle in$\overline{G^i(\bar{\mathbf{R }}^{-i})}$with respect to$R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$, then$x\in G^i(\bar{\mathbf{R }}^{-i})$.

Proof

We let $\hat{R}^i$ be an MMT of $R^i$ at x sufficiently close to $R^i$. See Momi (2013b, Lemma 4) for the existence of such an MMT. Then, by applying Lemma 1, we let $\check{R}^i$ be a preference that is identical to $\hat{R}^i$ in a neighborhood of x and identical to $R^i$ outside $Q^i$. Thus, $\check{R}^i$ is an MMT of $R^i$ at x and an element of ${{\mathcal{R}}}(\bar{R}^i,Q^{ic})$. We show $x=f^i(\check{R}^i, \bar{\mathbf{R }}^{-i})$. Since $\check{R}^i$ is an MMT of $R^i$ at x, $UC(x;\check{R}^i)\setminus x\subset P(x;R^i)$. Therefore, x is the unique most preferred consumption bundle in $\overline{G^i(\mathbf{R }^{-i})}$ with respect to $\check{R}^i$.

As $x\in \overline{G^i(\bar{\mathbf{R }}^{-i})}$, there exists a $\hat{x}\in G^i(\bar{\mathbf{R }}^{-i})$ arbitrarily close to x. Therefore, if x is strictly preferred to $f^i(\check{R}^i,\bar{\mathbf{R }}^{-i})$ with respect to $\check{R}^i$, then $\hat{x}$ is strictly preferred to $f^i(\check{R}^i,\bar{\mathbf{R }}^{-i})$. This contradicts the strategy-proofness of f. Therefore, $f^i(\check{R}^i, \bar{\mathbf{R }}^{-i})\in UC(x;\check{R}^i)$. Then, $f^i(\check{R}^i,\bar{\mathbf{R }}^{-i})\ne x$ contradicts that x is the most preferred consumption bundle in $\overline{G^i(\mathbf{R }^{-i})}$ with respect to $\check{R}^i$. Hence, $x=f^i(\check{R}^i,\bar{\mathbf{R }}^{-i})\in G^i(\bar{\mathbf{R }}^{-i})$. $\square$

An immediate consequence of Lemma A2 is that if x is the unique most preferred consumption bundle in $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ with respect to $R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$ for $i=1,2$, then $x=f^i(R^i,\bar{\mathbf{R }}^{-i})$. The next lemma, which is a counterpart of Lemma 4 in Momi (2017), shows that the most preferred consumption bundle is actually unique in a neighborhood of $f^i(\bar{\mathbf{R }})$. Thus, for any preference $R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$, the most preferred consumption bundle in $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ with respect to the preference is unique, and it is exactly the consumption bundle $f^i(R^i, \bar{\mathbf{R }}^{-i})$ allocated to the agent for the preference by the social choice function f.

Lemma A3

For$i=1,2$, in a neighborhood of$f^i (\bar{\mathbf{R }})$, $f^i(\bar{\mathbf{R }})$is the unique most preferred consumption bundle in$\overline{G^i (\bar{\mathbf{R }}^{-i})}$with respect to$\bar{R}^i$.

Fig. 6

Proof of Lemma A3: Case 1

Fig. 7

Proof of Lemma A3: Case 2

Proof

The proof is slightly different from the proof of Lemma 4 in Momi (2017). Without loss of generality, we prove the statement for agent 1. We write $f(\bar{\mathbf{R }})=\bar{\mathbf{x }} =(\bar{x}^1,\ldots ,\bar{x}^N)$ and $\bar{p}=p(\bar{\mathbf{R }},f)$. From the definitions, $\bar{x}^1$ is (one of) the most preferred consumption bundles in $G^1(\bar{\mathbf{R }}^{-1})$ with respect to $\bar{R}^1$, and no consumption bundle in $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ is strictly preferred to $\bar{x}^1$ with respect to $\bar{R}^1$.⁷ We suppose that there exists another consumption bundle $\tilde{x}^1$ in $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ that is indifferent to $\bar{x}^1$ with respect to $\bar{R}^1$, and show a contradiction. We define the gradient vector of $\bar{R}^1$ at $\tilde{x}^1$ as $\tilde{p}$: $\tilde{p}=p(\bar{R}^1, \tilde{x}^1)$.

Since we are interested in a neighborhood of $\bar{x}^1$, we assume that $\tilde{x}^1$ is sufficiently close to $\bar{x}^1$ so that $\tilde{x}^1\in Q^1$, $g^2(\bar{R}^2,\tilde{p})\in Q^2$, $g^3(\bar{R}^3,\tilde{p})\in Q$, and these vectors are different.

We let $\check{R}^1$ be a preference that is identical to $\bar{R}^1$ outside $Q^1$, $I(\bar{x}^1;\check{R}^1)\bigcap I(\bar{x}^1;\bar{R}^1)=\{\bar{x}^1,\tilde{x}^1\}$ and $UC(\bar{x}^1;R) \subset UC(\bar{x}^1;\bar{R}^1)$. That is, both $\bar{x}^1$ and $\tilde{x}^1$ are the most preferred consumption bundles in $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ with respect to $\check{R}^1$. Note that $f^i(\check{R}^1,\bar{\mathbf{R }}^{-1})$ should be $\bar{x}^1$ or $\tilde{x}^1$ for such a preference $\check{R}^1$. We assume that there exists such a preference $\check{R}^1$ satisfying $f^i(\check{R}^1,\bar{\mathbf{R }}^{-1}) =\bar{x}^1$. At the end of the proof, we see that this assumption does not lose generality.

Observe that the scalars a and b satisfying $\tilde{x}^1+ag(\bar{R}^2, \tilde{p})+bg(R^3;\tilde{p})=\Omega$ are determined uniquely because the consumption-direction vectors of agents 2 and 3 are independent. We define $\tilde{x}^2=ag(\bar{R}^2;\tilde{p})$ and $x=bg(\bar{R}^3;\tilde{p})$ using the uniquely determined scalars.

Figure 6 describes the situation in which the closure of agent 1’s option set has the two most preferred consumption bundles $\bar{x}^1$ and $\tilde{x}^1$ with respect to the preference $\bar{R}^1$. In the following proof, we observe that such an option set contradicts the pseudo-efficiency and strategy-proofness of the social choice function.

We construct agent 1’s new preferences $R_t^1$ as follows. We pick $\tilde{R}^1$, which is an MMT of $\bar{R}^1$ at $\tilde{x}^1$, and $R^{1\prime }$, which is an MMT of $\bar{R}^1$ at $\bar{x}^1$. Let $\epsilon >0$ be a sufficiently small scalar. For a parameter $t\in (1-\epsilon ,1)$ sufficiently close to one, we define $K_t$ as the convex hull of the set $UC(t\tilde{x}^1; \tilde{R}^1)\bigcup UC(\bar{x}^1; R^{1\prime })$:

$$\begin{aligned} K_t=co \left( UC(t\tilde{x}^1; \tilde{R}^1)\bigcup UC(\bar{x}^1; R^{1\prime }) \right) . \end{aligned}$$

We set $\epsilon$ sufficiently small so that for any $t\in (1-\epsilon ,1)$, $\tilde{p}$ and $\bar{p}$ are normal vectors of the supporting hyperplanes to $K_t$ at $t\tilde{x}^1$ and $\bar{x}^1$, respectively.

Applying Lemma A1 in Momi (2017) with $K_t$, we slightly modify the indifference set $I(\bar{x}^1;\bar{R}^1)$. We consider the hyperplane $(\tilde{x}^1)^\perp$ passing through the origin and perpendicular to the vector $\tilde{x}^1$. For each $y\in (\tilde{x}^1)^\perp$, we let L(y) denote the half line starting from y and extending in the direction of $\tilde{x}^1$: $L(y)=\{x\in R^L|x=y+r\tilde{x}^1, r\ge 0\}$. We fix a scalar $\bar{s}<1$ close to one and define the following:

$$\begin{aligned} I_t=\bigcup _{y\in (\tilde{x}^1)^\perp } \left\{ \bar{s} \left( L(y)\bigcap I(\bar{x}^1;\bar{R}^1)\right) +(1-\bar{s}) \left( L(y)\bigcap \partial K_t\right) \right\} . \end{aligned}$$

We let $\tilde{R}_t^{1}\in{{\mathcal{R}}}$ be agent 1’s preference that has $I_t$ as its indifference set. Note that $\tilde{R}_t^{1}$ can be arbitrarily close to $\bar{R}^1$ by setting $\bar{s}$ close to one. We pick a cone $Q'$ including both $\bar{x}$ and $\tilde{x}^1$ and $\overline{Q'}\setminus \{0\}\subset Q^1$. We set $\bar{s}$ sufficiently close to one and apply Lemma 1 so that we have a preference $R^1_t$ that is identical to $\tilde{R}^1_t$ in $Q'$ and identical to $\bar{R}^1$ outside $Q^1$ for any $t\in (1-\epsilon ,1)$.

Both $R_t^{1}$ and $\bar{R}^1$ have the same gradient vectors $\tilde{p}$ and $\bar{p}$ at $\tilde{x}^1$ and $\bar{x}^1$, respectively. Observe that $R_t^{1}$ is an MMT of $\bar{R}^1$ at $\tilde{x}^1$. Furthermore, observe that $UC(\bar{x}^1; R_t^{1})\bigcap LC(\bar{x}^1; \bar{R}^1)$ consists of the point $\bar{x}^1$ and a set in a neighborhood of $\tilde{x}^1$ that converges to $\tilde{x}^1$ as t converges to 1.

We write $f(R_t^{1},\bar{\mathbf{R }}^{-1})=\mathbf{x }_t =(x_t^{1},\ldots ,x_t^{N})$. These t in $x_t^i$ should not be confused with the subscripts labeling goods. Since $R_t^{1}$ is an MMT of $\bar{R}^1$ at $\tilde{x}^1$, $x_t^{1}=\tilde{x}^1$. Therefore $x_t^2=\tilde{x}^2$ and $\sum _{i=3}^Nx_t^i=x$, as shown in Lemma A1.

