# Help with a specific norm and convex space.

I am struggling with an exercice, maybe I am missing something or I don't understand something. Here's the exercice:

Let $$V \subset \mathbb{R}^n$$ open, convex, bounded and such that if $$x \in V$$ then $$-x \in V$$. Now consider the function defined by

$$\mid \mid x \mid \mid_V := \inf_{r>0} \{ \frac{x}{r} \in V \}.$$

I have to show that:

a) $$\mid \mid . \mid \mid_V$$ defines a norm on $$\mathbb{R}^n$$.

b) $$V=B(0,1)$$, where $$B(0,1)$$ is the open ball with respect to the norm $$\mid \mid . \mid \mid_V$$.

My attempt:

I am not even sure how to show that $$\mid \mid x \mid \mid_V = 0 \Leftrightarrow x =0.$$ Since $$V$$ is convex and since $$x$$ and $$-x$$ are in $$V$$, $$0$$ has to be in $$V$$. Then, if $$x=0$$, I should have $$\inf_{r>0} \{ \frac{0}{r} \in V \} = \inf_{r>0} \{ 0 \in V \} =0.$$ Is this correct ?

For the other direction I am unsure how to interpret

$$\inf_{r>0} \{ \frac{x}{r} \in V \} = 0$$

This mean that the smallest value of $$r>0$$ I can take such that $$\frac{x}{r}$$ is in $$V$$ is $$0$$. But how does this show that $$x=0$$ ?

For the second property of the norm, $$\mid \mid \lambda x \mid \mid_V = \mid \lambda \mid \cdot \mid \mid x \mid \mid_V$$, is this just the property of the infimum ? I mean something like $$\inf \{ \lambda X \} = \lambda \inf \{ X \}$$? And similarly for the triangle inequality ?

To be honest I have no clue on how to prouve the part b), any help and hints will be appreciated.

Thank you.

• Your first implication $x=0\implies \|x\|=0$ is correct. For the second implication, use the fact that $V$ is bounded. – TSF Apr 30 '19 at 7:32
• Thank you for your comment. So, if I use the fact that $V$ is bounded, it means that $V$ is contained in a Ball of fine radius, let's say $R$. I don't really understand how this will be of any help. If I take any $x \in \mathbb{R}^n$ then I can maybe "put" it in $V$ by making it smaller enough : $\frac{x}{r} \in V$ for a value of $r$ big enough, no ? – Alain Apr 30 '19 at 8:39
• Yes we can "put" it in $V$ by making that fraction small enough but that means that we are making $r$ big which means the inf will be big (and hence nonzero). Try this: assume that $\|x\| =0$. Then $\inf\{r>0: \frac{x}{r} \in V\}=0$. This means that $\lim\limits_{r\searrow 0}\frac{x}{r}\in V$ but we have $$\|\lim\limits_{r\searrow 0}\frac{x}{r}\|_2 = \lim\limits_{r\searrow 0}\frac{1}{r}\|x\|_2 = \begin{cases}\infty & x\neq 0,\\ 0 & x=0.\end{cases}$$ Since $V$ is bounded, we must have $x=0$. – TSF Apr 30 '19 at 8:40
• How I see now. This is very clever. Thank you. – Alain Apr 30 '19 at 12:31
• I think it's actually wrong. The step where I write that the limit is contained in $V$ is not true, since we don't know if $V$ is closed. Instead it should be something like this: Since $\inf\{r>0: \frac{x}{r}\in V\}=0$ we have that $\frac{x}{r}\in V$ for arbitrarily small $r$. Since $V$ is bounded (say, in a $2$-ball of radius $R$), pick $r$ such that $\frac{\|x\|_2}{r} > R$ (this is possible because of the infimum condition). Then the $2$-norm of $\frac{x}{r}$ is bigger than $R$ unless $\|x\|_2=0$. – TSF Apr 30 '19 at 12:41

Proof of b): if $$\|x\|<1$$ then there exists $$r<1$$ such that $$\frac x r \in V$$. Now $$x= r(\frac x r)+(1-r)0$$. This implies that $$x \in V$$ because $$V$$ is convex. [Note that $$0 \in V$$ because $$0=\frac {y+(-y)} 2$$ (where $$y$$ is any element of $$V$$) and $$V$$ is convex]. Conversely, suppose $$x \in V$$. Since $$V$$ is open there exists $$t>0$$ such that $$(1+t)x \in V$$. This implies $$\|x\|_V \leq r$$ where $$r= \frac 1 {1+t}$$. Hence $$\|x\|_V <1$$.
Triangle inequality: suppose $$\frac x {r_1} \in V$$ and $$\frac y {r_2} \in V$$. Then $$\frac {x+y} {r_1+r_2}=\frac {r_1} {r_1+r_2} (\frac x {r_1})+\frac {r_2} {r_1+r_2} (\frac y {r_2}) \in V$$. This gives $$\|x+y\|_V \leq r_1+r_2$$. Now take infimum over $$r_1$$ and $$r_2$$.
• Welcome. For triangle inequality you have to use convexity of $V$. – Kavi Rama Murthy Apr 30 '19 at 8:43
• May I ask you some hints on how to use the convexity ? I don't know how to "cut" the expression $\inf_{r>0} \{ (\frac{x}{r} + \frac{y}{r}) \in V \}$ in two pieces. Thank you – Alain Apr 30 '19 at 12:36