Consider an economy with a continuum of commodities, with one commodity for each point in $[0,1]$.

Suppose a consumer wants to maximise $$U = \int_0^1 c_i^\theta\,di\qquad 0<\theta<1$$ subject to $$\int_0^1 p_i c_i\,di = M$$ where $c_i$ is the amount of the $i$-th commodity consumed, $p_i$ its price and $M$ the consumer's money income.

This kind of problem arises for example in applying the Dixit-Stiglitz model to macroeconomics or international trade.

The solution to this problem is supposedly $$c_i = Ap_i^{1 \over {\theta-1}}$$ where $A$ is a constant chosen to ensure that the budget constraint is satisfied.

I am not very satisfied with derivations of this result which use Lagrange multipliers in analogy with the case of a finite number of commodities. What would be a completely mathematically rigorous method of deriving the above result?

It seems clear that there isn't a unique solution since arbitrarily changing the values of $c_i$ for a finite number of values of $i$ will leave the integrals in the utility function and budget constraint unchanged. I am expecting that a completely rigorous derivation would also correctly pinpoint this degree of nonuniqueness.

EDIT: In response to the comments by @BKay, @Ubiquitous. My problem with starting out with economies with $n$ commodities and taking the limit as $n \to \infty$ is that this needs to be accompanied by an argument which shows that the limit of optima is an optimum of the limit problem. I would appreciate a reference to a result which shows this either for this particular problem or a general result which is applicable to this problem.

In response to @AlecosPapadopoulos. The proofs of the Langrange multiplier method that is taught in math for economics courses is usually for a finite number of choice variables. I would appreciate a reference to where the method is justified for a continuum of choice variables. Also, the nonuniqueness I mention above shows that the method cannot be exactly right. Then what exactly are the qualifications required for its validity?

  • 1
    $\begingroup$ I agree with OP, a lot can potentially go wrong when the space becomes infinite-dimensional. To me it is not clear at all that the limit of the optimum is the optimum of the limit. $\endgroup$
    – FooBar
    Commented Nov 22, 2014 at 3:23

3 Answers 3


The completely rigorous thing would be to write the Euler lagrange equation of this calculus of variations problem, this will give you a strong solution that is what you have or a weak solution that is written with respect to a distribution.

  • $\begingroup$ But how do I incorporate my budget constraint into a calculus of variations formulation? $\endgroup$ Commented Nov 22, 2014 at 5:05
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    $\begingroup$ Check this link, math.stackexchange.com/questions/279518/…, a lagrange multiplier function!, is what you need, this gives you a strong solution that can be interpreted pointwise, although it must hold almost sure with the dominant measure $\endgroup$
    – user157623
    Commented Nov 22, 2014 at 5:12
  • $\begingroup$ Thanks. Following you hint of using the calculus of variations I found a Theorem 1 in section 12 of Kolomogorov and Fomin's Calculus of Variations does seem to handle constraints expressed as integrals. So in a sense one can use Langrange multipliers after all. $\endgroup$ Commented Nov 22, 2014 at 5:24
  • $\begingroup$ This is useful -but as a comment, not as an answer. $\endgroup$ Commented Nov 22, 2014 at 9:31
  • $\begingroup$ You are right Jyotirmoy Bhattacharya, maybe someone can edit it to be a full answer with the links that have been provided in the comments. $\endgroup$
    – user157623
    Commented Nov 22, 2014 at 13:31

As the OP noted in a comment, Theorem 1 in section 12 of Gelfand and Fomin's Calculus of Variations seems to provide some comfort that we can indeed use the Langrange Multiplier method when the number of our variables is infinite. Still, the authors do that in a footnote, writing "the reader will easily recognize the analogy with Langrange multipliers". So no, this does not rigorously show what we want.

I think that what we need is a paper like Craven, B. D. (1970). A generalization of Lagrange multipliers. Bulletin of the Australian Mathematical Society, 3(03), 353-362. which in its summary writes:

The method of Lagrange multipliers for solving a constrained stationary-value problem is generalized to allow the functions to take values in arbitrary Banach spaces (over the real field). The set of Lagrange multipliers in a finite-dimensional problem is shown to be replaced by a continuous linear mapping between the relevant Banach spaces.