We show that $\tilde{x}^2$ is not indifferent to $\bar{x}^2$ with respect to $\bar{R}^2$.⁸ We assume that $\tilde{x}^2$ is indifferent to $\bar{x}^2$ with respect to $R^2$ and show a contradiction. Since the preferences of agents $i\ge 3$ are all identical in Q, the consumption vectors $\bar{x}^i$, $i\ge 3$ are parallel and $x_t^i$, $i\ge 3$ are parallel. Therefore for $i\ge 3$, their consumption bundles are written as $\bar{x}^i=a^i\sum _{i=3}^N\bar{x}^i$ and $x_t^i=b^i\sum _{i=3}^N x_t^i$ using the scalars $a^i$ and $b^i$, $i=3,\ldots , N$, such that $0\le a^i\le 1$, $\sum _{i=3}^Na^i=1$, $0\le b^i\le 1$, and $\sum _{i=3}^Nb^i=1$. Using a scalar $s\in (0,1)$, we set a new allocation $x^{*}=(x^{1*},\ldots ,x^{N*})$ as follows: $x^{1*}=s\bar{x}^1+(1-s)\tilde{x}^1$, $x^{2*} =s\bar{x}^2+(1-s)\tilde{x}^2$, and for $i\ge 3$

$$\begin{aligned} x^{i*}=\left\{ \begin{array}{l} a^i\left( s\sum _{i=3}^N \bar{x}^i+(1-s)\sum _{i=3}^N x_t^i\right) \ \text{ if }\ \left( \sum _{i=3}^N x_t^i\right) \bar{R}^3\left( \sum _{i=3}^N\bar{x}^i\right) \\ b^i\left( s\sum _{i=3}^N \bar{x}^i+(1-s)\sum _{i=3}^N x_t^i\right) \ \text{ if }\ \left( \sum _{i=3}^N \bar{x}^i\right) P_{\bar{R}^3}\left( \sum _{i=3}^N x_t^i\right) \end{array}\right. . \end{aligned}$$

Because of the strict convexity of the preferences, $x^{1*}$ is preferred to either $\bar{x}^i$ or $x_t^i=\tilde{x}^i$ for agent $i=1,2$. For agent $i\ge 3$, as $\bar{R}^i$ is identical to $\bar{R}^3$ in Q, if $\sum _{i=3}^N x_t^i$ is preferred to $\sum _{i=3}^N\bar{x}^i$ with respect to $\bar{R}^3$, then $x^{i*}$ is preferred to $\bar{x}^i=a^i\sum _{i=3}^N \bar{x}^i$. If $\sum _{i=3}^N \bar{x}^i$ is preferred to $\sum _{i=3}^N x_t^i$ with respect to $\bar{R}^3$, then $x^{i*}$ is preferred to $x_t^i=b^i\sum _{i=3}^N x_t^i$. The feasibility of the allocation $\mathbf{x }^{*}$ is clear from the fact that $\sum _{i=3}^Nx^{*i}= s\sum _{i=3}^N \bar{x}^i+(1-s)\sum _{i=3}^N x_t^i$ and the feasibility of $\bar{\mathbf{x }}$ and $\mathbf{x }_t$. This contradicts the Pareto efficiency of the allocations $\bar{\mathbf{x }}$ and $\mathbf{x }_t$.

Now, we separate two cases depending on whether $\tilde{x}^2$ is strictly preferred or strictly less preferred to $\bar{x}^2$ with respect to $R^2$

Case 1. We consider the case in which $\tilde{x}^2$ is strictly preferred to $\bar{x}^2$ with respect to $\bar{R}^2$. That is, there exists a positive vector $\alpha \in R^{L}_{++}$ such that $\tilde{x}^2 \in P(\bar{x}^2+\alpha ;\bar{R}^2)$. Figure 6 illustrates this case.

We pick $\hat{R}^2$ such that (i) $\hat{R}^2$ is an MMT of $\bar{R}^2$ at $\bar{x}^2$ and equivalent to $\bar{R}^2$ outside $Q^2$, (ii) $\tilde{x}^{2} \in P(\bar{x}^2;\hat{R}^2)$, and (iii) $g(\hat{R}^2, \tilde{p})$ is different from $g(\bar{R}^2, \tilde{p})$.

Observe that $f^2(\bar{R}^1, \hat{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\bar{x}^2$ because $\hat{R}^2$ is an MMT of $\bar{R}^2$ at $\bar{x}^2$. Therefore, $f^1(\bar{R}^1, \hat{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\bar{x}^1$, as shown in Lemma A1.

We write $f(R_{t}^{1},\hat{R}^2,\bar{\mathbf{R }}^{-\{1,2\}}) =\hat{\mathbf{x }}_{t}=(\hat{x}_{t}^{1},\ldots ,\hat{x}_{t}^{N})$ and focus on $\hat{x}_{t}^{1}$ with t sufficiently close to 1. Facing the other agents’ preferences $(\hat{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})$, agent 1 can achieve $\bar{x}^1$ by reporting $\bar{R}^1$, as mentioned above. Therefore, $\hat{x}_{t}^{1}$ should be preferred to $\bar{x}^1$ with respect to $R_{t}^{1}$. On the contrary, $\hat{x}_{t}^{1}$ is not strictly preferred to $\bar{x}^1$ with respect to $\bar{R}^1$. Therefore, $\hat{x}_{t}^{1}\in UC(\bar{x}^1; R_{t}^{1})\bigcap LC(\bar{x}^1; \bar{R}^1)$. Recall that this set consists of the point $\bar{x}^1$ and a set in a neighborhood of $\tilde{x}^1$ that converges to $\tilde{x}^1$ as t converges to 1. We investigate these two cases.

We consider the case in which $\hat{x}_{t}^{1}=\bar{x}^1$. As $R_{t}^{1}$ and $\bar{R}^1$ have the same gradient vector at $\bar{x}^1$, and $\hat{R}^2$ and $\bar{R}^2$ have the same gradient vector at $\bar{x}^2$, this case implies that $\hat{x}_{t}^{2}=\bar{x}^2$. Recall that $f^2(R_{t}^{1}, \bar{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\tilde{x}^{2}$ and we have chosen $\hat{R}^2$ so that $\tilde{x}^2 \in P(\bar{x}^2; \hat{R}^2)$. If agent 2 has the preference $\hat{R}^2$ and faces the other agents’ preferences $(R_{t}^{1}, \bar{\mathbf{R }}^{-\{1,2\}})$, he can become better off by reporting $\bar{R}^2$ and achieving $\tilde{x}^2$ than reporting his true preference $\hat{R}^2$ and achieving $\hat{x}^2_{t}=\bar{x}^2$. This contradicts the strategy-proofness of f.

We now consider the case in which $\hat{x}_{t}^1$ is in a set in a neighborhood of $\tilde{x}^1$ that converges to $\tilde{x}^1$ as $t\rightarrow 1$. In this case, as $t\rightarrow 1$, the gradient vector of $R_{t}^{1}$ at $\hat{x}_{t}^{1}$ converges to $\tilde{p}$. Therefore, the price vector at $\hat{\mathbf{x }}_{t}$ converges to $\tilde{p}$. In particular, the gradient of $\hat{R}^2$ at $\hat{x}_{t}^{2}$ converges to $\tilde{p}$.

However, note the equation $\hat{x}_{t}^{2}=\tilde{x}^2+\sum _{i\ne 2}(\tilde{x}^i-\hat{x}_{t}^{i})$ obtained from $\sum _{i=1}^N\tilde{x}^i=\Omega$ and $\sum _{i=1}^N\hat{x}_t^{i}=\Omega$. On the right-hand side of this equation, $\tilde{x}^1-\hat{x}_{t}^{1}\rightarrow 0$ as $t\rightarrow 1$. For agent $i\ge 3$ all $\tilde{x}^i$ are parallel and on $[g(R^3,\tilde{p})]$ and all $x^i_{t}$ are also parallel and converge to consumptions on $[g(R^3,\tilde{p})]$. Since $\hat{x}_{t}^{2}$ is different from $\tilde{x}^2$ because of (iii), the limit of $\sum _{i=3}^N\tilde{x}^i-\sum _{i=3}^N\hat{x}_{t_k}$ is written as $ag(R^3,\tilde{p})$, where the scalar a is not zero.

If $a>0$, then $\hat{x}_{t}^2$ with sufficiently large t should be strictly preferred to $\tilde{x}^2$ with respect to any preference. Then, agent 2 with the preference $\bar{R}^2$ facing the other agents’ preferences $(R^1_{t}, \bar{\mathbf{R }}^{-\{1,2\}})$ can be better off by reporting $\hat{R}^2$ and achieving $\hat{x}_{t}^2$ than reporting his true preference $\bar{R}^2$ and achieving $\tilde{x}^2$. This contradicts the strategy-proofness of f.

If $a<0$, then $\tilde{x}^2$ should be strictly preferred to $\hat{x}_{t}^2$ with sufficiently large t. Then, agent 2 with the preference $\hat{R}^2$ facing the other agents’ preferences $(R^1_{t},\bar{\mathbf{R }}^{-\{1,2\}})$ can be better off by reporting $\bar{R}^2$ and achieving $\tilde{x}^2$ than reporting his true preference $\hat{R}^2$ and achieving $\hat{x}_{t}^2$. This contradicts the strategy-proofness of f.

Case 2. We consider the case in which $\bar{x}^2$ is strictly preferred to $\tilde{x}^2$ with respect to $\bar{R}^2$. That is, there exists a positive vector $\alpha \in R^{L}_{++}$ such that $\bar{x}^2 \in P(\tilde{x}^2+\alpha ;\bar{R}^2)$. Figure 7 depicts this case.

Recall that we have assumed the existence of a preference $\check{R}^1$ such that $\check{R}^1\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})$, $I(\bar{x}^1;\check{R}^1)\bigcap I(\bar{x}^1;\bar{R}^1)=\{\bar{x}^1,\tilde{x}^1\}$, $UC(\bar{x}^1;\check{R}^1)\subset UC(\bar{x}^1;\bar{R}^1)$, and $f^1(\check{R}^1,\bar{\mathbf{R }}^{-1})=\bar{x}^1$. Thus, $f^2 (\check{R}^1,\bar{\mathbf{R }}^{-1})=\bar{x}^2$ as shown in Lemma A1.

We consider $\check{R}^2$ such that (i) $\check{R}^2$ is an MMT of $\bar{R}^2$ at $\tilde{x}^2$ and identical to $\bar{R}^2$ outside $Q^2$, (ii)$\bar{x}^2\in P(\tilde{x}^2;\check{R}^2)$, and (iii) $g(\check{R}^2,\bar{p})\ne g(\bar{R}^2,\bar{p})$.

Since $f^2(R_t^1,\bar{\mathbf{R }}^{-1})=\tilde{x}^2$ and $\check{R}^2$ is an MMT of $\bar{R}^2$ at $\tilde{x}^2$, we have $f^2(R_t^1,\check{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\tilde{x}^2$, and hence $f^1(R_t^1,\check{R}^2, \bar{\mathbf{R }}^{-\{1,2\}}) =\tilde{x}^1$, as shown in Lemma A1. We write $f(\check{R}^1, \check{R}^2,\bar{\mathbf{R }}^{-\{1,2\}})=\check{\mathbf{x }} =(\check{x}^1,\ldots , \check{x}^N)$ and consider the location of $\check{x}^1$.

Since $f^1(R_t^1,\check{R}^2,\bar{\mathbf{R }}^{-\{1,2\}}) =\tilde{x}^1$ and $f^1( \check{R}^1, \check{R}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\check{x}^1$, $\check{x}^1$ should be preferred to $\tilde{x}^1$ with respect to $\check{R}^1$ and $\tilde{x}^1$ should be preferred to $\check{x}^1$ with respect to $R_t^1$. Therefore, we have $\check{x}^1\in UC(\tilde{x}^1;\check{R}^1)\bigcap LC(\tilde{x}^1;R^1_t)$ for any t. As $t\rightarrow 1$, $R^1_t\rightarrow \bar{R}^1$, and hence $UC(\tilde{x}^1;\check{R}^1)\bigcap LC(\tilde{x}^1;R^1_t)\rightarrow UC(\tilde{x}^1;\check{R}^1)\bigcap LC(\tilde{x}^1;\bar{R}^1)=\{\bar{x}^1,\tilde{x}^1\}$. Therefore, $\check{x}^1$ should be $\tilde{x}^1$ or $\bar{x}^1$.

