# Why does strictly Walrasian demand with quasi-concave utility function mean that the walrasian demand having only one single consumption bundle?

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?

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.
• 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)? Oct 20 '19 at 17:25
• 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. Oct 20 '19 at 21:17