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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.

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  • $\begingroup$ Your first implication $x=0\implies \|x\|=0$ is correct. For the second implication, use the fact that $V$ is bounded. $\endgroup$ – TSF Apr 30 '19 at 7:32
  • $\begingroup$ 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 ? $\endgroup$ – Alain Apr 30 '19 at 8:39
  • $\begingroup$ 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 $x=0$. $\endgroup$ – TSF Apr 30 '19 at 8:40
  • $\begingroup$ How I see now. This is very clever. Thank you. $\endgroup$ – Alain Apr 30 '19 at 12:31
  • $\begingroup$ 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 $\|x\|_2=0$. $\endgroup$ – TSF Apr 30 '19 at 12:41
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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 $r_2$.

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  • $\begingroup$ Thank you for your help. $\endgroup$ – Alain Apr 30 '19 at 8:39
  • $\begingroup$ Welcome. For triangle inequality you have to use convexity of $V$. $\endgroup$ – Kavi Rama Murthy Apr 30 '19 at 8:43
  • $\begingroup$ 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 $\endgroup$ – Alain Apr 30 '19 at 12:36
  • $\begingroup$ @Alain I have added a proof of triangle inequality. $\endgroup$ – Kavi Rama Murthy Apr 30 '19 at 13:18

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