Suppose $\check{x}^1=\tilde{x}^1$. Then, $\check{x}^2=\tilde{x}^2$, and agent 2 with the preference $\check{R}^2$ facing the other agents’ preferences $(\check{R}^1, \bar{\mathbf{R }}^{-\{1,2\}})$ can be better off by reporting $\bar{R}^2$ and achieving $\bar{x}^2$ than reporting his true preference $\check{R}^2$ and achieving $\check{x}^2=\tilde{x}^2$. This contradicts the strategy-proofness of f.

Suppose $\check{x}^1=\bar{x}^1$. We have $\check{x}^2 =\bar{x}^2+\sum _{i=3}^N\bar{x}^i-\sum _{i=3}^N\check{x}^i$. All $\bar{x}^i$ and $\check{x}^i$ for $i\ge 3$ are on the ray $[g(\bar{R}^3;\bar{p})]$. Furthermore, because $\check{x}^2\ne \bar{x}^2$, as ensured by (iii), $\sum _{i=3}^N\bar{x}^i -\sum _{i=3}^N\check{x}^i\ne 0$, Therefore, we can write $\sum _{i=3}^N\bar{x}^i-\sum _{i=3}^N\check{x}^i=ag(\bar{R}^3;\bar{p})$ with a non-zero scalar a.

If $a>0$, then $\check{x}^2$ is strictly preferred to $\bar{x}^2$ with any preference. Agent 2 with the preference $\bar{R}^2$ facing the other agents’ preferences $(\check{R}^1, \bar{\mathbf{R }}^{-\{1,2\}})$ can be better off by reporting $\check{R}^2$ and achieving $\check{x}^2$ than reporting his true preference $\bar{R}^2$ and achieving $\bar{x}^2$. This contradicts the strategy-proofness of f.

If $a<0$, then $\bar{x}^2$ is strictly preferred to $\check{x}^2$ with any preference. Agent 2 with the preference $\check{R}^2$ facing other agents’ preferences $(\check{R}^1, \bar{\mathbf{R }}^{-\{1,2\}})$ can be better off by reporting $\bar{R}^2$ and achieving $\bar{x}^2$ than reporting his true preference $\check{R}^2$ and achieving $\check{x}^2$. This contradicts the strategy-proofness of f. This ends Case 2.

We proved that a preference $\bar{R}^1$, such that $f^1(\bar{\mathbf{R }})=\bar{x}^1$, does not have another consumption bundle $\tilde{x}^1\in \overline{G^1(\bar{\mathbf{R }}^{-1})}$ indifferent to $\bar{x}^1$ with respect to $R^1$ under the assumption that there exists a preference $\check{R}^1$ such that $\check{R}^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$, $I(\bar{x}^1;\check{R}^1)\bigcap I(\bar{x}^1;\bar{R}^1) =\{\bar{x}^1,\tilde{x}^1\}$, $UC(\bar{x}^1;\check{R}^1)\subset UC(\bar{x}^1;\bar{R}^1)$, and $f^1(\check{R}^1, \bar{\mathbf{R }}^{-1})=\bar{x}^1$.

Suppose that there exists no such preference $\check{R}^1$. Then $f^1(R,\bar{\mathbf{R }}^{-1})=\tilde{x}^1$ for any preference R such that $R\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$, $I(\bar{x}^1; R)\bigcap I(\bar{x}^1;\bar{R}^1)=\{\bar{x}^1,\tilde{x}^1\}$, $UC(\bar{x}^1; R)\subset UC(\bar{x}^1;\bar{R}^1)$. We pick two such preferences $R^{\prime 1}$ and $R^{\prime \prime 1}$ satisfying $UC(\bar{x}^1; R^{\prime \prime 1})\subset UC(\bar{x}^1;R^{\prime 1})$.

Replacing $\bar{R}^1$, $\bar{x}^1$, and $\check{R}^1$ with $R^{\prime 1}$, $\tilde{x}^1$, and $R^{\prime \prime 1}$, respectively, what we proved establishes that a preference $R^{\prime 1}$, such that $f^1(R^{\prime 1}, \bar{\mathbf{R }}^{-1})=\tilde{x}^1$, does not have another consumption bundle $\bar{x}^1\in \overline{G^1 (\bar{\mathbf{R }}^{-1})}$ indifferent to $\tilde{x}^1$ with respect to $R^{\prime 1}$ because there exists the preference $R^{\prime \prime 1}$ such that $R^{\prime \prime 1}\in{{\mathcal{R}}}( R^{\prime 1},Q^{1c})$, $I(\tilde{x}^1; R^{\prime \prime 1})\bigcap I(\tilde{x}^1; R^{\prime 1})=\{\bar{x}^1,\tilde{x}^1\}$, $UC(\tilde{x}^1; R^{\prime \prime 1})\subset UC(\tilde{x}^1; R^{\prime 1})$, and $f^1( R^{\prime \prime 1},\bar{\mathbf{R }}^{-1})=\tilde{x}^1$. This implies that $\bar{x}^1$ and $\tilde{x}^1$ are not indifferent with respect to $\bar{R}^1$. Thus, a preference $\bar{R}^1$ such that $f^1(\bar{\mathbf{R }})=\bar{x}^1$ does not have another consumption bundle $\tilde{x}^1\in \overline{G^1(\bar{\mathbf{R }}^{-1})}$ indifferent to $\bar{x}^1$ with respect to $\bar{R}^1$ regardless of the existence of the preference $\check{R}^1$ we assumed. $\square$

The next proposition is a counterpart of Proposition 1 in Momi (2017) that showed that the social choice function $f(\cdot ,\bar{\mathbf{R }})$ is a continuous function of $R^i$ with the other agents’ preferences $\bar{\mathbf{R }}^{-i}$ fixed. In our setup, agent 1’s consumption determines only agent 2’s consumption and the sum of the other agents’ consumption. The consumption bundles for agents $i\ge 3$ themselves are indeterminate. Moreover, the efficiency and strategy-proofness of $f(\cdot ,\bar{\mathbf{R }}^{-i})$ are assumed on ${{\mathcal{R}}}(\bar{R}^{i},Q^{ic})$ for $i=1,2$, in our setup, while they are assumed in a neighborhood of $\bar{R}^i$ in Momi (2017).

Proposition A1

For$i,j\in \{1,2\}$, $f^j(\cdot ,\bar{\mathbf{R }}^{-i})$is a continuous function on the domain${{\mathcal{R}}} (\bar{R}^i,Q^{ic})$.

Proof

Without loss of generality, we prove the proposition for $i=1$. We only have to prove that if the consumption-direction vectors $g(\bar{R}^i,p(\bar{\mathbf{R }},f))$, $i=1,2,3$, are different at $\bar{\mathbf{R }}$ and if $f^1(\bar{\mathbf{R }})\in int A$, then $f^j(\cdot , \bar{\mathbf{R }}^{-i})$, $j\in \{1,2\}$ are continuous functions at $\bar{R}^1$. Suppose that the claim is true and let $R^1$ be a preference close to $\bar{R}^1$. Then, $f^1(R^i,\bar{\mathbf{R }}^{-i})$ is close to $f^1(\bar{\mathbf{R }})$ because of the supposed continuity, and hence the price vectors are close at the preference profiles $(R^1,\bar{\mathbf{R }}^{-1})$ and $\bar{\mathbf{R }}$. Then the consumption-direction vectors of agents 1,2,and 3 at the preference profile $( R^i,\bar{\mathbf{R }}^i)$ are still independent and $f^1( R^i,\bar{\mathbf{R }}^{-1})\in int A$. Then, $f^j(\cdot ,\bar{\mathbf{R }}^{-i})$, $j\in{1,2}$ are continuous at $R^1$ because of the supposed claim.

We first prove that $f^1(\cdot ,\bar{\mathbf{R }}^{-1})$ is a continuous function at $\bar{R}^1$. We let $\{R_n^{1}\}_{n=1}^\infty$ be a sequence of preferences identical to $\bar{R}^1$ outside $Q^1$ converging to $\bar{R}^i$ as $n\rightarrow \infty$. There exists a convergent subsequence $\{f_{n_k}^1(R^{1},\bar{\mathbf{R }}^{-1})\}_{k=1}^\infty$ because of the compactness of the feasible allocation set. We write $f^1(R_{n_k}^{1},\bar{\mathbf{R }}^{-1})\rightarrow x^{1*}$ as $k\rightarrow \infty$. All we have to show is that $x^{1*}=f^1(\bar{\mathbf{R }})$.

We observe that $x^{1*}$ is indifferent to $f^1(\bar{\mathbf{R }})$ with respect to $\bar{R}^1$. If $x^{1*}\in P(f^1(\bar{\mathbf{R }}); \bar{R}^1)$, then $f^1(R_{n_k}^{1},\bar{\mathbf{R }}^{-1})\in P(f^1(\bar{\mathbf{R }}); \bar{R}^1)$ for a sufficiently large k. This contradicts the strategy-proofness of f. If $f^1(\bar{\mathbf{R }})\in P(x^{1*};\bar{R}^1)$, then $f^1(\bar{\mathbf{R }})\in P(f^1(R_{n_k}^{1},\bar{\mathbf{R }}^{-1}); R_{n_k}^{1})$ for a sufficiently large k because $f^1(R_{n_k}^{1}, \bar{\mathbf{R }}^{-1})$ converges to $x^{1*}$ as $k\rightarrow \infty$ and $UC(x; R_{n_k}^{1})\bigcap K$ converges to $UC(x;\bar{R}^1)\bigcap K$ at any consumption x with a compact set K as $k\rightarrow \infty$ with respect to the standard metric in the set of compact subsets of $R^L$. Again, this contradicts the strategy-proofness of f. Thus, $x^{1*}$ is indifferent to $f^1(\bar{\mathbf{R }})$ with respect to $\bar{R}^1$. On the contrary, $x^{1*}\in \overline{G^1(\bar{\mathbf{R }}^{-1})}$ because $f^1(R_{n_k}^{1},\bar{\mathbf{R }}^{-1})\in G^1(\bar{\mathbf{R }}^{-1})$ for any k. Then, $x^{1*}=f^1(\bar{\mathbf{R }})$ by Lemma A3.

We now prove the continuity of $f^2(\cdot ,\bar{\mathbf{R }}^{-1})$ at $\bar{R}^1$. We have proved the continuity of $f^1(\cdot ,\bar{\mathbf{R }})$ at $\bar{R}^1$. That is, if $\{R_n^{1}\}_{n=1}^\infty$ is a sequence of preferences converging to $\bar{R}^1$ as $n\rightarrow \infty$, then $f^1(R_n^{1}, \bar{\mathbf{R }}^{-1})$ converges to $f^1(\bar{\mathbf{R }})$. According to this convergence, the gradient vector of $R_n^{1}$ at $f^1(R_n^{1},\bar{\mathbf{R }}^{-1})$ converges to the gradient vector of $\bar{R}^1$ at $f^1(\bar{\mathbf{R }})$; that is, $p((R_n^{1},\bar{\mathbf{R }}^{-1}),f)$ converges to $p(\bar{\mathbf{R }},f)$, and hence, $g^j((R_n^{1}, \bar{\mathbf{R }}^{-1}),f)$ converges to $g^j(\bar{\mathbf{R }},f)$ for any agent j .

