<p>By definition, a function is continuous if arbitrarily small changes in its value can be assured by choosing sufficiently small changes of its argument. Since <span class="math-container">$u(x')&gt;u(x^*)$</span> and the utility function <span class="math-container">$u$</span> is continuous, a small enough change from <span class="math-container">$x'$</span> from <span class="math-container">$x''$</span> will ensure that <span class="math-container">$u(x'')&gt;u(x^*)$</span> too.</p>
<hr />
<p>Formally, let <span class="math-container">$\varepsilon=\frac{1}{2}[u(x')-u(x^*)]$</span>. Then, by continuity of <span class="math-container">$u$</span>, there exists a <span class="math-container">$\delta&gt;0$</span> such that for all <span class="math-container">$x\in X$</span></p>
<p><span class="math-container">$$|x-x'|&lt;\delta \implies u(x')-\varepsilon&lt;u(x)&lt;u(x')+\varepsilon$$</span></p>
<p>In particular, there exists a <span class="math-container">$\delta&gt;0$</span> such that for all <span class="math-container">$x\in X$</span>:</p>
<p><span class="math-container">$$|x-x'|&lt;\delta \implies u(x)&gt;u(x')-\varepsilon=\frac{1}{2}[u(x')+u(x^*)]&gt;u(x^*)$$</span></p>
<p>In particular, we know that if <span class="math-container">$x''=\alpha x'$</span> with:
<span class="math-container">$$|\alpha x'-x'|&lt;\delta \iff (1-\alpha)&lt;\frac{\delta}{|x'|} \iff \alpha&gt;1-\frac{\delta}{|x'|}$$</span></p>
<p>then <span class="math-container">$u(x'')&gt;u(x^*)$</span>.</p>

CC BY-SA 4.0
2023-09-08T15:30:40.110Z
Thanks for your answer. How do you guarantee that $\alpha\in(0,1)$?
manifold
CC BY-SA 4.0
2023-09-08T16:44:12.127Z
Since $\delta>0$, it must be that $1-\frac{\delta}{|x'|}<1$, so there certainly exist  $\alpha\in(0,1)$ such that $\alpha>1-\frac{\delta}{|x'|}$.
smcc
CC BY-SA 4.0
2023-09-08T05:07:04.990Z
2023-09-08T10:41:59.643Z
2023-09-08T10:41:59.643Z
smcc
<p>In <a href="https://rads.stackoverflow.com/amzn/click/com/0195073401" rel="nofollow noreferrer" rel="nofollow noreferrer">Mas-Colell et al.'s Microeconomic Theory</a> Proposition 3.E.1(ii) (p. 58) states that if <span class="math-container">$\succsim$</span> is a rational (i.e. complete and transitive), continuous, and locally nonsatiated preference relation on (<span class="math-container">$X\equiv\mathbb{R}_{+}^L$</span>) represented by continuous utility function <span class="math-container">$u(\cdot)$</span>, then <span class="math-container">$x^*\in\text{arg}\min\{p\cdot x:u(x)\ge \underline{u}&gt;u(0)\}$</span> implies <span class="math-container">$x^*\in\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$</span> (in words, that under the assumptions the solution to the EMP also solves the UMP).</p>
<p>Now I have trouble deciphering a step of the proof (bolded). The proof goes as follows.</p>
<p>Proof. Suppose <span class="math-container">$x^*\in\text{arg}\min\{p\cdot x:u(x)\ge u(x^*)&gt;u(0)\}$</span> and <span class="math-container">$x^*\notin\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$</span>. First we note that <span class="math-container">$x^*\ne0$</span>. Now, by our hypothesis we have that there exists <span class="math-container">$x'\in X$</span> with <span class="math-container">$u(x')&gt;u(x^*)$</span> and <span class="math-container">$p\cdot x'\le p\cdot x^*$</span>. <strong>Consider a bundle <span class="math-container">$x''=\alpha x'$</span> where <span class="math-container">$\alpha\in(0,1)$</span> (<span class="math-container">$x''$</span> is a scaled version of <span class="math-container">$x'$</span>). By continuity of <span class="math-container">$u(\cdot)$</span>, if <span class="math-container">$\alpha$</span> is close enough to <span class="math-container">$1$</span>, then we will have <span class="math-container">$u(x'')&gt;u(x^*)$</span> and <span class="math-container">$p\cdot x''&lt;p\cdot x^*$</span></strong>. And the conclusion clearly follows from the previous argument. <span class="math-container">$\square$</span></p>
<p>I don't understand the argument bolded (which I transcribed exactly as in the book), I suppose it has something to do with continuity of preferences that in the limit (i.e. &quot;<span class="math-container">$\alpha$</span> close to <span class="math-container">$1$</span>&quot;) <span class="math-container">$x''\succ x'$</span> but I don't see how, I think I should build some sort of sequence like <span class="math-container">$(1-1/(n+1))x'$</span> that in the limit has that property, but not sure how to see this argument.</p>
<p>I would be very grateful if anyone can point me the underlying argument of the reasoning mentioned.</p>
<p>Thanks.</p>

