Show that $x(p,w)=w\cdot x(p,1)$ with homothetic preferences Someone gave me a proof of this, but I am not sure if it is correct. 

Let $B(p,w) = \{x: p\cdot x \leq w\}$ (the budget set). Then:

\begin{align}
x(p,w) &amp;= \arg \max_{x\in B(p,w)} u(x)\\
&amp;=\arg \max_{\alpha x\in \alpha B(p,w)}  u(\alpha x) \\
&amp;=\arg \max_{y\in B(p,\alpha w)} u(y) \\
&amp;=\frac{1}{\alpha} \arg \max_{y\in B(p,\alpha w)} u(y) \\
&amp;=\frac{1}{\alpha} x(p, \alpha w)
\end{align}
Where the result follows from taking $\alpha=\frac{1}{w}$.

Is this proof correct (I am not sure of the middle three equalities)? Where is homotheticity used? 

EDIT: A monotone preference relation $\succsim$ on $X= \mathbb{R}^{L}_{+}$ is homothetic if all indifference sets are related by proportional expansion along rays; that is, if $x \sim y$, then $\alpha x \sim \alpha y$ for any $\alpha \geq 0$. 

Also, recall a continuous $\succsim$ on $X = \mathbb{R}_{+}^{L}$ is homothetic iff it admits a utility function that is homogenous of degree one; $u(\alpha x) = \alpha u(x)$.
An indirect proof. Suppose
$$
x(p,w) = w\cdot x(p,1)
$$
does not hold. This is equivalent with stating
$$
U(x(p,w)) \neq U(w\cdot x(p,1)).
$$
(To be precise: $x(p,w)$ and $x(p,1)$ may be set valued. In this case we are talking about two elements at least one of which is not included in both sets.)

Case 1. 
$$
U(x(p,w)) &gt; U(w\cdot x(p,1))
$$
As $U$ is homothetic
$$
U(x(p,w)) =  U(w \cdot \frac{1}{w}\cdot x(p,w)) = w \cdot U(\frac{1}{w}\cdot x(p,w)).
$$
Using this we have
$$
w \cdot U(\frac{1}{w}\cdot x(p,w)) = U(x(p,w)) &gt; U(w\cdot x(p,1)) = w\cdot U(x(p,1))
$$
and thus
$$
U(\frac{1}{w}\cdot x(p,w)) &gt; U(x(p,1))
$$
However as $\frac{1}{w} \cdot x(p,w)$ is clearly an element of $B(p,1)$ this is impossible as $x(p,1)$ gives maximal utility in that budget set.

Case 2. 
$$
U(x(p,w)) &lt; U(w\cdot x(p,1))
$$
As $w \cdot x(p,1)$ is clearly an element of $B(p,w)$ this is impossible as $x(p,w)$ gives maximal utility in that budget set.

Thus we have proven that
$$
U(x(p,w)) = U(w\cdot x(p,1))
$$
which is equivalent with
$$
x(p,w) = w\cdot x(p,1).
$$