# Prove this indirect utility function is quasi-convex

The indirect utility function is as follows:

$$v(m,p) = \frac{m}{p_{1}^{1/2} p_{2}^{1/4} p_{3}^{1/4}}$$

I need to prove that it is quasi-convex. I tried both definition of a quasiconvex function and the property of a convex lower contour set but couldn't get it.

This is a question in my homework so you may just want to give hint.

• Could you please type in what you tried with the definition of quasiconvexity, so that we can see where you run into problems? – Giskard Nov 22 '19 at 19:55

If the other two methods did not work out for you, another possibility, albeit a more difficult one:

Set up the bordered Hessian:

$$\bar{H}=\begin{bmatrix}0 & \frac{\partial v}{\partial p_1} & \frac{\partial v}{\partial p_2} & \frac{\partial v}{\partial p_3} & \frac{\partial v}{\partial m}\\ \frac{\partial v}{\partial p_1} & \frac{\partial^2 v}{\partial p_1^2} & \frac{\partial^2 v}{\partial p_1\partial p_2} & \frac{\partial^2 v}{\partial p_1\partial p_3} & \frac{\partial^2 v}{\partial p_1\partial m}\\ \frac{\partial v}{\partial p_2} & \frac{\partial^2 v}{\partial p_2 \partial p_1} & \frac{\partial^2 v}{\partial p_2^2} & \frac{\partial^2 v}{\partial p_2\partial p_3} & \frac{\partial^2 v}{\partial p_2\partial m}\\ \frac{\partial v}{\partial p_3} & \frac{\partial^2 v}{\partial p_3 \partial p_1} & \frac{\partial^2 v}{\partial p_3 \partial p_2} & \frac{\partial^2 v}{\partial p_3^2} & \frac{\partial^2 v}{\partial p_3\partial m}\\ \frac{\partial v}{\partial m} & \frac{\partial^2 v}{\partial m \partial p_1} & \frac{\partial^2 v}{\partial m \partial p_2} & \frac{\partial^2 v}{\partial m \partial p_3} & \frac{\partial^2 v}{\partial m^2}\\ \end{bmatrix}$$

Define $$\bar{H}_k$$ as the $$k_{th}$$ order leading principal submatrix. We can apply the following theorem:

If the $$\text{Det}\left(\bar{H}_k\right) <0$$ for all $$k=2,3,4,5$$, then $$v$$ is quasiconvex in $$(\textbf{p},m)$$.

This is an application of the more general result:

Let $$X$$ be an open convex subset of $$\mathbb{R}^n$$ and let $$f:X \rightarrow \mathbb{R}$$ be a twice continuously differentiable function $$\left(f\in C^2\right)$$. Let $$\bar{H}$$ be the bordered Hessian and let $$\bar{H}_k$$ be the $$k_{th}$$ order leading principal submatrix. If the $$\text{Det}\left(\bar{H}_k\right) <0$$ for all $$k=2,3,\dots,n+1$$, then $$f$$ is quasiconvex on $$X$$.