As in the proof of Lemma A1, $f^j(\mathbf{R })=\parallel f^j(\mathbf{R }) \parallel g^j(\mathbf{R },f)$, and the sum of the agents’ consumption equals the total endowment. Thus, we have two equalities: $f^1(\bar{\mathbf{R }})+ \parallel f^2(\bar{\mathbf{R }}) \parallel g^2(\bar{\mathbf{R }},f)+\parallel \sum _{i=3}^Nf^i(\bar{\mathbf{R }}) \parallel g^3(\bar{\mathbf{R }},f)=\Omega$ and $f^1(R_n^{1}, \bar{\mathbf{R }}^{-1})+ \parallel f^2(R_n^{1},\bar{\mathbf{R }}^{-1}) \parallel g^2((R_n^{1},\bar{\mathbf{R }}^{-1}),f)+\parallel \sum _{i=3}^Nf^i(R_n^{1},\bar{\mathbf{R }}^{-1}) \parallel g^3((R_n^{1},\bar{\mathbf{R }}^{-1}),f)=\Omega$ for any n. Considering the difference between these equalities, we have:

$$\begin{aligned} 0= &{} f^1(\bar{\mathbf{R }})-f^1(R_n^{1},\bar{\mathbf{R }}^{-1})\\&+\parallel f^2(\bar{\mathbf{R }}) \parallel g^2(\bar{\mathbf{R }},f) -\parallel f^2(R_n^{1},\bar{\mathbf{R }}^{-1}) \parallel g^2((R_n^{i},\bar{\mathbf{R }}^{-i}),f)\\&+\left\| \sum _{i=3}^Nf^i(\bar{\mathbf{R }}) \right\| g^3(\bar{\mathbf{R }},f) -\left\| \sum _{i=3}^Nf^i(R_n^{1},\bar{\mathbf{R }}^{-1}) \right\| g^3((R_n^{i},\bar{\mathbf{R }}^{-i}),f)\\= &{} f^1(\bar{\mathbf{R }})-f^1(R_n^{1},\bar{\mathbf{R }}^{-1})\\&+(\parallel f^2(\bar{\mathbf{R }}) \parallel -\parallel f^2(R_n^{1},\bar{\mathbf{R }}^{-1}) \parallel ) g^2(\bar{\mathbf{R }},f)\\&+ \parallel f^2(R^{1}_n,\bar{\mathbf{R }}^{-1}) \parallel (g^2(\bar{\mathbf{R }},f)- g^2((R_n^{1},\bar{\mathbf{R }}^{-1}),f)).\\&+\left( \left\| \sum _{i=3}^Nf^i(\bar{\mathbf{R }}) \right\| -\left\| \sum _{i=3}^Nf^i(R_n^{1},\bar{\mathbf{R }}^{-1}) \right\| \right) g^3(\bar{\mathbf{R }},f)\\&+ \left\| \sum _{i=3}^Nf^i(R^{1}_n,\bar{\mathbf{R }}^{-1})\right\| (g^3(\bar{\mathbf{R }},f)- g^3((R_n^{1},\bar{\mathbf{R }}^{-1}),f)). \end{aligned}$$

As $n\rightarrow \infty$, the first, third, and fifth elements in the last equation converge to zeros. As $g^j(\bar{\mathbf{R }},f)$, $j=2,3$, are independent vectors, $\parallel f^2(R_n^{1},\bar{\mathbf{R }}^{-1}) \parallel$ converges to $\parallel f^2(\bar{\mathbf{R }}) \parallel$. This implies that $f^2(R_n^{1},\bar{\mathbf{R }}^{-1})$ converges to $f^2(\bar{\mathbf{R }})$ as $n\rightarrow \infty$. That is, $f^2(\cdot , \bar{\mathbf{R }}^{-1})$ is continuous at $\bar{R}^1$. $\square$

We can show that $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ is the surface of a convex set in the sense that in a neighborhood of $f^i(\bar{\mathbf{R }})$, $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ is in the lower left-hand side of the hyperplane $f^i (R^i,\bar{\mathbf{R }}^{-i})+p((R^i,\bar{\mathbf{R }}^{-i}),f)^\perp$ for any $R^i$ in a neighborhood of $\bar{R}^i$. The next proposition is a counterpart of Proposition 2 in Momi (2017).

Proposition A2

For any$R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$in a neighborhood of$\bar{R}^i$, $i=1,2$, $\overline{G^i(\bar{\mathbf{R }}^{-i})} \subset f^i(R^i, \bar{\mathbf{R }}^{-i})+p((R^i, \bar{\mathbf{R }}^{-i}),f))^\perp -R_{+}^L$holds in a neighborhood of$f^i(\bar{\mathbf{R }})$; that is, there exist positive scalars$\bar{\epsilon }$and$\epsilon '$such that

$$\begin{aligned} D_{\bar{\epsilon }}(f^i(\bar{\mathbf{R }}))\bigcap \overline{G^i(\bar{\mathbf{R }}^{-i})}\subset f^i (R^i,\bar{\mathbf{R }}^{-i})+p((R^i,\bar{\mathbf{R }}^{-i}),f)^\perp -R_+^L, \end{aligned}$$

(3)

for any$R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})\bigcap{\mathcal{B}}_{\epsilon '}(\bar{R}^i)$, where$D_{\bar{\epsilon }} (f^i(\bar{\mathbf{R }}))$is the open ball in$R_+^L$with center$f^i(\bar{\mathbf{R }})$and radius$\bar{\epsilon }$and${\mathcal{B}}_{\epsilon '}(\bar{R}^i)$is the open ball in${{\mathcal{R}}}$with center$\bar{R}^i$and radius$\epsilon '$

Proof

Without loss of generality, we show the result for $i=1$. We first show that for any $R^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ in a neighborhood of $\bar{R}^1$, $\overline{G^1(\bar{\mathbf{R }}^{-1})} \subset f^1(R^1, \bar{\mathbf{R }}^{-1})+p((R^1, \bar{\mathbf{R }}^{-1}),f))^\perp -R_{+}^L$ holds in a neighborhood of $f^1(R^1,\bar{\mathbf{R }}^{-1})$. That is, there exists a positive scalar $\epsilon _{R^1}$, depending on $R^1$, such that:

$$\begin{aligned} D_{\epsilon _{R^1}}(f^1(R^1,\bar{\mathbf{R }}^{-1}))\bigcap \overline{G^1(\bar{\mathbf{R }}^{-1})}\subset f^1(R^1, \bar{\mathbf{R }}^{-1})+p((R^1,\bar{\mathbf{R }}^{-1}),f)^\perp -R_+^L. \end{aligned}$$

(4)

All we have to show is the existence of an $\epsilon _{\bar{R}^1}$ satisfying (4) at $\bar{\mathbf{R }}$ where the consumption-direction vectors $g^i(\bar{\mathbf{R }},f)$, $i=1,2,3$, are different and $f^i(\bar{\mathbf{R }})\in int A$ for both $i=1,2$. Suppose this claim to be true. If $R^1\in{{\mathcal{R}}} (\bar{R}^i,Q^{1c})$ is sufficiently close to $\bar{R}^i$, then $f^1(R^i,\bar{\mathbf{R }}^{-i})$ is sufficiently close to $f(\bar{\mathbf{R }})$ by Proposition A1. Hence, the difference of consumption-direction vectors among agents 1,2, and 3 and positivity of consumption for agents 1 and 2 still hold at the preference profile $(R^1,\bar{\mathbf{R }}^{-1})$. Then, there exists an $\epsilon _{R^1}$ satisfying (4) by the supposed claim.

Contrary to the existence of an $\epsilon _{\bar{R}^1}$ satisfying (4), we suppose that $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ has an intersection with $f^1(\bar{\mathbf{R }})+ p(\bar{\mathbf{R }},f)^\perp +R_{++}^L$ in any neighborhood of $f^1(\bar{\mathbf{R }})$. Then, there exists a preference in a neighborhood of $\bar{R}^1$ that has the two most preferred consumption bundles in $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$. See Momi (2017, Proposition 2) for an example of such a preference $R^1_{\epsilon ,t(\epsilon )}$ that converges to $\bar{R}^1$ as $\epsilon \rightarrow 0$. Further, one of the two most preferred consumption bundles in $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ with respect to $R^1_{\epsilon ,t (\epsilon )}$ is $f^1(\bar{\mathbf{R }})$ and the other is in a neighborhood of $f^1(\bar{\mathbf{R }})$. Applying Lemma 1, we construct a new preference $\tilde{R}^1_{\epsilon ,t(\epsilon )}$ that is identical to $R^1_{\epsilon ,t(\epsilon )}$ in a neighborhood of $f^1(\bar{\mathbf{R }})$ and identical to $\bar{R}^1$ in $Q^{1c}$. Then, there exist the two most preferred consumption bundles in $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ with respect to $\tilde{R}^1_{\epsilon ,t(\epsilon )}$. This contradicts Lemma A3.

As for the choice of the scalars $\bar{\epsilon }$ and $\epsilon '$ satisfying (3), we follow Momi (2017, Proposition 2). We consider $\epsilon _{R^1}$ satisfying (4) for each $R^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ in a neighborhood of $\bar{R}^1$. For $\epsilon >0$, we consider the $\epsilon$-neighborhood of $\bar{R}^1$, ${\mathcal{B}}_{\epsilon } (\bar{R}^1)$, and define $\alpha (\epsilon )$ as the infimum of $\epsilon _{R^1}$ for $R^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})\bigcap{\mathcal{B}}_\epsilon (\bar{R}^1)$: $\alpha (\epsilon )=\inf _{ R^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})\bigcap{\mathcal{B}}_\epsilon (\bar{R}^1)} \epsilon _{R^1}$. Note that $\epsilon \mapsto \alpha (\epsilon )$ is a positive and decreasing function by the definition. On the other hand, we define $\beta (\epsilon )$ as the supremum of the distance between $f^1(R^1,\bar{\mathbf{R }}^{-1})$ and $f^1(\bar{\mathbf{R }})$ for $R^1\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})\bigcap{\mathcal{B}}_\epsilon (\bar{R}^1)$: $\beta (\epsilon )=\sup _{ R^1\in{{\mathcal{R}}}(\bar{R}^1, Q^{1c})\bigcap{\mathcal{B}}_\epsilon (\bar{R}^1)} \parallel f^1(R^1,\bar{\mathbf{R }}^{-1}) - f^1(\bar{\mathbf{R }}) \parallel$. Note that $\epsilon \mapsto \beta (\epsilon )$ is a positive and increasing function and $\beta (\epsilon )\rightarrow 0$ as $\epsilon \rightarrow 0$ because of the continuity of $f^1(\cdot ,\bar{\mathbf{R }}^{-1})$ shown in Proposition A1. We pick a scalar $\epsilon '$ satisfying $\beta (\epsilon ') <\frac{1}{2}\alpha (\epsilon ')$ and define $\bar{\epsilon }= \beta (\epsilon ')$. The existence of such a scalar $\epsilon '$ is ensured by the properties of the functions $\epsilon \mapsto \alpha (\epsilon )$ and $\epsilon \mapsto \beta (\epsilon )$ mentioned above.

