<p>Here is a simple proof by contradiction: If this were not the case, we could assume by passing to a subsequence that the sequence <span class="math-container">$(z(p_n))$</span> converges to some <span class="math-container">$z$</span>. The continuity of the inner product shows that <span class="math-container">$p\cdot z^i\leq 0$</span> and, therefore, <span class="math-container">$p\cdot x^i\leq p\cdot e^i$</span> with <span class="math-container">$x^i=z^i+e^i$</span> for all <span class="math-container">$i$</span>. Suppose <span class="math-container">$x^i$</span> were not optimal under <span class="math-container">$p$</span> for <span class="math-container">$i$</span>. Then there is some <span class="math-container">$y\succ x^i$</span> such that <span class="math-container">$p\cdot y\le p\cdot e^i&gt;0$</span>. By continuity of preferences, we can assume that <span class="math-container">$p\cdot y&lt;p\cdot e^i$</span> and <span class="math-container">$y\succ x_n^i$</span> for <span class="math-container">$n$</span> large enough. But <span class="math-container">$p_n\cdot y\leq p_n\cdot e^i$</span>, contradicting the optimality of <span class="math-container">$x^i_n$</span> at <span class="math-container">$p_n$</span>. So <span class="math-container">$x^i$</span> is optimal at <span class="math-container">$p$</span>. But this is impossible if <span class="math-container">$\succeq_i$</span> is strictly monotone and some price of <span class="math-container">$p$</span> equals zero; there are no optimal bundles in that case.</p>
<p>The argument is standard; see for example Theorem 1.3.6. of &quot;Existence and Optimality of Competitive Equilibria&quot; by Aliprantis, Brown, Burkinshaw.</p>

