<p>It might be worth stating that : i<strong>f $u$ is concave (as a multi-dimensional function), it will yield an indirect utility function that is concave in $m$.</strong></p>

<p>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 $v(p, m) = u(x^d(p, m))$.</p>

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

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

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

<p>Therefore,</p>

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

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

CC BY-SA 4.0
2019-07-25T10:21:08.087Z
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 $\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')$
Bertrand
CC BY-SA 4.0
2018-06-14T01:22:13.037Z
2018-06-14T01:22:13.037Z
Amit
<p>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 $w$.</p>

CC BY-SA 3.0
2017-11-02T11:56:44.507Z
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 $w$.
Theoretical Economist
CC BY-SA 3.0
2017-11-02T13:31:23.677Z
@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.
Giskard
CC BY-SA 3.0
2017-11-02T19:44:23.557Z
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?
user154729
CC BY-SA 3.0
2017-11-02T19:52:26.947Z
My proof is as follows: Let $x^*(w)$ be the optimal allocation at budget $w$.  Consider budgets $w_-<w_+$.  Then, at budget $w_{1/2}=(w_-+w_+)/2$, the allocations $x^*(w_-)/2+x^*(w_+)/2$ is a feasible allocation.   So, $U(w_{1/2})\geq u(x^*(w_-)/2+x^*(w_+)/2)\geq u(x^*(w_-))/2+u(x^*(w_+))/2=U(w_-)/2+U(w_+)/2$, where the second inequality is by concavity of $u$. 

Is this correct?
user154729
CC BY-SA 3.0
2017-11-02T20:08:29.193Z
@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.
user154729
CC BY-SA 3.0
2017-11-02T20:58:17.560Z
@user154729 Yes, you are indeed correct. W.r.t. your other question: Please post new questions as actual questions, not comments. But yes, your proof seems to have the right idea, though it is not rigorous.
Giskard
CC BY-SA 4.0
2019-07-25T11:26:27.943Z
@Bertrand Luckily it does not have to be, as the question specifies "$u(x_1,...,x_n)$ is concave in *each of its arguments*".
Giskard
CC BY-SA 4.0
2019-07-25T12:02:09.397Z
@Giskard. OK I see. So the apparent different conclusions between Amit and your answer are entirely due to different hypotheses regarding $u$ which is required to be concave in $x_1$ and in $x_2$ in your answer (and the initial question), and concave in $x_1,x_2$ by Amit. Again, the devil is in the details...
Bertrand
CC BY-SA 4.0
2019-07-25T12:50:49.963Z
@Bertrand Amit has made this quite apparent, he put his different assumption in bold font.
Giskard
CC BY-SA 4.0
2019-07-25T13:03:33.723Z
@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.
Bertrand
CC BY-SA 4.0
2019-07-25T10:09:06.267Z
I am afraid that $$
U(x_1,x_2) = (x_1x_2)^{\frac{2}{3}}
$$
is not concave in $(x_1,x_2)$
Bertrand
CC BY-SA 3.0
2017-11-02T10:03:29.057Z
2017-11-02T10:03:29.057Z
Giskard
<p>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?  </p>

<p>One can assume that $u$ is differential as many times as necessary.</p>

<p>Thanks.</p>

CC BY-SA 4.0
2019-07-24T22:25:07.827Z
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?
meh
<microeconomics><consumer-theory>
If the (direct) utility is concave in all goods, is the indirect utility necessarily concave in wealth?
user154729
# If the (direct) utility is concave in all goods, is the indirect utility necessarily concave in wealth?

(Asked by: username_0 on 2017-11-02)

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.

Thanks.

**Tags:** microeconomics, consumer-theory

> **Comments:**
> **username_1**: 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?  

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## 2 answers

### Answer by username_4 on 2017-11-02. Score: 4 | Accepted answer

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 $w$.

> **Comments:**
> **username_5**: 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 $w$.  
> **username_4**: @username_5 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.  
> **username_0**: 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?  
> **username_0**: My proof is as follows: Let $x^*(w)$ be the optimal allocation at budget $w$.  Consider budgets $w_-<w_+$.  Then, at budget $w_{1/2}=(w_-+w_+)/2$, the allocations $x^*(w_-)/2+x^*(w_+)/2$ is a feasible allocation.   So, $U(w_{1/2})\geq u(x^*(w_-)/2+x^*(w_+)/2)\geq u(x^*(w_-))/2+u(x^*(w_+))/2=U(w_-)/2+U(w_+)/2$, where the second inequality is by concavity of $u$.   Is this correct?  
> **username_0**: @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.  
> **username_4**: @username_0 Yes, you are indeed correct. W.r.t. your other question: Please post new questions as actual questions, not comments. But yes, your proof seems to have the right idea, though it is not rigorous.  
> **username_3**: I am afraid that $$ U(x_1,x_2) = (x_1x_2)^{\frac{2}{3}} $$ is not concave in $(x_1,x_2)$  
> **username_4**: @username_3 Luckily it does not have to be, as the question specifies "$u(x_1,...,x_n)$ is concave in *each of its arguments*".  
> **username_3**: @username_4. OK I see. So the apparent different conclusions between username_2 and your answer are entirely due to different hypotheses regarding $u$ which is required to be concave in $x_1$ and in $x_2$ in your answer (and the initial question), and concave in $x_1,x_2$ by username_2. Again, the devil is in the details...  
> **username_4**: @username_3 username_2 has made this quite apparent, he put his different assumption in bold font.  
> **username_3**: @username_4: 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.  

---

### Answer by username_2 on 2018-06-14. Score: 3

It might be worth stating that : i**f $u$ 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 $\lambda \in [0, 1]$,

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

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

Therefore,

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

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\*}

> **Comments:**
> **username_3**: 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 $\lambda x^d(p, m') + (1-\lambda)x^d(p, m'')$  

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