# About Characterization of Homothetic Function - Mathematics for Economists by Simon and Blume Chapter 20 Exercise 18(3)

I am studying homothetic function and got the following problem:

## Problem

Determine whether the function $$x^3y^6 + 3x^2y^4 + 6xy^2 + 9$$ is homothetic or not.

Here is my attempt.

## My Attempt

I did it in two different ways, but I got two opposite answers. First by using the following definition of a homothetic function, I showed that the above function is homothetic.

Definition$$\quad$$ A function $$v:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}$$ is called homothetic if it is a monotonic transformation of a homogeneous function, that is, if there is a monotonic transformation $$z \mapsto g(z)$$, where $$g:\mathbb{R}_{+} \to \mathbb{R}$$, and a homogeneous function $$u:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}_{+}$$ such that $$v(\mathbf{x}) = g(u(\mathbf{x}))$$ for all $$\mathbf{x}$$ in $$\mathbb{R}_{+}^{\mathbf{n}}$$.

Answer 1$$\quad$$ $$v(x,y) = x^3y^6 + 3x^2y^4 + 6xy^2 + 9 = (xy^2)^3 + 3(xy^2)^2 + 6xy^2 + 9$$ is a monotonic transformation of a homogeneous function. Define $$g:\mathbb{R}_{+} \to \mathbb{R}$$ by $$g(z) = z^3 + 3z^2 + 6z + 9$$. Since $$g'(z) = 3z^2 + 6z + 6 = 3(z+1)^2 + 3 > 0$$, $$g$$ is strictly increasing. Define $$u:\mathbb{R}_{+}^{2} \to \mathbb{R}_{+}$$ by $$u(x,y) = xy^2$$. Then $$u$$ is homogeneous of degree $$3$$. Thus, $$v = g \circ u$$ is homothetic.

However, if I use the following theorem, I would get that $$v$$ is not homothetic.

Theorem$$\quad$$ Let $$u$$ be a $$C^1$$ function on $$\mathbb{R}_{+}^{\mathbf{n}}$$. $$u$$ is homothetic if and only if, for all $$\mathbf{x}$$ in $$\mathbb{R}_{+}^{\mathbf{n}}$$, all $$t > 0$$, and all $$i$$, $$j$$, \begin{align*} \frac{\frac{\partial u}{\partial x_i}(t\mathbf{x})}{\frac{\partial u}{\partial x_j}(t\mathbf{x})} = \frac{\frac{\partial u}{\partial x_i}(\mathbf{x})}{\frac{\partial u}{\partial x_j}(\mathbf{x})}. \end{align*}

Answer 2$$\quad$$ Since for all $$x$$ in $$\mathbb{R}_{+}^{\mathbf{2}}$$, \begin{align*} \frac{\frac{\partial v}{\partial x}(tx,ty)}{\frac{\partial v}{\partial y}(tx,ty)} = \frac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)} \end{align*} which does not equal to \begin{align*} \frac{\frac{\partial v}{\partial x}(x,y)}{\frac{\partial v}{\partial y}(x,y)} = \frac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}, \end{align*} for all $$t > 0$$, we conclude that $$v$$ is not homothetic.

## My Question

Could someone please help me explain why would I get opposite conclusion in the above problem? What am I missing? I really appreciate it!

Proposition 1. For $$a>0,b>0,c>0,d>0$$, if $$\dfrac{a}{b}=\dfrac{c}{d}$$ then $$\dfrac{a+c}{b+d}=\dfrac{a}{b}$$.
Proposition 2. For $$a_i>0,b_i>0$$, where $$i\in\{1,2,\ldots,n\}$$, if $$\dfrac{a_i}{b_i}=\dfrac{a_1}{b_1}$$ (for all $$i\in\{1,2,\ldots,n\}$$), then $$\dfrac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}=\dfrac{a_1}{b_1}$$.
Using this proposition, it follows that $$\dfrac{3(tx)^2(ty)^6}{6(tx)^3(ty)^5} = \dfrac{6(tx)(ty)^4}{12(tx)^2(ty)^3} = \dfrac{6(ty)^2}{12(tx)(ty)}=\dfrac{3x^2y^6}{6x^3y^5}=\dfrac{6xy^4}{12x^2y^3} = \dfrac{6y^2}{12xy}$$ implies $$\dfrac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}=\dfrac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}$$ for all $$t>0$$.