<p>Consider any $x_2'$ and $x_2''$ in $\mathbb{R}_+$. Without loss of generality, let $x_2'' &gt; x_2'$. We can choose $x_1'=f(x_2'')-f(x_2') &gt; 0$ so that $U(0,x_2'')=U(x_1',x_2')$. Let $\lambda(x_1',x_2')+(1-\lambda)(0,x_2'')$ be a convex combination of $(x_1',x_2')$, and $(0,x_2'')$. Since $\succsim$ is strictly convex and $U(0,x_2'')=U(x_1',x_2')$,
        \begin{eqnarray*} 
 &amp;&amp; U(\lambda(x_1',x_2')+(1-\lambda)(0,x_2''))  &gt;  U(x_1',x_2') \\
 &amp;\Rightarrow &amp; U(\lambda x_1'+(1-\lambda)0,\lambda x_2'+(1-\lambda)x_2'')  &gt;  \lambda U(x_1',x_2')+(1-\lambda)U(0,x_2'')\ldots(\because U(0,x_2'')=U(x_1',x_2')) \\
 &amp;\Rightarrow &amp;  \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') &gt;  \lambda (x_1'+f(x_2'))+(1-\lambda)(0 + f(x_2''))\\
  &amp;\Rightarrow &amp;  \lambda x_1'+(1-\lambda)0 + f(\lambda x_2'+(1-\lambda)x_2'') &gt;  \lambda x_1'+(1-\lambda)0 + \lambda f(x_2')+(1-\lambda)f(x_2'') \\
    &amp;\Rightarrow &amp;  f(\lambda x_2'+(1-\lambda)x_2'') &gt;   \lambda f(x_2')+(1-\lambda)f(x_2'')\end{eqnarray*}
Therefore, $f(\cdot)$ is strict concave.</p>

CC BY-SA 3.0
2017-04-13T15:31:23.410Z
2017-04-13T15:31:23.410Z
Amit
<p>Let $\succsim$ be a strictly convex and quasilinear preference relation. It's defined over, say, $\mathbb{R}^2_{+}$ and is quasilinear on good 1.</p>

<p>So, $U(x_{1},x_{2}) = x_{1} + f(x_{2})$. How to prove that $f$ is a strictly concave function?</p>

<p>I'm solving problem 15.B.8 from MWG and I can't even understand what the solutions manual did! </p>

<p>Thanks a lot in advance for any hints and ideas!</p>

<microeconomics><preferences><proof>
How to prove convexity + quasilinear preferences imply concave utility?
Raul Guarini Riva
# How to prove convexity + quasilinear preferences imply concave utility?

(Asked by: username_0 on 2017-04-13)

Let $\succsim$ be a strictly convex and quasilinear preference relation. It's defined over, say, $\mathbb{R}^2\_{+}$ and is quasilinear on good 1.

So, $U(x\_{1},x\_{2}) = x\_{1} + f(x\_{2})$. How to prove that $f$ is a strictly concave function?

I'm solving problem 15.B.8 from MWG and I can't even understand what the solutions manual did!

Thanks a lot in advance for any hints and ideas!

**Tags:** microeconomics, preferences, proof


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## 1 answers

### Answer by username_1 on 2017-04-13. Score: 2 | Accepted answer

Consider any $x\_2'$ and $x\_2''$ in $\mathbb{R}\_+$. Without loss of generality, let $x\_2'' > x\_2'$. We can choose $x\_1'=f(x\_2'')-f(x\_2') > 0$ so that $U(0,x\_2'')=U(x\_1',x\_2')$. Let $\lambda(x\_1',x\_2')+(1-\lambda)(0,x\_2'')$ be a convex combination of $(x\_1',x\_2')$, and $(0,x\_2'')$. Since $\succsim$ is strictly convex and $U(0,x\_2'')=U(x\_1',x\_2')$,
\begin{eqnarray\*}
&& U(\lambda(x\_1',x\_2')+(1-\lambda)(0,x\_2'')) > U(x\_1',x\_2') \\
&\Rightarrow & U(\lambda x\_1'+(1-\lambda)0,\lambda x\_2'+(1-\lambda)x\_2'') > \lambda U(x\_1',x\_2')+(1-\lambda)U(0,x\_2'')\ldots(\because U(0,x\_2'')=U(x\_1',x\_2')) \\
&\Rightarrow & \lambda x\_1'+(1-\lambda)0 + f(\lambda x\_2'+(1-\lambda)x\_2'') > \lambda (x\_1'+f(x\_2'))+(1-\lambda)(0 + f(x\_2''))\\
&\Rightarrow & \lambda x\_1'+(1-\lambda)0 + f(\lambda x\_2'+(1-\lambda)x\_2'') > \lambda x\_1'+(1-\lambda)0 + \lambda f(x\_2')+(1-\lambda)f(x\_2'') \\
&\Rightarrow & f(\lambda x\_2'+(1-\lambda)x\_2'') > \lambda f(x\_2')+(1-\lambda)f(x\_2'')\end{eqnarray\*}
Therefore, $f(\cdot)$ is strict concave.


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