These are the desired scalars. If $R^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ is in the $\epsilon '$-neighborhood of $\bar{R}^1$, ${\mathcal{B}}_{\epsilon '}(\bar{R}^1)$, then in the neighborhood $D_{\alpha (\epsilon ')}(f^1(R^1,\bar{\mathbf{R }}^{-1}))$, the option set $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ is in the lower left-hand side of the hyperplane $f(R^1,\bar{\mathbf{R }}^{-1}) +p((R^1,\bar{\mathbf{R }}^{-1}),f)^\perp$:

$$\begin{aligned} D_{\alpha (\epsilon ')}(f^1(R^1,\bar{\mathbf{R }}^{-1})) \bigcap \overline{G^1(\bar{\mathbf{R }}^{-1})}\subset f^1(R^1,\bar{\mathbf{R }}^{-1})+p((R^1,\bar{\mathbf{R }}^{-1}),f)^\perp -R_+^L. \end{aligned}$$

As the distance between $f^1(R^i,\bar{\mathbf{R }}^{-1})$ and $f^1(\bar{\mathbf{R }})$ is at most $\bar{\epsilon }$ and $\alpha (\epsilon ')>2\bar{\epsilon }$, we have $D_{\bar{\epsilon }} (f^1(\bar{\mathbf{R }}))\subset D_{\alpha (\epsilon ')}(f^1(R^1, \bar{\mathbf{R }}^{-1}))$. Hence, we have (3). $\square$

This proposition asserts the convexity of the option set in the following sense. For $x'=f^i(R^{i\prime },\mathbf{R }^{-i})$ and $x''=f^i(R^{i\prime \prime },\mathbf{R }^{-i})$ in $\overline{G^i(\bar{\mathbf{R }}^{-i})}$, $i=1,2$, if a ray $[sx'+(1-s)x'']$ with $s\in (0,1)$ has an intersection with $\overline{G^i(\bar{\mathbf{R }}^{-i})}$, then the intersection is written as $t(sx'+(1-s)x'')$ with a scalar $t\ge 1$ greater than or equal to one. If the intersection is given as $t(sx'+(1-s)x'')$ with $t<1$, then there exists $R^i$ such that $f^i(R^i,\bar{\mathbf{R }}^{-i})$ is arbitrarily close to the intersection because of the definition of $\overline{G^i(\bar{\mathbf{R }}^{-i})}$. Then, at least one of $x'$ and $x''$ is in the upper right-hand side of the hyperplane $f^i(R^i,\bar{\mathbf{R }}^{-i})+ p((R^i, \bar{\mathbf{R }}^{-i}),f))^\perp$ regardless of the value of the price vector, which contradicts Proposition A2.

Next, we show that $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ is a surface of a strictly convex set; that is, the t in the previous paragraph is strictly greater than one. In other words, for any $R^i$ in a neighborhood of $\bar{R}^i$, $f^i(R^i, \bar{\mathbf{R }}^{-i})$ is the unique intersection between $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ and the hyperplane $f^i(R^i,\bar{\mathbf{R }}^{-i})+p((R^i,\bar{\mathbf{R }}^{-i}),f)^\perp$. The next proposition is a counterpart of Proposition 3 in Momi (2017).

Proposition A3

For any$R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$in a neighborhood of$\bar{R}^i$, $i=1,2$, $f^i(R^i,\bar{\mathbf{R }}^{-i})$is the unique intersection between$\overline{G^{i} (\bar{\mathbf{R }}^{-i})}$and$f^i(R^i,\bar{\mathbf{R }}^{-i}) +p((R^i,\bar{\mathbf{R }}^{-i}),f)^\perp$in a neighborhood of$f^i(R^i, \bar{\mathbf{R }})$.

Proof

As in the proof of Proposition A2, we only have to prove the statement of the proposition at the preference profile $\bar{\mathbf{R }}$.

Without loss of generality, we prove the proposition for agent 1. We write $f(\bar{\mathbf{R }})=\bar{\mathbf{x }}=(\bar{x}^1,\ldots , \bar{x}^N)$ and $p(\bar{\mathbf{R }},f)=\bar{p}$. We have to prove that $\bar{x}^1$ is the unique intersection between $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ and $\bar{x}^1+\bar{p}^\perp$ in a neighborhood of $\bar{x}^1$.

Contrary to the statement of the proposition, we suppose that for any scalar $\epsilon >0$, there exists $\tilde{x}^1_\epsilon$ in the $\epsilon$-neighborhood $D_\epsilon (\bar{x}^1)$ of $\bar{x}^1$ such that $\tilde{x}_\epsilon ^1$ is different from $\bar{x}^1$, and $\tilde{x}_\epsilon ^1$ is the intersection between $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ and $\bar{x}^1+\bar{p}^\perp$. We show a contradiction.

We first observe that when $\epsilon$ is sufficiently small, the hyperplane $\bar{x}^1+\bar{p}^\perp$ is tangent to $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ along the segment ${[}\bar{x}^1,\tilde{x}_\epsilon ^1]\equiv \{t\bar{x}^1+(1-t) \tilde{x}_\epsilon ^1\in R^L_+: 0\le t\le 1\}$. We pick an arbitrary consumption bundle $x^1\in (\bar{x}^1, \tilde{x}_\epsilon ^1)\equiv \{t\bar{x}^1+(1-t)\tilde{x}_\epsilon ^1\in R^L_+:0<t<1\}$ and consider a preference $R^1\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})$ in a neighborhood of $\bar{R}^1$ so that the gradient vector of $R^1$ at $x^1$ is $\bar{p}$. When $\epsilon$ is sufficiently small, $x^1$ is sufficiently close to $\bar{x}^1$, and we can have such a preference $R^1$ in a neighborhood of $\bar{R}^1$. We let $x^{1\prime }$ denote the most preferred consumption bundle in $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ with respect to $R^1$ and $p^\prime$ denote the gradient vector of $R^1$ at $x^{1\prime }$. As shown in Proposition A2, $x^{1\prime }$ is in the lower left-hand side of the hyperplane $\bar{x}^1+\bar{p}^\perp$. Therefore, $\bar{p} x^{1\prime }\le \bar{p}\bar{x}^1=\bar{p}\tilde{x}_\epsilon ^1$, and hence $\bar{p} x^{1\prime }\le \bar{p} x^1$. By the same reasoning, $\bar{x}^1$ and $\tilde{x}_\epsilon ^1$ are in the lower left-hand side of the hyperplane $x^{1\prime }+( p^\prime )^\perp$. Therefore, $p^\prime \bar{x}^1\le p^\prime x^{1\prime }$ and $p^\prime \tilde{x}_\epsilon ^1\le p^\prime x^{1\prime }$, and hence $p^\prime x^1\le \ p^\prime x^{1\prime }$. These two inequalities are satisfied for the two combinations $(\bar{p}, x^1)$ and $(p^\prime , x^{1\prime })$ of a gradient vector and a consumption bundle with the preference $R^1$ if and only if $\bar{p}= p^\prime$ and $x^{1\prime }=x^1$. As our choice of $x^1$ is arbitrary on $(\bar{x}^1,\tilde{x}_\epsilon ^1)$, this implies that $\bar{x}^1+\bar{p}^\perp$ is tangent to $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ along the segment $[\bar{x}^1,\tilde{x}_\epsilon ^1]$. Hereafter, we assume that $\epsilon$ is sufficiently small so that $\bar{x}^1+\bar{p}^\perp$ is tangent to $\overline{G^{1}(\bar{\mathbf{R }}^{-1})}$ along the segment $[\bar{x}^1,\tilde{x}_\epsilon ^1]$.

For each $\epsilon$, we pick a consumption bundle $\hat{x}_\epsilon ^1\in (\bar{x}^1,\tilde{x}_\epsilon ^1)$ and a preference $\hat{R}_\epsilon ^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ such that the gradient vector of $\hat{R}_\epsilon ^1$ at $\hat{x}_\epsilon ^1$ is $\bar{p}$ and $\hat{R}_\epsilon ^1$ converges to $\bar{R}^1$ as $\epsilon \rightarrow 0$. We can have such a preference because $\hat{x}^1_\epsilon \rightarrow \bar{x}$ as $\epsilon \rightarrow 0$. Similarly, we let $\tilde{R}_\epsilon ^1\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})$ be a preference such that the gradient vector of $\tilde{R}_\epsilon ^1$ at $\tilde{x}_\epsilon ^1$ is $\bar{p}$ and $\tilde{R}_\epsilon ^1$ converges to $R^1$ as $\epsilon \rightarrow 0$. It is clear that $f^1(\tilde{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1}) =\tilde{x}_\epsilon ^1$, $f^1(\hat{R}_\epsilon ^1, \bar{\mathbf{R }}^{-1})=\hat{x}_\epsilon ^1$, and $p((\tilde{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1}),f) =p((\hat{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1}),f)=\bar{p}$. We write $f(\tilde{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1})=\tilde{\mathbf{x }}_\epsilon =(\tilde{x}_\epsilon ^1,\ldots ,\tilde{x}_\epsilon ^N)$ and $f(\hat{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1}) =\hat{\mathbf{x }}_\epsilon =(\hat{x}_\epsilon ^1,\ldots ,\hat{x}_\epsilon ^N)$.

For agent $i\ne 1$, the consumption bundles, $\bar{x}^i$, $\hat{x}_\epsilon ^i$, and $\tilde{x}_\epsilon ^i$ are all on the same ray $[g(\bar{R}^i,\bar{p})]$ because of the same price vector $\bar{p}$. We observe that $\tilde{x}_\epsilon ^2\ne \bar{x}^2$. If $\tilde{x}_\epsilon ^2=\bar{x}^2$, then, by comparing the two equations $\tilde{x}^1_\epsilon +\tilde{x}_\epsilon ^2 +\sum _{i=3}^N\tilde{x}_\epsilon ^i=\Omega$ and $\bar{x}^1+\bar{x}^2+\sum _{i=3}^N\bar{x}^i=\Omega$, we have $\tilde{x}_\epsilon ^1-\bar{x}^1=\sum _{i=3}^N\bar{x}^i -\sum _{i=3}^N\tilde{x}_\epsilon ^i$. Considering the inner product with $\bar{p}$ we have a contradiction, $0=\bar{p}\cdot (\tilde{x}_\epsilon ^1-\bar{x}^1)=\bar{p}\cdot (\sum _{i=3}^N\bar{x}^i-\sum _{i=3}^N\tilde{x}_\epsilon ^i)\ne 0$, where the last inequality comes from the fact that $\bar{p}$ is a positive vector and $\sum _{i=3}^N\bar{x}^i-\sum _{i=3}^N$ is parallel to a positive vector $g(R^3,\bar{p})$ under the equality of the consumption-direction vectors for $i\ge 3$.

Then, $\hat{x}_\epsilon ^2$ is between $\bar{x}^2$ and $\tilde{x}_\epsilon ^2$ on the same ray, and either $\bar{x}^2< \hat{x}_\epsilon ^2< \tilde{x}_\epsilon ^2$ or $\tilde{x}_\epsilon ^2<\hat{x}_\epsilon ^2< \bar{x}^2$ holds.

Figure 8 describes the situation in which the closure of agent 1’s option set is tangent to the hyperplane $\bar{x}^1+\bar{p}^\perp$ along the segment $[\bar{x}^1,\tilde{x}_\epsilon ^1]$, contrary to the statement of the proposition. In the following proof, we show that agent 2’s option set is also flat and observe that such flat option sets contradict the efficiency and strategy-proofness of the social choice function.