CC BY-SA 4.0
2024-12-02T13:17:15.860Z
Perfect! Thank you. It's so simple in retrospect.
Joseph Basford
CC BY-SA 4.0
2024-12-02T00:12:42.363Z
2024-12-02T10:44:08.300Z
2024-12-02T10:44:08.300Z
Michael Greinecker
<p>Exercise 17.B.2 of MWG asks us to prove the following.</p>
<p>Suppose we have <span class="math-container">$N$</span> consumers, each with access to commodity space <span class="math-container">$X_i=\mathbb{R}_+^L$</span> and preferences <span class="math-container">$\succeq_i$</span> which are continuous, strictly convex, and strongly monotone. Further, suppose initial endowments <span class="math-container">$\{e^i\}$</span> satisfy <span class="math-container">$\sum_ie^i&gt;&gt;0$</span> (element-wise), then the aggregate excess demand function <span class="math-container">$z(p)$</span>, defined for <span class="math-container">$p&gt;&gt;0$</span> satisfies <span class="math-container">$$p^n\to p\neq 0 \text{ with }p_{\ell}=0 \text{ for some }\ell \implies \max\{z_1(p_n),\dots,z_L(p^n)\}\to \infty$$</span></p>
<hr />
<p>I have seen the following proof of this which assumes preferences are represented by a continuously differentiable utility function <span class="math-container">$u_i$</span>.</p>
<p>As <span class="math-container">$\sum_{j=1}^Ne^j_{\ell}&gt;0$</span> for all <span class="math-container">$\ell$</span> and <span class="math-container">$p\neq 0$</span>, for <span class="math-container">$n$</span> large <span class="math-container">$p^n\neq 0$</span>. Therefore, there is some consumer, indexed as <span class="math-container">$j\in I$</span>, which for such large <span class="math-container">$n$</span>, has a strictly positive amount of purchasing power. That is, for large <span class="math-container">$n$</span>, <span class="math-container">$\exists \varepsilon_1&gt;0$</span> such that <span class="math-container">$$w^j(p)=\sum_{\ell =1}^{\infty}e_{\ell}^jp_{\ell}^n\geq \varepsilon_1&gt;0$$</span> As <span class="math-container">$p\neq 0$</span>, there is <span class="math-container">$k$</span> such that <span class="math-container">$p_k &gt; 0$</span>.</p>
<p>If consumer <span class="math-container">$j$</span> is not purchasing any good which has strictly positive price, in the limit their expenditure is <span class="math-container">$$\lim_{n\to \infty} \sum_{\ell :p_{\ell}=0}x_{\ell}^j(p^n,w^j(p^n))p_{\ell}^n$$</span> as since by local non-satiation the Marshallian demand must be such that all purchasing power is utilised and <span class="math-container">$p_{\ell}^n\to 0$</span> in the above, it must be that <span class="math-container">$$\lim_{n\to \infty}\max_{\ell:p_{\ell}=0}\{x_{\ell}^j(p^n,w^j(p^n))\}=\infty $$</span> Otherwise, in the limit, total expenditure of consumer <span class="math-container">$j$</span> tends to <span class="math-container">$0$</span>: <span class="math-container">$$\lim_{n\to \infty} \sum_{\ell :p_{\ell}=0}x_{\ell}^j(p^n,w^j(p^n))p_{\ell}^n=0$$</span> Therefore, <span class="math-container">$$\max\{z_1(p^n),\dots , z_L (p^n)\}\xrightarrow{n\to \infty} \infty$$</span> as other agents can only sell up to their endowment amount so cannot counteract the <span class="math-container">$\infty$</span> demand by consumer <span class="math-container">$j$</span>.</p>
<p>If consumer <span class="math-container">$j$</span> instead does consume some of good <span class="math-container">$k$</span> at prices <span class="math-container">$p$</span> which has <span class="math-container">$p_k&gt;0$</span>, then there exists <span class="math-container">$\varepsilon_2&gt;0$</span> such that <span class="math-container">$x_k^j(p^n,w^k(p^n))$</span> for all <span class="math-container">$n$</span> large (because Marshallian demand is continuous). As <span class="math-container">$p_k&gt;0$</span>, we consider the case for <span class="math-container">$n$</span> sufficiently large than <span class="math-container">$p^n_k&gt;0$</span>. Consider the consumer's optimisation problem of picking between goods <span class="math-container">$\ell$</span> and <span class="math-container">$k$</span>, where <span class="math-container">$p_k&gt;0$</span> and <span class="math-container">$p_{\ell}=0$</span>. Hold the expenditure on all other goods constant to isolate the trade-off between <span class="math-container">$\ell$</span> and <span class="math-container">$k$</span>. Let <span class="math-container">$E^j_{-\ell,k}$</span> be the expenditure not used on goods <span class="math-container">$\ell$</span> and <span class="math-container">$k$</span>. The consumer then solves <span class="math-container">$$\max_{x_{\ell},x_k\geq 0}u^j(x_{\ell},x_k,x^j_{-\ell,k}) \text{  st. } p_k^nx_k+p_{\ell}^n x_{\ell} \leq w^j(p^n)-E^j_{-\ell,k}$$</span> The associated Lagrangian is <span class="math-container">$$\mathcal{L}=u^j(x_{\ell},x_k,x^j_{-\ell,k}) -\lambda  (p_k^nx_k+p_{\ell}^n x_{\ell}-w^j(p^n)+E^j_{\ell,k})$$</span> First order conditions (strict concavity used here) gives that <span class="math-container">$$\frac{\partial u}{\partial x_{\ell}}(x_{\ell},x_k,x^j_{-\ell,k})-\lambda p_{\ell}^n\leq 0$$</span> and <span class="math-container">$$\frac{\partial u}{\partial x_{k}}(x_{\ell},x_k,x^j_{-\ell,k})-\lambda p_{k}^n\leq 0$$</span> Dividing one by the other... <span class="math-container">$$\frac{\partial u}{\partial x_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))\leq \frac{p_{\ell}^n}{p_k^n}$$</span> Where we recover the Marshallian demands by setting <span class="math-container">$x^j_{-\ell,k}$</span> to be the optimal choices over goods not <span class="math-container">$\ell$</span> and <span class="math-container">$k$</span>. Because <span class="math-container">$p_k^n\to p_k&gt;0$</span> and <span class="math-container">$p_{\ell}^n\to p_{\ell}=0$</span>, this gives us that <span class="math-container">$$\lim_{n\to \infty}\frac{\partial u}{\partial x_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))\leq 0$$</span> Because utility is increasing, this is also greater than <span class="math-container">$0$</span>, so <span class="math-container">$$\lim_{n\to \infty}\frac{\partial u}{\partial x_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))= 0$$</span></p>
<p>So, either <span class="math-container">$\lim_{n\to \infty}\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))=\infty$</span> or <span class="math-container">$\lim_{n\to \infty}\frac{\partial u}{\partial x_{\ell}}(x^j(p^n,w^j(p^n)))=0$</span>. Since we assumed utility is strictly increasing, <span class="math-container">$\frac{\partial u}{\partial x_{\ell}}(\cdot)&gt;0$</span>. Assume, seeking contradiction, that the sequence of Marshallian demands <span class="math-container">$(x^j(p^n,w^j(p^n)))_{n\in \mathbb{N}}$</span> stay in a compact set <span class="math-container">$K$</span> for all <span class="math-container">$n$</span>. Then, by Weierstrass <span class="math-container">$\frac{\partial u}{\partial x_{\ell}}(\cdot)$</span> achieves it's minimum so <span class="math-container">$\varepsilon_3&gt;0$</span> such that <span class="math-container">$\frac{\partial u}{\partial x_{\ell}}(\cdot)&gt;\varepsilon_3&gt;0$</span> But then it cannot converge to <span class="math-container">$0$</span>. We can do exactly the same argument but with a maximum to show that <span class="math-container">$\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))$</span> cannot converge off to infinity. But this contradicts <span class="math-container">$$\lim_{n\to \infty}\frac{\partial u}{\partial x_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x_{k}}(x^j(p^n,w^j(p^n)))= 0$$</span> So it must be that the sequence <span class="math-container">$(x^j(p^n,w^j(p^n)))_{n\in \mathbb{N}}$</span> leaves the compact set <span class="math-container">$K$</span>. Therefore, <span class="math-container">$(x^j(p^n,w^j(p^n)))_{n\in \mathbb{N}}$</span> is unbounded. Thus, it must be that <span class="math-container">$x^j(p^n,w^j(p^n))\to \infty$</span>, so that <span class="math-container">$$\max\{z_1(p^n),\dots , z_L (p^n)\}\xrightarrow{n\to \infty} \infty$$</span></p>
<hr />
<p>How can I adjust this argument to not assume a continuously differentiable utility representation?</p>

