<p>Let <span class="math-container">$x,y,z \in \mathbb{R}_{+}^3$</span> be consumption bundles such that <span class="math-container">$y \geq x, z \geq x$</span>. Let <span class="math-container">$\lambda \in [0,1]$</span>.</p>
<p><span class="math-container">\begin{equation}
    u(\lambda y + (1-\lambda) z) = \min(\lambda (y_1,y_2,y_3) + (1 - \lambda) (z_1,z_2,z_3))
\end{equation}</span></p>
<p>Using the inequality &quot;the minimum of a sum <span class="math-container">$\geq$</span> the sum of the minimums&quot;</p>
<p><span class="math-container">\begin{equation}
    u(\lambda y + (1-\lambda) z) \geq \min(\lambda (y_1,y_2,y_3)) + \min( (1 - \lambda) (z_1,z_2,z_3) )
\end{equation}</span></p>
<p>Since <span class="math-container">$\lambda \in [0,1]$</span>, we have <span class="math-container">$\lambda \geq 0, 1 - \lambda \geq 0$</span>, we can factor each of them out of each min.</p>
<p><span class="math-container">\begin{equation}
    u(\lambda y + (1-\lambda) z) \geq \lambda \min(y_1,y_2,y_3) + (1-\lambda) \min(z_1,z_2,z_3)
\end{equation}</span></p>
<p>Recalling the definition of the function <span class="math-container">$u$</span></p>
<p><span class="math-container">\begin{equation}
    u(\lambda y + (1-\lambda) z) \geq \lambda u(y) + (1-\lambda) u(z)
\end{equation}</span></p>
<p>Since <span class="math-container">$y \geq x, z \geq x$</span>, we have that <span class="math-container">$u(y) \geq u(x), u(z) \geq u(x)$</span>. Therefore,</p>
<p><span class="math-container">\begin{equation}
u(\lambda y + (1-\lambda) z) \geq \lambda u(x) + (1-\lambda) u(x) = u(x) 
\end{equation}</span></p>
<p>From here we get <span class="math-container">$\lambda y + (1-\lambda) z \geq x$</span>.</p>
<p>Therefore, we conclude that the preference relation represented by <span class="math-container">$u$</span> is convex.</p>
<p>Answer: True</p>

CC BY-SA 4.0
2022-11-10T06:39:41.283Z
I guess the problem asks for the convexity of the represented preferences.
Michael Greinecker
CC BY-SA 4.0
2022-11-10T07:47:02.467Z
Isn't the same u(x)=min(x,y) convex in 2D? Shouldn't that extend to 3D also?
reindeer
CC BY-SA 4.0
2022-11-10T14:42:58.737Z
I tweaked it a bit to match the new question (OP edited it). Someone just decided to give me a -1.
Nicolas Torres
CC BY-SA 4.0
2022-11-10T14:58:13.483Z
@ Nicolas Torres  You, while editing your post, forgot to delete the sentence 'It is actually false' at the beginning'.
BakerStreet
CC BY-SA 4.0
2022-11-22T13:46:01.747Z
what is x here? is it just another bundle? how is it comparable to the convex combination?
reindeer
CC BY-SA 4.0
2022-11-22T14:31:28.347Z
@reindeer x is another bundle, which is defined to be weakly worse than both y and z. To prove that the preference relation is convex, I prove the convex combination of y and z is also weakly preferred to x. I know there are more equivalent definitions of convexity for preference relations, but this is the one I used here.
Nicolas Torres
CC BY-SA 4.0
2022-11-10T06:20:46.357Z
2022-11-10T15:01:25.460Z
2022-11-10T15:01:25.460Z
Nicolas Torres
<p>Taking two bundles <span class="math-container">$x$</span> and <span class="math-container">$y$</span>,
<span class="math-container">$x = (x_1, x_2, x_3)$</span> and  <span class="math-container">$y = (y_1, y_2, y_3)$</span>,
I understand I have to consider three cases to prove convexity, when <span class="math-container">$x\sim y$</span>, <span class="math-container">$x\succ y$</span>, <span class="math-container">$y\succ x$</span>.
I understand when <span class="math-container">$x\sim y$</span>, but I don't know how to prove the convex combination for the last two cases.</p>

<microeconomics><consumer-theory>
Is the preference relation represented by $u(x)=min(x_1, x_2, x_3)$ convex? True or False
reindeer
# Is the preference relation represented by $u(x)=min(x_1, x_2, x_3)$ convex? True or False

(Asked by: username_0 on 2022-11-10)

Taking two bundles $x$ and $y$,
$x = (x\_1, x\_2, x\_3)$ and $y = (y\_1, y\_2, y\_3)$,
I understand I have to consider three cases to prove convexity, when $x\sim y$, $x\succ y$, $y\succ x$.
I understand when $x\sim y$, but I don't know how to prove the convex combination for the last two cases.

**Tags:** microeconomics, consumer-theory

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## 1 answers

### Answer by username_1 on 2022-11-10. Score: 5 | Accepted answer

Let $x,y,z \in \mathbb{R}\_{+}^3$ be consumption bundles such that $y \geq x, z \geq x$. Let $\lambda \in [0,1]$.

\begin{equation}
u(\lambda y + (1-\lambda) z) = \min(\lambda (y\_1,y\_2,y\_3) + (1 - \lambda) (z\_1,z\_2,z\_3))
\end{equation}

Using the inequality "the minimum of a sum $\geq$ the sum of the minimums"

\begin{equation}
u(\lambda y + (1-\lambda) z) \geq \min(\lambda (y\_1,y\_2,y\_3)) + \min( (1 - \lambda) (z\_1,z\_2,z\_3) )
\end{equation}

Since $\lambda \in [0,1]$, we have $\lambda \geq 0, 1 - \lambda \geq 0$, we can factor each of them out of each min.

\begin{equation}
u(\lambda y + (1-\lambda) z) \geq \lambda \min(y\_1,y\_2,y\_3) + (1-\lambda) \min(z\_1,z\_2,z\_3)
\end{equation}

Recalling the definition of the function $u$

\begin{equation}
u(\lambda y + (1-\lambda) z) \geq \lambda u(y) + (1-\lambda) u(z)
\end{equation}

Since $y \geq x, z \geq x$, we have that $u(y) \geq u(x), u(z) \geq u(x)$. Therefore,

\begin{equation}
u(\lambda y + (1-\lambda) z) \geq \lambda u(x) + (1-\lambda) u(x) = u(x)
\end{equation}

From here we get $\lambda y + (1-\lambda) z \geq x$.

Therefore, we conclude that the preference relation represented by $u$ is convex.

Answer: True

> **Comments:**
> **username_2**: I guess the problem asks for the convexity of the represented preferences.  
> **username_0**: Isn't the same u(x)=min(x,y) convex in 2D? Shouldn't that extend to 3D also?  
> **username_1**: I tweaked it a bit to match the new question (OP edited it). Someone just decided to give me a -1.  
> **username_3**: @ Nicolas Torres  You, while editing your post, forgot to delete the sentence 'It is actually false' at the beginning'.  
> **username_0**: what is x here? is it just another bundle? how is it comparable to the convex combination?  
> **username_1**: @reindeer x is another bundle, which is defined to be weakly worse than both y and z. To prove that the preference relation is convex, I prove the convex combination of y and z is also weakly preferred to x. I know there are more equivalent definitions of convexity for preference relations, but this is the one I used here.  

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