We now show that when $\epsilon$ is sufficiently small, $\overline{G^2(\hat{R}^1_\epsilon ,\bar{\mathbf{R }}^{-\{1,2\}})}$ is flat in a neighborhood of $\hat{x}_\epsilon ^2$. We suppose that it is not flat in any neighborhood of $\hat{x}_\epsilon ^2$, as illustrated in Figure 8. We let $\epsilon '>0$ be a scalar and let $\check{R}_{\epsilon '}^2\in{{\mathcal{R}}}(\bar{R}^2,Q^{2c})$ be agent 2’s preference in the $\epsilon '$ neighborhood ${\mathcal{B}}_{\epsilon '}(\bar{R}^2)$ of $\bar{R}^2$ such that the gradient vector $\check{p}_{\epsilon ,\epsilon '}$ of $\check{R}_{\epsilon '}^2$ at the most preferred consumption bundle in $\overline{G^2(\hat{R}^1_\epsilon ,\bar{\mathbf{R }}^{-\{1,2\}})}$ with respect to $\check{R}_{\epsilon '}^2$ is different from $\bar{p}$, as illustrated in the figure. As the closure of the option set is not flat in any neighborhood of $\hat{x}^2_\epsilon$, we can have such a preference $\check{R}_{\epsilon '}^2 \in{{\mathcal{R}}}(\bar{R}^2,Q^{2c})$ in any $\epsilon '$-neighborhood of $\bar{R}^2$. We write $f(\hat{R}_\epsilon ^1,\check{R}_{\epsilon '}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\check{\mathbf{x }}_{\epsilon ,\epsilon '} =(\check{x}_{\epsilon ,\epsilon '}^1,\ldots ,\check{x}_{\epsilon , \epsilon '}^N)$.⁹ We have $\check{p}_{\epsilon ,\epsilon '} \rightarrow \bar{p}$ and $(\check{x}_{\epsilon ,\epsilon '}^1, \check{x}_{\epsilon ,\epsilon '}^2)\rightarrow (\hat{x}_\epsilon ^1, \hat{x}_\epsilon ^2)$ as $\epsilon '\rightarrow 0$.

As $\hat{x}_\epsilon ^1$ is between $\bar{x}^1$ and $\tilde{x}_\epsilon ^1$, either $\check{p}_{\epsilon ,\epsilon '} \bar{x}^1>\check{p}_{\epsilon ,\epsilon '}\hat{x}_\epsilon ^1 >\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$, $\check{p}_{\epsilon ,\epsilon '}\bar{x}^1<\check{p}_{\epsilon ,\epsilon '} \hat{x}_\epsilon ^1<\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$, or $\check{p}_{\epsilon ,\epsilon '} \bar{x}^1=\check{p}_{\epsilon , \epsilon '}\hat{x}_\epsilon ^1=\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$ holds. When $\check{p}_{\epsilon ,\epsilon '} \bar{x}^1>\check{p}_{\epsilon ,\epsilon '}\hat{x}_\epsilon ^1 >\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$, as illustrated in Fig. 8, or when $\check{p}_{\epsilon ,\epsilon '} \bar{x}^1=\check{p}_{\epsilon ,\epsilon '}\hat{x}_\epsilon ^1 =\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$, we focus on the two combinations $(\bar{x}^1,\bar{p})$ and $(\check{x}_{\epsilon ,\epsilon '}^1,\check{p}_{\epsilon ,\epsilon '})$ of a consumption bundle and a price vector. We consider a preference $R_{\epsilon ,\epsilon '}^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ in a neighborhood of $\bar{R}^1$ such that the gradient vector of $R_{\epsilon ,\epsilon '}^1$ at $\bar{x}^1$ is $\bar{p}$ and that at $\check{x}_{\epsilon ,\epsilon '}^1$ is $\check{p}_{\epsilon , \epsilon '}$: $\bar{p}=p(R_{\epsilon ,\epsilon '}^1,\bar{x}^1)$ and $\check{p}_{\epsilon ,\epsilon '}=p(R_{\epsilon ,\epsilon '}^1, \check{x}_{\epsilon ,\epsilon '}^1)$. Refer to Momi (2017) for an example of such a preference.

With the preference $R_{\epsilon ,\epsilon '}^1$, we have $f^2(R_{\epsilon ,\epsilon '}^1, \check{R}^2_{\epsilon , \epsilon '},\bar{\mathbf{R }}^{-\{1,2\}}) =\check{x}^2_{\epsilon , \epsilon '}$ and $f^2(R_{\epsilon ,\epsilon '}^1, \bar{\mathbf{R }}^{-2}) =\bar{x}^2$. This contradicts the strategy-proofness of f with respect to agent 2 because $\check{x}^2_{\epsilon ,\epsilon '}\rightarrow \hat{x}^2_\epsilon$ as $\epsilon '\rightarrow 0$, and $\hat{x}_\epsilon ^2$, which is on the same ray as $\bar{x}^2$, satisfies $\hat{x}_\epsilon ^2<\bar{x}^2$ or $\bar{x}^2<\hat{x}_\epsilon ^2$ as mentioned above.

The discussion is symmetric when $\check{p}_{\epsilon ,\epsilon '} \bar{x}^1<\check{p}_{\epsilon ,\epsilon '}\hat{x}_{\epsilon ,\epsilon '}^1 <\check{p}_{\epsilon ,\epsilon '}\tilde{x}_\epsilon ^1$. We focus on the two pairs $(\check{x}_{\epsilon ,\epsilon '}^1, \check{p}_{\epsilon , \epsilon '})$ and $(\tilde{x}_\epsilon ^1,\bar{p})$. Similar to the discussion above, we consider a preference $R_{\epsilon ,\epsilon '}^1 \in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ in a neighborhood of $\tilde{R}_\epsilon ^1$, and thus, in a neighborhood of $\bar{R}^1$, such that the gradient vectors of $R_{\epsilon ,\epsilon '}^1$ at $\tilde{x}_\epsilon ^1$ and $\check{x}_{\epsilon ,\epsilon '}^1$ are $\bar{p}$ and $\check{p}_{\epsilon ,\epsilon '}$, respectively. Then, we have $f^i(R_{\epsilon ,\epsilon '}^1, \check{R}_{\epsilon , \epsilon '}^2,\bar{\mathbf{R }}^{-\{1,2\}})=\check{x}^i_{\epsilon , \epsilon '}$ for $i=1,2$, which converges to $\hat{x}^i_\epsilon$ as $\epsilon '\rightarrow 0$, and $f^i(R_{\epsilon ,\epsilon '}^1, \bar{\mathbf{R }}^{-2})=\tilde{x}^i_\epsilon$, $i=1,2$. This again contradicts the strategy-proofness of f with respect to agent 2. This ends the proof that $\overline{G^2(\hat{R}^1_\epsilon , \bar{\mathbf{R }}^{-\{1,2\}})}$ is flat in a neighborhood of $\hat{x}^2_\epsilon$ when $\epsilon$ is sufficiently small.

In addition to $\hat{x}_\epsilon ^1$ and $\hat{R}_\epsilon ^1$, we pick another consumption bundle $\acute{x}_\epsilon ^1(\ne \hat{x}_\epsilon ^1)$ on $(\bar{x}^1,\tilde{x}_\epsilon ^1)$ and another preference $\acute{R}_\epsilon ^1\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})$ for each $\epsilon$ such that the gradient vector of $\acute{R}_\epsilon ^1$ at $\acute{x}_\epsilon ^1$ is $\bar{p}$ and $\acute{R}_\epsilon ^1$ converges to $\bar{R}^1$ as $\epsilon \rightarrow 0$. It is clear that $f^1(\acute{R}_\epsilon ^1, \bar{\mathbf{R }}^{-\{1,2\}})=\acute{x}_\epsilon ^1$ and $p((\acute{R}_\epsilon ^1,\bar{\mathbf{R }}^{-\{1,2\}}),f)=\bar{p}$. We write $f(\acute{R}_\epsilon ^1,\bar{\mathbf{R }}^{-\{1,2\}}) =\acute{\mathbf{x }}_\epsilon =(\acute{x}_\epsilon ^1,\ldots , \acute{x}_\epsilon ^N)$. Similar to the discussion above, $\overline{G^2 (\acute{R}_\epsilon ^1, \bar{\mathbf{R }}^{-\{1,2\}})}$ is flat in a neighborhood of $\acute{x}_\epsilon ^2$ when $\epsilon$ is sufficiently small. Under the difference of the consumption-direction vectors among agents 1,2, and 3, $\acute{x}_\epsilon ^2$ is on the same ray as $\hat{x}_\epsilon ^2$ and $\acute{x}_\epsilon ^2\ne \hat{x}_\epsilon ^2$ because of $\acute{x}_\epsilon ^1\ne \hat{x}_\epsilon ^1$.

Now, we exchange the roles of agents 1 and 2. The closure of the option set $\overline{G^2(\hat{R}_\epsilon ^1, \bar{\mathbf{R }}^{-\{1,2\}})}$ is flat in a neighborhood of $\hat{x}_\epsilon ^2$. For a sufficiently small scalar $\epsilon ''$, we let $\grave{R}_{\epsilon ''}^2\in{{\mathcal{R}}} (\bar{R}^2,Q^{2c})$ be agent 2’s preference in ${\mathcal{B}}_{\epsilon ''}(\bar{R}^2)$ such that $f^2(\hat{R}_\epsilon ^1,\grave{R}_{\epsilon ''}^2, \bar{\mathbf{R }}^{-\{1,2\}})$ is on the flat part. Note that $f^2(\hat{R}_\epsilon ^1,\grave{R}_{\epsilon ''}^2, \bar{\mathbf{R }}^{-\{1,2\}})-\hat{x}_\epsilon ^2 \notin [g(\bar{R}^3;\bar{p})]$.

We write $f(\hat{R}^1_\epsilon ,\grave{R}^2_{\epsilon ''}, \bar{\mathbf{R }}^{-\{1,2\}})=\grave{\mathbf{x }}_{\epsilon , \epsilon ''}=(\grave{x}_{\epsilon ,\epsilon ''}^1,\ldots , \grave{x}_{\epsilon ,\epsilon ''}^N)$. Similar to the discussion above, we now have that $\grave{x}_{\epsilon ,\epsilon ''}^1(\ne \hat{x}_\epsilon ^1)$ is on the same ray as $\hat{x}_\epsilon ^1$ and $\overline{G^1(\grave{R}_{\epsilon ''}^2,\bar{\mathbf{R }}^{-\{1,2\}})}$ is flat in a neighborhood of $\grave{x}_{\epsilon ,\epsilon ''}^1$ when $\epsilon ''$ is sufficiently small.

