# Doesnt convexity prevent thick indifference curves aswell?

It is standard in many micro textbooks when analyzing the relationship between preference axioms and the shape of the utility function (and consequently the shape of indifference curves), to attributed the "non-thickness" of indifference curves to "local non-satiation". While it is easily seen (and proven) that LNS preferences don't admit thick IC, my question is the following:

Thick IC violate as well strict convexity (notice that weak convexity appears to "survive"), so, isn't convexity alone another property that prevents thickness? Put it differently, can we find non monotonic, convex preferences that don't admit thick IC?

If any proof or reference to relevant literature would be very useful.

• Strict convexity at least implies that indifference curves have nonempty interior. – Michael Greinecker Aug 20 '16 at 22:03

My approach would be to define thick indifference curves in terms of a stronger form of local non-satiation:

Definition (thick indifference curves) Preferences are said to have thick indifference curves if there exists at least one bundle $$A\in\mathbb{R}^l$$ and an open ball $$\mathscr{B}(A)$$ around $$A$$ such that $$A'\sim A$$ for every $$A'\in\mathscr{B}(A)$$.

Let's set up a definition for strict convexity to be sure we are on the same page:

Definition (strict convexity) Preferences obey strict convexity if

$$A\sim B\implies \lambda A+(1-\lambda) B\succ A\sim B$$

for $$A,B\in\mathbb{R}^l$$ and any $$\lambda\in(0,1)$$.

Now we can state the desired result:

Proposition If $$\succsim$$ is strictly convex then $$\succsim$$ does not have thick indifference curves.

Proof Suppose that $$\succsim$$ has thick indifference curves. Then there exists an $$A$$ such that $$A'\sim A$$ for all $$A'\in\mathscr{B}(A)$$. Thus, fix two bundles $$A$$ and $$B$$ such that $$A\sim B$$ and $$B\in\mathscr{B}(A)$$. We know that $$\lambda A+(1-\lambda) B \in \mathscr{B}(A).$$ Thus, if the indifference curve is thick around point$$A$$ we must have $$\lambda A+(1-\lambda) B\sim A\sim B$$. But this contradicts the strict convexity hypothesis. QED

That should do the job for strict convexity. As frage_man already pointed out, the proposition fails to hold generally under weak convexity. Indifference everywhere is a very elegant example why.

• I thought about a definition for thick IC like yours and the thing I was unsure about was whether we needed every point in the neighborhood to be indifferent, Because a point on the boundary of the thick IC is only indifferent to a `half-ball' in a sense. Thoughts? – VCG Aug 21 '16 at 11:58
• @frage_man Yes, this is why I don't require my definition to hold at every point on the IC. Instead, I require only that we can find at least one point in the interior of the IC's area around which we can draw a ball of indifference. As soon as that's true for one point, we know the IC isn't 'thin'. I can't think of a way to construct a thick IC that would have no such points at all. I edited the wording of my answer to be slightly clearer on this point. – Ubiquitous Aug 21 '16 at 14:43
• Gotcha - great answer btw. – VCG Aug 21 '16 at 14:47

Now a simple example of weakly convex with thick IC is indfference everywhere. So $\forall x,y\in X, x\sim y$. In this case the IC is all of $\mathbb{R}$.