In the context of Walrasian demand:

Suppose u is continuous, satisfies local nonsatiation, and is strictly quasi-concave, each $w(p, x)$ contains a single consumption bundle.

The proof I got from a textbook is:

Let $x $~$ y$ with $p^T x=p^T y =w$, $z=\alpha x + (1- \alpha) y$ with $0 < \alpha < 1$. $p^T z =w$ because of the convexity of budget set.

Note: I understand why this makes sense graphically.

Case 1:

If $x \neq y$, strict quasi-concavity implies $u(z) > u(x) = u(y)$, thus $z$ is preferred to $x$ and $y$, hence $x, y$ are not elements of Walrasian demand.

Case 2:

Otherwise $x = y = z$.

For case 1, how can one be sure that the $z$ is unique?


1 Answer 1


For case 1, you can argue $z$ will be unique by contradiction:

Suppose ad absurdum there is another $z'$ that is feasible (i.e. $p^Tz' =w$) optimal and $z'\neq z$. Then you can consider a convex combination of $z$ and $z'$: $\bar z = \beta z + (1-\beta) z'$, for $\beta\in(0,1)$. Notice that $\bar z$ is still feasible (because it is a combination of two feasible bundles) and, because of strict quasi-concavity, $u(\bar z) > u(z')$ so that $z'$ cannot be optimal, a contradiction.

  • $\begingroup$ Thank you for your answer. Just to make sure: do you think the proof given in my question proved that z is unique? $\endgroup$
    – Aqqqq
    Oct 20, 2019 at 8:28
  • $\begingroup$ You're welcome. I'd say so, yes. Case 1 and 2 are the only two possible cases and they both imply uniqueness $\endgroup$
    – GabMac
    Oct 20, 2019 at 13:34
  • $\begingroup$ How does the proof given in my question for Case 1 imply that z is unique (that there can not be another $\alpha$ which can generate another z which also belongs to Walrasian demand)? $\endgroup$
    – Aqqqq
    Oct 20, 2019 at 17:25
  • $\begingroup$ Suppose that there is another $\alpha$ that generates an optimal solution $z'$, s.t. $z'\neq z$, then you can proceed with the argument I've outlined above to conclude that this is impossible. $\endgroup$
    – GabMac
    Oct 20, 2019 at 21:17
  • $\begingroup$ So the proof given in my question did not prove that z is unique for Case 1. $\endgroup$
    – Aqqqq
    Oct 21, 2019 at 7:40

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