<p>For case 1, you can argue <span class="math-container">$z$</span> will be unique by contradiction:</p>

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

CC BY-SA 4.0
2019-10-20T08:28:49.660Z
Thank you for your answer. Just to make sure: do you think the proof given in my question proved that z is unique?
Aqqqq
CC BY-SA 4.0
2019-10-20T13:34:17.130Z
You're welcome. I'd say so, yes. Case 1 and 2 are the only two possible cases and they both imply uniqueness
GabMac
CC BY-SA 4.0
2019-10-20T17:25:09.573Z
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)?
Aqqqq
CC BY-SA 4.0
2019-10-20T21:17:37.760Z
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.
GabMac
CC BY-SA 4.0
2019-10-21T07:40:07.993Z
So the proof given in my question did not prove that z is unique for Case 1.
Aqqqq
CC BY-SA 4.0
2019-10-19T23:32:53.997Z
2019-10-19T23:32:53.997Z
GabMac
<p>In the context of Walrasian demand:</p>

<p>Suppose u is continuous, satisfies local nonsatiation, and is strictly quasi-concave, each <span class="math-container">$w(p, x)$</span> contains a single consumption bundle.</p>

<p>The proof I got from a textbook is:</p>

<p>Let <span class="math-container">$x $</span>~<span class="math-container">$ y$</span> with <span class="math-container">$p^T x=p^T y =w$</span>, <span class="math-container">$z=\alpha x + (1- \alpha) y$</span> with <span class="math-container">$0 &lt; \alpha &lt; 1$</span>.
<span class="math-container">$p^T z =w$</span> because of the convexity of budget set.</p>

<p>Note: I understand why this makes sense graphically.</p>

<p>Case 1:</p>

<p>If <span class="math-container">$x \neq y$</span>, strict quasi-concavity implies <span class="math-container">$u(z) &gt; u(x) = u(y)$</span>, thus <span class="math-container">$z$</span> is preferred to <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, hence <span class="math-container">$x, y$</span> are not elements of Walrasian demand.</p>

<p>Case 2:</p>

<p>Otherwise <span class="math-container">$x = y = z$</span>.</p>

<p>For case 1, how can one be sure that the <span class="math-container">$z$</span> is unique?</p>

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

(Asked by: username_0 on 2019-10-19)

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?

**Tags:** microeconomics, utility, consumption, walrasian


---

## 1 answers

### Answer by username_1 on 2019-10-19. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: Thank you for your answer. Just to make sure: do you think the proof given in my question proved that z is unique?  
> **username_1**: You're welcome. I'd say so, yes. Case 1 and 2 are the only two possible cases and they both imply uniqueness  
> **username_0**: 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)?  
> **username_1**: 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.  
> **username_0**: So the proof given in my question did not prove that z is unique for Case 1.  

---

