Given a (profit) function of the form
$$ \pi(p) = \sup \{p.y:y \in Y\} $$, where $p \in R_+^k$ is a positive (price) vector and $Y \in R^k$ is a (production-possibility) set.
I need to proof that $\pi()$ is homogeneous of degree one i.e. $\pi(p \alpha)=\alpha \pi(p)$ for $\alpha \geq 0$.
According to my professor, the proof goes like this: For all $y \in Y$, $$ \pi(p) \geq p.y $$ $$ \alpha \pi(p) \geq \alpha p.y $$ $$ \alpha \pi(p) \geq \sup \alpha p.y $$ $$ \alpha \pi(p) \geq \pi(\alpha p)$$ On the other hand,
$$ \frac{1}{\alpha} \pi(\alpha p) \geq \pi(\frac{\alpha}{\alpha} p.y) $$ $$ \pi(\alpha p) \geq \alpha \pi(p.y) $$, thus $$ \alpha \pi(p) = \pi(p \alpha)$$
No problems with this proof, but before I saw it, I tried to make a proof on my own.
I got something like this,
$$\alpha \pi(p) = \alpha \sup \{p.y:y \in Y\} = \sup \{(\alpha p).y:y \in Y\} = \pi( p \alpha )$$ since I'm not confident with supremum, and my proof is much shorter, I'm suspecting that I have some mistake, or I'm disregarding some situation such as when $\pi(p)=\infty$.