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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!

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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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