# Proving that Marshallian demand is of the form: $x_i^*(p,I) = \hat{x}_i^*(p)I$ with certain conditions

Can I please have some feedback/help proving the following. My proof is below but I am quite uncertain as to whether my solution is efficient. Thank you.

If $$u(x)$$ is a homothetic utility, then show that Marshallian demand is of the form: $$x_i^*(p,I) = \hat{x}_i^*(p)I.$$

$$\textit{Proof.}$$ If $$u(x)$$ is homothetic then $$\forall \alpha \in \mathbb{R}_{+}, \forall x : \hskip 6pt \frac{\partial u(x)}{\partial x} = \frac{\partial u(\alpha \cdot x)}{\partial x}.$$ Now suppose $$x_i^*(p,I) = \hat{x}_i^*(p)I$$ does not hold, which is equivalent to $$u(x_i^*(p,I)) \ne u(\hat{x}_i^*(p)I).$$ Precisely, $$x_i^*(p,I)$$ and $$\hat{x}_i^*(p)I$$ may be set valued. In this case we refer to two elements, with at least one of which is not included in both sets.

Case 1. $$u(x_i^*(p,I)) > u(\hat{x}_i^*(p)I)$$ As $$u$$ is homothetic $$u(x_i^*(p,I)) = u(I \cdot \frac{1}{I}\cdot x_i^*(p,I)) = I \cdot u(\frac{1}{I}\cdot \hat{x}_i^*(p)I).$$ Using this we have $$I \cdot u(\frac{1}{I}\cdot x_i^*(p,I)) = u(x_i^*(p,I)) > u(I\cdot \hat{x}_i^*(p)I) = I\cdot u(x_i^*(p,I))$$ thus we have $$u(\frac{1}{I}\cdot x_i^*(p,I)) > \hat{x}_i^*(p)I)$$ However as $$\frac{1}{I} \cdot x_i^*(p,I)$$ is clearly an element of $$B(p)I$$ this is impossible as $$\hat{x}_i^*(p)I$$ gives maximal utility in that budget set.

Case 2. $$u(x_i^*(p,I)) < u(\hat{x}_i^*(p)I)$$ As $$\hat{x}_i^*(p)I$$ is clearly an element of $$B(p,I)$$ this is impossible as $$x_i^*(p,I)$$ gives maximal utility in that budget set.

Thus we have proven that $$u(x_i^*(p,I)) = u(\hat{x}_i^*(p)I)$$ which is equivalent with $$x_i^*(p,I) = \hat{x}_i^*(p)I.$$

• I am not entirely sure what the index $$i$$ is referring to. A specific good of the bundle? I don't think you need this.
• $$\hat x (p)$$ denotes an optimal bundle for an income equal to $$1$$, right? It is clearer to write $$\hat x (p, 1)$$ in order to be consistent with your notation.
• The claim which you want to prove is "If $$\hat x (p, 1)$$ is optimal at an income of 1, then $$I \hat x (p, 1)$$ is optimal at an income of $$I$$" . By the way, it is not generally true that every optimal bundle at $$I$$ is given by $$I \hat x (p, 1)$$.
• To prove the above claim, you only need to note that $$I \hat x (p, 1)$$ is feasible at $$I$$ and then argue that it is weakly preferred to every other $$y$$ that's feasible at $$I$$. You basically did this in your answer when you noted that the bundle $$y / I$$ is feasible at an income of 1, and then used optimality of $$\hat x (p, 1)$$.