<p>We consider the following problem:
<span class="math-container">$$
\max_{x} u(x) \text{ s.t. } p'x \le w.
$$</span>
Let us show that if <span class="math-container">$u$</span> is strictly concave, then the solution (if it exists) is unique.</p>
<p>We proof by contradiction. Assume that the solution is not unique. Then there should exist at least two solutions <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>. As both bundles are optimal, they should give the same utility, so <span class="math-container">$u(x_1) = u(x_2)$</span>.</p>
<p>Also, both of these bundles should be in the budget constraint: <span class="math-container">$p' x_1 \le w$</span> and <span class="math-container">$p' x_2 \le w$</span>.</p>
<p>Now, take any strict convex combination of these two bundles, say <span class="math-container">$\widetilde{x} = \frac{x_1 + x_2}{2}$</span>. As the budget set is convex, we have that <span class="math-container">$p' \widetilde{x} = \frac{1}{2} p'x_1 + \frac{1}{2} p'x_2 \le w$</span>, so the bundle <span class="math-container">$\widetilde{x}$</span> is also in the budget set.</p>
<p>But now, by strict concavity of <span class="math-container">$u$</span>, we have that:
<span class="math-container">$$
u(\widetilde{x}) = u(\frac{1}{2}x_1 + \frac{1}{2} x_2) &gt; \frac{1}{2}u(x_1) + \frac{1}{2} u(x_2) = u(x_1).
$$</span>
(the last equality follows as <span class="math-container">$u(x_1) = u(x_2)$</span>). However, this is impossible as <span class="math-container">$x_1$</span> was optimal (i.e. gives the highest utility for all bundles in the budget set).</p>

CC BY-SA 4.0
2023-09-04T14:15:15.280Z
2023-09-04T14:15:15.280Z
tdm
<p>In the basic problem of consumer's choice the budget set <span class="math-container">$B(p,w)$</span>, with <span class="math-container">$p$</span> the vector of prices belonging to <span class="math-container">$\mathbb{R}^N_{++}$</span> and <span class="math-container">$w &gt;0$</span> wealth, define a convex set. Following from that I don't understand why this implies that if the utility function <span class="math-container">$U: X \to \mathbb{R}$</span>  is strictly concave, then the solution <span class="math-container">$x^*(p,w)$</span> is unique for each <span class="math-container">$(p,w)&gt;0$</span>.</p>

<microeconomics>
Consumer problem
Francesco Favuzza
# Consumer problem

(Asked by: username_0 on 2023-09-04)

In the basic problem of consumer's choice the budget set $B(p,w)$, with $p$ the vector of prices belonging to $\mathbb{R}^N\_{++}$ and $w >0$ wealth, define a convex set. Following from that I don't understand why this implies that if the utility function $U: X \to \mathbb{R}$ is strictly concave, then the solution $x^\*(p,w)$ is unique for each $(p,w)>0$.

**Tags:** microeconomics


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## 1 answers

### Answer by username_1 on 2023-09-04. Score: 3 | Accepted answer

We consider the following problem:
$$
\max\_{x} u(x) \text{ s.t. } p'x \le w.
$$
Let us show that if $u$ is strictly concave, then the solution (if it exists) is unique.

We proof by contradiction. Assume that the solution is not unique. Then there should exist at least two solutions $x\_1$ and $x\_2$. As both bundles are optimal, they should give the same utility, so $u(x\_1) = u(x\_2)$.

Also, both of these bundles should be in the budget constraint: $p' x\_1 \le w$ and $p' x\_2 \le w$.

Now, take any strict convex combination of these two bundles, say $\widetilde{x} = \frac{x\_1 + x\_2}{2}$. As the budget set is convex, we have that $p' \widetilde{x} = \frac{1}{2} p'x\_1 + \frac{1}{2} p'x\_2 \le w$, so the bundle $\widetilde{x}$ is also in the budget set.

But now, by strict concavity of $u$, we have that:
$$
u(\widetilde{x}) = u(\frac{1}{2}x\_1 + \frac{1}{2} x\_2) > \frac{1}{2}u(x\_1) + \frac{1}{2} u(x\_2) = u(x\_1).
$$
(the last equality follows as $u(x\_1) = u(x\_2)$). However, this is impossible as $x\_1$ was optimal (i.e. gives the highest utility for all bundles in the budget set).


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