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.

  • $\begingroup$ If C is open, any point in C has a ball around it, but still in C. How can you choose $\alpha$? $\endgroup$
    – ryan221b
    May 17 '19 at 17:27
  • $\begingroup$ So if $x \in C $ means $x \in 1C $ and with $C$ open, we can say $p_C(x)<1$ ? $\endgroup$ May 17 '19 at 17:46
  • 1
    $\begingroup$ You can conclude that, yes. But as it's what you're trying to prove, you might want to explain more carefully why being open gives you an $\alpha < 1$ (so the infimum is < 1). The ball around $x\in C$ is convex... $\endgroup$
    – ryan221b
    May 17 '19 at 17:51
  • $\begingroup$ Do I need to define some $s$ and $t$ in the ball that somehow $\lambda s + (1-\lambda)t = \frac{x}{\alpha}$ with $\alpha < 1$? $\endgroup$ May 17 '19 at 19:33
  • $\begingroup$ Almost. You want to use that $C$ is convex too, and $0 \in C$. I think that if you draw yourself a diagram with $0, x, C$ and the ball you'll see the geometry. $\endgroup$
    – ryan221b
    May 17 '19 at 19:40

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?


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, $p_C(x) \leq \alpha < 1$.

  • $\begingroup$ So as I understand, the idea is to draw a "smaller" ball inside $B(x,\delta)$ hence, the "smaller" ball is in $C$, since $p_C(x)\le1$ as $x\in1C$ we now have a point $s \in C$ (as the center of "smaller ball") where $\alpha$ in this case is even smaller, so $\alpha < 1$. $\endgroup$ May 17 '19 at 21:33
  • $\begingroup$ You don't need any extra balls, as $B(x, \delta)$ is already contained in $C$. The point is that you want an $\alpha < 1$, so $\alpha^{-1} > 1$. This means that $x$ is “pushed further away from $0$”. I'll add more details. $\endgroup$
    – ryan221b
    May 17 '19 at 21:42
  • $\begingroup$ Thank you! I understand it now :) $\endgroup$ May 18 '19 at 15:33

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