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I wonder how a CES function over a continuum of goods,

$$\left(\int_1^\infty c(\theta)^\delta g(\theta) \mathrm{d}\theta\right)^\frac{1}{\delta}$$

where $c(\theta), g(\theta)>0 \forall \theta\in\left[1,\infty\right)$, collapses to a minimum function in the limit as $\delta \to -\infty$.

I have seen the answer to a very similar question and below is my adaption of it to this problem:

First, let $\theta' \in \arg\min_{\theta\in\left[1,\infty\right)}c(\theta)$. Then,

\begin{align} \lim_{\delta \to -\infty} \left(\int_1^\infty c(\theta)^\delta g(\theta) \mathrm{d}\theta\right)^\frac{1}{\delta} &= \lim_{\delta \to -\infty} \left(c(\theta')^\delta g(\theta')\int_1^\infty \left(\frac{c(\theta)}{c(\theta')}\right)^\delta \frac{g(\theta)}{g(\theta')} \mathrm{d}\theta\right)^\frac{1}{\delta} \\ &= \lim_{\delta \to -\infty} c(\theta') g(\theta')^\frac{1}{\delta}\left(\int_1^\infty \left(\frac{c(\theta)}{c(\theta')}\right)^\delta \frac{g(\theta)}{g(\theta')} \mathrm{d}\theta\right)^\dfrac{1}{\delta} \\ &= c(\theta') \lim_{\delta \to -\infty} g(\theta')^\frac{1}{\delta}\left(\int_1^\infty \left(\frac{c(\theta)}{c(\theta')}\right)^\delta \frac{g(\theta)}{g(\theta')} \mathrm{d}\theta\right)^\dfrac{1}{\delta} \\ &= c(\theta') \end{align}

It is the very last step that I can't really get my head around. I'm fine with that, as $\delta\to-\infty$, $g(\theta')^\frac{1}{\delta}\to 1$ and also that if $c(\theta)<c(\theta')$ then $\left(\frac{c(\theta)}{c(\theta')}\right)^\delta\to 0$ and if $c(\theta)=c(\theta')$ then $\left(\frac{c(\theta)}{c(\theta')}\right)^\delta\to 1$.

  1. How can I know that the entire integral goes to unity? This could perhaps be rephrased: how do I show that

$$\lim_{\delta\to-\infty} \left(\int_1^\infty \theta^\delta\mathrm{d}\theta\right)^\frac{1}{\delta}=1?$$

  1. Also, and more fundamentally, under what circumstances can a limit be factorized so that it is the product of the "factor limits"?
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Fredrik P
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  • Can you provide background for math people who don't know CES fcns?: 0) are c, g > 0? 1) Why does the integral exist? are there conditions on c and g that make this happen? 2) Why does the arg min exist? – cantorhead Apr 28 '15 at 10:45
  • If you define $f(\theta)=c(\theta)^{-1}$ and define $p=-\delta$ then this is very similar to the proof that the $L^p$-norm approaches the $\infty$-norm as $p\to\infty$. – cantorhead Apr 28 '15 at 10:55
  • Sorry about that. I have tried to include further details. Let's just assume that the arg min exists. The existence of the arg min is not really at the heart of the question, which is more about the limit of the integral. – Fredrik P Apr 28 '15 at 10:55
  • Could you supply a link to the proof you are referring to, @cantorhead ? – Fredrik P Apr 28 '15 at 10:57
  • is it not just about showing that the integral is bounded? – tired Apr 28 '15 at 12:20
  • Question is $c$ a differentiable function? – tired Apr 28 '15 at 12:28
  • Heres a link. You'll need to do something like $\int \cdot g(\theta)\ d \theta = \int \cdot \ dw$ first. Their proof follows different lines, though, I think because it requires fewer assumptions. – cantorhead Apr 28 '15 at 12:29
  • At top you say you want $\delta \to -\infty$ but in the equations you have $\lim_{\delta\to\infty}$ is this intended? – cantorhead Apr 28 '15 at 12:37
  • I don't seem to have the rep to add the functional analysis tag, but I think it would be appropriate. – cantorhead Apr 28 '15 at 12:39
  • @tired: $c$ is a differentiable function of $\theta$ but not of $\delta$, so I think that in this problem $c(\theta)$ for each $\theta$ can be viewed as a constant. – Fredrik P Apr 28 '15 at 13:07
  • @cantorhead: All the limits should be $\delta\to-\infty$. Thanks for spotting the typo! – Fredrik P Apr 28 '15 at 13:09
  • @tired: it could be that I would need to show that the integral is bounded, but I am afraid that I don't know. That's why I'm asking :) – Fredrik P Apr 28 '15 at 13:20
  • @cantorhead: Thanks for the link! I will have a look at it. – Fredrik P Apr 28 '15 at 13:21

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