My understanding of local non-satiation is that increasing your allocation of one good by a marginal amount increases utility. Suppose your utility takes the following form: $$U(x,y)=x^\alpha y^\beta$$ and your initial endowment is $(0,0)$. Now, if you increase either good without increasing the other, your utility does not increase. Does this mean that your utility is not locally non-satiated?

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    $\begingroup$ Actually no, wait. Local non-statiation does NOT mean that. Locally non satiated in common words mean that for every possible allocation there exists an allocation arbiitrarily close to it that gives strictly more utility. It could be increasing both goods. What you mean is strictly monotone, not locally non satiated. $\endgroup$
    – MathUser
    Jun 21, 2016 at 11:32

2 Answers 2



Cobb-Douglas utility is monotonic and monotonicity implies L.N.S.

The issue here is that you're only considering edge cases. You've correctly reasoned that edge points are not more desirable that the origin. However, LNS simply claims that there exists a more desirable bundle within the open epsilon ball of your allocation under consideration (and this is true for all allocations). So given any epsilon, an open epsilon ball centered at the origin will necessarily contain an allocation where both elements are strictly positive.

To see this, pick an epsilon and draw the epsilon ball around the origin. It should become fairly obvious what's going on.

Hope that helps.

  • $\begingroup$ Is the function strictly monotonic? I have a similar concern as the OP because the marginal utility of x and y each is 0 at (0,0): does this still mean the function is strictly monotonic? Or does the marginal utility definition of monotonicity doesn't apply here? $\endgroup$
    – PGupta
    Jul 9, 2020 at 8:07

The sequence of bundles $\big\{\frac{1}{n}\frac{1}{n}\big\}_{n\in\mathbb{N}}$ converge to $(0,0)$ and each bundle in the seuqence is strictly preferred to $(0,0)$.


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