<microeconomics><mathematical-economics>
MWG Exercise 17.B.2
Joseph Basford
# MWG Exercise 17.B.2

(Asked by: username_0 on 2024-12-01)

Exercise 17.B.2 of MWG asks us to prove the following.

Suppose we have $N$ consumers, each with access to commodity space $X\_i=\mathbb{R}\_+^L$ and preferences $\succeq\_i$ which are continuous, strictly convex, and strongly monotone. Further, suppose initial endowments $\{e^i\}$ satisfy $\sum\_ie^i>>0$ (element-wise), then the aggregate excess demand function $z(p)$, defined for $p>>0$ satisfies $$p^n\to p\neq 0 \text{ with }p\_{\ell}=0 \text{ for some }\ell \implies \max\{z\_1(p\_n),\dots,z\_L(p^n)\}\to \infty$$

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I have seen the following proof of this which assumes preferences are represented by a continuously differentiable utility function $u\_i$.

As $\sum\_{j=1}^Ne^j\_{\ell}>0$ for all $\ell$ and $p\neq 0$, for $n$ large $p^n\neq 0$. Therefore, there is some consumer, indexed as $j\in I$, which for such large $n$, has a strictly positive amount of purchasing power. That is, for large $n$, $\exists \varepsilon\_1>0$ such that $$w^j(p)=\sum\_{\ell =1}^{\infty}e\_{\ell}^jp\_{\ell}^n\geq \varepsilon\_1>0$$ As $p\neq 0$, there is $k$ such that $p\_k > 0$.

If consumer $j$ is not purchasing any good which has strictly positive price, in the limit their expenditure is $$\lim\_{n\to \infty} \sum\_{\ell :p\_{\ell}=0}x\_{\ell}^j(p^n,w^j(p^n))p\_{\ell}^n$$ as since by local non-satiation the Marshallian demand must be such that all purchasing power is utilised and $p\_{\ell}^n\to 0$ in the above, it must be that $$\lim\_{n\to \infty}\max\_{\ell:p\_{\ell}=0}\{x\_{\ell}^j(p^n,w^j(p^n))\}=\infty $$ Otherwise, in the limit, total expenditure of consumer $j$ tends to $0$: $$\lim\_{n\to \infty} \sum\_{\ell :p\_{\ell}=0}x\_{\ell}^j(p^n,w^j(p^n))p\_{\ell}^n=0$$ Therefore, $$\max\{z\_1(p^n),\dots , z\_L (p^n)\}\xrightarrow{n\to \infty} \infty$$ as other agents can only sell up to their endowment amount so cannot counteract the $\infty$ demand by consumer $j$.