We write $f(\acute{R}_{\epsilon }^1,\grave{R}_{\epsilon ''}^2, \bar{\mathbf{R }}^{-\{1,2\}})=\dot{\mathbf{x }}_{\epsilon ,\epsilon ''} =(\dot{x}_{\epsilon ,\epsilon ''}^1,\ldots ,\dot{x}_{\epsilon ,\epsilon ''}^N)$. As $\epsilon ''\rightarrow 0$, we have $\grave{R}_{\epsilon ''}^2\rightarrow \bar{R}^2$, and hence $(\dot{x}^1_{\epsilon ,\epsilon ''}, \dot{x}^2_{\epsilon ,\epsilon ''})\rightarrow (\acute{x}^1_{\epsilon , \epsilon ''}, \acute{x}^2_{\epsilon ,\epsilon ''})$. Therefore, when $\epsilon ''$ is sufficiently small, $\dot{x}_{\epsilon , \epsilon ''}^2$ is on the flat part of $\overline{G^2 (\acute{R}_{\epsilon }^1, \bar{\mathbf{R }}^{-\{1,2\}})}$ in a neighborhood of $\acute{x}_{\epsilon }^2$. As $\epsilon \rightarrow 0$, both $\hat{R}_\epsilon ^1$ and $\acute{R}_\epsilon ^1$ converge to $\bar{R}^1$, and hence both preferences become closer. Therefore, when $\epsilon$ is sufficiently small, $\dot{x}_{\epsilon , \epsilon ''}^1$ is sufficiently close to $\grave{x}_{\epsilon , \epsilon ''}^1$ and it is in the flat part of $\overline{G^1 (\grave{R}_{\epsilon ''}^2,\bar{\mathbf{R }}^{-\{1,2\}})}$ in a neighborhood of $\grave{x}_{\epsilon ,\epsilon ''}^1$.

Now, we consider four allocations, namely $\hat{\mathbf{x }}_\epsilon$, $\acute{\mathbf{x }}_{\epsilon }$, $\grave{\mathbf{x }}_{\epsilon ,\epsilon ''}$, and $\dot{\mathbf{x }}_{\epsilon ,\epsilon ''}$ with sufficiently small $\epsilon$ and $\epsilon ''$. Clearly, these satisfy four equations: $\sum _{i=1}^{N}\hat{x}_\epsilon ^i=\Omega$, $\sum _{i=1}^{N} \acute{x}_{\epsilon }^i=\Omega$, $\sum _{i=1}^{N}\grave{x}_{\epsilon , \epsilon ''}^i=\Omega$, and $\sum _{i=1}^{N}\dot{x}_{\epsilon , \epsilon ''}^i=\Omega$. As the price vector $\bar{p}$ is the same in all allocations and the preferences of agents $i\ge 3$ are unchanged, we have $\sum _{i=3}^N\acute{x}_{\epsilon }^i =a\sum _{i=3}^N\hat{x}_\epsilon ^i$, $\sum _{i=3}^N \grave{x}_{\epsilon ,\epsilon ''}^i=b\sum _{i=3}^N\hat{x}_\epsilon ^i$, and $\sum _{i=3}^N\dot{x}_{\epsilon ,\epsilon ''}^i=c\sum _{i=3}^N \hat{x}_\epsilon ^i$ using some scalars a, b, and c. As for agent 1, we have $\grave{x}_{\epsilon ,\epsilon ''}^1 =t\hat{x}_\epsilon ^1$ and $\dot{x}_{\epsilon ,\epsilon ''}^1 =t\acute{x}_{\epsilon }^1$ with a scalar t, because $\grave{x}_{\epsilon ,\epsilon ''}^1$ and $\hat{x}_\epsilon ^1$ are on the same ray $[g(\hat{R}_\epsilon ^1,\bar{p})]$, $\dot{x}_{\epsilon , \epsilon ''}^1$ and $\acute{x}_{\epsilon }^1$ are on the same ray $[g(\acute{R}_{\epsilon }^1,\bar{p})]$, and the segments $[\grave{x}_{\epsilon ,\epsilon ''}^1,\dot{x}_{\epsilon ,\epsilon ''}^1]$ and $[\hat{x}_\epsilon ^1,\acute{x}_{\epsilon }^1]$ are both perpendicular to $\bar{p}$. Similarly, we have $\acute{x}_{\epsilon }^2 =s \hat{x}_\epsilon ^2$ and $\dot{x}_{\epsilon ,\epsilon ''}^2=s\grave{x}_{\epsilon ,\epsilon ''}^2$ using a scalar s for agent 2.

Thus, we have

$$\begin{aligned}&\hat{x}_\epsilon ^1+\hat{x}_\epsilon ^2 +\sum _{i=3}^N\hat{x}_\epsilon ^i=\Omega , \end{aligned}$$

(5)

$$\begin{aligned}&\acute{x}_{\epsilon }^1+s\hat{x}_\epsilon ^2 +a\sum _{i=3}^N\hat{x}_\epsilon ^i=\Omega , \end{aligned}$$

(6)

$$\begin{aligned}&t\hat{x}_\epsilon ^1+\grave{x}_{\epsilon ,\epsilon '}^2 +b\sum _{i=3}^N\hat{x}_\epsilon ^i=\Omega , \end{aligned}$$

(7)

$$\begin{aligned}&t\acute{x}_{\epsilon }^1+s\grave{x}_{\epsilon ,\epsilon '}^2 +c\sum _{i=3}^N\hat{x}_\epsilon ^i=\Omega . \end{aligned}$$

(8)

From the first and second equations, we have:

$$\begin{aligned} s\hat{x}_\epsilon ^1-\acute{x}_{\epsilon }^1+(s-a) \sum _{i=3}^N\hat{x}_\epsilon ^i=(s-1)\Omega . \end{aligned}$$

(9)

From the third and fourth equations, we have:

$$\begin{aligned} t(s\hat{x}_\epsilon ^1-\acute{x}_{\epsilon }^1)+(sb-c) \sum _{i=3}^N\hat{x}_\epsilon ^i=(s-1)\Omega . \end{aligned}$$

(10)

From $\grave{x}_{\epsilon ,\epsilon ''}^1\ne \hat{x}_{\epsilon }^1$ and $\acute{x}_{\epsilon }^2\ne \hat{x}_{\epsilon }^2$, we have $s\ne 1$ and $t\ne 1$. Then, from (9) and (10), we have $(t(s-a)-(sb-c))\sum _{i=3}^N\hat{x}_{\epsilon }^i=(t-1)(s-1)\Omega$; that is, $\sum _{i=3}^N\hat{x}_{\epsilon }^i$ is parallel to $\Omega$. Then, from (5) and (6), we have that both $\hat{x}_{\epsilon }^1+\hat{x}_{\epsilon }^2$ and $\acute{x}_{\epsilon }^1 +s\hat{x}_{\epsilon }^2$ are parallel to $\Omega$, and then $\acute{x}_{\epsilon }^1=s\hat{x}_{\epsilon }^1$. This contradicts the fact that both $\acute{x}_{\epsilon }^1$ and $\hat{x}_{\epsilon }^1$ are on agent 1’s flat option set perpendicular to $\bar{p}$. $\square$

Fig. 8

Proof of Proposition A3

The next proposition, which is a counterpart of Proposition 4 in Momi (2017), shows that $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ is a smooth surface in the sense that at each point $f^i(R^i,\bar{\mathbf{R }}^{-i})$, the hyperplane tangent to $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ is unique.

Proposition A4

For any$R^i\in{{\mathcal{R}}}(\bar{R}^i,Q^{ic})$sufficiently close to$\bar{R}^i$, $i=1,2$, $f^i(R^i,\bar{\mathbf{R }}^{-i}) +p((R^i,\bar{\mathbf{R }}^{-i}), f)^\perp$is the unique hyperplane tangent to$\overline{G^i(\bar{\mathbf{R }}^{-i})}$at$f^i(R^i,\bar{\mathbf{R }}^{-i})$.

Proof

As in the proof of Proposition A2, we only have to prove the statement of the proposition at the preference profile $\bar{\mathbf{R }}$. Without loss of generality, we prove the statement for agent 1. We write $f(\bar{\mathbf{R }}) =\bar{\mathbf{x }}=(\bar{x}^1,\ldots ,\bar{x}^N)$ and $\bar{p}=p(\bar{\mathbf{R }},f)$.

As shown in Proposition A2, $\bar{x}^1+\bar{p}^\perp$ is a hyperplane tangent to $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ at $x^1$, and $\overline{ G^1(\bar{\mathbf{R }}^{-1})}$ is in the lower left-hand side of this hyperplane. We suppose that $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ has an edge at $\bar{x}^1$ and that there exists another hyperplane tangent to $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ at $\bar{x}^1$ with a normal vector $\tilde{p}$ different from $\bar{p}$. We show a contradiction.

As $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ is in the lower left-hand side of $\bar{x}^1+\bar{p}^\perp$, and in the lower left-hand side of $\bar{x}^1+\tilde{p}^\perp$, any hyperplane $\bar{x}^1+(t\bar{p}+(1-t)\tilde{p})^\perp$, $t\in [0,1]$, is tangent to $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ at $\bar{x}^1$

We let $\epsilon >0$ be a small scalar. For each $\epsilon$, we pick a preference $\tilde{R}_\epsilon ^1$ in the $\epsilon$-neighborhood ${\mathcal{B}}_\epsilon (\bar{R}^1)$ of $\bar{R}^1$ such that $\tilde{R}^1_\epsilon$ is identical to $\bar{R}^1$ outside $Q^1$, the gradient vector of $\tilde{R}_\epsilon ^1$ at $\bar{x}^1$ is different from $\bar{p}$ and $\bar{x}^1$ is the most preferred consumption bundle in $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ with respect to $\tilde{R}_\epsilon ^1$. The existence of such a preference should be clear from the above discussion. We let $\tilde{p}_\epsilon$ denote the gradient vector of $\tilde{R}_\epsilon ^1$ at $\bar{x}^1$: $\tilde{p}_\epsilon =p(\tilde{R}_\epsilon ^1,\bar{x}^1)$. We write $f(\tilde{R}_\epsilon ^1, \bar{\mathbf{R }}^{-1}) =\tilde{\mathbf{x }}_\epsilon =(\tilde{x}_\epsilon ^1,\ldots ,\tilde{x}_\epsilon ^N)$. It is clear that $\tilde{x}_\epsilon ^1=\bar{x}^1$ and $\tilde{p}_\epsilon =p((\tilde{R}_\epsilon ^1,\bar{\mathbf{R }}^{-1}),f)$. As $\epsilon \rightarrow 0$, $\tilde{R}^1_\epsilon \rightarrow \bar{R}^1$, $\tilde{p}_\epsilon \rightarrow \bar{p}$, and $(\tilde{x}_\epsilon ^1, \tilde{x}_\epsilon ^2) \rightarrow (\bar{x}^1,\bar{x}^2)$. Figure 9 describes the situation in which the closure of agent 1’s option set has an edge at $\bar{x}^1$, contrary to the statement of the proposition. In the following proof we observe that this induces allocations that contradict the pseudo-efficiency and strategy-proofness of the social choice function f.

As $\bar{R}^1$ and $\tilde{R}_\epsilon ^1$ have different gradient vectors at $\bar{x}^1$, $P(\bar{x}^1;\tilde{R}_\epsilon ^1)\setminus UC(\bar{x}^1;\bar{R}^1)\ne \emptyset$. Then, there exists a consumption bundle y in any neighborhood of $\bar{x}^1$ such that y is indifferent to $\bar{x}^1$ with respect to $\bar{R}^1$ and y is strictly preferred to $\bar{x}^1$ with respect to $\tilde{R}_\epsilon ^1$. We let $\{y_n\}_{n=1}^\infty$ be a sequence of such consumption bundles converging to $\bar{x}^1$ as $n\rightarrow \infty$: $y_n\in I(\bar{x}^1;\bar{R}^1)$, $y_n\in P(\bar{x}^1;\tilde{R}_\epsilon ^1)$, and $y_n\rightarrow \bar{x}^1$ as $n\rightarrow \infty$. We let $p_n'$ denote the gradient vector of $\bar{R}^1$ at $y_n$: $p_n'=p(\bar{R}^1,y_n)$.

