Suppose that the direct utility $u(x_1,...,x_n)$ is concave in each of its arguments. Does this imply that the indirect utility $U(w,p)$ is concave with respect to $w$? If all goods are normal than this can proved. e.g. using Lagrange multipliers. But, is it true in general?

One can assume that $u$ is differential as many times as necessary.


  • $\begingroup$ what you wrote is \begin{eqnarray*} u(x^d(p, \lambda m' + (1-\lambda)m'')) \geq u(\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')) \end{eqnarray*} assuming that $u(.)$ is non-decreasing that means ; \begin{eqnarray*} x^d(p, \lambda m' + (1-\lambda)m'') \geq \lambda x^d(p, m') + (1-\lambda)x^d(p, m'') \end{eqnarray*} how did you concluded that? I cannot easily that from $(1)$. Does that mean demand is concave function when utility is concave? $\endgroup$ – meh Jul 24 '19 at 22:25

Your statement does not seem to be true for normal goods either. A counterexample: $$ U(x_1,x_2) = (x_1x_2)^{\frac{2}{3}} $$ By the Cobb-Douglas property the optimum bundle given $w,p$ is $$ (x_1,x_2) = \left(\frac{w}{2p_1}, \frac{w}{2p_2}\right), $$ so both goods are normal. The indirect utility is $$ U(w,p) = \left(\frac{1}{4p_1p_2}\right)^{\frac{2}{3}}w^{\frac{4}{3}}. $$ Here $w$ is raised to a power higher than 1 so $U(w,p)$ is convex in $w$.

| improve this answer | |
  • 2
    $\begingroup$ It might be worth pointing out that this is at least partly due to your choice of utility representation. The utility function $\hat{U}(x_1,x_2) = (x_1 x_2)^{1/2}$ represents the same preferences, but has an indirect utility that is linear in $w$. $\endgroup$ – Theoretical Economist Nov 2 '17 at 11:56
  • $\begingroup$ @TheoreticalEconomist Yes, but also note that the choice of utility representation is not completely free. That is $U(x_1,x_2) = x_1^2x_2^2$ is not a good counterexample, because although it obviously has a convex indirect utility function, it is not concave in the goods individually. So the representation space was somewhat limited and my choice was in within limitations. The inverse question of "If the indirect utility function is concave in income does an individually concave in goods utility function representation always exist?" is interesting. $\endgroup$ – Giskard Nov 2 '17 at 13:31
  • $\begingroup$ Question: if $u$ is concave (as a multi-dimensional function), does this guarantee that the indirect utility is concave? I think it is true, but am I wrong again? $\endgroup$ – user154729 Nov 2 '17 at 19:44
  • 1
    $\begingroup$ @denesp To your question "If the indirect utility function is concave in income does an individually concave in goods utility function representation always exist?". It would seem to me that concave indirect utility only says something about the structure of the utility around the optimal allocation path. Away from this path, the utility can behave arbitrarily, so long as it does not give an optimal allocation. So, it seems to me that it would be hard to conclude any type of global concavity from concavity of the indirect utility. $\endgroup$ – user154729 Nov 2 '17 at 20:08
  • 1
    $\begingroup$ @Giskard: yes but still the subtlety was not clear (to me) because of the title of the question (in bold too) "If the (direct) utility is concave in all goods…". Anyway I think that both answers are interesting to see when then property holds and when not. $\endgroup$ – Bertrand Jul 25 '19 at 13:03

It might be worth stating that : if $u$ is concave (as a multi-dimensional function), it will yield an indirect utility function that is concave in $m$.

If $u:\mathbb{R}^n_+\rightarrow \mathbb{R}$ is concave, then indirect utility function $v:\mathbb{R}^n_+ \times \mathbb{R}_+\rightarrow \mathbb{R}$ defined as $v(p, m) := \displaystyle\max_{x\in B(p, m)} u(x)$ is also concave in $m$. Here $B(p, m) = \{x \in \mathbb{R}^n_+ : p\cdot x \leq m\}$ is the budget set. Let $x^d(p, m)$ denotes the solution to the maximization problem $\displaystyle\max_{x\in B(p, m)} u(x)$ so that $v(p, m) = u(x^d(p, m))$.

Consider any arbitrary $m'$, and $m''$ and a $\lambda \in [0, 1]$,

$p\cdot x^d(p, m') \leq m'$

$p\cdot x^d(p, m'') \leq m''$


$p\cdot (\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')) \leq \lambda m' + (1-\lambda) m'' \tag{1}$

Consequently, \begin{eqnarray*} v(p, \lambda m' + (1-\lambda)m'')& = & u(x^d(p, \lambda m' + (1-\lambda)m'')) & \\ & \geq & u(\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')) & \ \ [by \ (1)] \\ & \geq & \lambda u(x^d(p, m')) + (1-\lambda)u(x^d(p, m'')) & \ \ [by \ \text{concavity of } u] \\ & = & \lambda v(p, m') + (1-\lambda)v(p, m'') & \end{eqnarray*}

| improve this answer | |
  • 1
    $\begingroup$ You could precise that your first inequality follows from the fact that at income $\lambda m' + (1-\lambda)m''$, the bundle $x^d(p, \lambda m' + (1-\lambda)m'')$ maximizes utility, while this is not necessarily the case for the feasible alternative bundle $\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')$ $\endgroup$ – Bertrand Jul 25 '19 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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