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?


1 Answer 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 $[u(0), u^b)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.