<microeconomics><decision-theory>
Proving duality of UMP and EMP arguing with continuity of utility
manifold
# Proving duality of UMP and EMP arguing with continuity of utility

(Asked by: username_0 on 2023-09-08)

In [Mas-Colell et al.'s Microeconomic Theory](https://rads.stackoverflow.com/amzn/click/com/0195073401) Proposition 3.E.1(ii) (p. 58) states that if $\succsim$ is a rational (i.e. complete and transitive), continuous, and locally nonsatiated preference relation on ($X\equiv\mathbb{R}\_{+}^L$) represented by continuous utility function $u(\cdot)$, then $x^\*\in\text{arg}\min\{p\cdot x:u(x)\ge \underline{u}>u(0)\}$ implies $x^\*\in\text{arg}\max\{u(x):p\cdot x\le p\cdot x^\*\}$ (in words, that under the assumptions the solution to the EMP also solves the UMP).

Now I have trouble deciphering a step of the proof (bolded). The proof goes as follows.

Proof. Suppose $x^\*\in\text{arg}\min\{p\cdot x:u(x)\ge u(x^\*)>u(0)\}$ and $x^\*\notin\text{arg}\max\{u(x):p\cdot x\le p\cdot x^\*\}$. First we note that $x^\*\ne0$. Now, by our hypothesis we have that there exists $x'\in X$ with $u(x')>u(x^\*)$ and $p\cdot x'\le p\cdot x^\*$. **Consider a bundle $x''=\alpha x'$ where $\alpha\in(0,1)$ ($x''$ is a scaled version of $x'$). By continuity of $u(\cdot)$, if $\alpha$ is close enough to $1$, then we will have $u(x'')>u(x^\*)$ and $p\cdot x''<p\cdot x^\*$**. And the conclusion clearly follows from the previous argument. $\square$

I don't understand the argument bolded (which I transcribed exactly as in the book), I suppose it has something to do with continuity of preferences that in the limit (i.e. "$\alpha$ close to $1$") $x''\succ x'$ but I don't see how, I think I should build some sort of sequence like $(1-1/(n+1))x'$ that in the limit has that property, but not sure how to see this argument.

I would be very grateful if anyone can point me the underlying argument of the reasoning mentioned.

Thanks.

**Tags:** microeconomics, decision-theory


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## 1 answers

### Answer by username_1 on 2023-09-08. Score: 2 | Accepted answer

By definition, a function is continuous if arbitrarily small changes in its value can be assured by choosing sufficiently small changes of its argument. Since $u(x')>u(x^\*)$ and the utility function $u$ is continuous, a small enough change from $x'$ from $x''$ will ensure that $u(x'')>u(x^\*)$ too.

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Formally, let $\varepsilon=\frac{1}{2}[u(x')-u(x^\*)]$. Then, by continuity of $u$, there exists a $\delta>0$ such that for all $x\in X$

$$|x-x'|<\delta \implies u(x')-\varepsilon<u(x)<u(x')+\varepsilon$$

In particular, there exists a $\delta>0$ such that for all $x\in X$:

$$|x-x'|<\delta \implies u(x)>u(x')-\varepsilon=\frac{1}{2}[u(x')+u(x^\*)]>u(x^\*)$$

In particular, we know that if $x''=\alpha x'$ with:
$$|\alpha x'-x'|<\delta \iff (1-\alpha)<\frac{\delta}{|x'|} \iff \alpha>1-\frac{\delta}{|x'|}$$

then $u(x'')>u(x^\*)$.

> **Comments:**
> **username_0**: Thanks for your answer. How do you guarantee that $\alpha\in(0,1)$?  
> **username_1**: Since $\delta>0$, it must be that $1-\frac{\delta}{|x'|}<1$, so there certainly exist  $\alpha\in(0,1)$ such that $\alpha>1-\frac{\delta}{|x'|}$.  

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