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.

• If C is open, any point in C has a ball around it, but still in C. How can you choose $\alpha$? May 17 '19 at 17:27
• So if $x \in C$ means $x \in 1C$ and with $C$ open, we can say $p_C(x)<1$ ? May 17 '19 at 17:46
• 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... May 17 '19 at 17:51
• Do I need to define some $s$ and $t$ in the ball that somehow $\lambda s + (1-\lambda)t = \frac{x}{\alpha}$ with $\alpha < 1$? May 17 '19 at 19:33
• 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. 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?

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

• 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 $\alpha < 1$. May 17 '19 at 21:33
• 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. May 17 '19 at 21:42
• Thank you! I understand it now :) May 18 '19 at 15:33