I have this following to prove:
Let $X$ be normed space and $C \subset X$ be an open, convex subset of $X$ with $0\in C$. Show that:
$C= \{x \in X: p_C(x) < 1\}$
With $p_C$ defined as the Minkowski functional of $C$:
$p_C : X \to [0,\infty]$, $x \to \mathbb{inf}\{\alpha > 0: x \in \alpha C\}$
I could show that $\{x \in X: p_C(x) < 1\} \subset C$ but don't know how to show the other way. I suppose I need to use the openness of $C$. Any hint would be appreciated.
Best Answer
Hopefully this will clear up both the question and the last comment about $\alpha^{-1}x \in C$.
Because $0 \in C$ and $C$ is open, there is a ball $B(0, r) \subset C$ of radius $r > 0$. Then any point (non zero) $x\in X$ would satisfy $\alpha x \in C$ by taking $\alpha = \frac{r}{2\| x \|}$, as then $\alpha x \in B(0, r)$. Can you see how this implies that $p_C(x) = 0$ for all $x \in X$?
Now consider $x \in C$, so there is a $\delta > 0$ such that $x \in B(x, \delta) \subset C$. Look at the line joining $0$ to $x$ in $C$ (as $C$ is convex). You can extend this line further away from $x$ inside the ball $B(x, \delta)$. (This is the picture I meant.)
Now, the distance from $x$ to $0$ is $\| x \|$. Pick the point $\delta/2 + \| x \|$ away from $0$ along the line $0$ to $x$.
Can you see how to finish from here?
Edit
We want to pick a point in the same “direction” as $x$ but “further away” from $0$. Dividing by $\|x\|$ and multiplying by $\delta/2 + \|x\|$ we get the desired point. Why is it desired? Because this point is of the form $\alpha^{-1}x$ with
$$ \alpha^{-1} = \frac{\delta/2 + \|x\|}{\|x\|} > 1$$
so $\alpha < 1$. And this point is only $\delta/2$ away from $x$, so it's in $B(x, \delta) \subset C$. Hence, $p_C(x) \leq \alpha < 1$.