Homeomorphism via Minkowski functional?

Question

Suppose $E$ is an infinite dimensional topological vector space and $\Omega\subset E$ is open, convex and $0\in \Omega$. The Minkowski-functional of $\Omega$ is defined by: $$ p_\Omega:E\to \mathbb{R},\quad p_\Omega(v):=\inf\{\lambda>0\, |\, v\in \lambda \Omega\}. $$ I would like to show that $\Omega$ is homeomorphic to $E$.
Question: Is the map $$ \varphi:\Omega\to E, \quad \varphi(x):=\frac{x}{1-p_\Omega(x)} $$ a homeomorphism from $\Omega$ to $E$?

Answer 1

Yes, it is. Since $p_\Omega$ is homogeneous of degree $1$, we have $$p_\Omega(\varphi(x))= \frac{p_{\Omega}(x)}{1-p_\Omega(x)} \tag1$$ Let $y=\varphi(x)$. Rearrange (1) into $$p_\Omega(x) = \frac{p_{\Omega}(y)}{1+p_\Omega(y)} \tag2$$ and conclude that $$\varphi^{-1}(y) = \frac{y}{1+p_\Omega(y)}\tag2$$ It remains to show that $p_\Omega$ is a continuous function on $E$. To this end, observe that for any positive number $t$ the sets $\{x:p_\Omega(x)<t\}=t\Omega$ and $\{x:p_\Omega(x)>t\}=E\setminus \overline{t\Omega}$ are open.