This is math-speak but it says what we wanted to hear (one can also find a short exposition in wikipedia to the degree that it trusts the content).

Then, we can form the Lagrangean of the problem

$$\Lambda = \int_0^1 c_i^\theta\,\text{d}i +\lambda\left(M - \int_0^1 p_i c_i\,\text{d}i\right)$$

and calculate the first-order condition(s) by , informally speaking, "looking at the integral and seeing a sum",

$$\frac{\partial \Lambda}{\partial c_i} = 0 \Rightarrow \theta c_i^{\theta-1} = \lambda p_i, \;\; i \in [0,1] \tag{1}$$

...a continuum of conditions. For later use we define

$$\sigma \equiv 1/(1-\theta), \Rightarrow 1-\theta = 1/\sigma, \;\; \theta = \frac {\sigma-1}{\sigma}$$

The constant $\sigma$ can be shown to be the elasticity of substitution between any two goods.

Writing $(1)$ for commodity $j$ and equating through the common lagrange multiplier we arrive at

$$c_i = \left(\frac {p_i}{p_j}\right)^{-\sigma}c_j \tag{2}$$

Multiply both sides by $p_i$ and take the integral over the commmodity space with respect to $i$:

$$\int_0^1 p_i c_i\,\text{d}i = \int_0^1 p_i^{1-\sigma}p_j^{-\sigma}c_j\,\text{d}i$$

$$\Rightarrow M = p_j^{\sigma}c_j\int_0^1 p_i^{1-\sigma}\,\text{d}i$$

$$\Rightarrow c_j = p_j^{-\sigma} \cdot M \cdot \left(\int_0^1 p_i^{1-\sigma}\,\text{d}i\right)^{-1} \tag{3}$$

which is Marshallian demand for commodity $j$.

  • $\begingroup$ The Kolmogorov-Fomin result applied mechanically gives us a solution. So we need not appeal to the analogy with Lagrange multipliers. I am writing it out in a separate answer. $\endgroup$ Commented Nov 23, 2014 at 2:52

This is just an elaboration of the answer given by @user157623. I am posting it as a community wiki for convenience.

Theorem 1 of Section 12 of Kolmogorov and Fomin's Calculus of Variations says

Given the functional $$J[y] = \int_a^b F(x,y,y')\,dx,$$ let the admissible curves satisfy the conditions $$y(a)=A,\quad y(b)=b, \quad K[y] = \int_a^bG(x,y,y')\,dx=l,$$ where $K[y]$ is another functional, and let $J[y]$ have an extremum for $y=y(x)$. Then, if $y=y(x)$ is not an extremal of $K[y]$, there exists a constant $\lambda$ such that $y=y(x)$ is an extremal of the functional $$\int_a^b (F+\lambda G)\,dx,$$ i.e., $y=y(x)$ satisfies the differential equation $$F_y-\frac{d}{dx}F_{y'}+\lambda\left(G_y-\frac{d}{dx}G_{y'}\right)=0.$$

We can try to apply this theorem to our problem by taking $x$ to be $i$, $c$ to be $y$, $F(i,c,c')=c^\theta$ and $G(i,c,c')=pc$.

Then the final differential equation in the theorem becomes $$\theta c_i^{\theta-1} + \lambda p_i=0$$ which is precisely what we need.

Is the theorem applicable? Our $K[y]$ is linear, so cannot have an extremal, so the requirement of not having an extremal is easily satisfied. The boundary conditions on $y(a)$ and $y(b)$ do not matter since if a path of $c$, say $c^*(i)$, is extremal without any boundary conditions then it is extremal within the set $c(0)=c^*(0), c(1)=c^*(1)$.

The only catch is in the nature of the theorem itself. It gives necessary conditions for an optimum. Given that in our case the necessary condition gives a unique result, all that we need to make it sufficient is to argue that our problem has a solution.

The proofs in Kolmogorov-Fomin assume that the functions we are dealing with have continuous first derivatives. So we still need to show that the consumer's problem has an optimal in this class of functions but given that the problem is solved.


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