We focus on agent 2. With each sufficiently large n, we let $x_n^{2\prime }$ be a consumption vector on $\overline{G^2(\bar{\mathbf{R }}^{-2})}$ such that the hyperplane $x_n^{2\prime }+p_n^{\prime \perp }$ is tangent to $\overline{G^2(\bar{\mathbf{R }}^{-2})}$ at $x_n^{2\prime }$. Such a consumption vector $x_n^{2\prime }$ is obtained uniquely because $\bar{x}^2$ is the unique intersection between $\overline{G^2(\bar{\mathbf{R }}^{-2})}$ and $\bar{x}^1+\bar{p}^\perp$, as shown in Proposition A3, and $p_n^\prime \rightarrow \bar{p}$ as $n\rightarrow \infty$. We have $x_n^{2\prime }\rightarrow \bar{x}^2$ as $n\rightarrow \infty$.¹⁰

We let $R_n^{2\prime }$ be agent 2’s preference that is identical to $\bar{R}^2$ outside $Q^2$ and has gradient vector $p_n^\prime$ at $x_n^{2\prime }$, and converges to $\bar{R}^2$ as $n\rightarrow \infty$. Refer to Momi (2017, Proposition 4) for an example of such a preference. We write $f(R_n^{2\prime },\bar{\mathbf{R }}^{-2}) =\mathbf{x }_n^\prime =(x_n^{1\prime },\ldots ,x_n^{N\prime })$. Observe that $x_n^{1\prime }$ is on the ray $[y_n]$.

For each sufficiently small $\epsilon$, there exists a sufficiently large n, and we can have agent 2’s preference $\hat{R}_{\epsilon ,n}^2$ identical to $\bar{R}^2$ outside $Q^2$ such that (i) the gradient vector of $\hat{R}_{\epsilon ,n}^2$ at $\tilde{x}_\epsilon ^2$ is $\tilde{p}_\epsilon$ and (ii) the gradient vector of $\hat{R}_{\epsilon ,n}^2$ at $x_{n}^{2\prime }$ is $p_{n}'$. Refer to Momi (2017, Proposition 4) for an example of such a preference.

We write $f(\hat{R}_{\epsilon ,n}^2, \bar{\mathbf{R }}^{-2}) =\hat{\mathbf{x }}_{\epsilon ,n}=(\hat{x}_{\epsilon ,n}^1,\ldots , \hat{x}_{\epsilon ,n}^N)$ and $f(\tilde{R}_\epsilon ^1, \hat{R}_{\epsilon ,n}^2,\bar{\mathbf{R }}^{-\{1,2\}}) =\check{\mathbf{x }}_{\epsilon ,n}=(\check{x}_{\epsilon ,n}^1, \ldots ,\check{x}_{\epsilon ,n}^N)$. We have $\check{x}_{\epsilon ,n}^2=\tilde{x}_\epsilon ^2$ because of (i), and hence $\check{x}^1_{\epsilon ,n}=\tilde{x}^1_\epsilon =\bar{x}^1$ by Lemma A1. On the contrary, we have $\hat{x}_{\epsilon ,n}^2 =x_{n}^{2\prime }$ because of (ii), and hence $\hat{x}^1_{\epsilon ,n}={x}_{n}^{1\prime }$. Note that $\hat{x}_{\epsilon ,n}^1$ is on the ray $[y_n]$.

We first consider the case in which $\hat{x}_{\epsilon ,n}^1\in P(\bar{x}^1; \tilde{R}_\epsilon ^1)$. If agent 1 has the preference $\tilde{R}_\epsilon ^1$ and faces the other agents’ preferences $(\hat{R}_{\epsilon ,n}^2,\bar{\mathbf{R }}^{-\{1,2\}})$, he is better off by reporting $\bar{R}^1$ and achieving $\hat{x}_{\epsilon ,n}^1$ than reporting the true preference $\tilde{R}_\epsilon ^1$ and achieving $\check{x}_{\epsilon ,n}^1=\tilde{x}_\epsilon ^1=\bar{x}^1$. This contradicts the strategy-proofness of f.

Next, we consider the case $\hat{x}_{\epsilon ,n}^1\notin UC(\bar{x}^1;\bar{R}^1)$. If agent 1 has the preference $\bar{R}^1$ and faces the other agents’ preferences $(\hat{R}_{\epsilon ,n}^2, \bar{\mathbf{R }}^{-\{1,2\}})$, he is better off by reporting $\tilde{R}_\epsilon ^1$ and achieving $\check{x}_{\epsilon ,n}^1 =\tilde{x}_\epsilon ^1=\bar{x}^1$ than reporting his true preference $\bar{R}^1$ and achieving $\hat{x}_{\epsilon ,n}^1$. Again, this contradicts the strategy-proofness of f.

As $\hat{x}_{\epsilon ,n}^1\in [y_\epsilon ']$, where $y_{n}\in I(\bar{x}^1;\bar{R}^1)$ and $y_{n}\in P(\bar{x}^1; \tilde{R}_\epsilon ^{1})$, only these two cases need to be considered. $\square$

Fig. 9

Proof of Proposition A4

Finally we show that $\overline{G^i(\bar{\mathbf{R }}^{-i})}$ coincides with $G^i(\bar{\mathbf{R }}^{-i})$ and that it is an $L-1$-dimensional manifold in a neighborhood of $f^i(\bar{\mathbf{R }})\in int A$, for $i=1,2$.

Proposition A5

For$i=1,2$, in a neighborhood of$f^i (\bar{\mathbf{R }})$, $G^i(\bar{\mathbf{R }}^{-i})$coincides with$\overline{G^i(\bar{\mathbf{R }}^{-i})}$and it is an$L-1$-dimensional manifold.

Proof

Without loss of generality, we prove the proposition for agent 1. We write $f(\bar{\mathbf{R }})=\bar{\mathbf{x }}=(\bar{x}^1,\ldots , \bar{x}^N)$ and $p(\bar{\mathbf{R }},f)=\bar{p}$.

We consider the Cobb–Douglas utility functions $u^{\alpha }(x)=x_1^{\alpha _1}\cdots x_L^{\alpha _L}$ with a parameter $\alpha =(\alpha _1,\ldots ,\alpha _L)\in S_{++}^{L-1}$ and the preferences represented by these utility functions.

Observe that the gradient vector of the preferences represented by the utility function $u^\alpha (x)$ at a consumption bundle $x=(x_1,\ldots ,x_L)$ is given by the normalization of $(\frac{\alpha _1}{x_1},\ldots ,\frac{\alpha _L}{x_L})$. On the contrary, if a preference represented by a Cobb–Douglas utility function $u^\alpha (x)$ has the gradient vector $p=(p_1,\ldots ,p_L)$ at $x=(x_1,\ldots ,x_L)$, then the parameter $\alpha$ is the normalization of $({p_1}{x_1},\ldots ,{p_L}{x_L})$.

We let $u^{\alpha ^{*}}(x)$ be the Cobb–Douglas utility function such that agent 1’s preference $R^{1\alpha ^{*}}$ represented by $u^{\alpha ^{*}}(x)$ has the gradient vector $\bar{p}$ at $\bar{x}^1$.

We consider a preference $R^{1\alpha }$ represented by a Cobb–Douglas utility function $u^\alpha$ where $\alpha$ is in a neighborhood of $\alpha ^{*}$. Then, $R^{1\alpha }$ is in a neighborhood of $R^{1\alpha ^{*}}$.

As f is supposed to satisfy efficiency and strategy-proofness only on ${{\mathcal{R}}}(\bar{R}^1,Q^{1c})\times{{\mathcal{R}}} (\bar{R}^2,Q^{2c})\times \{\bar{R}^3\}\times \cdots \times \{\bar{R}^N\}$, we let $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$ denote the most preferred consumption bundle in $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ with respect to the preference $R^{1\alpha }$. If $R^{1\alpha }\in{{\mathcal{R}}} (\bar{R}^1,Q^{1c})$, then, $f^1(R^{1\alpha },\bar{\mathbf{R }}^{-1}) =\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$.

We have $\tilde{f}^1(R^{1\alpha ^{*}},\bar{\mathbf{R }}^{-i}) =\bar{x}^1$ and $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-i})$ is in a neighborhood of $\bar{x}^1$ when $\alpha$ is in a neighborhood of $\alpha ^{*}$ because of the properties of $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ shown in Propositions A2–A4. Observe that $\alpha \ne \alpha ^\prime$ implies that $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})\ne \tilde{f}^1(R^{1 \alpha ^\prime },\bar{\mathbf{R }}^{-1})$ because $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1}) =\tilde{f}^1(R^{1\alpha ^\prime },\bar{\mathbf{R }}^{-1})$ and $\alpha \ne \alpha ^\prime$ imply that $R^{1\alpha }$ and $R^{1\alpha ^\prime }$ have different gradient vectors at the same consumption bundle, which contradicts Proposition A4. Thus, each $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$ in a neighborhood of $\bar{x}^1$ is identified with the corresponding parameter $\alpha \in S_{++}^{L-1}$ in a neighborhood of $\alpha ^{*}$, and hence in a neighborhood of $\bar{x}^1$, $\bigcup _{\alpha } \tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$ is an $L-1$-dimensional manifold.

To end the proof, we prove that in a neighborhood of $\bar{x}^1$, $\bigcup _{\alpha }\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$, $G^1(\bar{\mathbf{R }}^{-1})$, and $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ coincide. We let $\tilde{p}_\alpha$ denote the gradient vector of $R^{1\alpha }$ at $\tilde{f}^1(R^{1\alpha }, \bar{\mathbf{R }}^{-1})$: $\tilde{p}_\alpha =p(R^{1\alpha }, \tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1}))$. When $\alpha$ is in a neighborhood of $\alpha ^{*}$ and $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$ is in a neighborhood of $\bar{x}^1$, we can construct a new preference $\tilde{R}^1\in{{\mathcal{R}}}(\bar{R}^1,Q^{1c})$ such that the gradient vector of $\tilde{R}^1$ at $\tilde{f}^1(R^{1\alpha }, \bar{\mathbf{R }}^{-1})$ is $\tilde{p}_\alpha$. Refer to Momi (2017, Proposition 5) for an example of such a preference.

Then, $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$ is the most preferred consumption bundle in $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ with respect to $\tilde{R}^1$, and $\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1}) =f(\tilde{R}^1,\bar{\mathbf{R }}^{-1}) \in G^1(\bar{\mathbf{R }}^{-1})$, as in Lemma A2. Therefore, we have $\bigcup _\alpha \tilde{f}(R^{1\alpha },\bar{\mathbf{R }}^{-1}) \subset G^1(\bar{\mathbf{R }}^{-1}) \subset \overline{G^1 (\bar{\mathbf{R }}^{-1})}$ in a neighborhood of $\bar{x}^1$. To observe that any element of $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ in a neighborhood of $\bar{x}^1$ is included in $\bigcup _{\alpha }\tilde{f}^1(R^{1\alpha },\bar{\mathbf{R }}^{-1})$, note that any ray [y] in a neighborhood of $[\bar{x}^1]$ intersects with $\overline{G^1(\bar{\mathbf{R }}^{-1})}$ once at most because of strategy-proofness, and hence $\overline{G^1 (\bar{\mathbf{R }}^{-1})}$ is at most $L-1$ dimensional. $\square$