<p>Here are some references for a somewhat unsatisfying picture.</p>

<p>Rubinstein's lecture notes in microeconomic theory discuss a necessary condition for differentiability.</p>

<p>Debreu provided necessary and sufficient conditions for differentiability of demand functions. Unfortunately, they are not very intuitive.</p>

<p>Debreu, G. (1972). Smooth Preferences. Econometrica, 40(4), 603-615. 
Debreu, G. (1976). Smooth Preferences: A Corrigendum. Econometrica, 44(4), 831. </p>

CC BY-SA 3.0
2017-09-18T03:30:44.713Z
2017-09-18T03:30:44.713Z
HRSE
<p>If you want to impose sufficient conditions on the preferences, sadly it seems to me that this is nearly impossible. If a utility function describes a preference ordering then so does every monotonic transformation of the same utility function. If a utility function $U$ takes at least two values, say $a$ and $b$, you could use the following monotonic function
$$
f(u) = \left\{
\begin{array}{ll}
u &amp; \mbox{ if } u \leq \frac{a+b}{2} \\
\\
u + 1 &amp; \mbox{ if } u &gt; \frac{a+b}{2}.
\end{array} \right.
$$
to get a new utility function $f(U)$ which would still describe the same preferences. However, this function will not be differentiable, even if $U$ was. The only way to guarantee that a utility function describing the preference ordering is always differentiable is if you do not allow any form of it to take two different values, that is if all the goods are neutral to the consumer.</p>

<p>If you want to impose conditions on the utility functions itself it seems to me you would have to require that the marginal utilities exist and are continuous which is kind of the same as saying that the utility function is differentiable. So this is not a very useful condition. A better condition may exist, but I don't think it does.</p>

CC BY-SA 3.0
2015-11-14T21:03:58.337Z
Just to note, if you consider von neumann morgenstern utility, these are defined up to only affine transformations.
jmbejara
CC BY-SA 3.0
2015-11-15T12:41:14.143Z
Thanks for your answer, however, the question asks about the existence of a continuous, differentiable representation, not that all representations need to be continuous and differentiable. See last sentence of first paragraph.
HRSE
CC BY-SA 3.0
2015-11-15T13:01:51.120Z
@HRSE I see. Then I think your question boils down to: Under which conditions will a continuous function have a monoton transformation that is also differentiable. It seems to me that if there are only countable many points were the original function is nondifferentiable, a suitable monotonic transformation can always be found by 'smoothing out' the original function at these points.
Giskard
CC BY-SA 3.0
2015-11-16T03:12:12.900Z
I am not so sure, since two nondifferentiable points may have the same utility value.
HRSE
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2015-11-16T10:03:49.537Z
@HRSE I apologize, my statement was not well formulated. What I meant to say is that there should only be finitely many utility values where the function is nondifferentiable. Multiple consumption bundles with the same nondifferentiable utility do not cause problems because the 'neighbouring' indifference curves would be the same ones because of the continuity of the utility function. And then smoothing out the utility values would work in all the bundles.
Giskard
CC BY-SA 3.0
2015-11-14T11:18:29.323Z
2015-11-14T13:23:37.967Z
2015-11-14T13:23:37.967Z
Giskard
<p>A continuous, transitive, complete, convex, and locally non-satiated preference on an open convex subset $V$ of some $\mathbb{R}^l$ has an $r$-times continuously differentiable utility representation with no critical points if and only if the space $I=\{(x,y)\in V\times V\mid x\sim y\}$ is a $C^r$-manifold.</p>

<p>This result is essentially Proposition 2.3.9. in the book "The Theory of General Economic Equilibrium: A Differentiable Approach" by Andreu Mas-Colell. Actually, the result there replaces convexity of $V$ and the preference relation to be represented by the weaker condition that all indifference curves are connected.</p>

<p>I'm not aware of any more elementary approach to the problem.</p>

CC BY-SA 3.0
2017-09-18T13:34:12.480Z
2017-09-18T13:34:12.480Z
Michael Greinecker
<p>We know that if we start with a connected, separable product space $V_1\times,...,\times V_n$ and a complete, transitive, and continuous preference relation $\succsim$ on this product space, that there exists a continuous! utility representation $u:V_1\times,...,\times V_n\rightarrow \mathbb{R}$. However, continuity $\neq$ differentiability. Thus, I am curious under which conditions there even exists a differentiable utility function.</p>

<p>My first idea would be to at least restrict the product space to $\mathbb{R}^n$, but there may be counterexamples. For example, if $n=1$ and $u$ is the Weierstrass function, $u$ is continuous, but not differentiable.</p>

<p>Since in Economics we constantly work with differentiable utility, I am wondering which assumptions are necessary to ensure differentiability.</p>

<utility>
When is a utility representation differentiable?
HRSE
# When is a utility representation differentiable?

