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.


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.


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.


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.

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    $\begingroup$ Just to note, if you consider von neumann morgenstern utility, these are defined up to only affine transformations. $\endgroup$ – jmbejara Nov 14 '15 at 21:03
  • $\begingroup$ 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. $\endgroup$ – HRSE Nov 15 '15 at 12:41
  • $\begingroup$ @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. $\endgroup$ – Giskard Nov 15 '15 at 13:01
  • $\begingroup$ I am not so sure, since two nondifferentiable points may have the same utility value. $\endgroup$ – HRSE Nov 16 '15 at 3:12
  • $\begingroup$ @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. $\endgroup$ – Giskard Nov 16 '15 at 10:03

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