If consumer $j$ instead does consume some of good $k$ at prices $p$ which has $p\_k>0$, then there exists $\varepsilon\_2>0$ such that $x\_k^j(p^n,w^k(p^n))$ for all $n$ large (because Marshallian demand is continuous). As $p\_k>0$, we consider the case for $n$ sufficiently large than $p^n\_k>0$. Consider the consumer's optimisation problem of picking between goods $\ell$ and $k$, where $p\_k>0$ and $p\_{\ell}=0$. Hold the expenditure on all other goods constant to isolate the trade-off between $\ell$ and $k$. Let $E^j\_{-\ell,k}$ be the expenditure not used on goods $\ell$ and $k$. The consumer then solves $$\max\_{x\_{\ell},x\_k\geq 0}u^j(x\_{\ell},x\_k,x^j\_{-\ell,k}) \text{ st. } p\_k^nx\_k+p\_{\ell}^n x\_{\ell} \leq w^j(p^n)-E^j\_{-\ell,k}$$ The associated Lagrangian is $$\mathcal{L}=u^j(x\_{\ell},x\_k,x^j\_{-\ell,k}) -\lambda (p\_k^nx\_k+p\_{\ell}^n x\_{\ell}-w^j(p^n)+E^j\_{\ell,k})$$ First order conditions (strict concavity used here) gives that $$\frac{\partial u}{\partial x\_{\ell}}(x\_{\ell},x\_k,x^j\_{-\ell,k})-\lambda p\_{\ell}^n\leq 0$$ and $$\frac{\partial u}{\partial x\_{k}}(x\_{\ell},x\_k,x^j\_{-\ell,k})-\lambda p\_{k}^n\leq 0$$ Dividing one by the other... $$\frac{\partial u}{\partial x\_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))\leq \frac{p\_{\ell}^n}{p\_k^n}$$ Where we recover the Marshallian demands by setting $x^j\_{-\ell,k}$ to be the optimal choices over goods not $\ell$ and $k$. Because $p\_k^n\to p\_k>0$ and $p\_{\ell}^n\to p\_{\ell}=0$, this gives us that $$\lim\_{n\to \infty}\frac{\partial u}{\partial x\_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))\leq 0$$ Because utility is increasing, this is also greater than $0$, so $$\lim\_{n\to \infty}\frac{\partial u}{\partial x\_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))= 0$$

So, either $\lim\_{n\to \infty}\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))=\infty$ or $\lim\_{n\to \infty}\frac{\partial u}{\partial x\_{\ell}}(x^j(p^n,w^j(p^n)))=0$. Since we assumed utility is strictly increasing, $\frac{\partial u}{\partial x\_{\ell}}(\cdot)>0$. Assume, seeking contradiction, that the sequence of Marshallian demands $(x^j(p^n,w^j(p^n)))\_{n\in \mathbb{N}}$ stay in a compact set $K$ for all $n$. Then, by Weierstrass $\frac{\partial u}{\partial x\_{\ell}}(\cdot)$ achieves it's minimum so $\varepsilon\_3>0$ such that $\frac{\partial u}{\partial x\_{\ell}}(\cdot)>\varepsilon\_3>0$ But then it cannot converge to $0$. We can do exactly the same argument but with a maximum to show that $\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))$ cannot converge off to infinity. But this contradicts $$\lim\_{n\to \infty}\frac{\partial u}{\partial x\_{\ell}}(x^j(p^n,w^j(p^n)))/\frac{\partial u}{\partial x\_{k}}(x^j(p^n,w^j(p^n)))= 0$$ So it must be that the sequence $(x^j(p^n,w^j(p^n)))\_{n\in \mathbb{N}}$ leaves the compact set $K$. Therefore, $(x^j(p^n,w^j(p^n)))\_{n\in \mathbb{N}}$ is unbounded. Thus, it must be that $x^j(p^n,w^j(p^n))\to \infty$, so that $$\max\{z\_1(p^n),\dots , z\_L (p^n)\}\xrightarrow{n\to \infty} \infty$$

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How can I adjust this argument to not assume a continuously differentiable utility representation?

**Tags:** microeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2024-12-02. Score: 3 | Accepted answer

Here is a simple proof by contradiction: If this were not the case, we could assume by passing to a subsequence that the sequence $(z(p\_n))$ converges to some $z$. The continuity of the inner product shows that $p\cdot z^i\leq 0$ and, therefore, $p\cdot x^i\leq p\cdot e^i$ with $x^i=z^i+e^i$ for all $i$. Suppose $x^i$ were not optimal under $p$ for $i$. Then there is some $y\succ x^i$ such that $p\cdot y\le p\cdot e^i>0$. By continuity of preferences, we can assume that $p\cdot y and $y\succ x\_n^i$ for $n$ large enough. But $p\_n\cdot y\leq p\_n\cdot e^i$, contradicting the optimality of $x^i\_n$ at $p\_n$. So $x^i$ is optimal at $p$. But this is impossible if $\succeq\_i$ is strictly monotone and some price of $p$ equals zero; there are no optimal bundles in that case.

The argument is standard; see for example Theorem 1.3.6. of "Existence and Optimality of Competitive Equilibria" by Aliprantis, Brown, Burkinshaw.

> **Comments:**
> **username_0**: Perfect! Thank you. It's so simple in retrospect.  

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