(Asked by: username_0 on 2015-11-14)

We know that if we start with a connected, separable product space $V\_1\times,...,\times V\_n$ and a complete, transitive, and continuous preference relation $\succsim$ on this product space, that there exists a continuous! utility representation $u:V\_1\times,...,\times V\_n\rightarrow \mathbb{R}$. However, continuity $\neq$ differentiability. Thus, I am curious under which conditions there even exists a differentiable utility function.

My first idea would be to at least restrict the product space to $\mathbb{R}^n$, but there may be counterexamples. For example, if $n=1$ and $u$ is the Weierstrass function, $u$ is continuous, but not differentiable.

Since in Economics we constantly work with differentiable utility, I am wondering which assumptions are necessary to ensure differentiability.

**Tags:** utility


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## 3 answers

### Answer by username_3 on 2017-09-18. Score: 5 | Accepted answer

A continuous, transitive, complete, convex, and locally non-satiated preference on an open convex subset $V$ of some $\mathbb{R}^l$ has an $r$-times continuously differentiable utility representation with no critical points if and only if the space $I=\{(x,y)\in V\times V\mid x\sim y\}$ is a $C^r$-manifold.

This result is essentially Proposition 2.3.9. in the book "The Theory of General Economic Equilibrium: A Differentiable Approach" by Andreu Mas-Colell. Actually, the result there replaces convexity of $V$ and the preference relation to be represented by the weaker condition that all indifference curves are connected.

I'm not aware of any more elementary approach to the problem.


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### Answer by username_0 on 2017-09-18. Score: 2

Here are some references for a somewhat unsatisfying picture.

Rubinstein's lecture notes in microeconomic theory discuss a necessary condition for differentiability.

Debreu provided necessary and sufficient conditions for differentiability of demand functions. Unfortunately, they are not very intuitive.

Debreu, G. (1972). Smooth Preferences. Econometrica, 40(4), 603-615.
Debreu, G. (1976). Smooth Preferences: A Corrigendum. Econometrica, 44(4), 831.


---

### Answer by username_1 on 2015-11-14. Score: 1

If you want to impose sufficient conditions on the preferences, sadly it seems to me that this is nearly impossible. If a utility function describes a preference ordering then so does every monotonic transformation of the same utility function. If a utility function $U$ takes at least two values, say $a$ and $b$, you could use the following monotonic function
$$
f(u) = \left\{
\begin{array}{ll}
u & \mbox{ if } u \leq \frac{a+b}{2} \\
\\
u + 1 & \mbox{ if } u > \frac{a+b}{2}.
\end{array} \right.
$$
to get a new utility function $f(U)$ which would still describe the same preferences. However, this function will not be differentiable, even if $U$ was. The only way to guarantee that a utility function describing the preference ordering is always differentiable is if you do not allow any form of it to take two different values, that is if all the goods are neutral to the consumer.

If you want to impose conditions on the utility functions itself it seems to me you would have to require that the marginal utilities exist and are continuous which is kind of the same as saying that the utility function is differentiable. So this is not a very useful condition. A better condition may exist, but I don't think it does.

> **Comments:**
> **username_2**: Just to note, if you consider von neumann morgenstern utility, these are defined up to only affine transformations.  
> **username_0**: Thanks for your answer, however, the question asks about the existence of a continuous, differentiable representation, not that all representations need to be continuous and differentiable. See last sentence of first paragraph.  
> **username_1**: @username_0 I see. Then I think your question boils down to: Under which conditions will a continuous function have a monoton transformation that is also differentiable. It seems to me that if there are only countable many points were the original function is nondifferentiable, a suitable monotonic transformation can always be found by 'smoothing out' the original function at these points.  
> **username_0**: I am not so sure, since two nondifferentiable points may have the same utility value.  
> **username_1**: @username_0 I apologize, my statement was not well formulated. What I meant to say is that there should only be finitely many utility values where the function is nondifferentiable. Multiple consumption bundles with the same nondifferentiable utility do not cause problems because the 'neighbouring' indifference curves would be the same ones because of the continuity of the utility function. And then smoothing out the utility values would work in all the bundles.  

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