<p>Even if <span class="math-container">$u^b$</span> is finite, it can never be achieved. This is what is meant by "does not attain a maximum". Rather, <span class="math-container">$u(x)$</span> <em>approaches</em> <span class="math-container">$u^b$</span> from below as <span class="math-container">$x \to \infty$</span>. This is because <span class="math-container">$u$</span> is <em>strictly increasing</em>. If we had <span class="math-container">$u(x_*) = u^b$</span> for some <span class="math-container">$x_*$</span>, then we would have <span class="math-container">$u(x_* + 1) &gt; u^b$</span> and <span class="math-container">$u^b$</span> could not be a bound. This is why <span class="math-container">$U$</span> is written as the half-open interval <span class="math-container">$[u(0), u^b)$</span>.</p>

CC BY-SA 4.0
2018-10-07T18:10:32.323Z
2018-10-07T18:10:32.323Z
nanoman
<p>I am trying to understand the fact that <span class="math-container">$e(p, v(p,y)) = y$</span>. There is a proof in the text Advanced Microeconomic Theory (Jehle and Reny) that states the following:</p>

<blockquote>
  <p>Because <span class="math-container">$u(·)$</span> is strictly increasing on <span class="math-container">$R_n^+$</span>, it attains a minimum at <span class="math-container">$x = 0$</span>, but does not attain a maximum. Moreover, because <span class="math-container">$u(·)$</span> is continuous, the set <span class="math-container">$U$</span> of attainable utility
  numbers must be an interval. Consequently, <span class="math-container">$U = [u(0), u^b)]$</span> for <span class="math-container">$u^b &gt; u(0)$</span>, and where <span class="math-container">$u^b$</span> may be either finite or <span class="math-container">$+∞$</span>.</p>
  
  <p>To prove, fix <span class="math-container">$(p, y) ∈ R^{++}_n × R^+$</span>. We know <span class="math-container">$e(p, v(p, y)) ≤ y$</span>. We would like to show in fact that equality must hold. So suppose not, that is, suppose <span class="math-container">$e(p, u)&lt;y$</span>, where <span class="math-container">$u = v(p, y)$</span>. Note that by definition of <span class="math-container">$v(·), u ∈ U$</span>, so that <span class="math-container">$u&lt;u^b$</span>. By the continuity of <span class="math-container">$e(·)$</span> from, we may therefore choose <span class="math-container">$ε &gt; 0$</span> small enough so that <span class="math-container">$u + ε&lt;u^b$</span>, and <span class="math-container">$e(p, u +
ε)&lt;y$</span>. </p>
</blockquote>

<p>I don't understand the point below:</p>

<blockquote>
  <p>Note that by definition of <span class="math-container">$v(·)$</span>, <span class="math-container">$u ∈ U$</span>, so that <span class="math-container">$u&lt;u^b$</span>. </p>
</blockquote>

<p><span class="math-container">$v(·)$</span> denotes the maximum utility that can be achieved for given prices and wealth, and I do not see why we cannot have <span class="math-container">$u = u^b$</span> such that <span class="math-container">$u ≤ u^b$</span> since <span class="math-container">$u^b$</span> may be finite. What am I missing here?  </p>

<microeconomics><utility><price><wealth>
The relationship between indirect utility and expenditure functions
Cola
# The relationship between indirect utility and expenditure functions

(Asked by: username_0 on 2018-10-07)

I am trying to understand the fact that $e(p, v(p,y)) = y$. There is a proof in the text Advanced Microeconomic Theory (Jehle and Reny) that states the following:

> Because $u(·)$ is strictly increasing on $R\_n^+$, it attains a minimum at $x = 0$, but does not attain a maximum. Moreover, because $u(·)$ is continuous, the set $U$ of attainable utility
> numbers must be an interval. Consequently, $U = [u(0), u^b)]$ for $u^b > u(0)$, and where $u^b$ may be either finite or $+∞$.
>
> To prove, fix $(p, y) ∈ R^{++}\_n × R^+$. We know $e(p, v(p, y)) ≤ y$. We would like to show in fact that equality must hold. So suppose not, that is, suppose $e(p, u)<y$, where $u = v(p, y)$. Note that by definition of $v(·), u ∈ U$, so that $u<u^b$. By the continuity of $e(·)$ from, we may therefore choose $ε > 0$ small enough so that $u + ε<u^b$, and $e(p, u +
> ε)<y$.

I don't understand the point below:

> Note that by definition of $v(·)$, $u ∈ U$, so that $u<u^b$.

$v(·)$ denotes the maximum utility that can be achieved for given prices and wealth, and I do not see why we cannot have $u = u^b$ such that $u ≤ u^b$ since $u^b$ may be finite. What am I missing here?

**Tags:** microeconomics, utility, price, wealth


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## 1 answers

### Answer by username_1 on 2018-10-07. Score: 1

Even if $u^b$ is finite, it can never be achieved. This is what is meant by "does not attain a maximum". Rather, $u(x)$ *approaches* $u^b$ from below as $x \to \infty$. This is because $u$ is *strictly increasing*. If we had $u(x\_\*) = u^b$ for some $x\_\*$, then we would have $u(x\_\* + 1) > u^b$ and $u^b$ could not be a bound. This is why $U$ is written as the half-open interval $[u(0), u^b)$.


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