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 $$ x_i^*(p,I) = \hat{x}_i^*(p)I. $$